[ { "text": "If the two foci of a hyperbola are $F_{1}(-3,0)$, $F_{2}(3,0)$, and one asymptote has equation $y=\\sqrt{2} x$, then the equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(G)={F1,F2};Expression(OneOf(Asymptote(G))) = (y = sqrt(2)*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/3-y^2/6=1", "fact_spans": "[[[2, 5], [67, 70]], [[13, 26]], [[28, 40]], [[13, 26]], [[28, 40]], [[2, 40]], [[2, 64]]]", "query_spans": "[[[67, 75]]]", "process": "" }, { "text": "An ellipse $\\frac{x^{2}}{k^{2}}+y^{2}=1$ $(k>0)$ has a focus at $(3 , 0)$, then $k=$?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2 + x^2/k^2 = 1);k: Number;k>0;Coordinate(OneOf(Focus(G))) = (3,0)", "query_expressions": "k", "answer_expressions": "sqrt(10)", "fact_spans": "[[[0, 36]], [[0, 36]], [[54, 57]], [[2, 36]], [[0, 51]]]", "query_spans": "[[[54, 59]]]", "process": "" }, { "text": "Given any point $P$ on the parabola $y^{2}=4 x$, let $d$ be the distance from point $P$ to the $y$-axis. For a given point $A(4,5)$, what is the minimum value of $|P A|+d$?", "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 5);PointOnCurve(P,G);Distance(P, yAxis) = d;d:Number", "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))", "answer_expressions": "sqrt(34)-1", "fact_spans": "[[[2, 16]], [[49, 57]], [[27, 31], [22, 25]], [[2, 16]], [[49, 57]], [[2, 25]], [[27, 43]], [[40, 43]]]", "query_spans": "[[[59, 74]]]", "process": "The directrix of the parabola is x = -1. Extending the distance from P to the y-axis one unit further to the left gives the distance from P to the directrix. According to the definition of a parabola, the distance from P to the directrix equals the distance from P to the focus F(1,0). Therefore, |PA| + d = |PA| + |PF| - 1. Since A lies outside the parabola, the minimum value occurs when A, P, and F are collinear (i.e., P is the intersection point of segment AF and the parabola). The minimum value is |AF| - 1 = \\sqrt{(9+25)} - 1 = \\sqrt{34} - 1^{n}" }, { "text": "The equation of the line on which the chord of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ lies, passing through the point $M(3,-1)$ and bisected by the point $M$, is?", "fact_expressions": "G: Hyperbola;H: LineSegment;M: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(M) = (3, -1);IsChordOf(H,G);MidPoint(H)=M;PointOnCurve(M,H)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "3*x+4*y-5=0", "fact_spans": "[[[21, 49]], [], [[1, 12], [14, 18]], [[21, 49]], [[1, 12]], [[21, 51]], [[13, 51]], [[0, 51]]]", "query_spans": "[[[21, 59]]]", "process": "" }, { "text": "The product of the slopes of the lines connecting a moving point $P$ to fixed points $A(-1,0)$ and $B(1,0)$ is $-1$. Then, what is the trajectory equation of point $P$?", "fact_expressions": "P: Point;Slope(LineSegmentOf(P, A))*Slope(LineSegmentOf(P, B)) = -1;A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0)", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x^2+y^2=1)&Negation(x=pm*1)", "fact_spans": "[[[2, 5], [43, 47]], [[2, 41]], [[8, 17]], [[8, 17]], [[20, 28]], [[20, 28]]]", "query_spans": "[[[43, 54]]]", "process": "Let P(x,y), then k_{PA}=\\frac{y-0}{x+1}, k_{PB}=\\frac{y-0}{x-1}. Since the product of the slopes of the lines joining the moving point P and the fixed points A(-1,0), B(1,0) is -1, \\therefore k_{PA} \\cdot k_{PB} = -1, \\therefore \\frac{y^{2}}{x^{2}-1} = -1, that is, x^{2} + y^{2} = 1, and x \\neq \\pm 1. In conclusion, the trajectory equation of point P is x^{2} + y^{2} = 1 (x \\neq \\pm 1)." }, { "text": "The eccentricity of the hyperbola $x^{2}-y^{2}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 18]], [[0, 18]]]", "query_spans": "[[[0, 24]]]", "process": "From the given conditions, we have a = b = 1, c = \\sqrt{a^{2} + b^{2}} = \\sqrt{2}, thus the eccentricity e = \\frac{c}{a} = \\sqrt{2}." }, { "text": "Let $P(x, y)$ be an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and let $F_{1}$, $F_{2}$ be the two foci of the ellipse such that $\\angle F_{1} P F_{2} \\leq 90^{\\circ}$. Find the range of values for the eccentricity of the ellipse.", "fact_expressions": "P: Point;PointOnCurve(P, G) = True;Coordinate(P) = (x1, y1);x1: Number;y1: Number;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};AngleOf(F1, P, F2) <= ApplyUnit(90, degree)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(0,sqrt(2)/2]", "fact_spans": "[[[0, 10]], [[0, 73]], [[0, 10]], [[1, 10]], [[1, 10]], [[11, 67], [91, 93], [141, 143]], [[11, 67]], [[13, 67]], [[13, 67]], [[13, 67]], [[13, 67]], [[74, 81]], [[83, 90]], [[74, 98]], [[100, 138]]]", "query_spans": "[[[141, 154]]]", "process": "" }, { "text": "If the hyperbola $C$ has the same foci as the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and one asymptote has the equation $y= \\sqrt{7} x$, then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(C) = Focus(G);Expression(OneOf(Asymptote(C))) = (y = sqrt(7)*x)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/2-y^2/14=1", "fact_spans": "[[[1, 7], [80, 83]], [[8, 46]], [[8, 46]], [[1, 52]], [[1, 78]]]", "query_spans": "[[[80, 88]]]", "process": "" }, { "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ ($m, n>0$) has eccentricity $\\frac{1}{2}$, and one focus coincides exactly with the focus of the parabola $y^{2}=8x$, then what is the standard equation of the ellipse?", "fact_expressions": "H: Ellipse;Expression(H) = (y^2/n + x^2/m = 1);m: Number;n: Number;m>0;n>0;Eccentricity(H) = 1/2;G: Parabola;Expression(G) = (y^2 = 8*x);OneOf(Focus(H)) = Focus(G)", "query_expressions": "Expression(H)", "answer_expressions": "x^2/16+y^2/12=1", "fact_spans": "[[[1, 46], [93, 95]], [[1, 46]], [[3, 46]], [[3, 46]], [[3, 46]], [[3, 46]], [[1, 64]], [[74, 88]], [[74, 88]], [[1, 91]]]", "query_spans": "[[[93, 102]]]", "process": "" }, { "text": "It is known that the vertex of the parabola is at the origin, the focus lies on the $x$-axis, and the point $P(-3 , m)$ on it is at a distance of $5$ from the focus. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;P: Point;O: Origin;m:Number;Coordinate(P) = (-3, m);Vertex(G) = O;PointOnCurve(Focus(G), xAxis);PointOnCurve(P, G);Distance(P, Focus(G))=5", "query_expressions": "Expression(G)", "answer_expressions": "y^2=-8*x", "fact_spans": "[[[2, 5], [48, 51], [21, 22]], [[24, 36]], [[9, 11]], [[25, 36]], [[24, 36]], [[2, 11]], [[2, 20]], [[21, 36]], [[24, 46]]]", "query_spans": "[[[48, 55]]]", "process": "Since the vertex of the parabola is at the origin, the focus is on the x-axis, and it passes through point p(-3, m), the equation of the parabola can be written as y^{2} = -2px (p > 0). By the definition of the parabola, we have 3 + \\frac{p}{2} = 5. \\therefore p = 4. \\therefore The equation of the parabola is y^{2} = -8x." }, { "text": "Given that $M$ is a point on the parabola $x^{2}=4 y$, $F$ is its focus, and point $A(1,5)$, then the minimum value of $|M F|+|M A|$ is?", "fact_expressions": "G: Parabola;A: Point;M: Point;F: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (1, 5);PointOnCurve(M, G);Focus(G) = F", "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))", "answer_expressions": "6", "fact_spans": "[[[6, 20], [28, 29]], [[32, 41]], [[2, 5]], [[24, 27]], [[6, 20]], [[32, 41]], [[2, 23]], [[24, 31]]]", "query_spans": "[[[43, 62]]]", "process": "The parabola $ x^{2} = 4y $ has focus $ (0,1) $ and directrix $ y = -1 $. As shown in the figure, by the definition of a parabola, $ |MP| = |MF| $. When points $ A $, $ M $, and $ P $ are collinear and $ AM \\perp x $-axis, $ |MF| + |MA| $ is minimized, so $ |MF| + |MA| = |AP| = 5 + 1 = 6 $." }, { "text": "The line $y = x - 1$ intersects the parabola $y^{2} = 4x$ at points $A$ and $B$. Then $|AB| =$?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = x - 1);Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[10, 24]], [[0, 9]], [[27, 30]], [[31, 34]], [[10, 24]], [[0, 9]], [[0, 36]]]", "query_spans": "[[[38, 48]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, the line $y=\\frac{\\sqrt{6}}{3} b$ intersects the ellipse $C$ at points $A$ and $B$. If $O A \\perp O B$, then the value of the ellipse's eccentricity is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;G: Line;Expression(G) = (y = b*sqrt(6)/3);Intersection(G, C) = {A, B};A: Point;B: Point;O: Origin;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B)) = True", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[2, 59], [87, 92], [122, 124]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[60, 86]], [[60, 86]], [[60, 103]], [[94, 97]], [[98, 101]], [[105, 120]], [[105, 120]]]", "query_spans": "[[[122, 132]]]", "process": "" }, { "text": "The parabola $y=2 x^{2}$ and the circle $x^{2}+(y-a)^{2}=1$ have two distinct common points; then the set of values of $a$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2);H: Circle;Expression(H) = (x^2 + (-a + y)^2 = 1);a: Number;NumIntersection(G, H) = 2", "query_expressions": "Range(a)", "answer_expressions": "(-1, 1)+{17/8}", "fact_spans": "[[[0, 14]], [[0, 14]], [[15, 35]], [[15, 35]], [[46, 49]], [[0, 44]]]", "query_spans": "[[[46, 56]]]", "process": "The parabola and the circle have two intersection points in the following cases: \n1. When they are tangent to each other, solving the system of equations of the parabola and the circle yields $ y^{2} + \\left(\\frac{1}{2} - 2a\\right)y + a^{2} - 1 = 0 $. At this point, $ \\Delta = \\left(\\frac{1}{2} - 2a\\right)^{2} - 4(a^{2} - 1) = 0 $, solving gives $ a = \\frac{17}{8} $. In the figure above, if the circle moves upward, the two curves have no intersection points. \n2. From the figure in case 1, if the circle moves downward, there are situations with 4 intersection points and 3 intersection points. \n3. When continuing to move the circle downward from the case of 3 intersection points, the number of intersection points decreases from 2 to 1, as shown in the figure below. $ \\therefore $ In this process, the range of $ a $ is $ -1 < a < 1 $. After this, as the circle continues to move downward, the two curves have no intersection points. \nIn summary, the set of values of $ a $ is $ (-1, 1) \\cup \\left(\\frac{17}{8}\\right) $." }, { "text": "Given that point $P(x, y)$ lies on the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{9}=1$, what is the maximum value of $3x+2y$?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/5 + y^2/9 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G);x1:Number;y1:Number", "query_expressions": "Max(3*x1 + 2*y1)", "answer_expressions": "9", "fact_spans": "[[[13, 50]], [[2, 12]], [[13, 50]], [[2, 12]], [[2, 51]], [[3, 12]], [[3, 12]]]", "query_spans": "[[[53, 68]]]", "process": "Point $ P(x,y) $ lies on the ellipse $ \\frac{x^{2}}{5} + \\frac{y^{2}}{9} = 1 $. Let $ x = \\sqrt{5}\\cos\\theta $, $ y = 3\\sin\\theta $. Therefore, $ 3x + 2y = 3\\sqrt{5}\\cos\\theta + 2 \\times 3\\sin\\theta = 9\\left( \\frac{\\sqrt{5}}{3}\\cos\\theta + \\frac{2}{3}\\sin\\theta \\right) $. Let $ \\sin\\varphi = \\frac{\\sqrt{5}}{3} $, then $ \\cos\\varphi = \\frac{2}{3} $. Then the original expression becomes $ 9\\sin(\\varphi + \\theta) $. When $ \\varphi + \\theta = \\frac{\\pi}{2} $, its maximum value is $ 9 $. Hence, the maximum value is $ 9 $. This problem examines finding the maximum value after trigonometric substitution, and when solving such problems, pay attention to the substitution form related to the ellipse equation. It is a medium-level problem." }, { "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 $\\overrightarrow{P Q}=2 \\overrightarrow{F_{1} O}$, $\\overrightarrow{F_{1} Q}=\\lambda\\left(\\frac{\\overrightarrow{F_{1} P}}{|\\overrightarrow{F_{1} P}|}+\\frac{\\overrightarrow{F_{1} O}}{|\\overrightarrow{F_{1} O}|}\\right)$ $(\\lambda>0)$, 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;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, 54], [80, 82], [89, 91], [316, 318]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 65]], [[2, 65]], [[66, 69]], [[75, 79]], [[75, 83]], [[84, 88]], [[84, 96]], [[98, 147]], [[149, 315]], [[149, 315]], [[149, 315]]]", "query_spans": "[[[316, 324]]]", "process": "Let the right focus of the ellipse be $ F_{2} $. From $ \\overrightarrow{PQ} = 2\\overrightarrow{F_{1}O} $, it follows that $ \\overrightarrow{PQ} = \\overrightarrow{F_{1}F_{2}} $, so quadrilateral $ F_{1}F_{2}QP $ is a parallelogram. Since $ \\overrightarrow{F_{1}Q} = \\lambda\\left( \\frac{\\overrightarrow{F_{1}P}}{|\\overrightarrow{F_{1}P}|} + \\frac{\\overrightarrow{F_{1}O}}{|\\overrightarrow{F_{1}O}|} \\right) $, the parallelogram $ F_{1}F_{2}QP $ is a rhombus. Let point $ P(x,y) $, $ Q\\left( \\frac{a^{2}}{c}, y \\right) $, then $ \\overrightarrow{PQ} = \\left( \\frac{a^{2}}{c} - x, 0 \\right) $, $ \\overrightarrow{F_{1}F_{2}} = (2c, 0) $. Since $ \\overrightarrow{PQ} = \\overrightarrow{F_{1}F_{2}} $, we have $ \\frac{a^{2}}{c} - x = 2c $, so $ x = \\frac{a^{2}}{c} - 2c = \\frac{a^{2} - 2c^{2}}{c} $, hence the x-coordinate of point $ P $ is $ \\frac{a^{2} - 2c^{2}}{c} $. Since parallelogram $ F_{1}F_{2}QP $ is a rhombus, $ |\\overrightarrow{PF_{2}}| = |\\overrightarrow{PQ}| $. By the second definition of the ellipse, $ |\\overrightarrow{PF_{2}}| = \\frac{c}{a} \\cdot \\left( \\frac{a^{2} - 2c^{2}}{c} + \\frac{a^{2}}{c} \\right) = \\frac{2(a^{2} - c^{2})}{a} = 2b $. Simplifying yields $ c^{2} + ac - a^{2} = 0 $. Dividing both sides by $ a^{2} $ gives $ e^{2} + e - 1 = 0 $. Since $ 0 < e < 1 $, solving yields $ e = \\frac{\\sqrt{5} - 1}{2} $. Therefore, the eccentricity of the ellipse is $ \\frac{\\sqrt{5} - 1}{2} $." }, { "text": "The distance from the right focus $F$ of the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{b^{2}}=1(b>0)$ to one of its asymptotes is $1$. The directrix of the parabola $y^{2}=2 p x(p>0)$ passes through the left focus of the hyperbola. Then, the minimum distance from a moving point $M$ on the parabola to the point $(5,0)$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2/b^2 = 1);b: Number;b>0;F: Point;RightFocus(G) = F;Distance(F, OneOf(Asymptote(G))) = 1;H: Parabola;Expression(H) = (y^2 = 2*p*x);p: Number;p>0;PointOnCurve(LeftFocus(G), Directrix(H)) = True;M: Point;PointOnCurve(M, H) = True;I: Point;Coordinate(I) = (5, 0)", "query_expressions": "Min(Distance(M, I))", "answer_expressions": "2*sqrt(6)", "fact_spans": "[[[0, 47], [55, 56], [94, 97]], [[0, 47]], [[3, 47]], [[3, 47]], [[51, 54]], [[0, 54]], [[51, 68]], [[69, 90], [103, 106]], [[69, 90]], [[72, 90]], [[72, 90]], [[69, 101]], [[110, 113]], [[103, 113]], [[114, 123]], [[114, 123]]]", "query_spans": "[[[110, 131]]]", "process": "From the given conditions we have: d=\\frac{|cd|}{\\sqrt{a^{2}+b^{2}}}=1 \\therefore b=1,c=2, then the parabola is y^{2}=8x. Let the moving point M(x,y), the distance is \\sqrt{(x-5)^{2}+y^{2}}=\\sqrt{x^{2}-10x+25+8x}=\\sqrt{(x-1)^{2}+24} \\therefore the minimum value is 2\\sqrt{6}" }, { "text": "$F$ is the focus of the parabola $y^{2}=2 p x (p>0)$. A line passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the parabola at points $A$ and $B$. The perpendicular bisector of segment $A B$ intersects the $x$-axis at point $E$. If $|E F|=4$, then $p=$?", "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;E: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(60, degree);Intersection(H, G) = {A, B};Intersection(PerpendicularBisector(LineSegmentOf(A,B)), xAxis) = E;Abs(LineSegmentOf(E, F)) = 4", "query_expressions": "p", "answer_expressions": "3", "fact_spans": "[[[4, 27], [57, 60]], [[107, 110]], [[54, 56]], [[61, 64]], [[65, 68]], [[90, 94]], [[0, 3], [32, 36]], [[7, 27]], [[4, 27]], [[0, 30]], [[31, 56]], [[37, 56]], [[54, 70]], [[71, 94]], [[96, 105]]]", "query_spans": "[[[107, 112]]]", "process": "This problem first allows us to write the line equation $ y = \\sqrt{3}\\left(x - \\frac{p}{2}\\right) $ according to the given conditions. Then, by solving the system of equations consisting of the line equation and the parabola equation, and using Vieta's formulas, we can find the coordinates of the midpoint of segment $ AB $, thus obtaining the equation of the perpendicular bisector of segment $ AB $. Next, we determine the coordinates of point $ E $ from the perpendicular bisector equation. Finally, using $ |EF| = 4 $, we can solve for the value of $ p $. \n**Solution:** Since $ F $ is the focus of the parabola $ y^2 = 2px $, we have $ F\\left(\\frac{p}{2}, 0\\right) $. Because the line passes through point $ F $ and has an inclination angle of $ 60^\\circ $, the equation of the line is $ y = \\sqrt{3}\\left(x - \\frac{p}{2}\\right) $. \nSolving the system: \n$$\n\\begin{cases}\ny = \\sqrt{3}\\left(x - \\frac{p}{2}\\right) \\\\\ny^2 = 2px\n\\end{cases}\n$$\nwe obtain the quadratic equation $ 3x^2 - 5px + \\frac{3p^2}{4} = 0 $. \nLet $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ x_1 + x_2 = \\frac{5p}{3} $. \nThe x-coordinate of the midpoint of segment $ AB $ is $ \\frac{5p}{6} $, and the y-coordinate is $ \\frac{\\sqrt{3}p}{2} $. \nThe equation of the perpendicular bisector of segment $ AB $ is $ y - \\frac{\\sqrt{3}p}{2} = -\\frac{\\sqrt{3}}{3}\\left(x - \\frac{5p}{6}\\right) $. \nSince the perpendicular bisector intersects the x-axis at point $ E $, we get $ E\\left(\\frac{11p}{6}, 0\\right) $. \nGiven $ |EF| = 4 $, we have $ \\frac{11p}{6} - \\frac{p}{2} = 4 $, solving this gives $ p = 3 $." }, { "text": "Given that point $M(t, 0)$ lies on the axis of symmetry of the parabola $y^{2}=2 p x$ $(p>0)$, and a line passing through this point intersects the parabola at points $A$ and $B$, then the product of the slopes of lines $O A$ and $O B$ is?", "fact_expressions": "G: Parabola;p: Number;H: Line;O: Origin;A: Point;M: Point;B: Point;t:Number;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(M,SymmetryAxis(G));Coordinate(M) = (t, 0);PointOnCurve(M, H);Intersection(H, G) = {A, B}", "query_expressions": "Slope(LineOf(O,A))*Slope(LineOf(O,B))", "answer_expressions": "-2*p/t", "fact_spans": "[[[12, 33], [48, 51]], [[15, 33]], [[45, 47]], [[66, 71]], [[53, 56]], [[2, 11], [43, 44]], [[57, 60]], [[2, 11]], [[15, 33]], [[12, 33]], [[2, 39]], [[2, 11]], [[41, 47]], [[45, 62]]]", "query_spans": "[[[64, 84]]]", "process": "Let the equation of the line passing through $ M(t,0) $ be: $ x = my + t $. Solving the system of equations \n$$\n\\begin{cases}\nx = my + t \\\\\ny^2 = 2px\n\\end{cases}\n$$ \nwe obtain $ y^2 - 2pmy - 2pt = 0 $. Let the coordinates of points $ A $ and $ B $ be $ (x_1, y_1) $, $ (x_2, y_2) $, then $ y_1 y_2 = -2pt $. Then $ k_{OA} \\cdot k_{OB} = \\frac{y_1}{x_1} \\cdot \\frac{y_2}{x_2} = \\frac{y_1 \\cdot y_2}{\\frac{y_1^2}{2p} \\cdot \\frac{y_2^2}{2p}} = \\frac{4p^2}{y_1 y_2} = \\frac{4p^2}{-2pt} = -\\frac{2p}{t} $." }, { "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{32}=1$, and let $M$, $N$ be points on the two circles $(x-2)^{2}+y^{2}=1$ and $(x+2)^{2}+y^{2}=\\frac{1}{4}$, respectively. Then the range of values of $|P M|+|P N|$ is?", "fact_expressions": "G: Ellipse;H: Circle;J: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/36 + y^2/32 = 1);Expression(H) = (y^2 + (x - 2)^2 = 1);Expression(J) = (y^2 + (x + 2)^2 = 1/4);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, J)", "query_expressions": "Range(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))", "answer_expressions": "[21/2, 27/2]", "fact_spans": "[[[5, 44]], [[61, 80]], [[81, 110]], [[1, 4]], [[48, 51]], [[52, 55]], [[5, 44]], [[61, 80]], [[81, 110]], [[1, 47]], [[48, 113]], [[48, 113]]]", "query_spans": "[[[115, 135]]]", "process": "\\because the centers of the two circles (2,0), (-2,0) are exactly the foci F_{1}, F_{2} of the ellipse \\frac{x^{2}}{36}+\\frac{y^{2}}{32}=1, \\therefore |PF_{1}|+|PF_{2}|=12, the radii of the two circles are respectively: 1, \\frac{1}{2}, \\therefore (|PM|+|PN|)\\min=|PF_{1}|+|PF_{2}|-1-\\frac{1}{2}=12-1-\\frac{1}{2}=\\frac{21}{2}, (|PM|+|PN|)\\max=|PF_{1}|+|PF_{2}|+1+\\frac{1}{2}=12+1+\\frac{1}{2}=\\frac{27}{2}, then the range of |PM|+|PN| is: [\\frac{21}{2},\\frac{27}{2}]. This is a medium-level problem." }, { "text": "Given that a line $ l $ with slope $ k $ passes through the focus of the parabola $ C: y^2 = 2px \\ (p > 0) $, and intersects the parabola $ C $ at points $ A $ and $ B $. A point $ M(-1, -1) $ on the directrix of the parabola $ C $ satisfies $ \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = 0 $. Then $ |AB| = $?", "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;A: Point;B: Point;k: Number;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (-1, -1);Slope(l)=k;PointOnCurve(Focus(C),l);Intersection(l, C) = {A,B};PointOnCurve(M,Directrix(C));DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "5", "fact_spans": "[[[9, 14]], [[15, 41], [47, 53], [65, 71]], [[22, 41]], [[77, 87]], [[55, 58]], [[59, 62]], [[5, 8]], [[22, 41]], [[15, 41]], [[77, 87]], [[2, 14]], [[9, 44]], [[9, 64]], [[65, 87]], [[89, 140]]]", "query_spans": "[[[142, 151]]]", "process": "" }, { "text": "The standard equation of the ellipse passing through the two points $A(0,2)$, $B\\left(\\frac{1}{2}, \\sqrt{3}\\right)$ is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;Coordinate(A) = (0, 2);Coordinate(B) = (1/2, sqrt(3));PointOnCurve(A,G) = True;PointOnCurve(B,G) = True", "query_expressions": "Expression(G)", "answer_expressions": "x^2+y^2/4=1", "fact_spans": "[[[42, 44]], [[4, 12]], [[15, 41]], [[4, 12]], [[15, 41]], [[0, 44]], [[0, 44]]]", "query_spans": "[[[42, 51]]]", "process": "Substitute A(0,2), B(\\frac{1}{2},\\sqrt{3}) into the standard equation of an ellipse, solve the system of equations to find a and b, thereby obtaining the ellipse equation. Since it is unknown on which axis the foci of the ellipse lie, we can assume the ellipse equation as \\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{n^{2}}=1. Substituting A(0,2), B(\\frac{1}{2},\\sqrt{3}) into the equation yields \\begin{cases}0+\\frac{4}{n^{2}}=1\\\\\\frac{1}{4m^{2}}+\\frac{3}{n^{2}}=1\\end{cases}, solving gives m=1, n=2, so the ellipse equation is x^{2}+\\frac{y^{2}}{4}=1" }, { "text": "The line $x - m y + m = 0$ passing through the focus of the parabola $y^2 = 2 p x$ ($p > 0$) intersects the parabola at points $A$ and $B$, and the area of $\\Delta O A B$ ($O$ being the origin) is $2 \\sqrt{2}$. Then $m^6 + m^4 = $?", "fact_expressions": "G: Parabola;p: Number;H: Line;m: Number;O: Origin;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (m - m*y + x = 0);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};Area(TriangleOf(O,A,B))=2*sqrt(2)", "query_expressions": "m^6 + m^4", "answer_expressions": "2", "fact_spans": "[[[1, 22], [40, 43]], [[4, 22]], [[26, 39]], [[99, 112]], [[71, 74]], [[45, 48]], [[49, 52]], [[4, 22]], [[1, 22]], [[26, 39]], [[0, 39]], [[26, 54]], [[56, 97]]]", "query_spans": "[[[99, 114]]]", "process": "Problem Analysis: First, find the coordinates of the focus based on the equation of the parabola, substitute into the equation of the line to obtain a relation between m and p. Then, solve the system of equations of the line and the parabola by eliminating y, and find the solutions of the resulting equation. Next, use the equation of the line to find y_{1} and y_{2}, respectively. The area of \\triangle OAB can be expressed as the sum of the areas of \\triangle OAP and \\triangle OBP. If OP is taken as the common base for \\triangle OAP and \\triangle OBP, then their heights are the absolute values of the y-coordinates of points A and B, respectively. Thus, express the area of the triangle, solve for p, and then obtain the value of m. Substitute into m^{6}+m^{4} to find the answer." }, { "text": "From a point $P$ on the parabola $y^{2}=4 x$, draw a perpendicular to the directrix of the parabola, with foot of perpendicular at $M$, and $|P M|=5$. Let $F$ be the focus of the parabola. Then $\\cos \\angle MPF$=?", "fact_expressions": "G: Parabola;P: Point;M: 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": "Cos(AngleOf(M, P, F))", "answer_expressions": "3/5", "fact_spans": "[[[1, 15], [22, 25], [50, 53]], [[18, 21]], [[34, 37]], [[57, 60]], [[1, 15]], [[1, 21]], [], [[0, 30]], [[0, 30]], [[0, 37]], [[39, 48]], [[50, 60]]]", "query_spans": "[[[62, 81]]]", "process": "" }, { "text": "Given that 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]], [[117, 120]], [[54, 57]], [[58, 61]], [[40, 51]], [[2, 39]], [[2, 63]], [[65, 116]]]", "query_spans": "[[[126, 155]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and a line passing through point $F_{1}$ intersects the ellipse at points $A$ and $B$, then the perimeter of $\\triangle A B F_{2}$ is?", "fact_expressions": "F2: Point;F1: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);H: Line;PointOnCurve(F1, H) = True;A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "20", "fact_spans": "[[[10, 17]], [[2, 9], [64, 72]], [[2, 62]], [[18, 57], [76, 78]], [[18, 57]], [[73, 75]], [[63, 75]], [[80, 83]], [[84, 87]], [[73, 89]]]", "query_spans": "[[[91, 117]]]", "process": "Since \\( F_{1}, F_{2} \\) are the two foci of the ellipse \\( \\frac{x^{2}}{25} + \\frac{y^{2}}{16} = 1 \\), \\( |AF_{1}| + |AF_{2}| = 10 \\), \\( |BF_{1}| + |BF_{2}| = 10 \\), the perimeter of \\( \\triangle AF_{2}B \\) is \\( |AB| + |AF_{2}| + |BF_{2}| = |AF_{1}| + |AF_{2}| + |BF_{1}| + |BF_{2}| = 10 + 10 = 20 \\);" }, { "text": "Given that the parabola $x^{2}=2 p y$ passes through the point $(2 , 1)$, then the coordinates of the focus of the parabola are?", "fact_expressions": "G: Parabola;p: Number;H: Point;Expression(G) = (x^2 = 2*(p*y));Coordinate(H) = (2, 1);PointOnCurve(H, G)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,1)", "fact_spans": "[[[1, 17], [31, 34]], [[4, 17]], [[19, 29]], [[1, 17]], [[19, 29]], [[1, 29]]]", "query_spans": "[[[31, 41]]]", "process": "Since the point lies on the parabola, substitute the point to find the equation of the parabola, and then find the coordinates of the focus. Because the parabola x^{2}=2py passes through the point (2,1), we have 2^{2}=2p\\times1, so 2p=4, that is, the equation of the parabola is: x^{2}=4y. Therefore, the coordinates of the focus are: (0,1)" }, { "text": "Given that one focus of the hyperbola $\\frac{y^{2}}{m}-x^{2}=1$ $(m>0)$ coincides with the focus of the parabola $y=\\frac{1}{8} x^{2}$, find the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;m>0;Expression(G) = (-x^2 + y^2/m = 1);Expression(H) = (y = x^2/8);OneOf(Focus(G)) = Focus(H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 35], [73, 76]], [[5, 35]], [[41, 65]], [[5, 35]], [[2, 35]], [[41, 65]], [[2, 70]]]", "query_spans": "[[[73, 82]]]", "process": "Analysis: Based on the properties of hyperbolas and parabolas, find the coordinates of the focus, then determine $ m = a^{2} = 3 $, and thus find the eccentricity of the hyperbola. \nSolution: Since one focus of the hyperbola $ \\frac{y^{2}}{m} - x^{2} = 1 $ ($ m > 0 $) coincides with the focus of the parabola $ y = \\frac{1}{8}x^{2} $, and the focus of the parabola $ y = \\frac{1}{8}x^{2} $ is at $ (0, 2) $, $ \\therefore c = 2 $, $ \\therefore 1 + m = 4 $, i.e., $ m = a^{2} = 3 $, $ \\therefore a = \\sqrt{3} $, $ \\therefore e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{3} $." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$, a line with slope $\\sqrt{2}$ is drawn through the left focus $F$ of the hyperbola $C$, intersecting the left branch of the hyperbola $C$ at points $A$ and $B$. If the circle with diameter $AB$ passes through the origin $O$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;a > 0;b > 0;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);F: Point;LeftFocus(C) = F;l: Line;Slope(l) = sqrt(2);PointOnCurve(F, l);A: Point;B: Point;Intersection(l, LeftPart(C)) = {A, B};E: Circle;IsDiameter(LineSegmentOf(A, B), E);O: Origin;PointOnCurve(O, E)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 66], [68, 74], [100, 106], [142, 148]], [[10, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[2, 66]], [[78, 81]], [[68, 81]], [[97, 99]], [[83, 99]], [[67, 99]], [[110, 113]], [[114, 117]], [[97, 119]], [[131, 132]], [[121, 132]], [[133, 140]], [[121, 140]]]", "query_spans": "[[[142, 154]]]", "process": "" }, { "text": "If the distance from point $P$ to the line $x = -1$ is $1$ less than its distance to the point $(2 , 0)$, then what is the equation of the trajectory of point $P$?", "fact_expressions": "G: Line;H: Point;P: Point;Expression(G) = (x = -1);Coordinate(H) = (2, 0);Distance(P, G) = Distance(P, H) - 1", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2 = 8*x", "fact_spans": "[[[6, 14]], [[20, 30]], [[1, 5], [18, 19], [39, 43]], [[6, 14]], [[20, 30]], [[1, 37]]]", "query_spans": "[[[39, 50]]]", "process": "\\because the distance from point P to the line x = -1 is 1 less than its distance to the point (2,0), \\therefore the distance from point P to the line x = -2 is equal to its distance to the point (2,0), \\therefore the locus of point P is a parabola with equation y^{2} = 8x" }, { "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have its right focus at $F$. A line passing through $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{A B}=5 \\overrightarrow{F B}$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;PointOnCurve(F, G);Slope(G) = sqrt(3);Intersection(G, C) = {A, B};VectorOf(A, B) = 5*(VectorOf(F, B))", "query_expressions": "Eccentricity(C)", "answer_expressions": "6/5", "fact_spans": "[[[1, 62], [93, 96], [156, 159]], [[9, 62]], [[9, 62]], [[90, 92]], [[97, 100]], [[101, 104]], [[67, 70], [72, 75]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 70]], [[71, 92]], [[76, 92]], [[90, 106]], [[108, 154]]]", "query_spans": "[[[156, 165]]]", "process": "Analysis: From $\\overrightarrow{AB}=5\\overrightarrow{FB}$, it follows that $\\overrightarrow{AF}=4\\overrightarrow{FB}$, $|\\overrightarrow{AF}|=4m$, $|\\overrightarrow{BF}|=m$. Therefore, in $\\triangle ABD$, using $|AD|=\\frac{1}{2}|AB|$, the result can be obtained. \nDetail: From $\\overrightarrow{AB}=5\\overrightarrow{FB}$, it follows that $\\overrightarrow{AF}=4\\overrightarrow{FB}$. Let $|\\overrightarrow{AF}|=4m$, $|\\overrightarrow{BF}|=m$. Draw perpendiculars from $A$ and $B$ to the directrix, with feet $A_{1}$ and $B_{1}$, respectively. By the definition of the hyperbola, $|AA_{1}|=\\frac{4m}{e}$, $|BB_{1}|=\\frac{m}{e}$. Draw $BD$ perpendicular to $AA_{1}$, with foot $D$. Since the slope of $AB$ is $\\sqrt{3}$, in $\\triangle ABD$, $\\angle BAD=60^{\\circ}$, so $|AD|=\\frac{1}{2}|AB|$, that is, $\\frac{4m}{e}-\\frac{m}{e}=\\frac{3m}{e}=\\frac{5m}{2}$. Solving gives $e=\\frac{6}{5}$. The eccentricity of $C$ is $\\frac{6}{5}$." }, { "text": "Given that points $A$ and $B$ lie on the parabola $y^{2}=2 p x(p>0)$ with focus $F$, if $|\\overrightarrow{A F}|+|\\overrightarrow{B F}|=4$ and the distance from the midpoint of segment $A B$ to the line $x=\\frac{p}{2}$ is $1$, then what is the value of $p$?", "fact_expressions": "G: Parabola;p: Number;H: Line;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*p*x);Expression(H) = (x = p/2);Focus(G)=F;PointOnCurve(A,G);PointOnCurve(B,G);Abs(VectorOf(A, F)) + Abs(VectorOf(B, F)) = 4;Distance(MidPoint(LineSegmentOf(A,B)),H)=1", "query_expressions": "p", "answer_expressions": "{1,3}", "fact_spans": "[[[20, 41]], [[131, 134]], [[105, 122]], [[6, 9]], [[2, 5]], [[16, 19]], [[23, 41]], [[20, 41]], [[105, 122]], [[13, 41]], [[2, 41]], [[2, 41]], [[44, 93]], [[94, 129]]]", "query_spans": "[[[131, 138]]]", "process": "Draw perpendiculars from A and B to the line x = \\frac{p}{2}, and let the projection of the midpoint M of AB onto the directrix be N. According to the definition of the parabola, |AF| + |BF| = |AC| + |BD| = 4. In trapezoid ACDB, the midline |MN| = \\frac{1}{2}(|AC| + |BD|). Given that the distance from the midpoint of segment AB to x = \\frac{p}{2} is 1, we have |x_{0} - \\frac{p}{2}| = 1, which allows us to solve the problem. Draw perpendiculars from A and B to the line x = \\frac{p}{2}, with feet C and D respectively. Let N be the projection of the midpoint M of AB onto the directrix, and connect MN. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), M(x_{0}, y_{0}). According to the definition of the parabola, |AF| + |BF| = |AC| + |BD| = 4. Thus, in trapezoid ACDB, the midline |MN| = \\frac{1}{2}(|AC| + |BD|) = 2. It follows that x_{0} + \\frac{p}{2} = 2, so x_{0} = 2 - \\frac{p}{2}. Since the distance from the midpoint of AB to x = \\frac{p}{2} is 1, we have |x_{0} - \\frac{p}{2}| = 1. Therefore, |2 - p| = 1, solving which gives p = 1 or p = 3." }, { "text": "The equations of the two asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(3/4)*x", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 50]]]", "process": "" }, { "text": "If the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, then its eccentricity is equal to?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/3", "fact_spans": "[[[1, 4], [45, 46]], [[1, 43]]]", "query_spans": "[[[45, 52]]]", "process": "" }, { "text": "Given that the equation $\\frac{x^{2}}{3+k}+\\frac{y^{2}}{2-k}=1$ represents an ellipse, what is the range of values for $k$?", "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(3+k)+y^2/(2-k)=1);k:Number", "query_expressions": "Range(k)", "answer_expressions": "{(-3,2)&Negation(x=-1/2)}", "fact_spans": "[[[45, 47]], [[2, 47]], [[49, 52]]]", "query_spans": "[[[49, 59]]]", "process": "Problem Analysis: From the problem: \n\\begin{cases}3+k>0\\\\2-k>0\\\\3+k+2-k\\end{cases}, \nsolving yields: {x|-3 0, which simplifies to m^{2} < 2k^{2}+1, that is, k^{2} > \\frac{m^{2}-1}{2}. By Vieta's formulas, we have x_{1}+x_{2} = -\\frac{4km}{2k^{2}+1} = -1, so 4km = 2k^{2}+1. Squaring both sides gives 16k^{2}m^{2} = (2k^{2}+1)^{2}, so m^{2} = \\frac{(2k^{2}+1)^{2}}{16k^{2}} = \\frac{k^{2}}{4} + \\frac{1}{16k^{2}} + \\frac{1}{4} \\geqslant 2\\sqrt{\\frac{k^{2}}{4} \\cdot \\frac{1}{16k^{2}}} + \\frac{1}{4} = \\frac{1}{2}. The equality holds if and only if k = \\pm\\frac{\\sqrt{2}}{2}. Since m^{2} \\geqslant \\frac{1}{2}, it follows that m \\geqslant \\frac{\\sqrt{2}}{2} or m \\leqslant -\\frac{\\sqrt{2}}{2}. Therefore, the range of the y-coordinate of the intersection point of line AB with the y-axis is (-\\infty, -\\frac{\\sqrt{2}}{2}] \\cup [\\frac{\\sqrt{2}}{2}, +\\infty)." }, { "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is $\\sqrt{5}$, then what is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(5)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[1, 57], [74, 75]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 72]]]", "query_spans": "[[[74, 82]]]", "process": "From $\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{5}$, we get $\\frac{b^{2}}{a^{2}}=4$, $\\therefore\\frac{b}{a}=2$, $\\therefore$ the asymptotes of the hyperbola are $y=\\pm2x$." }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has two foci $F_{1}$, $F_{2}$. If $P$ is a point on its right branch such that $|P F_{1}|=2|P F_{2}|$, then the range of the eccentricity of the 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);Focus(G) = {F1,F2};PointOnCurve(P,RightPart(G));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,3]", "fact_spans": "[[[0, 56], [84, 85], [116, 119]], [[3, 56]], [[3, 56]], [[80, 83]], [[62, 69]], [[71, 78]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 78]], [[80, 90]], [[92, 114]]]", "query_spans": "[[[116, 129]]]", "process": "" }, { "text": "It is known that the foci of hyperbola $C$ lie on the coordinate axes, and the equations of the asymptotes are $y = \\pm \\frac{\\sqrt{2}}{2} x$. If the point $(4,2)$ lies on $C$, then the focal distance of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (4, 2);Expression(Asymptote(C)) = (y = pm*(sqrt(2)/2)*X);PointOnCurve(G, C);PointOnCurve(Focus(C), axis)", "query_expressions": "FocalLength(C)", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[2, 8], [62, 65], [68, 74]], [[53, 61]], [[53, 61]], [[2, 51]], [[53, 66]], [[2, 16]]]", "query_spans": "[[[68, 79]]]", "process": "If the foci of hyperbola C lie on the x-axis, let the equation of hyperbola C be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Since the asymptotes are given by y=\\pm\\frac{\\sqrt{2}}{2}x and the point (4,2) lies on C, we have \\begin{cases}\\frac{b}{a}=\\frac{\\sqrt{2}}{2}\\\\\\frac{16}{a^{2}}-\\frac{4}{b^{2}}=1\\end{cases}, solving which yields \\begin{cases}a^{2}=8\\\\b^{2}=4\\end{cases}, thus c=\\sqrt{a^{2}+b^{2}}=2\\sqrt{3}, so the focal length of hyperbola C is 2c=4\\sqrt{3}; if the foci of hyperbola C lie on the y-axis, let the equation of hyperbola C be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 (a>0,b>0). Since the asymptotes are given by y=\\pm\\frac{\\sqrt{2}}{2}x and the point (4,2) lies on C, we have \\begin{cases}\\frac{a}{b}=\\frac{\\sqrt{2}}{2}\\\\\\frac{4}{a^{2}}-\\frac{16}{b^{2}}=1\\end{cases}, which has no solution, hence the foci of hyperbola C do not lie on the y-axis; in conclusion, 2c=4\\sqrt{3}." }, { "text": "It is known that a hyperbola centered at the origin with symmetry axes along the coordinate axes has an asymptote $y=\\frac{4}{3} x$. What is the eccentricity of this hyperbola?", "fact_expressions": "C: Hyperbola;O:Origin;Center(C)=O;Expression(OneOf(Asymptote(C)))=(y=4/3*x);SymmetryAxis(C)=axis", "query_expressions": "Eccentricity(C)", "answer_expressions": "{5/3,5/4}", "fact_spans": "[[[16, 19], [46, 49]], [[5, 7]], [[2, 19]], [[16, 43]], [[8, 19]]]", "query_spans": "[[[46, 55]]]", "process": "" }, { "text": "Given that point $P\\left(\\frac{3}{2},-1\\right)$ lies on the directrix of the parabola $E$: $x^{2}=2 p y$ ($p>0$). Draw a tangent from point $P$ to the parabola, and let the point of tangency $A$ be in the first quadrant. Let $F$ be the focus of the parabola $E$, point $M$ lies on the line $A F$, and point $N$ lies on the circle $C$: $(x+2)^{2}+(y+2)^{2}=1$. Then the minimum value of $|M N|$ is?", "fact_expressions": "E: Parabola;p: Number;C: Circle;G: Line;A: Point;F: Point;P: Point;M: Point;N: Point;p>0;Expression(E) = (x^2 = 2*(p*y));Expression(C) = ((x + 2)^2 + (y + 2)^2 = 1);Coordinate(P) = (3/2, -1);PointOnCurve(P, Directrix(E));TangentOfPoint(P , E) = G;TangentPoint(G, E) = A;Quadrant(A) = 1;Focus(E) = F;PointOnCurve(M, LineOf(A, F));PointOnCurve(N, C)", "query_expressions": "Min(Abs(LineSegmentOf(M, N)))", "answer_expressions": "1/3", "fact_spans": "[[[23, 49], [60, 63], [83, 89]], [[31, 49]], [[112, 140]], [], [[70, 73]], [[79, 82]], [[2, 22], [55, 59]], [[93, 97]], [[107, 111]], [[31, 49]], [[23, 49]], [[112, 140]], [[2, 22]], [[2, 53]], [[54, 66]], [[54, 73]], [[70, 78]], [[79, 92]], [[93, 106]], [[107, 141]]]", "query_spans": "[[[143, 156]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ is tangent to the line $l$: $x-2 y+6=0$, then $p$=?", "fact_expressions": "l: Line;C: Parabola;p: Number;p>0;Expression(C) = (y^2 = 2*p*x);Expression(l)=(x-2*y+6=0);IsTangent(l,C)", "query_expressions": "p", "answer_expressions": "3", "fact_spans": "[[[29, 47]], [[2, 28]], [[51, 54]], [[10, 28]], [[2, 28]], [[29, 47]], [[2, 49]]]", "query_spans": "[[[51, 56]]]", "process": "The system of equations \\begin{cases} y2=2px, \\\\ x-2y+6=0, \\end{cases} simplifies to y^{2}-4py+12p=0. Since C is tangent to l, it follows that A=16p^{2}-48p=0, solving which gives p=3 or p=0 (discarded)." }, { "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$ respectively, with eccentricity $e$. A line passing through $F_{1}$ and perpendicular to the line $y=-\\frac{b}{a} x$ intersects the right branch of the hyperbola at point $P$. If $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then $e^{2}$=?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F1: Point;P: Point;F2: Point;e:Number;L:Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = -x*b/a);LeftFocus(G)=F1;RightFocus(G)=F2;Eccentricity(G)=e;PointOnCurve(F1,L);IsPerpendicular(L,H);Intersection(L,RightPart(G))=P;AngleOf(F1, P, F2) = pi/3", "query_expressions": "e^2", "answer_expressions": "(7+2*sqrt(3))/3", "fact_spans": "[[[0, 56], [121, 124]], [[3, 56]], [[3, 56]], [[97, 117]], [[64, 71], [89, 96]], [[127, 131]], [[72, 79]], [[84, 87]], [], [[3, 56]], [[3, 56]], [[0, 56]], [[97, 117]], [[0, 79]], [[0, 79]], [[0, 87]], [[88, 120]], [[88, 120]], [[88, 131]], [[133, 169]]]", "query_spans": "[[[171, 180]]]", "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 $F_{1}$ and $F_{2}$ respectively. If there exists a point $P$ on the right directrix such that the perpendicular bisector of segment $P F_{1}$ passes exactly through $F_{2}$, 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;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightDirectrix(G)) = True;P: Point;PointOnCurve(F2, PerpendicularBisector(LineSegmentOf(P, F1))) = True", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(3)/3,1)", "fact_spans": "[[[2, 54], [122, 124]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70]], [[71, 78], [112, 119]], [[2, 78]], [[2, 78]], [[2, 90]], [[86, 90]], [[92, 119]]]", "query_spans": "[[[122, 135]]]", "process": "" }, { "text": "Let the line $y = kx$ intersect the hyperbola $C$: $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$ at points $A$ and $B$, and let $P$ be a point on $C$ distinct from $A$ and $B$. The slopes of lines $PA$ and $PB$ are $k_{1}$ and $k_{2}$, respectively. If the eccentricity of $C$ is $2$, then $k_{1} \\cdot k_{2} = $?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;k1: Number;k2:Number;A: Point;P: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = k*x);Intersection(G, C) = {A, B};PointOnCurve(P, C);Negation(P = A);Negation(P = B);Slope(LineOf(P,A)) = k1;Slope(LineOf(P,B))=k2;Eccentricity(C) = 2;k:Number", "query_expressions": "k1*k2", "answer_expressions": "3", "fact_spans": "[[[11, 73], [90, 93], [146, 149]], [[19, 73]], [[19, 73]], [[1, 10]], [[128, 135]], [[137, 144]], [[97, 100]], [[86, 89]], [[101, 104]], [[19, 73]], [[18, 73]], [[11, 73]], [[1, 10]], [[1, 85]], [[86, 107]], [[86, 107]], [[86, 107]], [[108, 144]], [[108, 144]], [[146, 157]], [[1, 10]]]", "query_spans": "[[[159, 180]]]", "process": "According to the problem, points A and B are symmetric with respect to the origin. Let point A(x_{0},y_{0}), B(-x_{0},-y_{0}), P(x,y), then \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, \\frac{x^{2}}{a^{2}}-\\frac{y_{0}^{2}}{b^{2}}=1. Subtracting these two equations gives \\frac{x^{2}-x_{0}^{2}}{a^{2}}=\\frac{y^{2}-y_{0}^{2}}{b^{2}}, then \\frac{y^{2}-y_{0}^{2}}{x^{2}-x_{0}^{2}}=\\frac{b^{2}}{a^{2}}. Since e=2, we have k_{1}\\cdot k_{2}=\\frac{y-y_{0}}{x-x_{0}}\\cdot\\frac{y+y_{0}}{x+x_{0}}=\\frac{y^{2}-y_{0}^{2}}{x^{2}-x_{0}^{2}}=\\frac{b^{2}}{a^{2}}=\\frac{c^{2}-a^{2}}{a^{2}}=e^{2}-1=3" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the foci of an ellipse, $P$ is a point on the ellipse, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then what is the minimum value of the eccentricity $e$?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;P: Point;e: Number;Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree);Eccentricity(G) = e", "query_expressions": "Min(e)", "answer_expressions": "1/2", "fact_spans": "[[[18, 20], [29, 31]], [[2, 9]], [[10, 17]], [[25, 28]], [[74, 77]], [[2, 24]], [[25, 34]], [[36, 69]], [[29, 77]]]", "query_spans": "[[[74, 83]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=6x$ with focus $F$, a line $l$ passing through $F$ intersects $C$ at points $A$ and $B$, and intersects the directrix of $C$ at point $D$. If $|AB|=8$, then what are the coordinates of point $D$?", "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;D: Point;Expression(C) = (y^2 = 6*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Intersection(l,Directrix(C))=D;Abs(LineSegmentOf(A, B)) = 8", "query_expressions": "Coordinate(D)", "answer_expressions": "(-3/2,pm*3*sqrt(3))", "fact_spans": "[[[34, 39]], [[2, 21], [40, 43], [56, 59]], [[45, 48]], [[49, 52]], [[25, 28], [30, 33]], [[64, 68], [81, 85]], [[2, 21]], [[2, 28]], [[29, 39]], [[34, 54]], [[34, 68]], [[70, 79]]]", "query_spans": "[[[81, 90]]]", "process": "From the problem, the focus is at $ F\\left(\\frac{3}{2}, 0\\right) $. If the slope of line $ l $ does not exist, then the equation of line $ l $ is $ x = \\frac{3}{2} $. Substituting into the parabola gives $ y = \\pm 3 $, so $ |AB| = 6 $, which does not satisfy the condition. Let the slope of the line be $ k $, then the equation is $ y = k\\left(x - \\frac{3}{2}\\right) $. Substituting into the parabola equation yields $ 4k^{2}x^{2} - (12k^{2} + 24)x + 9k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = \\frac{3k^{2} + 6}{k^{2}} $. By the definition of the parabola, $ |AB| = x_{1} + x_{2} + 3 = \\frac{3k^{2}}{k^{2}} + 6 + 3 = 8 $. Solving gives $ k = \\pm \\sqrt{3} $. Thus, the equation of the line is $ y = \\pm \\sqrt{3}\\left(x - \\frac{3}{2}\\right) $. When $ x = -\\frac{3}{2} $, $ y = \\pm 3\\sqrt{3} $, so the coordinates of point $ D $ are $ \\left(-\\frac{3}{2}, \\pm 3\\sqrt{3}\\right) $." }, { "text": "The line $ l $: $ 2x - 3y + 12 = 0 $ intersects the $ x $-axis and $ y $-axis at points $ A $ and $ B $, respectively. What is the standard equation of the ellipse with focus at $ A $ and passing through point $ B $?", "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;Expression(l)=(2*x - 3*y + 12 = 0);Intersection(l,xAxis) = A;Intersection(l, yAxis) = B;OneOf(Focus(G))=A;PointOnCurve(B, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/52+y^2/16=1", "fact_spans": "[[[0, 21]], [[61, 63]], [[35, 38], [47, 50]], [[39, 42], [56, 60]], [[0, 21]], [[0, 44]], [[0, 44]], [[46, 63]], [[54, 63]]]", "query_spans": "[[[61, 70]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$ has eccentricity greater than $\\frac{\\sqrt{2}}{2}$, $A$ is the top vertex of the ellipse, $B$ is a point on the ellipse, then the maximum value of $|A B|^{2}$ is $?$.", "fact_expressions": "G: Ellipse;a: Number;A: Point;B: Point;a>1;Expression(G) = (y^2 + x^2/a^2 = 1);Eccentricity(G)>sqrt(2)/2;UpperVertex(G)=A;PointOnCurve(B, G)", "query_expressions": "Max(Abs(LineSegmentOf(A, B))^2)", "answer_expressions": "a^2+1+1/(a^2-1)", "fact_spans": "[[[2, 38], [71, 73], [82, 84]], [[4, 38]], [[67, 70]], [[78, 81]], [[4, 38]], [[2, 38]], [[2, 65]], [[67, 77]], [[78, 87]]]", "query_spans": "[[[89, 106]]]", "process": "Since the upper vertex of the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$ is $A(0,1)$, by using the parametric equations of the ellipse, let $B(a\\cos\\theta,\\sin\\theta)$. Therefore, $|AB|^{2}=(a\\cos\\theta)^{2}+(1-\\sin\\theta)^{2}=a^{2}(1-\\sin^{2}\\theta)+1-2\\sin\\theta+\\sin^{2}\\theta=-(a^{2}-1)\\sin^{2}\\theta+2\\sin\\theta+a^{2}+1=-(a^{2}-1)(\\sin^{2}\\theta+\\frac{2}{a^{2}-1}\\sin\\theta)+a^{2}+=-(a^{2}-1)(\\sin\\theta+\\frac{1}{a^{2}-1})^{2}+\\frac{1}{a^{2}-1}+a^{2}+1$. Hence, when $\\sin\\theta+\\frac{1}{a^{2}-1}=0$, $|AB|^{2}$ attains its maximum value of $\\frac{1}{a^{2}-1}+a^{2}+1$." }, { "text": "A moving point $P$ travels on the curve $y = x^{2} + 1$, and $A(1,0)$. Then the equation of the locus of the midpoint of $PA$ is?", "fact_expressions": "G: Curve;P: Point;A: Point;Expression(G) = (y = x^2 + 1);Coordinate(A) = (1, 0);PointOnCurve(P, G)", "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P, A)))", "answer_expressions": "y = 2*x^2 - 2*x + 1", "fact_spans": "[[[6, 19]], [[2, 5]], [[23, 31]], [[6, 19]], [[23, 31]], [[2, 22]]]", "query_spans": "[[[33, 47]]]", "process": "Let M(x, y) be an arbitrary point on the desired locus, where P(x_{1}, y_{1}). Since M is the midpoint of PA, we have \\begin{cases}x=\\frac{1+x_{1}}{2}\\\\y=\\frac{0+y_{1}}{2}\\end{cases}, so \\begin{cases}x_{1}=2x-1\\\\y_{1}=2y\\end{cases}. Substituting into the curve y=x^{2}+1 gives 2y=(2x-1)^{2}+1. Simplifying yields y=2x^{2}-2x+1, which is the equation of the locus of the midpoint of PA." }, { "text": "Given the parabola $C$: $y^{2}=4x$ has focus $F$. A line passing through point $F$ intersects $C$ at points $A$ and $B$, and $|FA|=4$. Then $|AB|=$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);A: Point;B: Point;Intersection(G, C) = {A, B};Abs(LineSegmentOf(F, A)) = 4", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16/3", "fact_spans": "[[[2, 20], [37, 40]], [[2, 20]], [[24, 27], [29, 33]], [[2, 27]], [[34, 36]], [[28, 36]], [[42, 45]], [[46, 49]], [[34, 51]], [[53, 62]]]", "query_spans": "[[[64, 73]]]", "process": "Let the equation of the line passing through $ F(1,0) $ be $ x = my + 1 $, and let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Combining the equations of the line and the parabola \n\\[\n\\begin{cases}\nx = my + 1 \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nyields $ y^2 - 4my - 4 = 0 $. By Vieta's formulas, we obtain $ y_{1}y_{2} = -4 $, then $ x_{1}x_{2} = \\frac{y_{1}^{2}}{4} \\cdot \\frac{y_{2}^{2}}{4} = 1 $. Since by the property of the parabola, $ |FA| = x_{1} + 1 $, we have $ x_{1} = 3 $, $ x_{2} = \\frac{1}{3} $, thus $ |FB| = x_{2} + 1 = \\frac{4}{3} $, $ |AB| = 4 + \\frac{4}{3} = \\frac{16}{3} $." }, { "text": "The parabola $C$: $x^{2}=3 y$ has focus $F$, and its directrix intersects the $y$-axis at point $A$. Point $M$ lies on the parabola $C$. Then the range of $\\frac{M F}{|M A|}$ is?", "fact_expressions": "C: Parabola;M: Point;A: Point;F: Point;Expression(C) = (x^2 = 3*y);Focus(C)=F;Intersection(Directrix(C),yAxis)=A;PointOnCurve(M, C)", "query_expressions": "Range(LineSegmentOf(M, F)/Abs(LineSegmentOf(M, A)))", "answer_expressions": "[sqrt(2)/2,1]", "fact_spans": "[[[7, 26], [46, 52]], [[41, 45]], [[36, 40]], [[3, 6]], [[7, 26]], [[0, 26]], [[7, 40]], [[41, 53]]]", "query_spans": "[[[55, 81]]]", "process": "Draw MN perpendicular to the directrix at point N, $\\frac{|MF|}{|MA|} = \\frac{|MN|}{|MA|} = \\sin\\angle MAN$. Without loss of generality, take point M in the first quadrant. When MA is tangent to the parabola, $\\angle MAN$ is minimized. Let the point of tangency be $M(x_{0}, y_{0})$. Since $A(0, -\\frac{3}{4})$, let the equation of tangent line MA be $y = kx - \\frac{3}{4}$. Combining with $x^{2} = 3y$ gives $x^{2} - 3kx + \\frac{9}{4} = 0$. From $\\Delta = 9k^{2} - 4 \\times \\frac{9}{4} = 0$, we get $k = 1$. Therefore, the slope of tangent $k_{AM} = 1$, so $\\angle MAN = 45^{\\circ}$. Hence, $45^{\\circ} \\leqslant \\angle MAN \\leqslant 90^{\\circ}$, so $\\frac{|MF|}{|MA|} = \\frac{|MN|}{|MA|} = \\sin\\angle MAN \\in \\left[\\frac{\\sqrt{2}}{2}, 1\\right]$. The range of $\\frac{MF}{MA}$ is $\\left[\\frac{\\sqrt{2}}{2}, 1\\right]$." }, { "text": "Given the hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a tangent from the left focus $F_{1}$ of $C_{1}$ to the circle $x^{2}+y^{2}=a^{2}$. Let the point of tangency be $M$. Extend $F_{1}M$ to intersect the parabola $C_{2}$: $y^{2}=2px$ $(p>0)$ at point $N$. It is known that $C_{1}$ and $C_{2}$ share a common focus. If $|M F_{1}|=|M N|$, then the eccentricity of the hyperbola $C_{1}$ is?", "fact_expressions": "C1: Hyperbola;Expression(C1)=(x^2/a^2-y^2/b^2=1);a:Number;b:Number;a>0;b>0;F1:Point;LeftFocus(C1)=F1;H:Circle;M:Point;l:Line;Expression(H)=(x^2+y^2=a^2);TangentOfPoint(F1,H)=l;TangentPoint(l,H)=M;C2:Parabola;Expression(C2)=(y^2=2*p*x);p:Number;p>0;N:Point;Intersection(OverlappingLine(LineSegmentOf(F1,M)),C2)=N;OneOf(Focus(C1))=Focus(C2);Abs(LineSegmentOf(M, F1)) = Abs(LineSegmentOf(M, N))", "query_expressions": "Eccentricity(C1)", "answer_expressions": "(sqrt(5)+1)/2", "fact_spans": "[[[1, 65], [159, 166], [203, 213]], [[1, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[69, 76]], [[1, 76]], [[77, 97]], [[105, 108]], [], [[77, 97]], [[0, 100]], [[0, 108]], [[121, 151], [167, 174]], [[121, 151]], [[133, 151]], [[133, 151]], [[152, 156]], [[109, 156]], [[159, 182]], [[184, 201]]]", "query_spans": "[[[203, 219]]]", "process": "Since the line $ F_{1}M $ is tangent to the circle, the distance from the center of the circle to the line equals the radius, and $ |OF_{1}| = c $, so $ |F_{1}M| = |MN| = b $. Therefore, the slope of the line $ F_{1}M $ is $ \\frac{b}{a} $, hence the equation of the line $ F_{1}M $ is $ y = \\frac{a}{b}(x + c) $. Since $ O $ is the midpoint of $ |F_{1}F_{2}| $, $ |OM| $ is the midline of triangle $ NF_{1}F_{2} $, so $ |NF_{2}| = 2a $ and $ \\angle N = \\frac{\\pi}{2} $. By the area method, we obtain $ \\frac{1}{2} \\times 2c \\times y_{N} = \\frac{1}{2} \\times 2a \\times 2b $, solving gives $ y_{N} = \\frac{2ab}{c} $. Substituting into the equation of line $ F_{1}M $, we find $ x_{N} = \\frac{b^{2} - a^{2}}{c} $, thus $ N\\left( \\frac{b^{2} - a^{2}}{c}, \\frac{2ab}{c} \\right) $. Since the parabola and hyperbola have the same focus, $ \\frac{p}{2} = c $, $ p = 2c $, so the equation of the parabola is $ y^{2} = 4cx $. Substituting the coordinates of point $ N $ into the parabola equation and simplifying yields $ b^{4} - a^{2}b^{2} - a^{4} = 0 $, i.e., $ \\frac{b^{4}}{a^{4}} - \\frac{b^{2}}{a^{2}} - 1 = 0 $, solving gives $ \\frac{b^{2}}{a^{2}} = \\frac{\\sqrt{5} + 1}{2} $, hence the eccentricity of the hyperbola is $ e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{\\frac{3 + \\sqrt{5}}{2}} = \\frac{\\sqrt{5} + 1}{2} $." }, { "text": "Let the endpoints of the minor axis of the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1$ be $B_{1}$, $B_{2}$, and let $F_{1}$ be a focus of the ellipse. Then $\\angle B_{1} F_{1} B_{2}$ = ?", "fact_expressions": "G: Ellipse;B1: Point;F1: Point;B2: Point;Expression(G) = (x^2/12 + y^2/9 = 1);Endpoint(MinorAxis(G)) = {B1, B2};OneOf(Focus(G)) = F1", "query_expressions": "AngleOf(B1, F1, B2)", "answer_expressions": "ApplyUnit(120, degree)", "fact_spans": "[[[1, 39], [69, 71]], [[45, 52]], [[61, 68]], [[53, 60]], [[1, 39]], [[1, 60]], [[61, 76]]]", "query_spans": "[[[78, 106]]]", "process": "As shown in the figure, according to the problem, B_{1}(0,-3), B_{2}(0,3), F_{1}(-\\sqrt{3},0). In Rt\\triangle B_{2}F_{1}O, \\tan\\angle B_{2}F_{1}O = \\frac{|OB_{2}|}{|OF_{1}|} = \\frac{3}{\\sqrt{3}} = \\sqrt{3}, so \\angle B_{2}F_{1}O = 60^{\\circ}. By the symmetry of the ellipse, \\angle B_{1}FB_{2} = 2\\angle B_{2}F_{1}O = 120^{\\circ}." }, { "text": "The left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ are $F_{1}$ and $F_{2}$, respectively. Chord $AB$ passes through $F_{1}$. If the circumference of the incircle of $\\triangle ABF_{2}$ is $2 \\pi$, and the coordinates of points $A$ and $B$ are $(x_{1} , y_{1})$ and $(x_{2} , y_{2})$, respectively, then what is the value of $y_{2}-y_{1}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(F1, LineSegmentOf(A, B));Perimeter(InscribedCircle(TriangleOf(A, B, F2))) = 2*pi;x1: Number;x2: Number;y1: Number;y2: Number;A: Point;B: Point;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2)", "query_expressions": "-y1 + y2", "answer_expressions": "3", "fact_spans": "[[[0, 37]], [[0, 37]], [[46, 53], [68, 75]], [[54, 61]], [[0, 61]], [[0, 61]], [[0, 67]], [[63, 75]], [[77, 110]], [[126, 143]], [[144, 161]], [[126, 143]], [[144, 161]], [[111, 114]], [[115, 118]], [[111, 161]], [[111, 161]]]", "query_spans": "[[[163, 180]]]", "process": "" }, { "text": "If a point $M$ on the parabola $y^{2}=4x$ is at a distance of $4$ from the focus $F$, then what is the abscissa of point $M$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);M: Point;PointOnCurve(M, G);F: Point;Focus(G) = F;Distance(M, F) = 4", "query_expressions": "XCoordinate(M)", "answer_expressions": "3", "fact_spans": "[[[1, 15]], [[1, 15]], [[18, 21], [36, 40]], [[1, 21]], [[24, 27]], [[1, 27]], [[18, 34]]]", "query_spans": "[[[36, 46]]]", "process": "" }, { "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a chord $AB$. If $AF=2FB$, then the equation of the line containing the chord $AB$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, LineSegmentOf(A, B));LineSegmentOf(A, F) = 2*LineSegmentOf(F, B);IsChordOf(LineSegmentOf(A, B), G)", "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A, B)))", "answer_expressions": "y = pm*2*sqrt(2)*(x - 1)", "fact_spans": "[[[1, 15]], [[23, 28]], [[23, 28]], [[18, 21]], [[1, 15]], [[1, 21]], [[0, 28]], [[30, 41]], [[1, 28]]]", "query_spans": "[[[44, 58]]]", "process": "" }, { "text": "Through the point $P(-1,0)$, draw a line intersecting the parabola $y^{2}=8x$ at points $A$ and $B$, such that $2|PA|=|AB|$. Then, what is the distance from point $B$ to the focus of the parabola?", "fact_expressions": "P: Point;Coordinate(P) = (-1, 0);H: Line;PointOnCurve(P, H);G: Parabola;Expression(G) = (y^2 = 8*x);A: Point;B: Point;Intersection(H, G) = {A, B};2*Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(A, B))", "query_expressions": "Distance(B, Focus(G))", "answer_expressions": "5", "fact_spans": "[[[1, 11]], [[1, 11]], [[12, 14]], [[0, 14]], [[15, 29], [65, 68]], [[15, 29]], [[32, 35]], [[36, 39], [59, 63]], [[12, 41]], [[43, 57]]]", "query_spans": "[[[59, 75]]]", "process": "From the given conditions, the equation of the line passing through point P(-1,0) is y = k(x+1). Substituting into the parabola equation and simplifying yields k^{2}x^{2} + (2k^{2}-8)x + k^{2} = 0. Let the intersection points be A(x_{1},y_{1}), B(x_{2},y_{2}). From the given condition \\overrightarrow{AP} = 3\\overrightarrow{PB}, we have x_{2}+1 = 3(x_{1}+1), so x_{2} = 3x_{1}+2. Substituting into x_{1}x_{2} = 1 gives x_{1}(3x_{1}+2) = 1, or 3x_{1}^{2} + 2x_{1} - 1 = 0. Solving yields x_{1} = \\frac{1}{3}, x_{1} = -1 (discarded). Then x_{2} = 1+2 = 3. Hence |BF| = x_{2}+2 = 5. The answer is 5." }, { "text": "Given $A(2,0)$, and point $B$ lies on the parabola $y^{2}=x$, then the minimum value of $|AB|$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = x);PointOnCurve(B,G);Coordinate(A)=(2,0)", "query_expressions": "Min(Abs(LineSegmentOf(A, B)))", "answer_expressions": "sqrt(7)/2", "fact_spans": "[[[17, 29]], [[2, 10]], [[13, 16]], [[17, 29]], [[13, 33]], [[2, 10]]]", "query_spans": "[[[35, 48]]]", "process": "Let point B(x, y), then x = y² ≥ 0. Since |AB| = √((x−2)² + y²) = √((x−2)² + x) = √(x² − 3x + 4) = √((x−3/2)² + 7/4), therefore when x = 3/2, |AB| attains the minimum value, and |AB|ₘᵢₙ = √7/2 √7/(7/4)" }, { "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 $F_{1}$ and $F_{2}$ respectively. A line $l_{1}$ passing through $F_{1}$ with an inclination angle of $30^{\\circ}$ intersects a line $l_{2}$ passing through $F_{2}$ at point $P$, where $P$ lies on the ellipse and $\\angle F_{1} P F_{2}=90^{\\circ}$. Find the eccentricity $e$ of the ellipse $C$.", "fact_expressions": "C: Ellipse;b: Number;a: Number;l1: Line;l2: Line;F1: Point;P: Point;F2: Point;e: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, l1);Inclination(l1) = ApplyUnit(30, degree);PointOnCurve(F2, l2);Intersection(l1, l2) = P;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(90, degree);Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "sqrt(3) - 1", "fact_spans": "[[[2, 59], [149, 151], [189, 194]], [[9, 59]], [[9, 59]], [[109, 118]], [[128, 137]], [[67, 74], [84, 91]], [[144, 148], [139, 143]], [[120, 127], [75, 82]], [[198, 201]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 82]], [[2, 82]], [[83, 118]], [[92, 118]], [[119, 137]], [[109, 143]], [[144, 152]], [[154, 187]], [[189, 201]]]", "query_spans": "[[[198, 203]]]", "process": "From the given conditions, we know that $\\triangle PF_{1}F_{2}$ is a right triangle and $\\angle PF_{1}F_{2}=30^{\\circ}$. Therefore, $PF_{2}=c$, $F_{1}F_{2}=2c$, so $PF_{1}=\\sqrt{3}c$. Since $PF_{1}+PF_{2}=2a$, that is, $\\sqrt{3}c+c=2a$, thus $e=\\sqrt{3}-1$." }, { "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 $B$. Then, what is the maximum value of the perimeter of $\\triangle F A B$?", "fact_expressions": "G: Ellipse;a: Number;H: Line;m: Number;F: Point;A: Point;B: Point;a>0;Expression(G) = (x^2/(4*a^2) + y^2/(3*a^2) = 1);Expression(H) = (x = m);LeftFocus(G) = F;Intersection(H, G) = {A, B}", "query_expressions": "Max(Perimeter(TriangleOf(F, A, B)))", "answer_expressions": "8*a", "fact_spans": "[[[0, 54], [71, 73]], [[2, 54]], [[63, 70]], [[65, 70]], [[59, 62]], [[76, 80]], [[81, 84]], [[2, 54]], [[0, 54]], [[63, 70]], [[0, 62]], [[63, 84]]]", "query_spans": "[[[86, 112]]]", "process": "As shown in the figure, let the right focus of the ellipse be M, and the major axis of the ellipse be $2 \\times 2a$. The perimeter of $\\triangle FAB$, $AF + FB + AB \\leqslant FA + AM + FB + BM = 2$>" }, { "text": "A hyperbola shares the same asymptotes as $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $A(-3, 2\\sqrt{3})$. The distance from one focus of this hyperbola to one of its asymptotes is?", "fact_expressions": "G: Hyperbola;A: Point;C:Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (-3, 2*sqrt(3));Asymptote(G)=Asymptote(C);PointOnCurve(A,C)", "query_expressions": "Distance(OneOf(Focus(C)),OneOf(Asymptote(C)))", "answer_expressions": "2", "fact_spans": "[[[1, 40]], [[50, 69]], [[70, 73]], [[1, 40]], [[50, 69]], [[0, 73]], [[48, 73]]]", "query_spans": "[[[70, 89]]]", "process": "According to the problem, let the hyperbola equation be $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=\\lambda$. Substituting the point $(-3,2\\sqrt{3})$ into the hyperbola equation gives $\\lambda=\\frac{1}{4}$. Therefore, the hyperbola equation passing through point $A(-3,2\\sqrt{3})$ is: $\\frac{4x^{2}}{9}-\\frac{y^{2}}{4}=1$. Hence, a focus coordinate of $\\frac{4x^{2}}{9}-\\frac{y^{2}}{4}=1$ is $(\\frac{5}{2},0)$, and an asymptote equation is $y=\\frac{4}{3}x$, that is, $4x-3y=0$. Therefore, the distance from the focus to an asymptote is $\\frac{10}{\\sqrt{9+16}}=2$. The answer is: $2$." }, { "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0)$ are $y=\\pm \\frac{4}{3} x$, 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(Asymptote(G)) = (y = pm*(4/3)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/3", "fact_spans": "[[[2, 60], [91, 94]], [[5, 60]], [[5, 60]], [[5, 60]], [[5, 60]], [[2, 60]], [[2, 88]]]", "query_spans": "[[[91, 100]]]", "process": "" }, { "text": "Given that the tangent line $l$ at point $A(x_{0}, y_{0})$ on the parabola $y^{2}=2 p x$ ($p>0$) is tangent to the circle $M$: $(x+2)^{2}+y^{2}=4$ at another point $B$, find the minimum value of the distance $|A F|$ between the focus $F$ of the parabola and the point of tangency $A$.", "fact_expressions": "G: Parabola;p: Number;M: Circle;A: Point;B:Point;F: Point;x0:Number;y0:Number;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(M) = (y^2 + (x + 2)^2 = 4);Coordinate(A) = (x0, y0);PointOnCurve(A,G);TangentOnPoint(A,G)=l;TangentPoint(l,M)=B;Focus(G)=F;Distance(F,A)=Abs(LineSegmentOf(A,F));l:Line", "query_expressions": "Min(Abs(LineSegmentOf(A,F)))", "answer_expressions": "8", "fact_spans": "[[[2, 23], [86, 89]], [[5, 23]], [[51, 75]], [[26, 43], [97, 100]], [[81, 84]], [[91, 94]], [[26, 43]], [[26, 43]], [[5, 23]], [[2, 23]], [[51, 75]], [[26, 43]], [[2, 43]], [[2, 50]], [[47, 84]], [[86, 94]], [[91, 109]], [[47, 50]]]", "query_spans": "[[[102, 115]]]", "process": "Since both the parabola and the circle are symmetric about the x-axis, assume point A is above the x-axis, then y = \\sqrt{2px}, y' = \\frac{\\sqrt{2p}}{2\\sqrt{x}}, y_{0}^{2} = 2px_{0}. The slope of the tangent line at point A is k = y'|_{x=x_{0}} = \\frac{\\sqrt{2p}}{2\\sqrt{x_{0}}} = \\frac{p}{y_{0}}. The equation of the tangent line l is: y - y_{0} = \\frac{p}{y_{0}}(x - x_{0}), i.e., l: y_{0}y - px - \\frac{y_{0}^{2}}{2} = 0. Since the tangent line l is tangent to the circle M: (x + 2)^{2} + y^{2} = 4 at another point B, we have: \\frac{|2p - \\frac{y_{0}^{2}}{2}|}{\\sqrt{p^{2} + y_{0}^{2}}} = 2. Squaring both sides and simplifying yields y_{0}^{4} - (16 + 8p)y_{0}^{2} = 0. Solving gives y_{0}^{2} = 0 or y_{0}^{2} = 8p + 16. When y_{0}^{2} = 0, points A and B coincide, which is invalid; thus y_{0}^{2} = 8p + 16, x_{0} = \\frac{y_{0}^{2}}{2p} = \\frac{8p + 16}{2p} = 4 + \\frac{8}{p}. Therefore, |AF| = \\frac{p}{2} + x_{0} = \\frac{p}{2} + 4 + \\frac{8}{p} \\geqslant 2\\sqrt{\\frac{p}{2} \\cdot \\frac{8}{p}} + 4 = 8, with equality if and only if \\frac{p}{2} = \\frac{8}{p}, i.e., p = 4, where the minimum value is attained." }, { "text": "The coordinates of the focus of the parabola $C$: $y=-\\frac{x^{2}}{8}$ are?", "fact_expressions": "C: Parabola;Expression(C) = (y = -x^2/8)", "query_expressions": "Coordinate(Focus(C))", "answer_expressions": "(0, -2)", "fact_spans": "[[[0, 27]], [[0, 27]]]", "query_spans": "[[[0, 34]]]", "process": "" }, { "text": "If a line passing through the point $P(8,1)$ intersects the hyperbola $x^{2}-4 y^{2}=4$ at points $A$ and $B$, and $P$ is the midpoint of segment $AB$, then the equation of line $AB$ is?", "fact_expressions": "G: Hyperbola;H: Line;B: Point;A: Point;P: Point;Expression(G) = (x^2 - 4*y^2 = 4);Coordinate(P) = (8, 1);PointOnCurve(P,H);Intersection(H,G) = {A, B};MidPoint(LineSegmentOf(A,B))=P", "query_expressions": "Expression(LineOf(A,B))", "answer_expressions": "2*x-y-15=0", "fact_spans": "[[[15, 35]], [[12, 14]], [[42, 45]], [[38, 41]], [[2, 11], [49, 52]], [[15, 35]], [[2, 11]], [[1, 14]], [[12, 47]], [[49, 63]]]", "query_spans": "[[[65, 77]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=16, y_{1}+y_{2}=2. Since x_{1}^{2}-4y_{1}=4, x_{2}^{2}-4y_{2}=4, \\therefore (x_{1}+x_{2})(x_{1}-x_{2})-4(y_{1}+y_{2})(y_{1}-y_{2})=0, \\therefore 16(x_{1}-x_{2})-8(y_{1}-y_{2})=0, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{16}{8}=2, \\therefore k_{AB}=2. \\therefore The equation of the line is y-1=2(x-8), i.e., 2x-y-15=0." }, { "text": "Given the point $A(2,2)$ on the parabola $x=2 p y^{2}$, then the distance from $A$ to the directrix is?", "fact_expressions": "G: Parabola;p: Number;A: Point;Expression(G) = (x = 2*(p*y^2));Coordinate(A) = (2, 2);PointOnCurve(A,G)", "query_expressions": "Distance(A, Directrix(G))", "answer_expressions": "5/2", "fact_spans": "[[[2, 18]], [[5, 18]], [[20, 29], [31, 34]], [[2, 18]], [[20, 29]], [0, 29]]", "query_spans": "[[[31, 42]]]", "process": "From the point A(2,2) on the parabola x=2py^{2}, we get p=\\frac{1}{4}, so the equation of the parabola is y^{2}=2x. The equation of the directrix is x=-\\frac{1}{2}, then the distance from A to the directrix is \\frac{5}{2}." }, { "text": "Let $O$ be the origin, the parabola $C$: $y^{2}=4x$ has directrix $l$ and focus $F$. A line passing through $F$ with slope $\\sqrt{3}$ intersects the parabola $C$ at points $A$ and $B$, with $|AF| > |BF|$. If the line $AO$ intersects $l$ at $D$, then $\\frac{|OF|}{|BD|}=$?", "fact_expressions": "C: Parabola;G: Line;l: Line;A: Point;O: Origin;F: Point;B: Point;D: Point;Expression(C) = (y^2 = 4*x);Directrix(C) = l;Focus(C) = F;PointOnCurve(F, G);Slope(G) = sqrt(3);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) > Abs(LineSegmentOf(B, F));Intersection(LineOf(A,O), l) = D", "query_expressions": "Abs(LineSegmentOf(O, F))/Abs(LineSegmentOf(B, D))", "answer_expressions": "3/4", "fact_spans": "[[[10, 29], [66, 72]], [[63, 65]], [[33, 36], [108, 111]], [[74, 77]], [[1, 4]], [[40, 43], [45, 48]], [[78, 81]], [[114, 117]], [[10, 29]], [[10, 36]], [[10, 43]], [[44, 65]], [[49, 65]], [[63, 83]], [[85, 98]], [[100, 117]]]", "query_spans": "[[[119, 142]]]", "process": "The line passing through F with slope $\\sqrt{3}$ has equation $y=\\sqrt{3}(x-1)$. Combined with the parabola $C: y^2=4x$, it yields $x^{2}-10x+3=0$. $A(3,2\\sqrt{3})$, $B\\left(\\frac{1}{3},\\frac{-2\\sqrt{3}}{3}\\right)$. Then the equation of line AO is $y=\\frac{3\\sqrt{3}}{3}x$, and its intersection point with $l: x=-1$ is $D\\left(-1,-\\frac{2\\sqrt{3}}{3}\\right)$. Therefore, $\\frac{|OF|}{BD}=\\frac{1}{4}=\\frac{3}{4}$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$. If $\\angle A F_{2} B=60^{\\circ}$, then the inradius of $\\triangle A F_{2} B$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;L: Line;PointOnCurve(F1, L);A: Point;B: Point;Intersection(L, LeftPart(G)) = {A, B};AngleOf(A, F2, B) = ApplyUnit(60, degree)", "query_expressions": "Radius(InscribedCircle(TriangleOf(A, F2, B)))", "answer_expressions": "(4*sqrt(3))/3", "fact_spans": "[[[2, 40], [77, 80]], [[2, 40]], [[49, 56], [66, 73]], [[57, 64]], [[2, 64]], [[2, 64]], [[74, 76]], [[65, 76]], [[85, 88]], [[89, 92]], [[74, 94]], [[96, 125]]]", "query_spans": "[[[127, 156]]]", "process": "Let the incenter be M, and let the points where the incircle of $\\triangle AF_{2}B$ touches its sides be connected to the center M by line segments. By the properties of the incircle, we have $|AF_{2}| - |AQ| = |BF_{2}| - |BQ|$. Furthermore, by the definition of a hyperbola: $|AF_{2}| - |AF_{1}| = |BF_{2}| - |BF_{1}| = 2a$, it follows that Q and $F_{1}$ coincide. Then, given $\\angle AF_{2}B = 60^{\\circ}$, we can find the radius of the incircle. Let the incenter be $M(x, y)$, and let the incircle touch the sides of the triangle at points T, Q, S respectively, as shown in the figure. Connect MS, MT, MQ. By the properties of the incircle, we obtain: $|F_{2}T| = |F_{2}S|$, $|AT| = |AQ|$, $|BS| = |BQ|$. Therefore, $|AF_{2}| - |AQ| = |AF_{2}| - |AT| = |F_{2}T|$, and $|BF_{2}| - |BQ| = |BF_{2}| - |BS| = |F_{2}S|$, so $|AF_{2}| - |AQ| = |BF_{2}| - |BQ|$. By the definition of the hyperbola: $|AF_{2}| - |AF_{1}| = |BF_{2}| - |BF_{1}| = 2a$, thus Q and $F_{1}$ coincide. Hence $|TF_{2}| = 2a = 4$, so the radius of the circle is $r = |MT| = |TF_{2}|\\tan\\frac{\\angle AF_{2}B}{2} = \\frac{4\\sqrt{3}}{3}$." }, { "text": "The endpoints of a line segment $AB$ of fixed length $3$ move along the parabola $y^2 = x$, and $M$ is the midpoint of segment $AB$. Then the minimum distance from $M$ to the $y$-axis is?", "fact_expressions": "Length(LineSegmentOf(A, B)) = 3;A: Point;B: Point;Endpoint(LineSegmentOf(A, B)) = {A, B};PointOnCurve(A, G);PointOnCurve(B, G);G: Parabola;Expression(G) = (y^2 = x);M: Point;MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Min(Distance(M, yAxis))", "answer_expressions": "5/4", "fact_spans": "[[[0, 14]], [[9, 14]], [[9, 14]], [[7, 18]], [[7, 34]], [[7, 34]], [[19, 31]], [[19, 31]], [[35, 38], [51, 54]], [[35, 49]]]", "query_spans": "[[[51, 67]]]", "process": "By the definition of a parabola, we can obtain |AF| + |BF| = x_{1} + x_{2} + \\frac{1}{2}. Using |AF| + |BF| \\geqslant |AB|, we can find the minimum value of x_{1} + x_{2}. Since the distance we are seeking is \\frac{x_{1} + x_{2}}{2}, we can determine the minimum value of the required distance. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}). From the equation of the parabola, the focus is F(\\frac{1}{4}, 0) and the directrix is x = -\\frac{1}{4}. By the definition of a parabola, |AF| = x_{1} + \\frac{1}{4}, |BF| = x_{2} + \\frac{1}{4}. Since |AF| + |BF| \\geqslant |AB| = 3 (equality holds if and only if points A, F, B are collinear), it follows that x_{1} + x_{2} + \\frac{1}{2} \\geqslant 3, solving gives x_{1} + x_{2} \\geqslant \\frac{5}{2}. Since the distance from the midpoint M of AB to the y-axis is d = \\frac{x_{1} + x_{2}}{2} \\geqslant \\frac{5}{4}, the required minimum value is \\frac{5}{4}." }, { "text": "Given that the eccentricity of the ellipse $y^{2}+m x^{2}=1$ $(m>0)$ is $\\frac{1}{2}$, then $m=$?", "fact_expressions": "G: Ellipse;m: Number;m>0;Expression(G) = (m*x^2 + y^2 = 1);Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "{3/4,4/3}", "fact_spans": "[[[2, 26]], [[46, 49]], [[4, 26]], [[2, 26]], [[2, 44]]]", "query_spans": "[[[46, 51]]]", "process": "Transform the ellipse equation y^{2}+mx^{2}=1 (m>0) into standard form: y^{2}+\\frac{x^{2}}{m}=1. If \\frac{1}{m}>1, that is, 01, then the eccentricity of the ellipse is e=\\frac{c}{a}=\\frac{\\sqrt{1-\\frac{1}{m}}}{1}=\\frac{1}{2}, solving gives: m=\\frac{4}{3}." }, { "text": "Given that points $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, and point $A$ is an intersection point of the line $x=\\frac{4}{3}a$ and the hyperbola $C$. If point $A$ lies on the circle with diameter $F_{1}F_{2}$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;b: Number;a: Number;a>0;b>0;A: Point;H: Line;Expression(H) = (x = (4/3)*a);OneOf(Intersection(H,C)) = A;G: Circle;IsDiameter(LineSegmentOf(F1, F2), G);PointOnCurve(A, G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "3*sqrt(2)/2", "fact_spans": "[[[21, 82], [113, 119], [153, 159]], [[21, 82]], [[2, 10]], [[11, 18]], [[2, 88]], [[2, 88]], [[29, 82]], [[29, 82]], [[29, 82]], [[29, 82]], [[89, 92], [126, 130]], [[93, 112]], [[93, 112]], [[89, 124]], [[149, 150]], [[131, 150]], [[126, 151]]]", "query_spans": "[[[153, 165]]]", "process": "Find the point A(\\frac{4a}{3},y), then since point A lies on the circle with diameter F_{1}F_{2}, we have F_{1}A\\bot F_{2}A; next, using the dot product of vectors equal to 0, obtain an equation in terms of a and c, thereby finding the eccentricity. Let A(\\frac{4a}{3},y), substitute into \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 and simplify to get y^{2}=\\frac{7}{9}b^{2}. From the given condition F_{1}A\\bot F_{2}A, it follows that \\overrightarrow{F_{1}A}\\cdot\\overrightarrow{F_{2}A}=0. Since \\overrightarrow{F_{1}A}=(\\frac{4a}{3}+c,y), \\overrightarrow{F_{2}A}=(\\frac{4a}{3}-c,y), then (\\frac{4}{3}a+c)(\\frac{4}{3}a-c)+y^{2}=(\\frac{4}{3}a+c)(\\frac{4}{3}a-c)+\\frac{7}{9}b^{2}=0. Also, a^{2}+b^{2}=c^{2}, simplifying yields: a^{2}=\\frac{2}{9}c^{2} \\Rightarrow \\frac{c^{2}}{a^{2}}=\\frac{9}{2} \\Rightarrow e=\\frac{3\\sqrt{2}}{2}." }, { "text": "Given that one asymptote of the hyperbola $x^{2}+m y^{2}=1$ is $y=2 x$, then the focal length of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (m*y^2 + x^2 = 1);m: Number;Expression(OneOf(Asymptote(G))) = (y = 2*x)", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 22], [41, 44]], [[2, 22]], [[5, 22]], [[2, 38]]]", "query_spans": "[[[41, 49]]]", "process": "Since $ x^{2} + m y^{2} = 1 $ is a hyperbola, we have $ a^{2} = 1 $, $ b^{2} = -\\frac{1}{m} $. Also, since one asymptote of the hyperbola $ x^{2} + m y^{2} = 1 $ is $ y = 2x $, it follows that $ \\frac{b^{2}}{a^{2}} = -\\frac{1}{m} = 2^{2} = 4 $, so $ b^{2} = 4 $. Therefore, $ c^{2} = a^{2} + b^{2} = 5 $, $ c = \\sqrt{5} $, hence the focal distance $ 2c = 2\\sqrt{5} $. The answer is $ 2\\sqrt{5} $." }, { "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ passing through the focus $F$ intersects the parabola $C$ at points $A$ and $B$. If the projections on the $y$-axis of the midpoints of segments $AF$ and $BF$ are $P$ and $Q$ respectively, and $|PQ|=4$, then the equation of line $l$ is?", "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};P: Point;Q: Point;Projection(MidPoint(LineSegmentOf(A, F)), yAxis) = P;Projection(MidPoint(LineSegmentOf(B, F)), yAxis) = Q;Abs(LineSegmentOf(P, Q)) = 4", "query_expressions": "Expression(l)", "answer_expressions": "{y=((sqrt(3)/3)*x)-sqrt(3)/3, y=((-sqrt(3)/3)*x)+sqrt(3)/3}", "fact_spans": "[[[2, 21], [35, 41]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 34], [103, 108]], [[22, 34]], [[43, 46]], [[47, 50]], [[29, 52]], [[83, 86]], [[87, 90]], [[54, 90]], [[54, 90]], [[92, 101]]]", "query_spans": "[[[103, 113]]]", "process": "\\because parabola C: y^{2} = 4x, \\therefore the focus of the parabola is F(1,0). Let A(x_{1},y_{1}), B(x_{2},y_{2}). Let the equation of line l be x = my + 1. Solving the system of equations:\n\\begin{cases}\nx = my + 1 \\\\\ny^{2} = 4x\n\\end{cases}\nSimplifying and rearranging yields y^{2} - 4my - 4 = 0. By Vieta's formulas, we get y_{1} + y_{2} = 4m, y_{1} \\cdot y_{2} = -4. \\because the projections on the y-axis of the midpoints of segments AF and BF are P and Q respectively, \\therefore P(0, \\frac{1}{2}y_{1}), Q(0, \\frac{1}{2}y_{2}), \\therefore |PQ| = \\frac{1}{2}|y_{1} - y_{2}| = \\frac{1}{2}\\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\frac{1}{2}\\sqrt{16m^{2} + 16} = 4. Solving gives m = \\pm\\sqrt{3}. Hence, the equation of line l is y = \\frac{\\sqrt{3}}{3}x - \\frac{\\sqrt{3}}{3} or y = -\\frac{\\sqrt{3}}{3}x + \\frac{\\sqrt{3}}{3}" }, { "text": "Given that the directrix of the parabola $y^{2}=8 x$ passes through a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and the eccentricity of the hyperbola is $2$, then the equation of the 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);PointOnCurve(OneOf(Focus(G)), Directrix(H));Eccentricity(G) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[20, 76], [84, 87], [99, 102]], [[23, 76]], [[23, 76]], [[2, 16]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 16]], [[2, 81]], [[84, 95]]]", "query_spans": "[[[99, 107]]]", "process": "The directrix of $ y^{2} = 8x $ is $ x = -2 $. Then in the hyperbola, $ c = 2 $, $ \\frac{c}{a} = \\frac{2}{a} = 2 $, $ a = 1 $, $ b = \\sqrt{3} $, so the equation of the hyperbola is $ x^{2} - \\frac{y^{2}}{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 $F_{1}$ and $F_{2}$ respectively. A tangent line drawn from point $F_{1}$ to the circle $x^{2}+y^{2}=a^{2}$ intersects the right branch of the hyperbola at point $M$. If $\\angle F_{1} M F_{2}=\\frac{\\pi}{4}$, 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 > b;b > 0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;H: Circle;Expression(H) = (x^2 + y^2 = a^2);M: Point;AngleOf(F1, M, F2) = pi/4;L:Line;TangentOfPoint(F1,H)=L;Intersection(L,RightPart(G))=M", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 55], [114, 117], [164, 167]], [[2, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[64, 71], [81, 89]], [[72, 79]], [[2, 79]], [[2, 79]], [[90, 110]], [[90, 110]], [[120, 124]], [[126, 162]], [-1, -1], [78, 110], [78, 121]]", "query_spans": "[[[164, 173]]]", "process": "Let the point of tangency be N, connect ON, and draw F_{2}A\\botMN from F_{2}, with foot at A, as shown in the figure: by the tangent property of a circle, we have ON\\botF_{1}M, ON=a; by the midline theorem of a triangle, we obtain AF_{2}=2a, AF_{2}\\botF_{1}M. In right triangle AF_{1}F_{2}, AF_{1}=\\sqrt{F_{1}F_{2}^{2}-AF_{2}^{2}}=2b. In right triangle AF_{2}M, \\angleF_{1}MF_{2}=\\frac{\\pi}{4}, so MA=2a, F_{2}M=2\\sqrt{2}a. By the definition of a hyperbola: F_{1}M-F_{2}M=2a, that is, 2b+2a-2\\sqrt{2}a=2a, thus b=\\sqrt{2}a. Since c=\\sqrt{a^{2}+b^{2}}, we have c=\\sqrt{a^{2}+b^{2}}=\\sqrt{3}a. Therefore, e=\\frac{c}{a}=\\sqrt{3}, that is, the eccentricity of the hyperbola is \\sqrt{3}." }, { "text": "Given point $A(0 , 2)$, the focus of the parabola $y^{2}=2 px(p>0)$ is $F$, the directrix is $l$, the line segment $FA$ intersects the parabola at point $B$, from $B$ a perpendicular is drawn to $l$, with foot of perpendicular at $M$. If $A M \\perp M F$, then $p$=?", "fact_expressions": "A: Point;Coordinate(A) = (0, 2);G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;B: Point;Intersection(LineSegmentOf(F, A), G) = B;Z: Line;PointOnCurve(B, Z);IsPerpendicular(Z, l);M: Point;FootPoint(Z, l) = M;IsPerpendicular(LineSegmentOf(A, M), LineSegmentOf(M, F))", "query_expressions": "p", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 13]], [[2, 13]], [[13, 33], [55, 58]], [[13, 33]], [[103, 106]], [[16, 33]], [[37, 40]], [[13, 40]], [[44, 47], [69, 72]], [[13, 47]], [[59, 63], [65, 68]], [[48, 63]], [], [[64, 75]], [[64, 75]], [[79, 82]], [[64, 82]], [[85, 100]]]", "query_spans": "[[[103, 108]]]", "process": "" }, { "text": "Given that $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the circle with diameter $OF$ ($O$ being the origin) intersects the hyperbola at a point $A$, if the angle of inclination of $OA$ is $\\frac{\\pi}{6}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;O: Origin;F: Point;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;IsDiameter(LineSegmentOf(O, F), H);OneOf(Intersection(H, G)) = A;Inclination(LineSegmentOf(O, A)) = pi/6", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(13) + 1)/3", "fact_spans": "[[[6, 62], [90, 93], [132, 135]], [[9, 62]], [[9, 62]], [[88, 89]], [[74, 77]], [[2, 5]], [[99, 102]], [[9, 62]], [[9, 62]], [[6, 62]], [[2, 66]], [[67, 89]], [[88, 102]], [[104, 129]]]", "query_spans": "[[[132, 141]]]", "process": "Let $ F_{1} $ be the left focus of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $, then $ |AF| = \\frac{c}{2} $, $ |AF_{1}| = 2a + \\frac{c}{2} $. In triangle $ AF_{1}F $, $ \\angle OFA = \\frac{\\pi}{3} $, by the cosine law we get $ \\left(2a + \\frac{c}{2}\\right)^{2} = (2c)^{2} + \\left(\\frac{c}{2}\\right)^{2} - 2 \\cdot 2c \\cdot \\frac{c}{2} \\cdot \\cos\\frac{\\pi}{3} $, then $ 2a + \\frac{c}{2} = \\frac{\\sqrt{13}c}{2} $, i.e., $ e = \\frac{c}{a} = \\frac{\\sqrt{13} + 1}{b} $." }, { "text": "The line passing through the right focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ and perpendicular to the $x$-axis intersects the two asymptotes of this hyperbola at points $A$ and $B$. Then the length of $AB$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);H: Line;PointOnCurve(RightFocus(G), H);IsPerpendicular(H, xAxis);A: Point;B: Point;L1: Line;L2: Line;Asymptote(G) = {L1, L2};Intersection(H, L1) = A;Intersection(H, L2) = B", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[1, 29], [47, 50]], [[1, 29]], [[42, 44]], [[0, 44]], [[34, 44]], [[57, 60]], [[61, 64]], [], [], [[47, 56]], [[42, 66]], [[42, 66]]]", "query_spans": "[[[68, 78]]]", "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ are $y=\\pm\\frac{\\sqrt{3}}{3}x$. The right focus is $(2,0)$. Then the points are $A(2,\\frac{2\\sqrt{3}}{3})$, $B(2,-\\frac{2\\sqrt{3}}{3})$. Therefore, $|AB|=\\frac{2\\sqrt{3}}{3}-(-\\frac{2\\sqrt{3}}{3})=\\frac{4\\sqrt{3}}{3}$." }, { "text": "Given that point $P$ is any point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ except the vertices, $F_{1}$ and $F_{2}$ are the left and right foci respectively, $c$ is the semi-focal length, and the incircle of $\\angle P F_{1} F_{2}$ touches $F_{1} F_{2}$ at point $M$, then $|F_{1} M| \\cdot|F_{2} M|$=?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;M: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Negation(P=Vertex(G));LeftFocus(G) = F1;RightFocus(G) = F2;HalfFocalLength(G) = c;c:Number;TangentPoint(InscribedCircle(TriangleOf(P,F1,F2)),LineSegmentOf(F1,F2))=M", "query_expressions": "Abs(LineSegmentOf(F1, M))*Abs(LineSegmentOf(F2, M))", "answer_expressions": "b^2", "fact_spans": "[[[7, 53]], [[10, 53]], [[10, 53]], [[64, 71]], [[72, 79]], [[2, 6]], [[138, 142]], [[7, 53]], [[2, 63]], [[2, 63]], [[7, 87]], [[7, 87]], [[7, 95]], [[88, 91]], [[96, 142]]]", "query_spans": "[[[144, 172]]]", "process": "Let the point of tangency between the circle and $ PF_{1} $ be point $ S $, and the point of tangency with $ PF_{2} $ be point $ T $. According to the property that the lengths of two tangent segments drawn from an external point to a circle are equal, we have: $ |F_{1}M| = |F_{1}S| $, $ |F_{2}M| = |F_{2}T| $, $ |PS| = |PT| $\\textcircled{1}. When $ P $ is on the right branch of the hyperbola, by the definition of a hyperbola, we know $ |F_{1}M| - |F_{2}M| = |F_{1}P| - |F_{2}P| = 2a $\\textcircled{1}; and since $ |F_{1}M| + |MF_{2}| = |F_{1}F_{2}| = 2c $\\textcircled{2}, solving \\textcircled{1} and \\textcircled{2} together gives: $ |F_{1}M| = a + c $, $ |F_{2}M| = c - a $, so $ |F_{1}M| \\cdot |F_{2}M| = (a + c)(c - a) = c^{2} - a^{2} = b^{2} $; \\textcircled{2} When $ P $ is on the left branch of the hyperbola, by the definition of a hyperbola, we know $ |F_{2}M| - |F_{1}M| = |F_{2}P| - |F_{1}P| = 2a $\\textcircled{3}; and since $ |F_{1}M| + |MF_{2}| = |F_{1}F_{2}| = 2c $\\textcircled{4}, solving \\textcircled{3} and \\textcircled{4} together gives: $ |F_{2}M| = a + c $, $ |F_{1}M| = c - a $, $ |F_{1}M| \\cdot |F_{2}M| = (a + c)(c - a) = c^{2} - a^{2} = b^{2} $. In summary, $ |F_{1}M| \\cdot |F_{2}M| = b^{2} $." }, { "text": "Given that point $P(4,4)$ lies on the parabola $C$: $y^{2}=2 p x$, $F$ is the focus of parabola $C$, and fixed point $M(-1,4)$, then the area of the circumcircle of $\\Delta M P F$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;P: Point;Coordinate(P) = (4, 4);PointOnCurve(P, C);F: Point;Focus(C) = F;M: Point;Coordinate(M) = (-1, 4)", "query_expressions": "Area(CircumCircle(TriangleOf(M, P, F)))", "answer_expressions": "(125*pi)/16", "fact_spans": "[[[11, 32], [41, 47]], [[11, 32]], [[19, 32]], [[2, 10]], [[2, 10]], [[2, 36]], [[37, 40]], [[37, 50]], [[53, 62]], [[53, 62]]]", "query_spans": "[[[64, 87]]]", "process": "Point P(4,4) is a point on the parabola C: y^{2}=2px, so 16=8p. Solving gives p=2, thus the equation of the parabola is y^{2}=4x. Given F(1,0), M(-1,4), P(4,4), we have MP=5, PF=5, MF=2\\sqrt{5}\\cos\\angleMPF=\\frac{5^{2}+5^{2}-(2\\sqrt{5})^{2}}{2\\times5\\times5}=\\frac{3}{5}, then \\sin\\angleMPF=\\sqrt{1-\\frac{9}{25}}=\\frac{4}{5}. Let R be the circumradius of \\triangleMPF, then 2R=\\frac{2\\sqrt{5}}{4}=\\frac{5\\sqrt{5}}{2}, solving gives R=\\frac{5}{5}\\sqrt{5}. Thus, the area of the circumcircle of \\triangleMPF is \\pi\\times\\frac{125}{6}=\\frac{125\\pi}{6}" }, { "text": "If point $Q(4,1)$ is the midpoint of chord $AB$ of the parabola $y^{2}=8x$, then the equation of line $AB$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;Q: Point;Expression(G) = (y^2 = 8*x);IsChordOf(LineSegmentOf(A,B),G);Coordinate(Q) = (4, 1);MidPoint(LineSegmentOf(A,B))=Q", "query_expressions": "Expression(LineOf(A, B))", "answer_expressions": "4*x - y - 15 = 0", "fact_spans": "[[[11, 25]], [[39, 44]], [[39, 44]], [[1, 10]], [[11, 25]], [[11, 32]], [[1, 10]], [[1, 35]]]", "query_spans": "[[[37, 49]]]", "process": "Since the midpoint Q(4,1) is given, the slope of the line exists. Using the point difference method, the slope is found, and then the equation of line AB can be determined. According to the problem, when AB is perpendicular to the x-axis, it does not satisfy the condition, so the slope of line AB exists. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}=8x_{1}\\textcircled{1}, y_{2}=8x_{2}\\textcircled{2} and x_{1}+x_{2}=8, y_{1}+y_{2}=2. \\textcircled{1}-\\textcircled{2} gives (y_{1}+y_{2})(y_{1}-y_{2})=8(x_{1}-x_{2}), that is, 2(y_{1}-y_{2})=8(x_{1}-x_{2}), that is, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=4. Hence, the slope of line AB is k=4. Therefore, the equation of line AB is y=4(x-4)+1, that is, 4x-y-15=0." }, { "text": "Given that $O$ is the coordinate origin, the parabola $C$: $y^{2}=2 p x$ ($p>0$) has focus $F$, $P$ is a point on $C$, $PF$ is perpendicular to the $x$-axis, $Q$ is a point on the $x$-axis, and $PQ \\perp OP$. If $|FQ|=6$, then the equation of the directrix of $C$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;P: Point;F: Point;Q: Point;O: Origin;p>0;Focus(C)=F;PointOnCurve(P,C);IsPerpendicular(LineSegmentOf(P,F),xAxis);PointOnCurve(Q, xAxis);IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(O, P));Abs(LineSegmentOf(F, Q)) = 6", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "x=-3/2", "fact_spans": "[[[11, 36], [48, 51], [110, 113]], [[11, 36]], [[18, 36]], [[44, 47]], [[40, 43]], [[68, 71]], [[2, 5]], [[18, 36]], [[11, 43]], [[44, 54]], [[55, 67]], [[68, 79]], [[81, 96]], [[99, 108]]]", "query_spans": "[[[110, 120]]]", "process": "From the problem, we have |OF| = \\frac{p}{2}, |PF| = p, \\angle OPF = \\angle PQF, so \\tan\\angle OPF = \\tan\\angle PQF. Therefore, \\frac{|OF|}{|PF|} = \\frac{|PF|}{|FQ|}, that is, \\frac{\\frac{p}{2}}{p} = \\frac{p}{6}. Solving gives p = 3. Thus, the equation of the directrix of C is x = -\\frac{3}{2}." }, { "text": "The hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0, b>0 $) has eccentricity $ 2 $. Its asymptotes are tangent to the circle $ (x-a)^{2}+y^{2}=3 $. Then the equation of the hyperbola $ C $ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (-a + x)^2 = 3);Eccentricity(C) = 2;IsTangent(Asymptote(C),G)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4-y^2/12=1", "fact_spans": "[[[0, 61], [70, 71], [99, 105]], [[7, 61]], [[7, 61]], [[75, 95]], [[7, 61]], [[7, 61]], [[0, 61]], [[75, 95]], [[0, 69]], [[70, 97]]]", "query_spans": "[[[99, 110]]]", "process": "Given the eccentricity $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = 2 $, that is, $ b^{2} = 3a^{2} $; also, the asymptote $ bx + ay = 0 $ is tangent to the circle $ (x-a)^{2} + y^{2} = 3 $, yielding $ \\frac{ab}{\\sqrt{b^{2}+a^{2}}} = \\sqrt{3} $. Solving these equations simultaneously gives $ a^{2} = 4 $, $ b^{2} = 12 $, so the equation of the hyperbola is: $ \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 $." }, { "text": "The line $ l $ passing through the origin intersects the hyperbola $ \\frac{x^{2}}{4} - \\frac{y^{2}}{3} = -1 $ at two points. Then, the range of the slope of $ l $ is?", "fact_expressions": "l: Line;O: Origin;PointOnCurve(O, l) = True;G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = -1);NumIntersection(l, G) = 2", "query_expressions": "Range(Slope(l))", "answer_expressions": "{(sqrt(3)/2, +oo), (-oo, -sqrt(3)/2)}", "fact_spans": "[[[4, 9], [55, 58]], [[1, 3]], [[0, 9]], [[10, 49]], [[10, 49]], [[4, 53]]]", "query_spans": "[[[55, 68]]]", "process": "Let the equation of the line $ l $ passing through the origin be $ y = kx $. From \n\\[\n\\begin{cases}\n\\frac{x^2}{4} - \\frac{y^2}{3} = -1 \\\\\ny = kx\n\\end{cases}\n\\] \nwe obtain $ (3 - 4k^2)x^2 + 12 = 0 $. Since the line $ l $ intersects the hyperbola $ \\frac{x^2}{4} - \\frac{y^2}{3} = -1 $ at two points, it follows that $ \\Delta = 0 - 4 \\times (3 - 4k^2) \\times 12 > 0 $, that is, $ 4k^2 > 3 $. Solving gives: $ k > \\frac{\\sqrt{3}}{2} $ or $ k < -\\frac{\\sqrt{3}}{2} $." }, { "text": "Given point $M(-1 , 1)$ and parabola $C$: $y^{2}=4 x$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\angle A M B=90^{\\circ}$, then $k$=?", "fact_expressions": "M: Point;Coordinate(M) = (-1, 1);C: Parabola;Expression(C) = (y^2 = 4*x);G: Line;PointOnCurve(Focus(C),G) = True;Slope(G) = k;k: Number;Intersection(G, C) = {A, B};A: Point;B: Point;AngleOf(A, M, B) = ApplyUnit(90, degree)", "query_expressions": "k", "answer_expressions": "2", "fact_spans": "[[[2, 14]], [[2, 14]], [[15, 34], [36, 39], [53, 56]], [[15, 34]], [[50, 52]], [[35, 52]], [[43, 52]], [[46, 49], [97, 100]], [[50, 67]], [[58, 61]], [[62, 65]], [[70, 95]]]", "query_spans": "[[[97, 102]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Then \n\\begin{cases}y_{1}^{2}=4x_{1}\\\\y_{2}^{2}=4x_{2}\\end{cases} \nSo y_{1}^{2}-y_{2}^{2}=4x_{1}-4x_{2}. \nThus k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}}. \nTake the midpoint M(x_{0},y_{0}) of AB. Draw perpendiculars from points A and B to the directrix x=-1, with feet of perpendiculars A' and B' respectively. \nSince \\angle AMB=90^{\\circ}, \\therefore |MM'|=\\frac{1}{2}|AB|=\\frac{1}{2}(|AF|+|BF|)=\\frac{1}{2}(|AA'|+|BB'|). \nSince M' is the midpoint of A'B', MM' is parallel to the x-axis. \nGiven M(-1,1), so y_{0}=1, then y_{1}+y_{2}=2, hence k=2." }, { "text": "A point $M(1, m)$ $(m>0)$ on the parabola $y^{2}=2 p x$ $(p>0)$ is at a distance of $5$ from its focus. The left vertex of the hyperbola $\\frac{x^{2}}{a}-y^{2}=1$ is $A$. If one asymptote of the hyperbola is parallel to the line $AM$, then the real number $a$=?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;m: Number;m>0;Coordinate(M) = (1, m);PointOnCurve(M, H);Distance(M, Focus(H)) = 5;G: Hyperbola;Expression(G) = (-y^2 + x^2/a = 1);a: Real;A: Point;LeftVertex(G) = A;IsParallel(OneOf(Asymptote(G)), LineOf(A, M))", "query_expressions": "a", "answer_expressions": "", "fact_spans": "[[[0, 21], [39, 40]], [[0, 21]], [[3, 21]], [[3, 21]], [[24, 38]], [[24, 38]], [[24, 38]], [[24, 38]], [[0, 38]], [[24, 49]], [[50, 78], [89, 92]], [[50, 78]], [[110, 115]], [[83, 86]], [[50, 86]], [[89, 108]]]", "query_spans": "[[[110, 117]]]", "process": "\\frac{1}{3} Test Analysis: According to the definition of a parabola, the distance from point M(1,m) (m>0) to the focus of the parabola is equal to its distance to the directrix of the parabola x=-\\frac{p}{2}, that is: 1+\\frac{p}{2}=5. Solving gives: p=8, so the equation of the parabola is: y^{2}=16x. At this time M(1,4). The left vertex of the hyperbola \\frac{x^{2}}{a}-y^{2}=1 is A(-\\sqrt{a},0), and its asymptotes are: y=\\pm\\frac{\\sqrt{a}}{a}x'. Since one asymptote of the hyperbola is parallel to the line AM, it must hold that \\frac{4}{1+\\sqrt{a}}=\\frac{\\sqrt{a}}{a}. Solving gives: a=\\frac{1}{9}. Therefore, the answer is: \\frac{1}{3}." }, { "text": "Given points $M(-5,0)$, $N(5,0)$, and the perimeter of $\\triangle M N P$ is $36$, then the equation of the locus of vertex $P$ of $\\triangle M N P$ is?", "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-5, 0);Coordinate(N) = (5, 0);Perimeter(TriangleOf(M, N, P)) = 36", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x^2/169 + y^2/144 = 1)&Negation(y = 0)", "fact_spans": "[[[2, 12]], [[13, 22]], [[70, 73]], [[2, 12]], [[13, 22]], [[23, 48]]]", "query_spans": "[[[70, 80]]]", "process": "Since point P satisfies |PM| + |PN| = 36 - 10 = 26 > 10, the locus of point P is an ellipse with foci at M and N, and 2a = 26 (since P, M, and N are not collinear, y ≠ 0). ∴ a = 13, and c = 5, ∴ b² = a² - c² = 13² - 5² = 144. Hence, the equation of the locus of vertex P of △MNP is \\frac{x^{2}}{169} + \\frac{y^{2}}{144} = 1 (y ≠ 0)." }, { "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ are $\\sqrt{3} x \\pm y=0$, then $b=$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/b^2 = 1);b: Number;b>0;Expression(Asymptote(G)) = (sqrt(3)*x + pm*y = 0)", "query_expressions": "b", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 49]], [[2, 49]], [[78, 81]], [[5, 49]], [[2, 76]]]", "query_spans": "[[[78, 83]]]", "process": "Use the hyperbola equation to write the asymptote equations and solve for b. That is, [Detailed Solution] The asymptote equations of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1 (b>0) are: bx+2y=0. Since the asymptote equations of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1 (b>0) are \\sqrt{3}x \\pm y=0, it follows that \\frac{b}{2}=\\sqrt{3}, so b=2\\sqrt{3}." }, { "text": "Given points $F_{1}(-\\sqrt{2}, 0)$, $F_{2}(\\sqrt{2}, 0)$, a moving point $P$ satisfies $|P F_{2}|-|P F_{1}|=2$. When the $y$-coordinate of point $P$ is $\\frac{1}{2}$, what is the distance from point $P$ to the origin?", "fact_expressions": "F1: Point;Coordinate(F1) = (-sqrt(2), 0);F2: Point;Coordinate(F2) = (sqrt(2), 0);P: Point;-Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 2;YCoordinate(P) = 1/2;O: Origin", "query_expressions": "Distance(P, O)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[2, 24]], [[2, 24]], [[26, 46]], [[26, 46]], [[48, 51], [78, 82], [102, 106]], [[53, 76]], [[78, 100]], [[107, 111]]]", "query_spans": "[[[102, 116]]]", "process": "Test analysis: Since |PF₂| - |PF₁| = 2, and 2 < |F₁F₂| = 2√2, the locus of point P is the left branch of a hyperbola with foci F₁(-√2, 0) and F₂(√2, 0). Here, c = √2, a = 1, then b² = c² - a² = 1. Therefore, the equation of the locus of point P is x² - y² = 1 (x ≤ -1). Let P(x₀, 1/2) be substituted into the above equation, yielding x₀ = √5/2. Then |PO| = √(x₀² + (1/2)²) = √(5/4 + 1/4) = √6/2." }, { "text": "$F$ is the focus of the parabola $y^{2}=4x$, $P$ is any point on the parabola, $A(3, 1)$ is a fixed point. Then the minimum value of $|PF|+|PA|$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (3, 1);Focus(G) = F;PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "4", "fact_spans": "[[[4, 18], [26, 29]], [[34, 44]], [[22, 25]], [[0, 3]], [[4, 18]], [[34, 44]], [[0, 21]], [[22, 33]]]", "query_spans": "[[[49, 66]]]", "process": "" }, { "text": "Given points $A(-\\sqrt{2}, 0)$, $B(\\sqrt{2}, 0)$, a moving point $P$ satisfies $\\angle A P B=\\theta$ and $|P A| \\cdot|P B| \\cdot \\cos 2 \\frac{\\theta}{2}=1$, then the equation of the trajectory of point $P$ is?", "fact_expressions": "A: Point;Coordinate(A) = (-sqrt(2), 0);B: Point;Coordinate(B) = (sqrt(2), 0);P: Point;AngleOf(A, P, B) = theta;theta: Number;Abs(LineSegmentOf(P, A))*Abs(LineSegmentOf(P, B))*Cos(theta/2)^2 = 1", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/3+y^2=1", "fact_spans": "[[[2, 20]], [[2, 20]], [[22, 38]], [[22, 38]], [[41, 44], [120, 124]], [[46, 67]], [[46, 67]], [[68, 118]]]", "query_spans": "[[[120, 131]]]", "process": "According to the problem, |AB| = 2\\sqrt{2}. Given |PA| \\cdot |PB| \\cdot \\cos^{2}\\frac{\\theta}{2} = 1, \\theta \\in (0,\\pi), then |PA| \\cdot |PB| \\cdot \\frac{1+\\cos\\theta}{2} = 1, \\therefore |PA||PB| + |PA||PB|\\cos\\theta = 2. In \\Delta ABP, \\cos\\theta = \\frac{|PA|^{2} + |PB|^{2} - |AB|^{2}}{2|PA||PB|} = \\frac{|PA|^{2} + |PB|^{2} - 8}{2|PA||PB|}, \\therefore 2|PA||PB|\\cos\\theta = |PA|^{2} + |PB|^{2} - 8, \\begin{matrix} {|PA| & |PB| \\end{matrix} + \\frac{|PA|^{2} + |PB|^{2}}{2} - 4 = 2, that is, 2|PA||PB| + |PA|^{2} + |PB|^{2} = 12, \\therefore (|PA| + |PB|)^{2} = 12, so |PA| + |PB| = 2\\sqrt{3} is a constant and greater than |AB|. Thus, the locus of P is an ellipse with major axis length 2a = 2\\sqrt{3}, focal distance 2c = 2\\sqrt{2}, foci on the x-axis, center at the origin. Hence a = \\sqrt{3}, c = \\sqrt{2}, so b^{2} = a^{2} - c^{2} = 1. Therefore, the equation of the locus of P is: \\frac{x^{2}}{3} + y^{2} = 1" }, { "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $M$ is a point on the left branch of the hyperbola $C$, the perimeter of $\\Delta M F_{1} F_{2}$ is $9$, and point $P$ moves on the right branch of the hyperbola $C$. Then the range of the area of $\\Delta M F_{1} P$ is?", "fact_expressions": "C: Hyperbola;M: Point;F1: Point;F2: Point;P: Point;Expression(C) = (x^2 - y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, LeftPart(C));Perimeter(TriangleOf(M, F1, F2)) = 9;PointOnCurve(P, RightPart(C))", "query_expressions": "Range(Area(TriangleOf(M, F1, P)))", "answer_expressions": "(3*sqrt(3)/4, +oo)", "fact_spans": "[[[2, 35], [64, 70], [112, 118]], [[60, 63]], [[44, 51]], [[52, 59]], [[108, 111]], [[2, 35]], [[2, 59]], [[2, 59]], [[60, 75]], [[76, 105]], [[108, 122]]]", "query_spans": "[[[124, 151]]]", "process": "By the given condition and the definition of a hyperbola, we obtain |MF₁| = 3/2, |MF₂| = 7/2, then we can find M(-5/4, 3√3/4). Since the line MF₁ is parallel to the asymptote y = √3x, it follows that S△MF₁P > (1/2)|MF₁|·√3 = 3√3/4. ∵ M is a point on the left branch of hyperbola C, ∴ |MF₂| − |MF₁| = 2. ∵ the perimeter of ΔMF₁F₂ is 9, ∴ |MF₂| + |MF₁| + |F₁F₂| = 9. ∵ |F₁F₂| = 4, ∴ |MF₁| = 3/2, |MF₂| = 7/2. Solving gives x₀ = −5/4, y₀ = ±3√3/4. By symmetry of the hyperbola, without loss of generality take y₀ = 3√3/4, then M(−5/4, 3√3/4). ∴ kMF₁ = √3, ∴ the equation of line MF₁ is y = √3(x + 2). ∵ line MF₁ is parallel to the asymptote y = √3x, ∴ any point on the right branch of hyperbola C has distance to line MF₁ greater than the distance between two parallel lines, i.e., greater than √3, ∴ S△MF₁P > (1/2)|MF₁|·√3 = 3√3/4." }, { "text": "The foci of an ellipse are $F_{1}(-4,0)$ and $F_{2}(4,0)$. Point $P$ lies on the ellipse. If the maximum area of $\\triangle P F_{1} F_{2}$ is $12$, then what is the equation of the ellipse?", "fact_expressions": "G: Ellipse;Focus(G) = {F1, F2};F1: Point;Coordinate(F1) = (-4, 0);F2: Point;Coordinate(F2) = (4, 0);P: Point;PointOnCurve(P, G) = True;Max(Area(TriangleOf(P, F1, F2))) = 12", "query_expressions": "Expression(G)", "answer_expressions": "x^2/25+y^2/9=1", "fact_spans": "[[[0, 2], [41, 43], [83, 85]], [[0, 35]], [[7, 20]], [[7, 20]], [[23, 35]], [[23, 35]], [[36, 40]], [[36, 44]], [[46, 81]]]", "query_spans": "[[[83, 89]]]", "process": "When point P is the vertex on the minor axis of the ellipse, the area of triangle PF₁F₂ is maximized. At this time, the area of triangle PF₁F₂ is S = \\frac{1}{2} \\times 8 \\times b = 12, solving gives b = 3. Also, a^{2} = b^{2} + c^{2} = 25, so the equation of the ellipse is \\frac{x^{2}}{25} + \\frac{y^{2}}{9} =" }, { "text": "If a focus of the hyperbola $x^{2}-ky^{2}=1$ is $(3 , 0)$, then the real number $k$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (-k*y + x^2 = 1);k: Real;H: Point;Coordinate(H) = (3, 0);OneOf(Focus(G)) = H", "query_expressions": "k", "answer_expressions": "1/8", "fact_spans": "[[[1, 20]], [[1, 20]], [[37, 42]], [[26, 35]], [[26, 35]], [[1, 35]]]", "query_spans": "[[[37, 44]]]", "process": "" }, { "text": "Given that $O$ is the coordinate origin, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$, $A$ is the right vertex of the ellipse, and a moving point $P(x_{0}, y_{0})$ satisfies $\\overrightarrow{O P} \\cdot \\overrightarrow{O A}=9$, then when $\\angle F_{1} P F_{2}$ is maximized, $y_{0}^{2}$=?", "fact_expressions": "G: Ellipse;P: Point;O: Origin;A: Point;F1: Point;F2: Point;x0: Number;y0: Number;Expression(G) = (x^2/9 + y^2/8 = 1);Coordinate(P) = (x0, y0);LeftFocus(G) = F1;RightFocus(G) = F2;RightVertex(G) = A;DotProduct(VectorOf(O, P), VectorOf(O, A)) = 9;WhenMax(AngleOf(F1, P, F2))", "query_expressions": "y0^2", "answer_expressions": "8", "fact_spans": "[[[29, 66], [77, 79]], [[86, 103]], [[2, 5]], [[73, 76]], [[11, 18]], [[19, 26]], [[86, 103]], [[185, 196]], [[29, 66]], [[86, 103]], [[11, 72]], [[11, 72]], [[73, 83]], [[105, 156]], [[158, 184]]]", "query_spans": "[[[185, 198]]]", "process": "From $\\overrightarrow{OP}\\cdot\\overrightarrow{OA}=9$, $x_{0}$ can be determined. According to $\\tan\\angle F_{1}PF_{2}=\\frac{k_{PF_{2}}-k_{PF_{1}}}{1+k_{PF_{2}}\\cdot k_{PF_{1}}}$, express $\\tan\\angle F_{1}PF_{2}$ in terms of $y_{0}$, and combining with the basic inequality, the maximum value of $\\tan\\angle F_{1}PF_{2}$ can be found. By the monotonicity of the tangent function, $\\angle F_{1}PF_{2}$ is maximized at this point. The result follows from the equality condition of the basic inequality. [Solution] From the ellipse equation: $A(3,0)$, $F_{1}(-1,0)$, $F_{2}(1,0)$. Then $\\overrightarrow{OP}\\cdot\\overrightarrow{OA}=(x_{0},y_{0})\\cdot(3,0)=3x_{0}=9$, solving gives $x_{0}=3$. Without loss of generality, assume $y_{0}>0$. From $k_{PF_{1}}=\\frac{y_{0}}{4}$, $k_{PF_{2}}=\\frac{y_{0}}{2}$, we get $\\tan\\angle F_{1}PF_{2}=\\tan(\\angle PF_{2}A - \\angle PF_{1}A)=\\frac{k_{PF_{2}}-k_{PF_{1}}}{1+k_{PF_{2}}\\cdot k_{PF_{1}}}=\\frac{\\frac{y_{0}}{2}-\\frac{y_{0}}{4}}{1+\\frac{y_{0}}{2}\\cdot\\frac{y_{0}}{4}}=\\frac{\\frac{y_{0}}{4}}{1+\\frac{y_{0}^{2}}{8}}=\\frac{2}{\\frac{8}{y_{0}}+y_{0}}\\leq\\frac{2}{2\\sqrt{\\frac{8}{y_{0}}\\cdot y_{0}}}=\\frac{\\sqrt{2}}{4}$ (equality holds if and only if $\\frac{8}{y_{0}}=y_{0}$). Since $\\angle F_{1}PF_{2}\\in[0,\\frac{\\pi}{2})$, by the monotonicity of the tangent function, when $\\tan\\angle F_{1}PF_{2}$ is maximized, $\\angle F_{1}PF_{2}$ is maximized. At this time, $\\frac{8}{y_{0}}=y_{0}$, i.e., $y_{0}^{2}=8$." }, { "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has left and right foci $F_{1}$, $F_{2}$ ($|F_{1} F_{2}|=2 c$). A circle $A$ is drawn with the origin $O$ as center and radius $c$. The circle $A$ intersects the hyperbola $C$ at a point $P$. If the area of triangle $F_{1} P F_{2}$ is $a^{2}$, then what is the eccentricity of $C$?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;c:Number;A: Circle;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;Abs(LineSegmentOf(F1,F2))=2*c;O:Origin;Center(A)=O;Radius(A)=c;OneOf(Intersection(A, C)) = P;Area(TriangleOf(F1,P,F2)) = a^2", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 62], [136, 142], [184, 187]], [[8, 62]], [[8, 62]], [[119, 122]], [[126, 130], [131, 135]], [[68, 75]], [[76, 84]], [[148, 151]], [[8, 62]], [[8, 62]], [[0, 62]], [[0, 84]], [[0, 84]], [[85, 104]], [[107, 114]], [[106, 130]], [[118, 130]], [[131, 151]], [[153, 182]]]", "query_spans": "[[[184, 193]]]", "process": "Let P be a point on the right branch, and let |PF₁| = m, |PF₂| = n. By the definition of the hyperbola, we have m - n = 2a. According to the problem, triangle PF₁F₂ is a right triangle with ∠F₁PF₂ = 90°, so m² + n² = 4c², and (1/2)mn = a². From (m - n)² = m² + n² - 2mn = 4c² - 4a² = 4a², it follows that c = √2 a." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$. If point $P$ lies on the hyperbola and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, \nthen $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|$=?", "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;Expression(G) = (x^2 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "Abs(VectorOf(P, F1) + VectorOf(P, F2))", "answer_expressions": "2*sqrt(10)", "fact_spans": "[[[19, 47], [61, 64]], [[56, 60]], [[9, 16]], [[1, 8]], [[19, 47]], [[1, 53]], [[1, 53]], [[56, 65]], [[67, 127]]]", "query_spans": "[[[130, 185]]]", "process": "" }, { "text": "If a hyperbola passes through the point $(-\\sqrt{3}, 6)$ and its two asymptotes are given by $y = \\pm 3x$, then what is the equation of the hyperbola?", "fact_expressions": "E: Hyperbola;F: Point;Coordinate(F) = (-sqrt(3), 6);PointOnCurve(F, E);Expression(Asymptote(E)) = (y = pm*3*x)", "query_expressions": "Expression(E)", "answer_expressions": "y^2/9 - x^2 = 1", "fact_spans": "[[[1, 4], [49, 52], [26, 27]], [[6, 24]], [[6, 24]], [[1, 24]], [[26, 47]]]", "query_spans": "[[[49, 57]]]", "process": "" }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{8+m}+\\frac{y^{2}}{9}=1$ is $\\frac{1}{2}$. Then, what is the distance between the two directrices?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(m + 8) + y^2/9 = 1);Eccentricity(G) = 1/2;l1:Line;l2:Line;Directrix(G)={L1,L2}", "query_expressions": "Distance(l1,l2)", "answer_expressions": "{8*sqrt(3),12}", "fact_spans": "[[[0, 39]], [[2, 39]], [[0, 39]], [[0, 57]], [], [], [[0, 62]]]", "query_spans": "[[[0, 68]]]", "process": "" }, { "text": "Given that the point $P(t, 3)$ lies on the parabola $C$: $y^{2}=2 x$, and $F$ is the focus of $C$, then $|P F|$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);P: Point;t: Number;Coordinate(P) = (t, 3);PointOnCurve(P, C);F: Point;Focus(C) = F", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "5", "fact_spans": "[[[13, 32], [40, 43]], [[13, 32]], [[2, 12]], [[3, 12]], [[2, 12]], [[2, 35]], [[36, 39]], [[36, 46]]]", "query_spans": "[[[48, 57]]]", "process": "Point P(t,3) lies on the parabola C: y^{2}=2x, so 2t=9, solving gives t=\\frac{9}{2}. By the definition of the parabola: \\frac{9}{2}-(-\\frac{1}{2})=5, therefore |PF|=5." }, { "text": "In the Cartesian coordinate plane, $A(-5,0)$, $B(5,0)$. If $||P A|-| P B||=8$, then the equation of the locus of point $P$ is?", "fact_expressions": "A: Point;P: Point;B: Point;Coordinate(A) = (-5, 0);Coordinate(B)=(5,0);Abs(Abs(LineSegmentOf(P,A))-Abs(LineSegmentOf(P,B)))=8", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/16-y^2/9=1", "fact_spans": "[[[10, 19]], [[52, 55]], [[21, 30]], [[10, 20]], [[22, 30]], [[32, 50]]]", "query_spans": "[[[52, 63]]]", "process": "Since ||PA| - |PB|| = 8, the locus of point P is a hyperbola with A and B as foci and a major axis length of 8. That is, 2a = 8, 2c = 10, so a = 4, b = 3, therefore the equation of the hyperbola is \\frac{x^{2}}{16} - \\frac{y^{2}}{9} = 1" }, { "text": "The distance from the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1(b>0)$ to its asymptote is half the distance from the right focus to the left vertex. Then $b$=?", "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/9 - y^2/b^2 = 1);Distance(RightFocus(G),Asymptote(G))=(1/2)*Distance(RightFocus(G),LeftVertex(G))", "query_expressions": "b", "answer_expressions": "4", "fact_spans": "[[[0, 47], [59, 60]], [[71, 74]], [[3, 47]], [[0, 47]], [[0, 69]]]", "query_spans": "[[[71, 76]]]", "process": "According to the problem: b = \\frac{a+c}{2}, and since a = 3, and 9 + b^{2} = c^{2}, it follows that c = 5, b = 4." }, { "text": "The equation $(x^{2}+y^{2}-2) \\sqrt{x-3}=0$ represents which curve?", "fact_expressions": "G: Curve;Expression(G)=((x^2+y^2-2)*sqrt(x-3)=0)", "query_expressions": "Expression(G)", "answer_expressions": "x=3", "fact_spans": "[[[35, 37]], [[0, 37]]]", "query_spans": "[[[35, 41]]]", "process": "(x^{2}+y^{2}-2)\\sqrt{x-3}=0 is meaningful only if x-3\\geqslant0, and either x^{2}+y^{2}-2=0 or x-3=0, which gives x=3." }, { "text": "If the equation $\\frac{x^{2}}{2+k}+\\frac{y^{2}}{3-k}=1$ in terms of $x$ and $y$ represents an ellipse with foci on the $x$-axis, then what is the range of values for $k$?", "fact_expressions": "G: Ellipse;k:Number;PointOnCurve(Focus(G),xAxis);Expression(G)=(x^2/(2+k)+y^2/(3-k)=1)", "query_expressions": "Range(k)", "answer_expressions": "(1/2,3)", "fact_spans": "[[[64, 66]], [[68, 71]], [[55, 66]], [[12, 66]]]", "query_spans": "[[[68, 78]]]", "process": "" }, { "text": "If points $P$ and $Q$ move on the ellipse $T$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-1}=1$ $(a>1)$, and $F_{1}$, $F_{2}$ are the left and right foci of ellipse $T$, then what is the maximum value of $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}-2 \\overrightarrow{P Q}|$?", "fact_expressions": "T: Ellipse;Expression(T) = (y^2/(a^2 - 1) + x^2/a^2 = 1);a: Number;a>1;P: Point;Q: Point;PointOnCurve(P, T) = True;PointOnCurve(Q, T) = True;F1: Point;F2: Point;LeftFocus(T) = F1;RightFocus(T) = F2", "query_expressions": "Max(Abs(VectorOf(P, F1) + VectorOf(P, F2) - 2*VectorOf(P, Q)))", "answer_expressions": "2*a", "fact_spans": "[[[11, 68], [88, 93]], [[11, 68]], [[18, 68]], [[18, 68]], [[1, 5]], [[6, 9]], [[1, 71]], [[1, 71]], [[72, 79]], [[80, 87]], [[72, 99]], [[72, 99]]]", "query_spans": "[[[101, 183]]]", "process": "By vector addition and subtraction, we have |\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}-2\\overrightarrow{PQ}|=|2\\overrightarrow{PO}-2\\overrightarrow{PQ}|=2|\\overrightarrow{QO}|, and the maximum value of |\\overrightarrow{QO}| is clearly the semi-major axis length a. Since O is the midpoint of F_{1}F_{2}, then \\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}=2\\overrightarrow{PO}. Therefore, |\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}-2\\overrightarrow{PQ}|=|2\\overrightarrow{PO}-2\\overrightarrow{PQ}|=|2(\\overrightarrow{PO}-\\overrightarrow{PQ})|=|2\\overrightarrow{QO}|, whose maximum value is 2a." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has two foci $F_{1}$, $F_{2}$. Point $P$ lies on the ellipse $C$, and $P F_{1} \\perp F_{1} F_{2}$, $|P F_{1}|=\\frac{4}{3}$, $|P F_{2}|=\\frac{14}{3}$. Then the equation of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: 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);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(F1, F2));Abs(LineSegmentOf(P, F1)) = 4/3;Abs(LineSegmentOf(P, F2)) = 14/3", "query_expressions": "Expression(C)", "answer_expressions": "x^2/9 + y^2/4 = 1", "fact_spans": "[[[0, 57], [84, 89], [172, 177]], [[7, 57]], [[7, 57]], [[79, 83]], [[63, 70]], [[71, 78]], [[7, 57]], [[7, 57]], [[0, 57]], [[0, 78]], [[79, 90]], [[92, 119]], [[121, 144]], [[146, 170]]]", "query_spans": "[[[172, 182]]]", "process": "\\because|PF_{1}|+|PF_{2}|=2a,|PF_{1}|=\\frac{4}{3},|PF_{2}|=\\frac{14}{3}\\therefore2a=6,a=3, Also PF_{1}\\botF_{1}F_{2},\\therefore|F_{1}F_{2}|^{2}+|PF_{1}|^{2}=|PF_{2}|^{2},4c^{2}+(\\frac{4}{3})^{2}=(\\frac{14}{3})^{2}\\thereforec^{2}=5,\\thereforeb^{2}=a^{2}-c^{2}=9-5=4\\therefore the equation of ellipse C is \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=" }, { "text": "There is a chord $PQ$ passing through the right focus $F_2$ of the hyperbola $x^2 - y^2 = 8$, with $|PQ| = 7$, and $F_1$ is the left focus. What is the perimeter of $\\Delta F_1PQ$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 8);RightFocus(G) = F2;F2: Point;P: Point;Q: Point;PointOnCurve(F2, LineSegmentOf(P, Q)) = True;IsChordOf(LineSegmentOf(P, Q), G) = True;Abs(LineSegmentOf(P, Q)) = 7;F1: Point;LeftFocus(G) = F1", "query_expressions": "Perimeter(TriangleOf(F1, P, Q))", "answer_expressions": "14+8*sqrt(2)", "fact_spans": "[[[1, 19]], [[1, 19]], [[1, 30]], [[23, 30]], [[34, 39]], [[34, 39]], [[0, 39]], [[1, 39]], [[41, 50]], [[52, 59]], [[1, 63]]]", "query_spans": "[[[66, 89]]]", "process": "" }, { "text": "Given the equation of the ellipse is $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, $P$ is a point on the ellipse in the first quadrant, $I$ is the incenter of $\\triangle P F_{1} F_{2}$, the line $P I$ intersects the $x$-axis at point $Q$, the eccentricity of the ellipse is $\\frac{1}{3}$, if $\\overrightarrow{P Q}=\\lambda \\overrightarrow{I Q}$, then the value of $\\lambda$ is?", "fact_expressions": "G: Ellipse;I: Point;P: Point;F1: Point;F2: Point;Q: Point;lambda:Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G);Quadrant(P)=1;Incenter(TriangleOf(P, F1, F2)) = I;Intersection(LineOf(P,I), xAxis) = Q;Eccentricity(G) = 1/3;VectorOf(P, Q) = lambda*VectorOf(I, Q);a:Number;b:Number;a>b;b>0", "query_expressions": "lambda", "answer_expressions": "4", "fact_spans": "[[[2, 4], [78, 80], [90, 92], [154, 156]], [[102, 105]], [[86, 89]], [[60, 67]], [[70, 77]], [[149, 153]], [[229, 238]], [[2, 58]], [[60, 85]], [[60, 85]], [[86, 101]], [[86, 101]], [[102, 134]], [[135, 153]], [[154, 174]], [[176, 227]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]]]", "query_spans": "[[[229, 242]]]", "process": "As shown in the figure, connect IF₁, IF₂, I is the incenter of △PF₁F₂, so IF₁, IF₂ are the angle bisectors of ∠PF₁F₂ and ∠PF₂F respectively. Since the line passing through point P and the incenter I of △PF₁F₂ intersects the x-axis at point Q, then PQ is the angle bisector of ∠F₁PF₂, thus the distances from Q to lines PF₁, PF₂ are equal. Therefore, $\\frac{S_{\\triangle PF_{1}Q}}{S_{\\triangle PF_{2}Q}} = \\frac{|PF_{1}|}{|PF_{2}|} = \\frac{|C|}{|C|}$. Similarly, we obtain $\\frac{1}{1}$. From the property of proportional relations, $\\frac{|PI|}{|IQ|}$. Also, the eccentricity of the ellipse $e = \\frac{c}{a} = \\frac{|\\overrightarrow{IQ}|}{|\\overrightarrow{PI}|} = \\frac{1}{3}$. Therefore, $\\overrightarrow{PI} = ?\\overrightarrow{IQ}$, so $\\overrightarrow{PQ} = 4\\overrightarrow{IQ}$, hence $\\lambda = 4$." }, { "text": "The line $ l $ passes through the focus $ F $ of the parabola $ y^{2} = 4x $, and intersects the parabola at points $ A $ and $ B $. If $ \\overrightarrow{A F} = 5 \\overrightarrow{F B} $, then the slope of line $ l $ is?", "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l,G)={A,B};VectorOf(A, F) = 5*VectorOf(F, B)", "query_expressions": "Slope(l)", "answer_expressions": "pm*sqrt(5)/2", "fact_spans": "[[[0, 5], [93, 98]], [[7, 21], [30, 33]], [[35, 38]], [[24, 27]], [[39, 42]], [[7, 21]], [[7, 27]], [[0, 27]], [[0, 44]], [[46, 91]]]", "query_spans": "[[[93, 103]]]", "process": "According to the problem, the focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $. Let the equation of line $ l $ be $ 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} - 2(k^{2} + 2)x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. \n$ \\therefore x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $, $ x_{1} \\cdot x_{2} = 1 $. \n$ \\overrightarrow{FA} = 5\\overrightarrow{FB} \\Rightarrow x_{1} - 1 = 5x_{2} - 5 $, i.e., $ 5x_{2} + x_{1} - 6 = 0 $. \nSince $ x_{1} = \\frac{1}{x_{2}} $, $ \\therefore 5x_{2} + \\frac{1}{x_{2}} - 6 = 0 $, solving gives $ x_{2} = 1 $ or $ x_{2} = \\frac{1}{5} $. \n$ \\therefore x_{1} = 1 $ or $ x_{1} = 5 $. Also $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} > 2 $, substituting $ x_{1} = 5 $ gives $ k = \\pm \\frac{\\sqrt{5}}{2} $." }, { "text": "The equation of the circle whose center is at the right focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and which is tangent to its asymptotes is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);H: Circle;RightFocus(G) = Center(H);IsTangent(Asymptote(G), H)", "query_expressions": "Expression(H)", "answer_expressions": "(x-5)^2+y^2=16", "fact_spans": "[[[1, 40], [50, 51]], [[1, 40]], [[57, 58]], [[0, 58]], [[49, 58]]]", "query_spans": "[[[57, 63]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$ has eccentricity $\\frac{\\sqrt{3}}{3}$, then the real number $a$=?", "fact_expressions": "C: Ellipse;a: Real;a>1;Expression(C) = (y^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(3)/3", "query_expressions": "a", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[2, 43]], [[70, 75]], [[9, 43]], [[2, 43]], [[2, 68]]]", "query_spans": "[[[70, 77]]]", "process": "Given the ellipse equation $ C: \\frac{x^{2}}{a^{2}} + y^{2} = 1 $ $ (a > 1) $, and the eccentricity is $ \\frac{\\sqrt{3}}{3} $, thus we have $ \\sqrt{1 - \\left( \\frac{1}{a} \\right)^{2}} = \\frac{\\sqrt{3}}{2} $. Solving gives $ a = \\pm 2 $, and since $ a > 1 $, it follows that $ a = 2 $." }, { "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 the value of $| PF_{1}|+| PF_{2} |$ 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], [38, 39], [49, 52]], [[44, 48]], [[21, 29]], [[30, 37]], [[2, 20]], [[21, 43]], [[44, 55]], [[57, 80]]]", "query_spans": "[[[82, 108]]]", "process": "" }, { "text": "If the equation $x^2+k y^2=2$ represents an ellipse with foci on the $x$-axis and focal distance $\\sqrt{3}$, then what is the length of the minor axis of the ellipse?", "fact_expressions": "G: Ellipse;Expression(G)=(x^2+k*y^2=2);PointOnCurve(Focus(G),xAxis);FocalLength(G)=sqrt(3);k:Number", "query_expressions": "Length(MinorAxis(G))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[43, 45], [47, 49]], [[2, 45]], [[19, 45]], [[29, 45]], [[2, 45]]]", "query_spans": "[[[47, 55]]]", "process": "The equation x^{2}+ky^{2}=2 can be rewritten as \\frac{x^{2}}{2}+\\frac{y^{2}}{k}=1, then (\\frac{\\sqrt{3}}{2})^{2}+\\frac{2}{k}=2=\\frac{2}{k}=\\frac{5}{4}, \\therefore the length of the minor axis is 2x\\frac{\\sqrt{5}}{2}=\\sqrt{5}" }, { "text": "Through the left focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, draw a perpendicular line to one of the asymptotes, intersecting the two asymptotes at points $A$ and $B$, respectively. If $\\frac{|A F|}{|B F|}=\\frac{1}{2}$, 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 > b;b > 0;F: Point;LeftFocus(G) = F;L: Line;PointOnCurve(F, L) ;IsPerpendicular(L, OneOf(Asymptote(G))) ;Asymptote(G) = {l1, l2};Intersection(L, l1) = A;Intersection(L, l2) = B;A: Point;B: Point;l1: Line;l2: Line;Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F)) = 1/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[1, 54], [127, 130]], [[1, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[58, 61]], [[1, 61]], [], [[0, 70]], [[0, 70]], [[1, 78]], [[0, 90]], [[0, 90]], [[81, 84]], [[85, 88]], [], [], [[92, 125]]]", "query_spans": "[[[127, 136]]]", "process": "When $ a > b > 0 $, from $ \\frac{|AF|}{|BF|} = \\frac{1}{2} $, by the angle bisector theorem and the given condition, we obtain $ \\frac{|OA|}{|OB|} = \\frac{|AF|}{|BF|} = \\frac{1}{2} $. In right triangle $ \\triangle OAB $, $ \\angle AOB = \\frac{\\pi}{3} $. The slope of asymptote $ OB $ is $ k = \\frac{\\sqrt{3}}{3} $, so the eccentricity is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{1}{3}} = \\frac{2\\sqrt{3}}{3} $." }, { "text": "$F(c , 0)$ is a focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$. The maximum distance from $F$ to a point on the ellipse is $m$, and the minimum distance is $n$. What is the point on the ellipse whose distance to $F$ is $\\frac{m+n}{2}$?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F: Point;c: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (c, 0);OneOf(Focus(G)) = F;P: Point;PointOnCurve(P, G);Max(Distance(F, P)) = m ;m: Number;n: Number;Min(Distance(F, P)) = n;P0: Point;PointOnCurve(P0, G);Distance(F, P0) = (m + n)/2", "query_expressions": "Coordinate(P0)", "answer_expressions": "(0, pm*b)", "fact_spans": "[[[11, 63], [73, 75], [98, 100]], [[13, 63]], [[13, 63]], [[0, 10], [69, 72], [102, 106]], [[0, 10]], [[13, 63]], [[13, 63]], [[11, 63]], [[0, 10]], [[0, 68]], [[76, 77]], [[73, 77]], [[69, 88]], [[85, 88]], [[93, 96]], [[69, 96]], [[125, 126]], [[98, 126]], [[98, 126]]]", "query_spans": "[[[125, 128]]]", "process": "" }, { "text": "Given the equation of a circle is $x^{2}+y^{2}=4$. If a parabola passes through points $A(-1,0)$ and $B(1,0)$, and has a tangent line of the circle as its directrix, then what is the equation of the locus of the focus of the parabola?", "fact_expressions": "G: Parabola;H: Circle;A: Point;B: Point;Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Expression(H) = (x^2 + y^2 = 4);PointOnCurve(A, G);PointOnCurve(B, G);IsTangent(H, Directrix(G))", "query_expressions": "LocusEquation(Focus(G))", "answer_expressions": "(x^2/4 + y^2/3 = 1)&Negation(y = 0)", "fact_spans": "[[[24, 27], [61, 64]], [[2, 3], [52, 53]], [[28, 38]], [[41, 49]], [[28, 38]], [[41, 49]], [[2, 22]], [[24, 38]], [[24, 49]], [[24, 59]]]", "query_spans": "[[[61, 73]]]", "process": "According to the problem: the sum of the distances from the focus to A and B equals the sum of the distances from A and B to the directrix; and the sum of these distances is twice the distance from the midpoint O of A and B to the directrix, that is, 2r = 4. Therefore, the locus C of the focus is an ellipse with A and B as foci, from which we can find the locus equation of the focus F of the parabola. Let the focus of the parabola be F, and draw perpendiculars AA_{1}, BB_{1}, OO_{1} from A, B, O to the directrix. Then |AA_{1}| + |BB_{1}| = 2|OO_{1}| = 4; by the definition of a parabola, |AA_{1}| + |BB_{1}| = |FA| + |FB|, therefore |FA| + |FB| = 4. Hence, the locus of point F is an ellipse with A and B as foci and major axis length 4 (excluding the endpoints of the major axis). Therefore, the locus equation of the focus of the parabola is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 (y\\neq0)." }, { "text": "If the line $y = kx + 1$ ($k \\in \\mathbb{R}$) has one common point with the hyperbola $x^{2} - y^{2} = 1$, find the set of real values of $k$?", "fact_expressions": "H: Line;Expression(H) = (y = k*x + 1);k: Real;G: Hyperbola;Expression(G) = (x^2 - y^2 = 1);NumIntersection(H, G) = 1", "query_expressions": "Range(k)", "answer_expressions": "{sqrt(6)/2, -sqrt(6)/2, sqrt(2)/2, -sqrt(2)/2}", "fact_spans": "[[[1, 21]], [[1, 21]], [[48, 53]], [[22, 40]], [[22, 40]], [[1, 46]]]", "query_spans": "[[[48, 59]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ has eccentricity $e=\\frac{\\sqrt{10}}{5}$, then the value of $m$ is?", "fact_expressions": "G: Ellipse;m: Number;e: Number;Expression(G) = (x^2/5 + y^2/m = 1);Eccentricity(G) = e ;e = sqrt(10)/5", "query_expressions": "m", "answer_expressions": "3, 25/3", "fact_spans": "[[[2, 39]], [[68, 71]], [[43, 66]], [[2, 39]], [[2, 66]], [[43, 66]]]", "query_spans": "[[[68, 75]]]", "process": "" }, { "text": "Given a fixed point $B(1 , 1)$ on the parabola $y^{2}=x$ and two moving points $P$, $Q$, when $P$ moves along the parabola, $B P \\perp P Q$. Then the range of the $y$-coordinate of point $Q$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x);B: Point;Coordinate(B) = (1, 1);PointOnCurve(B, G) = True;PointOnCurve(P, G) = True;PointOnCurve(Q, G) = True;P: Point;Q: Point;IsPerpendicular(LineSegmentOf(B, P), LineSegmentOf(P, Q)) = True", "query_expressions": "Range(YCoordinate(Q))", "answer_expressions": "(-oo,-1]+[3,+oo)", "fact_spans": "[[[2, 14], [46, 49]], [[2, 14]], [[18, 28]], [[18, 28]], [[2, 28]], [[42, 50]], [[2, 40]], [[33, 36], [42, 45]], [[37, 40], [71, 75]], [[54, 69]]]", "query_spans": "[[[71, 86]]]", "process": "Problem Analysis: Let P(t^{2},t), Q(s^{2},s). BP \\perp PQ, so \\overrightarrow{BP} \\cdot \\overrightarrow{PQ} = 0, that is (t^{2}-1,t-1) \\cdot (s^{2}-t^{2},s-t) = (t^{2}-1) \\cdot (s^{2}-t^{2}) + (t-1) \\cdot (s-t) = 0, which implies t^{2} + (s+1)t + s+1 = 0. Since t \\in \\mathbb{R}, it must hold that \\Delta = (s+1)^{2} - 4(s+1) > 0, i.e., s^{2} - 2s - 3 \\geqslant 0. Solving gives s < -1 or s > 3. The range of the y-coordinate of point Q is (-\\infty,-1) \\cup [3,+\\infty). Properties of parabola." }, { "text": "Given the parabola equation $x^{2}=-8 y$, then the focus coordinates of the parabola are?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -8*y)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,-2)", "fact_spans": "[[[2, 5], [22, 25]], [[2, 20]]]", "query_spans": "[[[22, 31]]]", "process": "Since the parabola equation is x^{2}=-8y, the focus lies on the y-axis, opens downward, and the focus coordinates are (0,-2)." }, { "text": "Given the parabola equation $y^{2}=4x$, and the line $l$ with equation $x-y+4=0$. On the parabola, there is a moving point $P$ such that the distance from $P$ to the $y$-axis is $d_{1}$, and the distance from $P$ to the line $l$ is $d_{2}$. Then the minimum value of $d_{1}+d_{2}$ is?", "fact_expressions": "l: Line;G: Parabola;P: Point;d1:Number;d2:Number;Expression(G) = (y^2 = 4*x);Expression(l) = (x - y + 4 = 0);PointOnCurve(P, G);Distance(P, yAxis) = d1;Distance(P, l) = d2", "query_expressions": "Min(d1 + d2)", "answer_expressions": "5*sqrt(2)/2-1", "fact_spans": "[[[20, 25], [74, 79]], [[2, 5], [40, 43]], [[48, 51], [70, 73]], [[60, 67]], [[83, 90]], [[2, 19]], [[20, 38]], [[40, 51]], [[48, 68]], [[69, 90]]]", "query_spans": "[[[92, 111]]]", "process": "Let the focus of the parabola be $ F(1,0) $, then $ d_{1} = |PF| - 1 $. Draw a perpendicular from $ P $ to the line $ x - y + 4 = 0 $, with foot $ M $. Draw a perpendicular from $ F $ to the line $ x - y + 4 = 0 $, with foot $ N $. Let $ Q $ be the intersection point of the line segment $ FN $ and the parabola. As shown in the figure, $ d_{1} + d_{2} = |PF| - 1 + |PM| = |FP| + |PM| - 1 $. From the graph, it is clear that when points $ M $, $ P $, $ F $ are collinear, i.e., $ M $ coincides with $ N $ and $ P $ coincides with $ Q $, $ |PF| + |PM| $ attains its minimum value $ |FN| = \\frac{|1 - 0 + 4|}{\\sqrt{2}} = \\frac{5\\sqrt{2}}{2} $. Therefore, the minimum value of $ d_{1} + d_{2} $ is $ \\frac{5\\sqrt{2}}{2} - 1 $." }, { "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, find the minimum value of $\\frac{1}{|P F_{1}|^{2}}+\\frac{1}{|P F_{2}|^{2}}$.", "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, F2))^2) + 1/(Abs(LineSegmentOf(P, F1))^2))", "answer_expressions": "", "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, 144]]]", "process": "By the definition of an ellipse, we have |PF₁| + |PF₂| = 2a. Then from |PF₁| + |PF₂| ≥ 2√(|PF₁|·|PF₂|), we obtain |PF₁||PF₂| ≤ a² (equality holds if and only if |PF₁| = |PF₂| = a). Also, by 1/|PF₁| + 1/|PF₂| ≥ 2√(1/a⁴) = 2/a² (equality holds if and only if |PF₁| = |PF₂| = a), so the minimum value of 1/|PF₁| + 1/|PF₂| is 2/a²." }, { "text": "Let $M$ and $N$ be any two points on the parabola $C$: $y^{2}=4x$. The coordinates of point $E$ are $(-\\lambda, 0)$ where $\\lambda \\geq 0$. If the minimum value of $\\overrightarrow{E M} \\cdot \\overrightarrow{E N}$ is $0$, then the value of the real number $\\lambda$ is?", "fact_expressions": "C: Parabola;E: Point;M: Point;N: Point;lambda: Real;lambda>=0;Expression(C) = (y^2 = 4*x);Coordinate(E) = (-lambda, 0);PointOnCurve(M, C);PointOnCurve(N, C);Min(DotProduct(VectorOf(E, M), VectorOf(E, N))) = 0", "query_expressions": "lambda", "answer_expressions": "1", "fact_spans": "[[[9, 27]], [[33, 37]], [[1, 4]], [[5, 8]], [[132, 143]], [[41, 71]], [[9, 27]], [[33, 71]], [[1, 32]], [[1, 32]], [[73, 130]]]", "query_spans": "[[[132, 147]]]", "process": "The angle between \\overrightarrow{EM} and \\overrightarrow{EN} is at most \\frac{\\pi}{2}. From point E, two tangent lines are drawn to the parabola, with points of tangency M and N respectively. At this moment, the angle between \\overrightarrow{EM} and \\overrightarrow{EN} is maximized, thus obtaining the equation of line EM. By solving the system of equations consisting of the line EM and the parabola, we set A=0 and can solve for \\lambda. According to the problem, the angle between \\overrightarrow{EM} and \\overrightarrow{EN} is at most \\frac{\\pi}{2}, from point E two tangent lines are drawn to the parabola, with points of tangency M and N respectively, at which moment the angle between \\overrightarrow{EM} and \\overrightarrow{EN} is maximized, and the slope k of line EM is k=\\tan\\frac{\\pi}{4}=1, so the equation is y=x+\\lambda. Solving the system \\begin{cases}y2=4x\\\\v=x+2\\end{cases}, eliminating x gives y^{2}-4y+4\\lambda=0, then _{A}=(-4)^{2}-4\\times4\\lambda=0, solving yields \\lambda=1." }, { "text": "Let the parabola $C$: $y^{2}=12x$ have focus $F$ and directrix $l$. Point $M$ lies on $C$, point $N$ lies on $l$, and $FN = \\lambda FM$ ($\\lambda > 0$). If $|MF| = 4$, then the value of $\\lambda$ is?", "fact_expressions": "C: Parabola;F: Point;N: Point;M: Point;l: Line;lambda: Number;lambda > 0;Expression(C) = (y^2 = 12*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(M, C);PointOnCurve(N, l);LineSegmentOf(F, N) = lambda*LineSegmentOf(F, M);Abs(LineSegmentOf(M, F)) = 4", "query_expressions": "lambda", "answer_expressions": "3", "fact_spans": "[[[1, 21], [41, 44]], [[25, 28]], [[46, 50]], [[36, 40]], [[32, 35], [51, 54]], [[98, 107]], [[57, 85]], [[1, 21]], [[1, 28]], [[1, 35]], [[36, 45]], [[46, 55]], [[57, 85]], [[87, 96]]]", "query_spans": "[[[98, 111]]]", "process": "Since point M lies on C and |MF| = 4, we have x_{M} + 3 = 4, so x_{M} = 1. Without loss of generality, assume M is in the first quadrant, then M(1, 2\\sqrt{3}). Thus, k_{FM} = \\frac{2\\sqrt{3}}{1 - 3} = -\\sqrt{3}, so the equation of line FM is y = -\\sqrt{3}(x - 3). Since FN = \\lambda FM (\\lambda > 0), and the directrix l: x = -3, substituting into y = -\\sqrt{3}(x - 3) gives y_{N} = 6\\sqrt{3}. Therefore, \\lambda = \\frac{|FN|}{|FM|} = \\frac{y_{N}}{y_{M}} = 3." }, { "text": "What is the standard equation of a parabola passing through the point $(-2,3)$?", "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (-2, 3);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=-(9/2)*x,x^2=(4/3)*y}", "fact_spans": "[[[11, 14]], [[1, 10]], [[1, 10]], [[0, 14]]]", "query_spans": "[[[11, 21]]]", "process": "According to the problem, the standard equation of the parabola can be set as y^{2} = -2px, (p > 0) or x^{2} = 2p'y, (p' > 0). Substituting (-2, 3) into the equations, we obtain p = \\frac{9}{4} or p' = \\frac{2}{3}. Hence, the standard equations of the parabola are y^{2} = -\\frac{9}{2}x or x^{2} = \\frac{4}{3}y." }, { "text": "If point $P$ moves on the line $l_{1}$: $x - y - 2 = 0$, and point $Q$ moves on the line $l_{2}$: $x - y - 6 = 0$, let the midpoint of segment $PQ$ be $M(x_{0}, y_{0})$, and $(x_{0} - 2)^{2} + (y_{0} + 2)^{2} \\leq 8$, then what is the range of values of $x_{0}^{2} + y_{0}^{2}$?", "fact_expressions": "P: Point;l1: Line;Expression(l1) = (x - y - 2 = 0);PointOnCurve(P, l1);Q: Point;l2: Line;Expression(l2) = (x - y - 6 = 0);PointOnCurve(Q, l2);M: Point;Coordinate(M) = (x0, y0);MidPoint(LineSegmentOf(P, Q)) = M;(x0 - 2)^2 + (y0 + 2)^2 <= 8;x0: Number;y0: Number", "query_expressions": "Range(x0^2 + y0^2)", "answer_expressions": "[8, 16]", "fact_spans": "[[[3, 6]], [[7, 27]], [[7, 27]], [[3, 28]], [[31, 34]], [[35, 55]], [[35, 55]], [[31, 56]], [[69, 86]], [[69, 86]], [[58, 86]], [[88, 124]], [[69, 86]], [[69, 86]]]", "query_spans": "[[[126, 154]]]", "process": "" }, { "text": "The coordinates of the point on the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{25}=1$ where the product of the distances to the two foci is maximized are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/25 = 1);P:Point;F1:Point;F2:Point;PointOnCurve(P,G);Focus(G)={F1,F2};WhenMax(Distance(P,F1)*Distance(P,F2))", "query_expressions": "Coordinate(P)", "answer_expressions": "(pm*3,0)", "fact_spans": "[[[0, 38]], [[0, 38]], [[51, 52]], [], [], [[0, 52]], [[0, 44]], [[0, 52]]]", "query_spans": "[[[51, 57]]]", "process": "" }, { "text": "The asymptotes of the hyperbola $C$: $\\frac{x^{2}}{25}-\\frac{y^{2}}{9}=1$ are given by?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/25 - y^2/9 = 1)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(3/5)*x", "fact_spans": "[[[0, 44]], [[0, 44]]]", "query_spans": "[[[0, 52]]]", "process": "According to the hyperbola equation, find a, b, and solve with the focus on the x-axis. Since the hyperbola C: \\frac{x^2}{25}-\\frac{y^{2}}{9}=1, then a=5, b=3, and the focus is on the x-axis, so its asymptote equations are y=\\pm\\frac{3}{5}x." }, { "text": "Given that $M$ is a point on the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then the sum of the distances from point $M$ to the two foci is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2/5 = 1);M: Point;PointOnCurve(M, C) = True;F1: Point;F2: Point;Focus(C) = {F1, F2}", "query_expressions": "Distance(M,F1) + Distance(M,F2)", "answer_expressions": "6", "fact_spans": "[[[6, 48]], [[6, 48]], [[2, 5], [54, 58]], [[2, 52]], [], [], [[6, 62]]]", "query_spans": "[[[54, 69]]]", "process": "Ellipse C: \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1, then a^{2}=9, a=3, the sum of the distances from point M to the two foci is 2a=6" }, { "text": "The vertex of a parabola coincides with the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and the focus of the parabola coincides with the right focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);Vertex(G) = Center(H);Focus(G) = RightFocus(H)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=16*x", "fact_spans": "[[[0, 3], [51, 54], [104, 107]], [[7, 45], [58, 96]], [[7, 45]], [[0, 50]], [[51, 102]]]", "query_spans": "[[[104, 112]]]", "process": "Find the right focus of the ellipse $ F_{2}(4,0) $, and solve by $ \\frac{p}{2}=4 $. According to the problem, the right focus of the ellipse is $ F_{2}(4,0) $. Let the equation of the parabola be $ y^{2}=2px $ ($ p>0 $), then $ \\frac{p}{2}=4 $, $ \\therefore p=8 $. $ \\therefore $ The equation of the parabola is: $ y^{2}=16x $. The answer is: $ y^{2}=16x $." }, { "text": "If a point $P$ on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is at a distance of $4$ from the right focus, then what is the distance from point $P$ to the left focus?", "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(P, G);Distance(P, RightFocus(G)) = 4", "query_expressions": "Distance(P, LeftFocus(G))", "answer_expressions": "10", "fact_spans": "[[[1, 40]], [[43, 46], [59, 63]], [[1, 40]], [[1, 46]], [[1, 57]]]", "query_spans": "[[[1, 72]]]", "process": "Let the left and right foci of the hyperbola be $ F_1 $ and $ F_2 $, respectively. From the given condition, $ |PF_2| = 4 $. When point $ P $ lies on the left branch of the hyperbola, then $ |PF_2| - |PF_1| = 6 $, which does not satisfy the condition. When point $ P $ lies on the right branch of the hyperbola, then $ |PF| - |PF_2| = 6 $, so $ |PF| = |PF_2| + 6 = 10 $, which satisfies the condition." }, { "text": "Given that a focus of the hyperbola $x^{2}-\\frac{y^{2}}{8}=\\frac{1}{k}$ is $(0,-3)$, find the value of $k$.", "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2 - y^2/8 = 1/k);Coordinate(OneOf(Focus(G))) = (0,-3)", "query_expressions": "k", "answer_expressions": "-1", "fact_spans": "[[[2, 40]], [[56, 59]], [[2, 40]], [[2, 54]]]", "query_spans": "[[[56, 63]]]", "process": "The hyperbola x^{2}-\\frac{y^{2}}{8}=\\frac{1}{k} is rewritten as \\frac{x^{2}}{k}-\\frac{y^{2}}{8k}=1. Since one focus of the hyperbola is (0,-3), it follows that -\\frac{8}{k}-\\frac{1}{k}=3^{2}, solving gives k=-1." }, { "text": "In an ellipse $\\Gamma$, let $A$ be one endpoint of the major axis, $B$ be one endpoint of the minor axis, and $F_{1}$, $F_{2}$ be the two foci. If $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{A F_{2}}+\\overrightarrow{B F_{1}} \\cdot \\overrightarrow{B F_{2}}=0$, then what is the value of $\\frac{|A B|}{|F_{1} F_{2}|}$?", "fact_expressions": "Gamma: Ellipse;A: Point;OneOf(Endpoint(MajorAxis(Gamma))) = A;OneOf(Endpoint(MinorAxis(Gamma))) = B;B: Point;F1: Point;F2: Point;Focus(Gamma) = {F1, F2};DotProduct(VectorOf(A, F1), VectorOf(A, F2)) + DotProduct(VectorOf(B, F1), VectorOf(B, F2)) = 0", "query_expressions": "Abs(LineSegmentOf(A, B))/Abs(LineSegmentOf(F1, F2))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[1, 11]], [[13, 16]], [[1, 24]], [[1, 36]], [[25, 28]], [[37, 44]], [[45, 52]], [[1, 57]], [[60, 175]]]", "query_spans": "[[[177, 210]]]", "process": "Assume the equation of T is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), A(a,0), B(0,b), F_{1}(-c,0), F_{2}(c,0), where \\frac{c=\\sqrt{a^{2}-b^{2}}}{\\overrightarrow{AF}_{1}\\cdot\\overrightarrow{AF_{2}}+\\overrightarrow{BF_{1}}\\cdot\\overrightarrow{BF_{2}}}=\\overrightarrow{AF_{1}}\\cdot\\overrightarrow{AF_{2}}+\\overrightarrow{BF_{1}}\\cdot\\overrightarrow{BF_{2}}=(-c-a)(c- so \\frac{|AB|}{|F_{1}F_{2}|}=\\frac{\\sqrt{a^{2}+b^{2}}}{2c}=\\frac{\\sqrt{2c^{2}}}{2c}=\\frac{\\sqrt{2}}{2}-c-a)(c-a)+(-c^{2}+b^{2})=a^{2}+b^{2}-2c^{2}=0." }, { "text": "Let a point $P$ on the parabola $y = -2x^2$ be at a distance of $4$ from the $x$-axis. Then, what is the distance from point $P$ to the focus of this parabola?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y = -2*x^2);PointOnCurve(P, G);Distance(P, xAxis) = 4", "query_expressions": "Distance(P, Focus(G))", "answer_expressions": "33/8", "fact_spans": "[[[1, 16], [42, 45]], [[19, 22], [36, 40]], [[1, 16]], [[1, 22]], [[19, 34]]]", "query_spans": "[[[36, 52]]]", "process": "The standard form of the parabola equation is: $x^{2}=-\\frac{y}{2}$, and the equation of the directrix is $y=\\frac{1}{8}$. By the definition of a parabola, the distance from point P to the focus of the parabola equals the distance $d$ from point P to the directrix $y=\\frac{1}{8}$. Since the distance from point P to the x-axis is 4, it follows that $d=4+\\frac{1}{8}=\\frac{33}{8}$. Therefore, fill in: $\\frac{33}{8}$. This question examines the standard form of the parabola equation, the focal radius of a parabola, and basic computational solving ability." }, { "text": "Given a parabola $C$: $y^{2}=2 p x(p>0)$, a point $A(4, m)$ on it is at a distance of $\\frac{17}{4}$ from its focus. Then the value of $p$ is?", "fact_expressions": "C: Parabola;p: Number;A: Point;p>0;Expression(C) = (y^2 = 2*p*x);m:Number;Coordinate(A) = (4, m);PointOnCurve(A, C);Distance(A, Focus(C)) = 17/4", "query_expressions": "p", "answer_expressions": "p=1/2", "fact_spans": "[[[41, 42], [2, 28]], [[64, 67]], [[31, 40]], [[10, 28]], [[2, 28]], [[31, 40]], [[31, 40]], [[2, 40]], [[31, 62]]]", "query_spans": "[[[64, 71]]]", "process": "Test Analysis: According to the definition of a parabola, for a point A(4, m) on the parabola C: y^{2}=2px (p>0), the distance from A to its focus is equal to the distance from A to its directrix, that is, \\frac{17}{4}=\\frac{p}{2}+4, so p=\\frac{1}{2}." }, { "text": "If the line $y = x + t$ intersects the ellipse $\\frac{x^{2}}{4} + y^{2} = 1$ at points $A$ and $B$, then as $t$ varies, the maximum value of $AB$ is?", "fact_expressions": "G: Ellipse;H: Line;t: Number;A: Point;B: Point;Expression(G) = (x^2/4 + y^2 = 1);Expression(H) = (y = t + x);Intersection(H, G) = {A, B}", "query_expressions": "Max(LineSegmentOf(A, B))", "answer_expressions": "4*sqrt(10)/5", "fact_spans": "[[[11, 38]], [[1, 10]], [[52, 55]], [[41, 44]], [[45, 48]], [[11, 38]], [[1, 10]], [[1, 50]]]", "query_spans": "[[[59, 69]]]", "process": "" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=16 x$, $M$ is a point on $C$, $O$ is the origin, and the extension of $F M$ intersects the $y$-axis at point $N$. If $|F N|=2|O M|$, then the ordinate of point $M$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 16*x);F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C);O: Origin;N: Point;Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = N;Abs(LineSegmentOf(F, N)) = 2*Abs(LineSegmentOf(O, M))", "query_expressions": "YCoordinate(M)", "answer_expressions": "pm*4*sqrt(2)", "fact_spans": "[[[6, 26], [34, 37]], [[6, 26]], [[2, 5]], [[2, 29]], [[30, 33], [87, 91]], [[30, 40]], [[41, 44]], [[65, 69]], [[50, 69]], [[71, 85]]]", "query_spans": "[[[87, 97]]]", "process": "Since triangle FON is a right triangle, and |FN| = 2|OM|, it follows that M is the midpoint of FN. Let N be (0, b), and since F is (4, 0), then M is (2, \\frac{b}{2}). Substituting into the parabola equation gives (\\frac{b}{2})^{2} = 32, \\frac{b}{2} = \\pm4\\sqrt{2}, so the y-coordinate of point M is \\pm4\\sqrt{2}. This problem examines the geometric property that in a right triangle, the median to the hypotenuse is half the length of the hypotenuse. First, draw the figure according to the given conditions. The key lies in the given condition |FN| = 2|OM|, which suggests the geometric property that in a right triangle, the median to the hypotenuse equals half the hypotenuse, thus M is the midpoint of F and N. Assign coordinates and substitute into the parabola equation to obtain the desired result." }, { "text": "If an ellipse with foci on the $x$-axis passes through the point $P(3,0)$, and the length of the major axis is 3 times the length of the minor axis, then its standard equation is?", "fact_expressions": "G: Ellipse;P: Point;Coordinate(P) = (3, 0);PointOnCurve(Focus(G),xAxis);PointOnCurve(P,G);Length(MajorAxis(G))=3*Length(MinorAxis(G))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9+y^2=1", "fact_spans": "[[[9, 11], [37, 38]], [[12, 21]], [[12, 21]], [[1, 11]], [[9, 21]], [[9, 35]]]", "query_spans": "[[[37, 44]]]", "process": "Test analysis: From the properties of the ellipse, we know that a=3, a=3b ∴ b=1, so the equation of the ellipse is \\frac{x^{2}}{9}+y^{2}=1" }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{k+4}+\\frac{y^{2}}{4}=1$ is $0.5$, then $k$=?", "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/(k + 4) + y^2/4 = 1);Eccentricity(G) = 0.5", "query_expressions": "k", "answer_expressions": "{4/3,-1}", "fact_spans": "[[[1, 40]], [[52, 55]], [[1, 40]], [[1, 50]]]", "query_spans": "[[[52, 57]]]", "process": "① When the foci of the ellipse lie on the x-axis, it follows from the given condition that $\\frac{\\sqrt{k}}{\\sqrt{k+4}}=0.5$, solving gives $k=\\frac{4}{3}$. ② When the foci of the ellipse lie on the y-axis, it follows from the given condition that $\\frac{\\sqrt{-k}}{\\sqrt{4}}=\\frac{1}{2}$, solving gives $k=-1$. In conclusion, $k=\\frac{4}{3}$ or $-1$." }, { "text": "If $P(x_{0}, y_{0})$ is a point on the left branch of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{4}=1$, then what is the range of values for $x_{0}$?", "fact_expressions": "P: Point;Coordinate(P) = (x0, y0);x0: Number;y0: Number;G: Hyperbola;Expression(G) = (x^2/2 - y^2/4 = 1);PointOnCurve(P, LeftPart(G))", "query_expressions": "Range(x0)", "answer_expressions": "(-oo, -sqrt(2)]", "fact_spans": "[[[1, 18]], [[1, 18]], [[64, 71]], [[1, 18]], [[19, 57]], [[19, 57]], [[1, 62]]]", "query_spans": "[[[64, 78]]]", "process": "The hyperbola equation is: \\frac{x^2}{2}-\\frac{y^{2}}{4}=1, its foci lie on the x-axis, and a=\\sqrt{2}. Since point P(x_{0},y_{0}) lies on the left branch of the hyperbola, it follows that x_{0}\\leqslant-\\sqrt{2}." }, { "text": "Given that line $l$ passes through the right focus $F_{2}$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1$, intersecting the ellipse at points $A$ and $B$, and $F_{1}$ is its left focus, then the perimeter of $\\Delta A F_{1} B$ is?", "fact_expressions": "l: Line;PointOnCurve(F2, l);G: Ellipse;Expression(G) = (x^2/16 + y^2/8 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;Intersection(l, G) = {A, B}", "query_expressions": "Perimeter(TriangleOf(A, F1, B))", "answer_expressions": "16", "fact_spans": "[[[2, 7]], [[2, 57]], [[8, 46], [59, 61], [81, 82]], [[8, 46]], [[73, 80]], [[50, 57]], [[73, 86]], [[8, 57]], [[63, 66]], [[67, 70]], [[2, 72]]]", "query_spans": "[[[88, 111]]]", "process": "" }, { "text": "If the focus of a parabola lies on the line $x-2 y-4=0$, then the standard equation of this parabola is?", "fact_expressions": "G: Parabola;Z: Line;Expression(Z) = (x - 2*y - 4 = 0);PointOnCurve(Focus(G), Z)", "query_expressions": "Expression(G)", "answer_expressions": "{y^2 = 16*x, x^2 = -8*y}", "fact_spans": "[[[1, 4], [25, 28]], [[8, 21]], [[8, 21]], [[1, 22]]]", "query_spans": "[[[25, 35]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$. The line $l$ passing through point $F$ with slope $\\sqrt{3}$ intersects $C$ at points $A$ and $B$. The circle with diameter $AB$ intersects the $y$-axis at points $M$ and $N$. Let $Q$ be the midpoint of segment $AB$. If the distance from point $F$ to the directrix of $C$ is $3$, then the value of $\\sin \\angle Q M N$ is?", "fact_expressions": "l: Line;C: Parabola;p: Number;G: Circle;B: Point;A: Point;Q: Point;M: Point;N: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Slope(l) = sqrt(3);Intersection(l, C) = {A, B};IsDiameter(LineSegmentOf(A, B), G);Intersection(G, yAxis) = {M, N};MidPoint(LineSegmentOf(A, B)) = Q;Distance(F, Directrix(C)) = 3", "query_expressions": "Sin(AngleOf(Q, M, N))", "answer_expressions": "5/8", "fact_spans": "[[[56, 61]], [[2, 28], [62, 65], [127, 130]], [[10, 28]], [[88, 89]], [[70, 73]], [[66, 69]], [[117, 120]], [[95, 98]], [[99, 102]], [[32, 35], [37, 41], [122, 126]], [[10, 28]], [[2, 28]], [[2, 35]], [[36, 61]], [[42, 61]], [[56, 75]], [[76, 89]], [[88, 104]], [[106, 120]], [[122, 140]]]", "query_spans": "[[[142, 165]]]", "process": "Given that $ p = 3 $, the equation of the parabola and the equation of line $ AB $ can be obtained. By solving the system of equations consisting of the line $ AB $ and the parabola, and applying Vieta's formulas and the midpoint coordinate formula, the coordinates of the midpoint $ Q $ of $ AB $ and the chord length $ |AB| $ can be found. Then the radius of circle $ Q $ can be determined. In $ \\triangle QMN $, using the definition of acute angle trigonometric functions, the required value is obtained. The parabola $ C: y^2 = 2px $ ($ p > 0 $) has focus $ F\\left(\\frac{p}{2}, 0\\right) $ and directrix $ x = -\\frac{p}{2} $. Given $ p = 3 $, the equation of the parabola is $ y^2 = 6x $, $ F\\left(\\frac{3}{2}, 0\\right) $, and the equation of line $ AB $ is $ y = \\sqrt{3}\\left(x - \\frac{3}{2}\\right) $. From\n$$\n\\begin{cases}\ny = \\sqrt{3}\\left(x - \\frac{3}{2}\\right),\n\\end{cases}\n$$\nwe obtain $ 3x^2 - 15x + \\frac{27}{4} = 0 $. Let the abscissae of points $ A $ and $ B $ be $ x_1 $ and $ x_2 $, respectively. Then $ x_1 + x_2 = 5 $. Therefore, the coordinates of the midpoint $ Q $ of $ AB $ are $ \\left(\\frac{5}{2}, \\sqrt{3}\\right) $, and $ |AB| = x_1 + x_2 + p = 5 + 3 = 8 $. Thus, the radius of circle $ Q $ is 4. In $ \\triangle QMN $, $ \\sin\\angle QMN = \\frac{\\frac{5}{2}}{4} = \\frac{5}{8} $." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), a line passing through the left focus $F$ with slope $\\frac{1}{4}$ intersects an asymptote of $C$ at point $A$, and $A$ lies in the first quadrant. If $|O A|=|O F|$ ($O$ is the origin), then the equation of the asymptotes of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;O: Origin;A: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;Slope(G) = 1/4;PointOnCurve(F, G);Intersection(G, OneOf(Asymptote(C))) = A;Quadrant(A) = 1;Abs(LineSegmentOf(O, A)) = Abs(LineSegmentOf(O, F))", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y = pm*x*8/15", "fact_spans": "[[[2, 63], [92, 95], [143, 146]], [[10, 63]], [[10, 63]], [[89, 91]], [[132, 135]], [[102, 106], [108, 111]], [[68, 71]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[72, 91]], [[64, 91]], [[89, 106]], [[108, 116]], [[118, 131]]]", "query_spans": "[[[143, 154]]]", "process": "The system of equations \\begin{cases}y=\\frac{1}{4}(x+c)\\\\y=\\frac{b}{a}x\\end{cases} has the solution \\begin{cases}x=\\frac{ac}{4b-a}\\\\y=\\frac{bc}{4b-a}\\end{cases}, so point A is \\left(\\frac{ac}{4b-a},\\frac{bc}{4b-a}\\right). Therefore, the asymptotes of hyperbola C are given by the equation y=\\pm\\frac{8}{15}x." }, { "text": "What is the eccentricity of the hyperbola $16 x^{2}-9 y^{2}=144$?", "fact_expressions": "G: Hyperbola;Expression(G) = (16*x^2 - 9*y^2 = 144)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/3", "fact_spans": "[[[0, 25]], [[0, 25]]]", "query_spans": "[[[0, 31]]]", "process": "" }, { "text": "The standard equation of an ellipse with foci on the $x$-axis, eccentricity $e=\\frac{1}{2}$, and passing through $(2 \\sqrt{2}, \\sqrt{3})$ is?", "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (2*sqrt(2), sqrt(3));PointOnCurve(Focus(G), xAxis);PointOnCurve(H, G);Eccentricity(G)=e;e:Number;e=1/2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/12+y^2/9=1", "fact_spans": "[[[55, 57]], [[30, 54]], [[30, 54]], [[0, 57]], [[29, 57]], [[9, 57]], [[12, 27]], [[12, 27]]]", "query_spans": "[[[55, 64]]]", "process": "Let the equation of the ellipse be $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Since the eccentricity is $e=\\frac{1}{2}$ and the ellipse passes through the point $(2\\sqrt{2},\\sqrt{3})$, we have $\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{1}{4}$, $\\frac{8}{a^{2}}+\\frac{3}{b^{2}}=1$. Solving these equations gives $a^{2}=12$, $b^{2}=9$. Therefore, the standard equation of the ellipse is $\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1$." }, { "text": "The coordinates of the focus of the parabola $y^{2}+8 x=0$ are?", "fact_expressions": "G: Parabola;Expression(G) = (8*x + y^2 = 0)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(-2, 0)", "fact_spans": "[[[0, 16]], [[0, 16]]]", "query_spans": "[[[0, 23]]]", "process": "" }, { "text": "Given that the line $l$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, intersects the directrix at point $C$, $F$ is the focus of the parabola, and $\\overrightarrow{B C}=3 \\overrightarrow{F B}$, then the value of $|A F|-|B F|$ is?", "fact_expressions": "l: Line;G: Parabola;B: Point;C: Point;F: Point;A: Point;Expression(G) = (y^2 = 4*x);Intersection(l, G) = {A, B};Intersection(l,Directrix(G))=C;Focus(G) = F;VectorOf(B, C) = 3*VectorOf(F, B)", "query_expressions": "Abs(LineSegmentOf(A, F)) - Abs(LineSegmentOf(B, F))", "answer_expressions": "3/2", "fact_spans": "[[[2, 7]], [[8, 22], [48, 51]], [[28, 31]], [[39, 43]], [[44, 47]], [[24, 27]], [[8, 22]], [[2, 33]], [[2, 43]], [[44, 54]], [[56, 101]]]", "query_spans": "[[[103, 120]]]", "process": "According to $\\overrightarrow{BC}=3\\overrightarrow{FB}$, construct the following: draw a perpendicular line $BB_{1}$ from $B$ to the directrix, with foot $B_{1}$; draw a perpendicular line $AE$ from $A$ to the directrix, with foot $E$. By the geometric property of the parabola: $BB_{1}=BF$. So the inclination angle of line $AB$ is $\\alpha$, $\\cos\\alpha=\\cos\\angle CBB_{1}=\\frac{1}{3}\\frac{|BB_{1}|}{|DF|}=\\frac{|BC|}{|CF|}$, thus $\\frac{|BB_{1}|}{2}=\\frac{3}{4}$, so $|BF|=|BB_{1}|=\\frac{3}{2}$, $|FC|=6$. Also, since $\\frac{|BB_{1}|}{|BC|}=\\frac{|EA|}{|AC|}$, i.e., $\\frac{1}{3}=\\frac{|EA|}{|AC|}$, $\\frac{1}{3}=\\frac{|FA|}{|AF|+|FC|}$, $\\frac{1}{3}=\\frac{|FA|}{|AF|+6}$. Solving gives: $|FA|=3$, so $|AF|-|BF|=\\frac{3}{2}$." }, { "text": "Given the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ has its upper focus at $F$, and the lines $x+y+1=0$ and $x+y-1=0$ intersect the ellipse at points $A$, $B$, $C$, $D$, then $A F+B F+C F+D F$=?", "fact_expressions": "G: Ellipse;H: Line;A: Point;l:Line;B: Point;C: Point;D: Point;Expression(G) = (x^2/3 + y^2/4 = 1);Expression(H) = (x + y + 1 = 0);Expression(l) = (x+y-1=0);UpperFocus(G) = F;Intersection(H,G)={A,B};Intersection(l,G)={C,D};F:Point", "query_expressions": "LineSegmentOf(A,F)+LineSegmentOf(B,F)+LineSegmentOf(C,F)+LineSegmentOf(D,F)", "answer_expressions": "8", "fact_spans": "[[[2, 39], [70, 72]], [[48, 59]], [[75, 79]], [[60, 69]], [[80, 83]], [[84, 87]], [[88, 91]], [[2, 39]], [[48, 59]], [[60, 69]], [[2, 47]], [[48, 91]], [[48, 91]], [[44, 47]]]", "query_spans": "[[[93, 112]]]", "process": "" }, { "text": "Given the ellipse $\\frac{y^{2}}{9}+x^{2}=1$, a line passing through the point $P\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$ intersects the ellipse at points $A$ and $B$, and the chord $AB$ is bisected by the point $P$. Then the equation of the line $AB$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/9 = 1);P: Point;Coordinate(P) = (1/2, 1/2);L: Line;PointOnCurve(P,L);Intersection(P,G) = {A,B};A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B)) = P", "query_expressions": "Expression(LineOf(A, B))", "answer_expressions": "", "fact_spans": "[[[2, 29], [65, 67]], [[2, 29]], [[31, 61], [88, 92]], [[31, 61]], [[62, 64]], [[30, 64]], [[62, 79]], [[70, 73]], [[74, 77]], [[65, 87]], [[82, 94]]]", "query_spans": "[[[96, 108]]]", "process": "" }, { "text": "Given the line $ l $: $ x - y + 2 = 0 $ intersects the $ x $-axis at point $ A $, and point $ P $ lies on line $ l $. There is exactly one point $ B $ on the circle $ C $: $ (x - 2)^2 + y^2 = 2 $ such that $ AB \\perp BP $. Then the set of possible values for the $ x $-coordinate of point $ P $ is?", "fact_expressions": "l: Line;C: Circle;A: Point;B: Point;P: Point;Expression(C) = (y^2 + (x - 2)^2 = 2);Expression(l) = (x - y + 2 = 0);Intersection(l, xAxis) = A;PointOnCurve(P, l);PointOnCurve(B, C);IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, P))", "query_expressions": "Range(XCoordinate(P))", "answer_expressions": "{5, 1/3}", "fact_spans": "[[[2, 18], [35, 40]], [[42, 67]], [[25, 29]], [[74, 78]], [[30, 34], [97, 101]], [[42, 67]], [[2, 18]], [[2, 29]], [[30, 41]], [[42, 78]], [[80, 95]]]", "query_spans": "[[[97, 112]]]", "process": "From AB⊥BP, point B lies on circle D with AP as diameter. Since circle D is tangent to circle C, D lies on an ellipse with foci A and C, thus its trajectory equation can be obtained. D lies on line l; solving simultaneously gives coordinates of D, and finally coordinates of point P. From AB⊥BP, point B lies on circle D with AP as diameter, so circle D is tangent to circle C. From the given conditions, A(-2,0), C(2,0). If circle D is externally tangent to circle C, then DC−DA=√2; if circle D is internally tangent to circle C, then DA−DC=√2. Therefore, the center D lies on the hyperbola x²/2−y²/2=1 with foci A and C, i.e., 14x²−2y²=7. Also, point D lies on line l. From { y=x+2, 14x²−2y²=7 }, we get 12x²−8x−15=0, solving gives x_D=3/2 or x_D=−5/6. Thus x_P=2x_D−x_A=2x_D+2=5 or x_P=1/2." }, { "text": "For the curve $C$: $\\frac{x^{2}}{4-k}+\\frac{y^{2}}{k-1}=1$ to represent an ellipse with foci on the $x$-axis, what is the range of values for $k$?", "fact_expressions": "E: Ellipse;C: Curve;k: Number;Expression(C) = (x^2/(4 - k) + y^2/(k - 1) = 1);E = C;PointOnCurve(Focus(E), xAxis)", "query_expressions": "Range(k)", "answer_expressions": "(1, 5/2)", "fact_spans": "[[[58, 60]], [[2, 48]], [[62, 65]], [[2, 48]], [[2, 60]], [[50, 60]]]", "query_spans": "[[[62, 71]]]", "process": "According to the problem, we obtain the following system of inequalities in $ k $: \n$$\n\\begin{cases}\n4 - k > 0, \\\\\nk - 1 > 0, \\\\\n4 - k > k - 1,\n\\end{cases}\n$$\nsolving this system gives the answer. \n$\\because$ the curve $ C: \\frac{x^2}{4 - k} + \\frac{y^{2}}{k - 1} = 1 $ represents an ellipse with foci on the $ x $-axis, \n$ k - 1 > 0 $, solving yields: $ k \\in (1, \\frac{5}{2}) $, \n$ 4 - k > k - 1 $." }, { "text": "Given that the equation $\\frac{x^{2}}{k-7}-\\frac{y^{2}}{k-12}=1$ represents an ellipse with foci on the $y$-axis, what is the range of real values for $k$?", "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(k - 7) - y^2/(k - 12) = 1);PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(k)", "answer_expressions": "(7,19/2)", "fact_spans": "[[[55, 57]], [[59, 64]], [[2, 57]], [[46, 57]]]", "query_spans": "[[[59, 71]]]", "process": "" }, { "text": "Through one focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), draw two tangent lines to the circle $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$. If $\\angle A O B=120^{\\circ}$ ($O$ is the 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;G: Circle;Expression(G) = (x^2 + y^2 = a^2);L1: Line;L2: Line;TangentOfPoint(OneOf(Focus(C)), G) = {L1, L2};A: Point;B: Point;TangentPoint(L1, G) = A;TangentPoint(L2, G) = B;AngleOf(A, O, B) = ApplyUnit(120, degree);O: Origin", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[1, 62], [146, 152]], [[1, 62]], [[8, 62]], [[8, 62]], [[8, 62]], [[8, 62]], [[68, 88]], [[68, 88]], [], [], [[0, 93]], [[99, 102]], [[103, 106]], [[0, 106]], [[0, 106]], [[108, 134]], [[135, 138]]]", "query_spans": "[[[146, 158]]]", "process": "\\because\\angleAOB=120^{\\circ}\\Rightarrow\\angleAOF=60^{\\circ}\\Rightarrow\\angleAFO=30^{\\circ}\\Rightarrowc=2a,\\thereforee=\\frac{c}{a}=2." }, { "text": "Given that $P$ and $Q$ are two points on the parabola $x^{2}=2 y$, with the x-coordinates of points $P$ and $Q$ being $4$ and $-2$ respectively, tangents to the parabola are drawn at $P$ and $Q$, and these two tangents intersect at point $A$. Then the x-coordinate of point $A$ is?", "fact_expressions": "P: Point;Q: Point;PointOnCurve(P,G) = True;PointOnCurve(Q,G) = True;G: Parabola;Expression(G) = (x^2 = 2*y);XCoordinate(P) = 4;XCoordinate(Q) = -2;L1: Line;L2: Line;TangentOfPoint(P,G) = L1;TangentOfPoint(Q,G) = L2;Intersection(L1,L2) = A;A:Point", "query_expressions": "XCoordinate(A)", "answer_expressions": "1", "fact_spans": "[[[2, 5], [28, 32], [54, 57]], [[6, 9], [33, 36], [58, 61]], [[2, 27]], [[2, 27]], [[10, 24], [64, 67]], [[10, 24]], [[28, 52]], [[28, 52]], [], [], [[53, 70]], [[53, 70]], [[53, 80]], [[76, 80], [82, 86]]]", "query_spans": "[[[82, 92]]]", "process": "Since points P and Q lie on the parabola, P(4,8), Q(-2,2). According to the problem, the slopes of the two tangent lines exist. Let the tangent line through point P be y = k_{1}(x - 4) + 8. Then \n\\begin{cases} y = k_{1}(x - 4) + 8 \\\\ y = \\frac{1}{2}x^{2} \\end{cases} \nThus, \\frac{1}{2}x^{2} - k_{1}x + 4k_{1} - 8 = 0. \nSince the line is tangent to the parabola, A = k_{1}^{2} - 4 \\times \\frac{1}{2} \\times (4k_{1} - 8) = 0. \nSolving gives k_{1} = 4. Therefore, the tangent line through point P is y = 4x - 8. \nSimilarly, let the tangent line through point Q be y = k_{2}(x + 2) + 2. Then \n\\begin{cases} y = k_{2}(x + 2) + 2 \\\\ y = \\frac{1}{2}x^{2} \\end{cases}, \nThus, \\frac{1}{2}x^{2} - k_{2}x - 2k_{2} - 2 = 0. \nSince the line is tangent to the parabola, A = k_{2}^{2} + 4k_{2} + 4 = 0. \nSolving gives k_{2} = -2. Therefore, the tangent line through point Q is y = -2x - 2. \nSolving the system of equations \n\\begin{cases} y = 4x - 8 \\\\ y = -2x - 2 \\end{cases}, \ngives x = 1. Thus, the x-coordinate of the intersection point of the two tangent lines is 1." }, { "text": "There exists a point $M$ on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ such that its distance to the left focus is twice its distance to the right directrix. What is the minimum value of the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;M: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(M, G);Distance(M,LeftFocus(G))=2*Distance(M,RightDirectrix(G))", "query_expressions": "Min(Eccentricity(G))", "answer_expressions": "(sqrt(17)-3)/2", "fact_spans": "[[[0, 52], [84, 86]], [[2, 52]], [[2, 52]], [[57, 60], [61, 62], [70, 71]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 60]], [[0, 82]]]", "query_spans": "[[[84, 95]]]", "process": "" }, { "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ passes through the point $(4,3)$, and one focus of the hyperbola lies on the directrix of the parabola $y^{2}=20x$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;C:Parabola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (4, 3);PointOnCurve(H,OneOf(Asymptote(G)));Expression(C)=(y^2=20*x);PointOnCurve(OneOf(Focus(G)),Directrix(C))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16-y^2/9=1", "fact_spans": "[[[2, 48], [65, 68], [95, 98]], [[5, 48]], [[5, 48]], [[55, 63]], [[74, 89]], [[2, 48]], [[55, 63]], [[2, 63]], [[74, 89]], [[65, 93]]]", "query_spans": "[[[95, 103]]]", "process": "The directrix of the parabola y^{2}=20x is x=-5. Since a focus of the hyperbola lies on the directrix of the parabola y^{2}=20x, we have c=5. 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. Since one asymptote of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 passes through the point (4,3), the point (4,3) lies on the line y=\\frac{b}{a}x, i.e., 4\\cdot\\frac{b}{a}=3, so 4b=3a, b=\\frac{3}{4}a. Squaring both sides gives b^{2}=\\frac{9}{16}a^{2}=c^{2}-a^{2}=25-a^{2}. Then \\frac{25}{16}a^{2}=25. Thus a^{2}=16, b^{2}=25-16=9. Therefore, the equation of the hyperbola is \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1." }, { "text": "Let $A$ and $B$ be the two endpoints of the major axis of the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{m}=1$. If there exists a point $M$ on $C$ such that $\\angle A M B=120^{\\circ}$, then the range of values for $m$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/3 + y^2/m = 1);m: Number;A: Point;B: Point;Endpoint(MajorAxis(C)) = {A, B};M: Point;PointOnCurve(M, C);AngleOf(A, M, B) = ApplyUnit(120, degree)", "query_expressions": "Range(m)", "answer_expressions": "(0, 1]+[9, +oo)", "fact_spans": "[[[9, 51], [60, 63]], [[9, 51]], [[100, 103]], [[1, 4]], [[5, 8]], [[1, 58]], [[66, 70]], [[60, 70]], [[72, 98]]]", "query_spans": "[[[100, 110]]]", "process": "Suppose the foci of the ellipse lie on the x-axis, then when 0 < m < 3, assuming M is at an endpoint of the minor axis, ∠AMB takes its maximum value. To have a point M on the ellipse C such that ∠AMB = 120°, ∠AMB ≥ 120°, ∠AMO ≥ 60°, tan∠AMO = √3 / √m ≥ tan60° = √3, solving gives: 0 < m ≤ 1; when the foci of the ellipse lie on the y-axis, m > 3, assuming M is at an endpoint of the minor axis, ∠AMB takes its maximum value. To have a point M on the ellipse C such that ∠AMB = 120°, ∠AMB ≥ 120°, ∠AMO ≥ 60°, tan∠AMO = √m / √3 ≥ tan60° = √3, solving gives: m ≥ 9, ∴ the range of m is (0,1] ∪ [9,+∞)" }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2/3", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 43]]]", "process": "According to the problem, a=3, c=\\sqrt{a^{2}-b^{2}}=2, so the eccentricity is \\frac{2}{3}." }, { "text": "If the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ are $F_{1}$, $F_{2}$, and $P$ is a point on the hyperbola such that $|P F_{1}|=3|P F_{2}|$, then the range of values for 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;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, 2]", "fact_spans": "[[[1, 58], [84, 87], [117, 120]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[64, 71]], [[72, 79]], [[1, 79]], [[80, 83]], [[80, 90]], [[92, 114]]]", "query_spans": "[[[117, 130]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. If a point $P$ on the hyperbola satisfies $\\angle F_{1} P F_{2}=90^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "9", "fact_spans": "[[[2, 41], [67, 70]], [[2, 41]], [[50, 57]], [[58, 65]], [[2, 65]], [[2, 65]], [[73, 76]], [[67, 76]], [[77, 110]]]", "query_spans": "[[[112, 139]]]", "process": "" }, { "text": "The equation of a hyperbola that shares a common asymptote with the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{3}=1$ and has a directrix given by $x=\\frac{5}{2}$ is?", "fact_expressions": "G: Hyperbola;Z: Hyperbola;Expression(G) = (x^2/5 - y^2/3 = 1);Asymptote(Z) = Asymptote(G);Expression(OneOf(Directrix(Z))) = (x = 5/2)", "query_expressions": "Expression(Z)", "answer_expressions": "x^2/10 - y^2/6 = 1", "fact_spans": "[[[1, 39]], [[70, 73]], [[1, 39]], [[0, 73]], [[47, 73]]]", "query_spans": "[[[70, 77]]]", "process": "" }, { "text": "Let the focus of the parabola $y^{2}=8x$ be $F$, the directrix be $l$, and let $P$ be a point on the parabola. Let $PA \\perp l$, where $A$ is the foot of the perpendicular. If the angle of inclination of $AF$ is $\\frac{2\\pi}{3}$, then $|PF|=$?", "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;l: Line;Expression(G) = (y^2 = 8*x);Focus(G) = F;Directrix(G)=l;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,A),l);FootPoint(LineSegmentOf(P,A),l)=A;Inclination(LineSegmentOf(A, F)) = (2*pi)/3", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "8", "fact_spans": "[[[1, 15], [36, 39]], [[58, 62]], [[19, 22]], [[31, 35]], [[26, 29]], [[1, 15]], [[1, 22]], [[1, 29]], [[32, 42]], [[43, 57]], [[43, 65]], [[68, 95]]]", "query_spans": "[[[97, 106]]]", "process": "Test analysis: Since the parabola equation is y^{2}=8x, the focus F is (2,0), and the directrix l has equation x=-2. Since the inclination angle of line AF is \\frac{2\\pi}{3}, the equation of line AF is y=-\\sqrt{3}(x-2). Solving \\begin{cases}x=-2\\\\y=-\\sqrt{3}(x-2)\\end{cases}, the coordinates of point A are (-2,4\\sqrt{3}). Since PA\\bot l and A is the foot of the perpendicular, the y-coordinate of point P is 4\\sqrt{3}. Substituting into the parabola equation gives the coordinates of point P as (6,4\\sqrt{3}). |PF|=|PA|=6-(-2)=8." }, { "text": "From a moving point $P$, two tangent lines $PA$ and $PB$ are drawn to the circle $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$ respectively. If $\\angle APB = 120^{\\circ}$, then the equation of the locus of the moving point $P$ is?", "fact_expressions": "G: Circle;Expression(G) = (x^2 + y^2 = 1);P: Point;TangentOfPoint(P,G) = {LineSegmentOf(P,A),LineSegmentOf(P,B)};TangentPoint(LineSegmentOf(P,A),G) = A;TangentPoint(LineSegmentOf(P,B),G) = B;A: Point;B: Point;AngleOf(A, P, B) = ApplyUnit(120, degree)", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2+y^2=4/3", "fact_spans": "[[[7, 23]], [[7, 23]], [[3, 6], [85, 88]], [[0, 41]], [[6, 53]], [[6, 53]], [[46, 49]], [[50, 53]], [[55, 81]]]", "query_spans": "[[[85, 95]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through point $F_{1}$ intersects the ellipse $C$ at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{2}$?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/16 + y^2/8 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, l) ;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "16", "fact_spans": "[[[2, 45], [86, 91]], [[2, 45]], [[54, 61], [71, 79]], [[62, 69]], [[2, 69]], [[2, 69]], [[70, 85]], [[80, 85]], [[80, 101]], [[92, 95]], [[96, 99]]]", "query_spans": "[[[103, 129]]]", "process": "Since a=4, according to the definition of an ellipse, |AF_{1}|+|AF_{2}|=2a, |BF_{1}|+|BF_{2}|=2a, therefore the perimeter of \\triangle ABF_{2} is 4a=16" }, { "text": "Given that $P$ is a point on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $O$ is the origin, $F_{1}$ and $F_{2}$ are the left and right foci of the curve $C$, respectively. If $|O P|=|O F_{2}|$ and $\\tan \\angle P F_{2} F_{1}=3$, 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;P: Point;PointOnCurve(P, C);O: Origin;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F2));Tan(AngleOf(P, F2, F1)) = 3", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[6, 67], [96, 101], [160, 163]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5]], [[2, 70]], [[71, 74]], [[80, 87]], [[88, 95]], [[80, 106]], [[80, 106]], [[108, 125]], [[129, 158]]]", "query_spans": "[[[160, 169]]]", "process": "\\because|OP|=|OF_{2}|.\\thereforeO is the center of the circumcircle of \\trianglePF_{1}F_{2}.\\therefore\\angleF_{1}PF_{2}=90^{\\circ} Also \\tan\\anglePF_{2}F_{1}=3,\\therefore|PF_{1}|=3|PF_{2}|, by the definition of hyperbola |PF_{1}|-|PF_{2}|=2a, solving |PF_{1}|=3a,|PF_{2}|=a, from |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}, i.e., (3a)^{2}+a^{2}=4c^{2}, i.e., c^{2}=\\frac{5}{2}a^{2}, so _{e}=\\sqrt{10}" }, { "text": "Given that $A$ and $B$ are the two endpoints of the major axis of the ellipse $C$: $\\frac{x^{2}}{m+1}+\\frac{y^{2}}{m}=1$, and $P$ is a moving point on the ellipse $C$, and the maximum value of $\\angle A P B$ is $\\frac{2 \\pi}{3}$, then the value of the real number $m$ is?", "fact_expressions": "C: Ellipse;m: Real;A: Point;P: Point;B: Point;Expression(C) = (x^2/(m + 1) + y^2/m = 1);Endpoint(MajorAxis(C))={A,B};PointOnCurve(P, C);Max(AngleOf(A, P, B)) = 2*pi/3", "query_expressions": "m", "answer_expressions": "1/2", "fact_spans": "[[[10, 54], [67, 72]], [[116, 121]], [[2, 5]], [[63, 66]], [[6, 9]], [[10, 54]], [[2, 62]], [[63, 76]], [[78, 114]]]", "query_spans": "[[[116, 125]]]", "process": "When point P is at the endpoint of the minor axis, ∠APB reaches its maximum value, from which the relationship between a and b can be obtained. Using the standard equation of the ellipse, the value of m can be determined. As shown in the figure, when point P is at the endpoint of the minor axis, ∠APB reaches its maximum value. Since the maximum value of ∠APB is \\frac{2\\pi}{3}, it follows that \\tan\\frac{\\angleAPB}{2}=\\tan\\frac{\\pi}{3}=\\sqrt{3}, so \\frac{a}{b}=\\sqrt{3}. Therefore, \\frac{\\sqrt{m+1}}{\\sqrt{m}}=\\sqrt{3}. Solving gives m=\\frac{1}{2}." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively. A line $l$ passing through point $F_{1}$ intersects the ellipse $C$ at points $A$, $B$. Given that $(\\overrightarrow{A F_{2}}+ \\overrightarrow{F_{1} F_{2}}) \\cdot \\overrightarrow{A F_{1}}=0$ and $\\overrightarrow{A F_{1}}=\\frac{4}{3} \\overrightarrow{F_{1}B}$, find the eccentricity of the ellipse $C$.", "fact_expressions": "l: Line;C: Ellipse;A: Point;F2: Point;F1: Point;B: Point;LeftFocus(C) = F1;RightFocus(C)=F2;Expression(C)=(x^2/a^2 + y^2/b^2 = 1);PointOnCurve(F1, l);Intersection(l, C) = {A, B};DotProduct((VectorOf(A,F2)+VectorOf(F1,F2)),VectorOf(A,F1))=0;VectorOf(A,F1)=4/3*VectorOf(F1,B);a:Number;b:Number;a>b;b>0", "query_expressions": "Eccentricity(C)", "answer_expressions": "5/7", "fact_spans": "[[[91, 96]], [[0, 57], [97, 102], [276, 281]], [[103, 106]], [[73, 80]], [[65, 72], [82, 90]], [[107, 110]], [[0, 80]], [[0, 80]], [[0, 57]], [[81, 96]], [[91, 112]], [[115, 206]], [[209, 274]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]]]", "query_spans": "[[[276, 287]]]", "process": "\\because(\\overrightarrow{AF}_{2}+\\overrightarrow{F_{1}F_{2}})\\cdot\\overrightarrow{AF_{1}}=(\\overrightarrow{AF}_{2}+\\overrightarrow{F_{1}F_{2}})\\cdot(\\overrightarrow{AF_{2}}-\\overrightarrow{F_{1}F_{2}})=\\overrightarrow{AF_{2}}^{2}-F_{1}F_{2}^{2}=0 and |\\overrightarrow{F_{1}F_{2}}|=2c, |\\overrightarrow{a}|=|\\overrightarrow{F_{1}F_{2}}|=2c\\therefore|\\overrightarrow{AF}|=2a-\\because\\overrightarrow{AF_{1}}=\\frac{4}{3}\\overrightarrow{F_{1}B}=2a-2c\\therefore|\\overrightarrow{BF}_{1}|=\\frac{3}{2}(a-c),|\\overrightarrow{BF_{2}}|=\\frac{a}{2}+\\frac{3c}{2} Let the midpoint of AF_{1} be H, then F_{2}H\\botAB, |\\overrightarrow{AH}|=a-c,|\\overrightarrow{BH}|=\\frac{5}{2}(a-c).|\\overrightarrow{F_{2}A}|^{2}-|\\overrightarrow{AH}|^{2}=|\\overrightarrow{F_{2}B}|^{2}-|\\overrightarrow{BH}|^{2} Substituting the data and simplifying: 7c^{2}-12ac+5a^{2}=0, dividing both sides of the equation by a^2: 7e^{2}-12e+5=0, solving e=\\frac{5}{7} or e=1 (rejected). So the eccentricity of ellipse C is e=\\frac{5}{7}" }, { "text": "The equation of the directrix of the parabola $x^{2}=\\frac{1}{2} y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y/2)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y = -1/8", "fact_spans": "[[[0, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "Since the parabola equation is: $x^{2}=\\frac{1}{2}y$, so $2p=\\frac{1}{2}$, that is, $p=\\frac{1}{4}$. The directrix equation is $y=-\\frac{p}{2}=-\\frac{1}{8}$. [Note] This question examines finding the directrix of a parabola, which is a simple problem." }, { "text": "Let $F_{1}$, $F_{2}$ be the two 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 hyperbola such that $|\\overrightarrow{O P}|=|\\overrightarrow{O F_{1}}|$, where $O$ is the origin. If the area of $\\Delta F_{1} F_{2} P$ is $4$, then $b=$?", "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;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);O: Origin;Abs(VectorOf(O, P)) = Abs(VectorOf(O, F1));Area(TriangleOf(F1, F2, P)) = 4", "query_expressions": "b", "answer_expressions": "2", "fact_spans": "[[[17, 73], [83, 86]], [[17, 73]], [[187, 190]], [[20, 73]], [[20, 73]], [[20, 73]], [[1, 8]], [[9, 16]], [[1, 78]], [[79, 82]], [[79, 90]], [[146, 149]], [[93, 144]], [[156, 185]]]", "query_spans": "[[[187, 192]]]", "process": "Let P(x_{P}, y_{P}). If |\\overrightarrow{OP}| = |\\overrightarrow{OF_{1}}|, then the trajectory equation of point P is: x^{2} + y^{2} = c. Solving simultaneously the circle equation x^{2} + y^{2} = c and the hyperbola equation \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1, we obtain: y_{p} = \\pm\\frac{b^{2}}{c}. Then the area of \\triangle F_{1}F_{2}P is: S = \\frac{1}{2} \\times 2c \\times |y_{p}| = \\frac{1}{2} \\times 2c \\times \\frac{b^{2}}{c} = b^{2} = 4. Combining with b > 0, we get b = 2." }, { "text": "Given the parabola $ C $: $ y^{2} = 2 p x $ ($ p > 0 $) with focus $ F $, a line $ l $ with slope $ \\sqrt{3} $ passing through $ F $ intersects the parabola $ C $ at points $ A $ and $ B $. If the ordinate of the midpoint of segment $ A B $ is $ \\sqrt{3} $, then the equation of the parabola $ C $ is?", "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;B: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;PointOnCurve(F, l);Slope(l) = sqrt(3);Intersection(l, C) = {A, B};YCoordinate(MidPoint(LineSegmentOf(A,B))) = sqrt(3)", "query_expressions": "Expression(C)", "answer_expressions": "y^2=6*x", "fact_spans": "[[[54, 59]], [[2, 27], [60, 66], [104, 110]], [[9, 27]], [[67, 70]], [[71, 74]], [[31, 34], [36, 39]], [[9, 27]], [[2, 27]], [[2, 34]], [[35, 59]], [[40, 59]], [[54, 76]], [[78, 102]]]", "query_spans": "[[[104, 115]]]", "process": "" }, { "text": "Let a chord $AB$ of the parabola $y^2 = 4x$ have point $P(1, 1)$ as its midpoint. Then the length of chord $AB$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (1, 1);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=P", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "sqrt(15)", "fact_spans": "[[[1, 15]], [[19, 24]], [[19, 24]], [[25, 36]], [[1, 15]], [[25, 36]], [[1, 24]], [[19, 39]]]", "query_spans": "[[[42, 51]]]", "process": "" }, { "text": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has two foci $F_{1}$ and $F_{2}$. Point $P$ lies on the hyperbola. If $PF_{1} \\perp PF_{2}$, then what is the distance from point $P$ to the $x$-axis?", "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);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2))", "query_expressions": "Distance(P, xAxis)", "answer_expressions": "16/5", "fact_spans": "[[[0, 39], [66, 69]], [[0, 39]], [[45, 52]], [[53, 60]], [[0, 60]], [[61, 65], [95, 99]], [[61, 70]], [[72, 93]]]", "query_spans": "[[[95, 109]]]", "process": "" }, { "text": "It is known that the center of the hyperbola is at the origin, and its vertices are the foci of the ellipse $x^{2}+4 y^{2}=64$, with a focal length of $16$. Then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;H: Ellipse;O:Origin;Expression(H) = (x^2 + 4*y^2 = 64);Center(G)=O;Vertex(G)=Focus(H);FocalLength(G) = 16", "query_expressions": "Expression(G)", "answer_expressions": "x^2/48-y^2/16=1", "fact_spans": "[[[2, 5], [2, 5]], [[13, 33]], [[8, 10]], [[13, 33]], [[2, 10]], [[2, 39]], [[2, 48]]]", "query_spans": "[[[50, 59]]]", "process": "(Convert the ellipse equation to standard form, find the coordinates of the foci, thus obtain a; then from the hyperbola's focal distance obtain c, then find b^{2}, and thus obtain the solution. The ellipse x^{2}+4y^{2}=64 is converted to the standard equation \\frac{x^{2}}{64}+\\frac{y^{2}}{16}=1; then the foci coordinates of the ellipse are (\\pm4\\sqrt{3},0). Let the standard equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, then a^{2}=48. According to the problem, the hyperbola's focal distance 2c=16, so c=8, thus b^{2}=c^{2}-a^{2}=16. Therefore, the standard equation of the hyperbola is \\frac{x^{2}}{48}-\\frac{y^{2}}{16}=1)" }, { "text": "If the distance from the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ to its asymptote is equal to $\\frac{\\sqrt{3}}{4}$ times the focal length, 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;Distance(RightFocus(G), Asymptote(G)) = sqrt(3)/4*FocalLength(G)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 57], [96, 99]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 94]]]", "query_spans": "[[[96, 105]]]", "process": "" }, { "text": "A line $ l $ passing through the focus of the parabola $ y^{2} = 4x $ intersects the parabola at points $ P(x_{1}, y_{1}) $ and $ Q(x_{2}, y_{2}) $. If $ x_{1} + x_{2} = 6 $, then $ |PQ| $ equals?", "fact_expressions": "l: Line;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (x1,y1);Coordinate(Q) = (x2,y2);PointOnCurve(Focus(G),l);Intersection(l, G) = {P,Q};x1:Number;x2:Number;y1:Number;y2:Number;x1+x2=6;G:Parabola", "query_expressions": "Abs(LineSegmentOf(P, Q))", "answer_expressions": "8", "fact_spans": "[[[19, 24]], [[29, 46]], [[47, 64]], [[1, 15]], [[29, 46]], [[47, 64]], [[0, 24]], [[19, 66]], [[29, 46]], [[47, 64]], [[29, 46]], [[47, 64]], [[69, 84]], [[1, 15], [25, 28]]]", "query_spans": "[[[86, 96]]]", "process": "From the given conditions, F(1,0), p=2, |PF|=x_{1}+\\frac{p}{2}, |QF|=x_{2}+\\frac{p}{2}, so |PQ|=|PF|+|QF|=x_{1}+x_{2}+p=6+2=8." }, { "text": "Point $A(x_{0}, y_{0})$ lies on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{32}=1$. If the distance from point $A$ to the right focus is equal to $2 x_{0}$, then $x_{0}=$?", "fact_expressions": "G: Hyperbola;A: Point;x0:Number;y0:Number;Expression(G) = (x^2/4 - y^2/32 = 1);Coordinate(A) = (x0, y0);PointOnCurve(A, RightPart(G));Distance(A,RightFocus(G)) = 2*x0", "query_expressions": "x0", "answer_expressions": "2", "fact_spans": "[[[19, 58]], [[0, 18], [64, 68]], [[88, 95]], [[1, 18]], [[19, 58]], [[0, 18]], [[0, 62]], [[19, 86]]]", "query_spans": "[[[88, 97]]]", "process": "From the hyperbola $\\frac{x^2}{4}-\\frac{y^{2}}{32}=1$, we have $a=2$, $b=4\\sqrt{2}$, then $c=\\sqrt{a^{2}+b^{2}}=6$, so the right focus of the hyperbola is $F(6,0)$. Substituting point $A(x_{0},y_{0})$ $(x_{0}\\geqslant2)$ into the equation of the hyperbola, we get $y_{0}^{2}=8x_{0}^{2}-32$. Therefore, $|AF|=\\sqrt{(x_{0}-6)^{2}+y_{0}^{2}}=\\sqrt{(x_{0}-6)^{2}+8x_{0}^{2}-32}=2x_{0}$. Rearranging yields $5x_{0}^{2}-12x_{0}+4=0$, solving gives $x_{0}=2$ or $x_{0}=\\frac{2}{5}$ (discarded). So $x_{0}=2$." }, { "text": "Given that a line passing through the focus $F$ of the parabola $y^{2} = -4x$ with slope $\\sqrt{3}$ intersects the parabola at points $A$ and $B$, then $\\frac{|A F| \\cdot |B F|}{|A B|}$ = ?", "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = -4*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Slope(H) = sqrt(3)", "query_expressions": "(Abs(LineSegmentOf(A, F))*Abs(LineSegmentOf(B, F)))/Abs(LineSegmentOf(A, B))", "answer_expressions": "1", "fact_spans": "[[[3, 18], [44, 47]], [[41, 43]], [[49, 52]], [[21, 24]], [[53, 56]], [[3, 18]], [[3, 24]], [[2, 43]], [[41, 58]], [[27, 43]]]", "query_spans": "[[[60, 94]]]", "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(-1,0) $, the directrix equation is $ x = 1 $. The line passing through $ F $ with slope $ \\sqrt{3} $ has equation $ y = \\sqrt{3}(x+1) $. Substituting into the parabola equation $ y^{2} = 4x $, we obtain $ 3x^{2} + 10x + 3 = 0 $, solving gives $ x = -3 $ or $ x = -\\frac{1}{3} $. By the definition of the parabola, $ |AF| = 1 + 3 = 4 $, $ |BF| = 1 + \\frac{1}{3} = \\frac{4}{3} $, $ |AB| = 2 - (-3 - \\frac{1}{3}) = \\frac{16}{3} $, then $ \\frac{|AF| \\cdot |BF|}{|AB|} = \\frac{4 \\times \\frac{4}{3}}{\\frac{16}{3}} = $" }, { "text": "The standard equation of the hyperbola with foci at $(\\pm 3 , 0)$ and asymptotes $y=\\pm \\sqrt{2} x$ is?", "fact_expressions": "G: Hyperbola;Coordinate(Focus(G)) = (pm*3, 0);Expression(Asymptote(G)) = (y=pm*(sqrt(2)*x))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/3-y^2/6=1", "fact_spans": "[[[43, 46]], [[0, 46]], [[20, 46]]]", "query_spans": "[[[43, 52]]]", "process": "" }, { "text": "Given the equation of parabola $C$ is $y=-\\frac{1}{4} x^{2}$, then its directrix equation is?", "fact_expressions": "C: Parabola;Expression(C) = (y = -x^2/4)", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "y = 1", "fact_spans": "[[[2, 8], [35, 36]], [[2, 33]]]", "query_spans": "[[[35, 42]]]", "process": "The standard equation of parabola C is x^{2}=-4y, and the directrix is y=1." }, { "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{8}=1$, $P$ is a point on the left branch of $C$, and the coordinates of point $A$ are $(0,4)$, then the minimum perimeter of $\\triangle A P F$ is?", "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2/8 = 1);Coordinate(A) = (0, 4);RightFocus(C)= F;PointOnCurve(P,LeftPart(C))", "query_expressions": "Min(Perimeter(TriangleOf(A, P, F)))", "answer_expressions": "12", "fact_spans": "[[[6, 39], [48, 51]], [[58, 62]], [[44, 47]], [[2, 5]], [[6, 39]], [[58, 73]], [[2, 43]], [[44, 57]]]", "query_spans": "[[[75, 100]]]", "process": "Let the left focus be F_{1}(-3,0). By the definition of the hyperbola, the perimeter of \\triangle APF is transformed into |AP|+|PF_{1}|+2+|AF|, and thus the solution can be obtained. From the hyperbola equation x^{2}-\\frac{y^{2}}{8}=1, we have a=1, c=3, so F(3,0), left focus F_{1}(-3,0). When point P moves on the left branch of the hyperbola, by the definition of the hyperbola, |PF|-|PF_{1}|=2, hence |PF|=|PF_{1}|+2. Thus, the perimeter of \\triangle APF is |AP|+|PF|+|AF|=|AP|+|PF_{1}|+2+|AF|. Since |AF|=\\sqrt{3^{2}+4^{2}}=5 is constant, the perimeter of \\triangle APF is minimized when (|AP|+|PF_{1}|) is minimized. At this time, point P lies at the intersection of line segment AF_{1} and the hyperbola, as shown in the figure. At this point, (|AP|+|PF_{1}|)=|AF_{1}|=\\sqrt{3^{2}+4^{2}}=5. Therefore, the minimum perimeter of \\triangle APF is 12. Hence, the answer is: 12" }, { "text": "If the foci of the ellipse $\\frac{x^{2}}{3 m}+\\frac{y^{2}}{2 m+1}=1$ lie on the $x$-axis, then the range of real values for $m$ is?", "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/(2*m + 1) + x^2/((3*m)) = 1);PointOnCurve(Focus(G), xAxis)", "query_expressions": "Range(m)", "answer_expressions": "(1,+oo)", "fact_spans": "[[[1, 44]], [[55, 60]], [[1, 44]], [[1, 53]]]", "query_spans": "[[[55, 67]]]", "process": "Since the foci of the ellipse $\\frac{x^{2}}{3m}+\\frac{y^{2}}{2m+1}=1$ lie on the $x$-axis, it follows that $3m>2m+1>0$. Solving this inequality yields $m>1$. Therefore, the range of the real number $m$ is $(1,+\\infty)$." }, { "text": "The asymptotes of a hyperbola are given by $y=\\pm \\frac{3}{4} x$. What is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3/4)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{5/4, 5/3}", "fact_spans": "[[[0, 3], [33, 36]], [[0, 31]]]", "query_spans": "[[[33, 42]]]", "process": "Let the hyperbola equation be: \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=\\lambda. When \\lambda>0, a^{2}=16\\lambda, b^{2}=9\\lambda, c^{2}=25\\lambda, \\therefore e=\\frac{c}{a}=\\frac{5}{4}; when \\lambda<0, b^{2}=-16\\lambda, a^{2}=-9\\lambda, c^{2}=-25\\lambda, \\therefore e=\\frac{c}{a}=\\frac{5}{3};" }, { "text": "The equation of the hyperbola that has the same asymptotes as $\\frac{x^{2}}{2}-y^{2}=1$ and passes through the point $(-2,-3)$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);G1: Hyperbola;Asymptote(G) = Asymptote(G1);H: Point;Coordinate(H) = (-2, -3);PointOnCurve(H, G1) = True", "query_expressions": "Expression(G1)", "answer_expressions": "y^2/7-x^2/14=1", "fact_spans": "[[[1, 29]], [[1, 29]], [[51, 54]], [[0, 54]], [[40, 50]], [[40, 50]], [[39, 54]]]", "query_spans": "[[[51, 58]]]", "process": "" }, { "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. If $F_{1}$ and $F_{2}$ are the two foci of the ellipse, then $PF_{1}+PF_{2}$=?", "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G) = True;F1: Point;F2: Point;Focus(G) = {F1, F2}", "query_expressions": "LineSegmentOf(P, F1) + LineSegmentOf(P, F2)", "answer_expressions": "10", "fact_spans": "[[[1, 4]], [[5, 44], [66, 68]], [[5, 44]], [[1, 47]], [[50, 57]], [[58, 65]], [[50, 73]]]", "query_spans": "[[[75, 92]]]", "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 $F_{1}$ and $F_{2}$ respectively, the length of the real axis is $2$, the asymptotes are $y=\\pm \\frac{1}{2} x$, and $|M F_{1}|-|M F_{2}|=2$. Point $N$ lies on the circle $\\Omega$: $x^{2}+y^{2}-2 y=0$. Then the minimum value of $|M N|+|M F_{1}|$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;Omega: Circle;M: Point;F1: Point;F2: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Omega) = (-2*y + x^2 + y^2 = 0);LeftFocus(C) = F1;RightFocus(C) = F2;Length(RealAxis(C)) = 2;Expression(Asymptote(C)) = (y = pm*(1/2)*x);Abs(LineSegmentOf(M, F1)) - Abs(LineSegmentOf(M, F2)) = 2;PointOnCurve(N, Omega)", "query_expressions": "Min(Abs(LineSegmentOf(M, F1)) + Abs(LineSegmentOf(M, N)))", "answer_expressions": "5/2", "fact_spans": "[[[2, 63]], [[10, 63]], [[10, 63]], [[152, 182]], [[123, 146]], [[72, 79]], [[80, 87]], [[147, 151]], [[10, 63]], [[10, 63]], [[2, 63]], [[152, 182]], [[2, 87]], [[2, 87]], [[2, 95]], [[2, 121]], [[123, 146]], [[147, 183]]]", "query_spans": "[[[185, 208]]]", "process": "From the given condition, we have 2a = 2, so a = 1. The asymptotes of hyperbola C are y = \\pm\\frac{b}{a}x = \\pm\\frac{1}{2}x, hence b = \\frac{1}{2}. Therefore, the equation of hyperbola C is x^{2} - \\frac{y^{2}}{4} = 1, and then c = \\sqrt{a^{2} + b^{2}} = \\frac{\\sqrt{5}}{2}. Since |MF_{1}| - |MF_{2}| = 2 < |F_{1}F_{2}| = \\sqrt{5}, point M lies on the right branch of hyperbola C. The standard equation of circle Q is x^{2} + (y - 1)^{2} = 1, with center Q(0, 1) and radius 1, as shown in the figure below: Because |MF_{1}| = |MF_{2}| + 2, therefore |MN| + |MF| = |MN| + |MF_{2}| + 2 \\geqslant |NF_{2}| + 2 \\geqslant |OF_{2}| - 1 + 2 = \\sqrt{(}\\sqrt{(0 - \\frac{\\sqrt{5}}{2})^{2} + 1^{2}} + 1 = \\frac{5}{5}. The minimum value of |MN| + |MF| is \\frac{5}{5}, achieved if and only if points N, O, F_{2} are collinear and point N lies on segment OF_{2}." }, { "text": "Given that a line with slope $2$ passes through the focus $F$ of the parabola $y^{2}=p x$ ($p>0$) and intersects the $y$-axis at point $A$. If the area of $\\triangle OAF$ ($O$ is the origin) is $1$, then $p=$?", "fact_expressions": "G: Parabola;p: Number;H: Line;O: Origin;A: Point;F:Point;p>0;Expression(G) = (y^2 = p*x);Slope(H)=2;Focus(G)=F;PointOnCurve(F,H);Intersection(H,yAxis)=A;Area(TriangleOf(O,A,F))=1", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[12, 31]], [[90, 93]], [[9, 11]], [[70, 73]], [[47, 51]], [[34, 37]], [[15, 31]], [[12, 31]], [[2, 11]], [[12, 37]], [[9, 37]], [[9, 51]], [[54, 88]]]", "query_spans": "[[[90, 95]]]", "process": "" }, { "text": "Given that the focus $F$ of the parabola $y^{2}=2 p x(p>0)$ is a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$, and the line passing through the intersection points of the two curves goes exactly through the point $F$, then the eccentricity of the hyperbola is?", "fact_expressions": "D: Parabola;Expression(D) = ( y^2 = 2*(p*x) );p: Number;p>0;F: Point;F = Focus(D);Focus(D) = OneOf(Focus(C));C: Hyperbola;Expression(C) = (x^2/a^2 - y^2/b^2 = 1 );a: Number;b: Number;a>0;b>0;l: Line;PointOnCurve(F,l);PointOnCurve(Intersection(C,D),l)", "query_expressions": "Eccentricity(C)", "answer_expressions": "1+sqrt(2)", "fact_spans": "[[[2, 22]], [[2, 22]], [[5, 22]], [[5, 22]], [[25, 28], [105, 109]], [[2, 28]], [[2, 91]], [[29, 86], [112, 115]], [[29, 86]], [[32, 86]], [[32, 86]], [[32, 86]], [[32, 86]], [[101, 103]], [[101, 109]], [[92, 103]]]", "query_spans": "[[[112, 121]]]", "process": "" }, { "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{n}-y^{2}=1$ $(n>1)$ are $F_{1}$ and $F_{2}$ respectively, and a point $P$ lies on the hyperbola such that $|P F_{1}|+|P F_{2}|=2 \\sqrt{n+2}$, then the area of $\\Delta P F_{1} F_{2}$ is?", "fact_expressions": "G: Hyperbola;n: Number;P: Point;F1: Point;F2: Point;n>1;Expression(G) = (-y^2 + x^2/n = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 2*sqrt(n + 2)", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "1", "fact_spans": "[[[1, 34], [64, 67]], [[4, 34]], [[59, 63]], [[43, 50]], [[51, 58]], [[4, 34]], [[1, 34]], [[1, 58]], [[1, 58]], [[59, 68]], [[72, 106]]]", "query_spans": "[[[108, 135]]]", "process": "Using the properties of the focal triangle combined with the given conditions, we obtain |PF_{1}| = \\sqrt{n+2} + \\sqrt{n}, |PF_{2}| = \\sqrt{n+2} - \\sqrt{n}, thus the focal triangle is a right triangle, and its area can be found. [Detailed Solution] Suppose point P lies on the right branch of the hyperbola, then |PF_{1}| - |PF_{2}| = 2\\sqrt{n}, |PF_{1}| + |PF_{2}| = 2\\sqrt{n+2}. Solving gives |PF_{1}| = \\sqrt{n+2} + \\sqrt{n}, |PF_{2}| = \\sqrt{n+2} - \\sqrt{n}. Also, |F_{1}F_{2}| = 2\\sqrt{n+1}. Hence, |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2}, so PF_{1} \\bot PF_{2}, therefore \\triangle PF_{1}F_{2} is a right triangle. Also, |PF_{1}| \\cdot |PF_{2}| = 2, thus S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2} \\cdot |PF_{1}| \\cdot |PF_{2}| = \\frac{1}{2} \\cdot 2 = 1" }, { "text": "It is known that hyperbola $C$ shares the same asymptotes with the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ and passes through the point $(2,2)$. Then the equation of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;H:Hyperbola;Expression(H)=(x^2-y^2/4=1);G: Point;Coordinate(G) = (2, 2);Asymptote(C)=Asymptote(H);PointOnCurve(G,C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/3-y^2/12=1", "fact_spans": "[[[2, 8], [57, 63]], [[9, 37]], [[9, 37]], [[47, 55]], [[47, 55]], [[2, 44]], [[2, 55]]]", "query_spans": "[[[57, 68]]]", "process": "Since hyperbola C has the same asymptotes as the hyperbola \\( x^{2} - \\frac{y^{2}}{4} = 1 \\), the equation of hyperbola C can be written as \\( x^{2} - \\frac{y^{2}}{4} = k \\) (\\( k \\neq 0 \\)). Since hyperbola C passes through the point \\( (2,2) \\), we have \\( 2^{2} - \\frac{2^{2}}{4} = k \\), so \\( k = 3 \\). Therefore, \\( \\frac{x^{2}}{3} - \\frac{y^{2}}{12} = 1 \\) is the required equation." }, { "text": "If the focus of the parabola $y^{2}=-2 p x(p>0)$ coincides with the left 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);p>0;Expression(H) = (y^2 = -2*p*x);Focus(H) = LeftFocus(G)", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[27, 55]], [[1, 23]], [[63, 66]], [[27, 55]], [[4, 23]], [[1, 23]], [[1, 61]]]", "query_spans": "[[[63, 69]]]", "process": "" }, { "text": "Given two points $A(-1,0)$, $B(1,0)$, if there exists a point $P(x, y)$ on the line $x - y + a = 0$ such that $\\angle APB = 90^{\\circ}$, then the range of real values for $a$ is?", "fact_expressions": "G: Line;a: Real;A: Point;B: Point;P: Point;Expression(G) = (a + x - y = 0);Coordinate(A) = (-1, 0);Coordinate(B) = (1, 0);Coordinate(P) = (x1, y1);PointOnCurve(P, G);AngleOf(A, P, B) = ApplyUnit(90, degree);x1:Number;y1:Number", "query_expressions": "Range(a)", "answer_expressions": "[-sqrt(2),sqrt(2)]", "fact_spans": "[[[25, 36]], [[81, 86]], [[4, 13]], [[15, 23]], [[39, 51]], [[25, 36]], [[4, 13]], [[15, 23]], [[39, 51]], [[25, 51]], [[54, 79]], [[40, 51]], [[40, 51]]]", "query_spans": "[[[83, 93]]]", "process": "Since there exists a point P(x,y) on the line x−y+a=0 such that ∠APB=90°, it follows that AP⊥BP, i.e., $\\overrightarrow{AP}\\bot\\overrightarrow{BP}$, $\\overrightarrow{AP}=(x+1,y)$, $\\overrightarrow{BP}=(x-1,y)$, so $\\overrightarrow{AP}\\cdot\\overrightarrow{BP}=(x+1)(x-1)+y^{2}=0$, simplifying gives: $x^{2}+y^{2}=1$. From the condition, the line $x−y+a=0$ intersects the circle $x^{2}+y^{2}=1$, so the distance from the center $(0,0)$ to the line $x−y+a=0$ is $d=\\frac{|a|}{\\sqrt{1^{2}+1^{2}}}\\leqslant1$, solving yields: $-\\sqrt{2}\\leqslant a \\leqslant \\sqrt{2}$" }, { "text": "A hyperbola has an asymptote given by $y = \\sqrt{3} x$ and one focus at $(0, 2)$. What is the standard equation of this hyperbola?", "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Point;Coordinate(H) = (0, 2);OneOf(Focus(G)) = H", "query_expressions": "Expression(G)", "answer_expressions": "y^2/3-x^2=1", "fact_spans": "[[[41, 44]], [[0, 44]], [[26, 35]], [[26, 35]], [[25, 44]]]", "query_spans": "[[[41, 50]]]", "process": "" }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$, $A B$ is a chord passing through the focus $F$, $A A_{1} \\perp$ directrix $l$ at point $A_{1}$, $B B_{1} \\perp$ directrix $l$ at point $B_{1}$, $N$ is the midpoint of $A_{1} B_{1}$, if $A A_{1}=4$, $B B_{1}=2$, then the length of segment $N F$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2=2*p*x);p: Number;p>0;A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),G) = True;Focus(G) = F;F: Point;PointOnCurve(F,LineSegmentOf(A,B)) = True;l: Line;Directrix(G) = l;IsPerpendicular(LineSegmentOf(A,A1),l) = True;A1: Point;FootPoint(LineSegmentOf(A,A1),l) = A1;IsPerpendicular(LineSegmentOf(B,B1),l) = True;B1: Point;FootPoint(LineSegmentOf(B,B1),l) = B1;N: Point;MidPoint(LineSegmentOf(A1,B1)) = N;LineSegmentOf(A,A1) = 4;LineSegmentOf(B,B1) = 2", "query_expressions": "Length(LineSegmentOf(N,F))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[2, 23]], [[2, 23]], [[5, 23]], [[5, 23]], [[26, 31]], [[26, 31]], [[2, 42]], [[2, 38]], [[35, 38]], [[26, 42]], [[60, 63], [90, 93]], [[2, 63]], [[43, 63]], [[64, 72]], [[44, 72]], [[73, 93]], [[94, 102]], [[73, 102]], [[103, 106]], [[103, 122]], [[124, 135]], [[138, 149]]]", "query_spans": "[[[151, 162]]]", "process": "Let the coordinates of point A and point B be given. By the definition of the parabola, the x-coordinate of point A is expressed as $x_{1}=4-\\frac{p}{2}$, and the x-coordinate of point B is $x_{2}=2-\\frac{p}{2}$. Substitute the equation of line AB into the parabola equation, and use Vieta's formulas to obtain $x_{1}\\cdot x_{2}=\\frac{p^{2}}{4}$, solve to get $p=\\frac{8}{3}$, thus the coordinates of focus F and point N can be found, and the distance NF can be calculated using the distance formula between two points. By the symmetry of the parabola, assume point $A(x_{1},y_{1})$ with $y_{1}>0$, $A_{1}(-\\frac{p}{2},y_{1})$, point $B(x_{2},y_{2})$ with $y_{2}<0$, $B_{1}(-\\frac{p}{2},y_{2})$. By the definition of the parabola, $AA_{1}=x_{1}+\\frac{p}{2}=4$, so $x_{1}=4-\\frac{p}{2}$; $BB_{1}=x_{2}+\\frac{p}{2}=2$, so $x_{2}=2-\\frac{p}{2}$. The slope of line AB exists. Let the line AB be: $y=k(x-\\frac{p}{2})$. Substitute into the parabola equation and simplify to get: $k^{2}x^{2}-(k^{2}p+2p)x+\\frac{p^{2}k^{2}}{4}=0$. Therefore, $(4-\\frac{p}{2})\\cdot(2-\\frac{p}{2})=\\frac{p^{2}}{4}$, solving gives $p=\\frac{8}{3}$. By Vieta's formulas, $x_{1}\\cdot x_{2}=\\frac{p^{2}}{4}$, so $(4-\\frac{p}{2})\\cdot(2-\\frac{p}{2})=\\frac{p^{2}}{4}$, solving gives $p=\\frac{8}{3}$. Thus the focus coordinate is $F(\\frac{4}{8},0)$, $x_{1}=4-\\frac{p}{2}=4-\\frac{4}{3}=\\frac{8}{3}$, $x_{2}=2-\\frac{p}{2}=2-\\frac{4}{3}=\\frac{2}{3}$, so $y_{1}=\\sqrt{2\\times\\frac{8}{3}\\times\\frac{8}{3}}=\\frac{\\sqrt{82}}{3}$, $y_{2}=-\\sqrt{2\\times\\frac{8}{3}\\times}$ so point N $-\\frac{4}{3},\\sqrt{3}$. By the distance formula, $|NF|=|(-\\frac{4}{3}-\\frac{4}{3})^{2}+\\frac{2\\sqrt{2}}{3}|=2\\sqrt{2}$" }, { "text": "Given two points $M(-3,0)$, $N(3,0)$, and a point $P$ in the coordinate plane such that $|\\overrightarrow{M N}| \\cdot|\\overrightarrow{M P}|+\\overrightarrow{M N} \\cdot \\overrightarrow{M P}=0$, find the minimum value of the distance from the moving point $P(x, y)$ to the point $M(-3,0)$.", "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-3, 0);Coordinate(N) = (3, 0);Coordinate(P) = (x1, y1);x1:Number;y1:Number;DotProduct(VectorOf(M, N), VectorOf(M, P)) + DotProduct(Abs(VectorOf(M, N)),Abs(VectorOf(M, P))) = 0", "query_expressions": "Min(Distance(P, M))", "answer_expressions": "3", "fact_spans": "[[[4, 13], [156, 166]], [[16, 24]], [[145, 155], [25, 29]], [[4, 13]], [[16, 24]], [[145, 155]], [[145, 155]], [[145, 155]], [[39, 141]]]", "query_spans": "[[[145, 175]]]", "process": "Since M(-3,0), N(3,0), it follows that \\overrightarrow{MN}=(6,0), |\\overrightarrow{MN}|=6, \\overrightarrow{MP}=(x+3,y), \\overrightarrow{NP}=(x-3,y). From |\\overrightarrow{MN}|\\cdot|\\overrightarrow{MP}|+\\overrightarrow{MN}\\cdot\\overrightarrow{NP}=0, we obtain 6\\sqrt{(x+3)^{2}+y^{2}}+6(x-3)=0. Simplifying yields y^{2}=-12x. Thus, M is the focus of the parabola y^{2}=-12x. Therefore, the minimum distance from point P to point M is the distance from the origin to M(-3,0), which is 3." }, { "text": "If the line $x + y - 2 = 0$ passes through the focus of the parabola $y = m x^{2}$, then $m =$?", "fact_expressions": "G: Parabola;m: Number;H: Line;Expression(G) = (y = m*x^2);Expression(H) = (x + y - 2 = 0);PointOnCurve(Focus(G),H)", "query_expressions": "m", "answer_expressions": "1/8", "fact_spans": "[[[14, 28]], [[33, 36]], [[1, 12]], [[14, 28]], [[1, 12]], [[1, 31]]]", "query_spans": "[[[33, 38]]]", "process": "The equation of the parabola can be written as x^{2}=\\frac{1}{m}y, so the focus lies on the y-axis. Since the line x+y-2=0 passes through the focus, the focus is (0,2). Therefore, \\frac{1}{4m}=2, solving gives m=\\frac{1}{8}." }, { "text": "Given $A(2,-1)$, $B(-1,1)$, and $O$ as the origin, a moving point $M$ satisfies $\\overrightarrow{O M}=m \\overrightarrow{O A}+n \\overrightarrow{O B}$, where $m, n \\in \\mathbb{R}$ and $2 m^{2}-n^{2}=2$. Then the equation of the trajectory of $M$ is?", "fact_expressions": "A: Point;B: Point;O: Origin;M: Point;m: Real;n: Real;Coordinate(A) = (2, -1);Coordinate(B) = (-1, 1);VectorOf(O, M) = m*VectorOf(O, A) + n*VectorOf(O, B);2*m^2 - n^2 = 2", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2/2 - y^2 = 1", "fact_spans": "[[[2, 11]], [[13, 23]], [[25, 28]], [[36, 39], [145, 148]], [[112, 124]], [[112, 124]], [[2, 11]], [[13, 23]], [[41, 109]], [[126, 143]]]", "query_spans": "[[[145, 155]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4x$ has focus $F$, a line $l$ with inclination angle $\\frac{\\pi}{3}$ passes through point $F$ and intersects the parabola $C$ at points $A$ and $B$. Then the area of $\\triangle AOB$ is?", "fact_expressions": "l: Line;C: Parabola;A: Point;O: Origin;B: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Inclination(l)=pi/3;PointOnCurve(F,l);Intersection(l,C)={A,B}", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[49, 54]], [[2, 21], [62, 68]], [[70, 73]], [[81, 98]], [[74, 77]], [[25, 28], [55, 59]], [[2, 21]], [[2, 28]], [[29, 54]], [[49, 59]], [[49, 79]]]", "query_spans": "[[[81, 103]]]", "process": "From the given conditions, the focus of the parabola is F(1,0), and the equation of line l is y=\\sqrt{3}(x-1). Substituting into y^{2}=4x and simplifying yields 3x^{2}-10x+3=0, or (3x-1)(x-3)=0, giving solutions x=\\frac{1}{3} or 3. Let points A(x_{1},y_{1}) and B(x_{2},y_{2}), then |x_{1}-x_{2}|=\\frac{8}{3}, |y_{1}-y_{2}|=\\sqrt{3}|x_{1}-x_{2}|=\\frac{8\\sqrt{3}}{3}. Therefore, S_{\\triangleAOB}=\\frac{1}{2}|OF|\\cdot|y_{1}-y_{2}|=\\frac{1}{2}\\times1\\times\\frac{8\\sqrt{3}}{3}=\\frac{4\\sqrt{3}}{3}" }, { "text": "The curve of the equation $4 x^{2}+k y^{2}=1$ is an ellipse with foci on the $y$-axis. What is the range of values for $k$?", "fact_expressions": "G: Ellipse;H: Curve;k: Number;Expression(H) = (4*x^2 + k*y^2 = 1);PointOnCurve(Focus(G), yAxis);H = G", "query_expressions": "Range(k)", "answer_expressions": "(0, 4)", "fact_spans": "[[[34, 36]], [[22, 24]], [[38, 41]], [[0, 24]], [[25, 36]], [[22, 36]]]", "query_spans": "[[[38, 48]]]", "process": "The ellipse equation $4x^{2}+ky^{2}=1$ is rewritten as $\\frac{x^{2}}{\\frac{1}{4}}+\\frac{y^{2}}{\\frac{1}{k}}=1$. Since the foci of the ellipse lie on the $y$-axis, $\\frac{1}{k}>\\frac{1}{4}$, that is, $00, b>0)$, if the eccentricity of the hyperbola is $2$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2=2*p*x);Coordinate(A) = (2, 4);PointOnCurve(OneOf(Focus(G)), Directrix(H));Eccentricity(G) = 2;p:Number;PointOnCurve(A,H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2 - y^2/3 = 1", "fact_spans": "[[[38, 95], [102, 105], [116, 119]], [[41, 95]], [[41, 95]], [[12, 28], [31, 34]], [[2, 11]], [[41, 95]], [[41, 95]], [[38, 95]], [[12, 28]], [[2, 11]], [[31, 100]], [[102, 113]], [[15, 28]], [2, 28]]", "query_spans": "[[[116, 123]]]", "process": "Test analysis: Since point A(2,4) lies on the parabola y^{2}=2px, \\therefore 16=4p, so p=4. \\therefore The directrix of the parabola is x=-2. Since the directrix of the parabola passes through a focus of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, (a>0,b>0), then c=2, and e=\\frac{c}{a}=2, \\therefore a=1, then b^{2}=c^{2}-a^{2}=4-1=3. \\therefore The equation of the hyperbola is x^{2}-\\frac{y^{2}}{3}=1." }, { "text": "Given a hyperbola $E$ centered at the origin with eccentricity $2$, and right vertex $A$. A line $l$ perpendicular to the $x$-axis is drawn through the left focus $F$ of $E$, intersecting $E$ at points $M$ and $N$. If the area of $\\Delta A M N$ is $9$, then the standard equation of $E$ is?", "fact_expressions": "E: Hyperbola;A: Point;M: Point;N: Point;l:Line;O:Origin;Center(E)=O;Eccentricity(E)=2;RightVertex(E)=A;LeftFocus(E)=F;PointOnCurve(F,l);IsPerpendicular(l,xAxis);Intersection(l, E) = {M,N};Area(TriangleOf(A, M, N)) = 9;F:Point", "query_expressions": "Expression(E)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[8, 14], [32, 35], [59, 62], [98, 101]], [[27, 30]], [[64, 67]], [[68, 71]], [[50, 53], [55, 58]], [[5, 7]], [[2, 14]], [[8, 22]], [[8, 30]], [[32, 42]], [[31, 53]], [[43, 53]], [[55, 73]], [[75, 96]], [[39, 42]]]", "query_spans": "[[[98, 108]]]", "process": "Let the standard equation of the hyperbola be $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$. Let $x=-c$, then $\\frac{c^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, solving gives $y=\\pm\\frac{b^{2}}{a}$, so $|MN|=\\frac{2b^{2}}{a}$. It is clear that $|AF|=a+c$, thus $S_{\\triangle AMN}=\\frac{1}{2}|MN||AF|=\\frac{1}{2}\\times\\frac{2b^{2}}{a}\\times(a+c)$. Also $a^{2}+b^{2}=c^{2}\\cdots\\textcircled{2}$, $\\frac{c}{a}=2\\cdots\\textcircled{3}$. Solving equations $\\textcircled{1}$, $\\textcircled{2}$, $\\textcircled{3}$ simultaneously gives the hyperbola equation: $x^{2}-\\frac{y^{2}}{3}=1$." }, { "text": "Draw a line $l$ through the left focus $F_{1}$ of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, intersecting the left branch of the hyperbola at points $P$ and $Q$. If $|P Q|=4$ and $F_{2}$ is the right focus of the hyperbola, then the perimeter of $\\Delta P F_{2} Q$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F1, l);P: Point;Q: Point;Intersection(l, LeftPart(G)) = {P, Q};Abs(LineSegmentOf(P, Q)) = 4", "query_expressions": "Perimeter(TriangleOf(P, F2, Q))", "answer_expressions": "12", "fact_spans": "[[[1, 29], [49, 52], [86, 89]], [[1, 29]], [[33, 40]], [[78, 85]], [[1, 40]], [[78, 93]], [[43, 48]], [[0, 48]], [[55, 58]], [[59, 62]], [[43, 64]], [[66, 75]]]", "query_spans": "[[[95, 118]]]", "process": "According to the problem, draw the figure as follows: |PF₁| = 2, |QF₂| - |QF₁| = 2. Hence |QF₂| + |PF₂| = 4 + |PF₁| + |QF₁| = 4 + |PQ| = 8, therefore the perimeter of △PF₂O is |OF₂| + |PF₂| + |PO| = 8 + 4 = 12" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with foci $F_{1}$ and $F_{2}$. A point $P$ in the plane satisfies $P F_{2} \\perp F_{1} F_{2}$ and $|P F_{2}|=|F_{1} F_{2}|$. The segments $P F_{1}$ and $P F_{2}$ intersect the ellipse at points $A$ and $B$, respectively. If $|P A|=|A F_{1}|$, then $\\frac{|B F_{2}|}{|P F_{2}|}$=?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;A: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(C) ={F1,F2};IsPerpendicular(LineSegmentOf(P, F2), LineSegmentOf(F1, F2));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Intersection(LineSegmentOf(P, F1), C) = A;Intersection(LineSegmentOf(P, F2), C) =B;Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(A, F1))", "query_expressions": "Abs(LineSegmentOf(B, F2))/Abs(LineSegmentOf(P, F2))", "answer_expressions": "sqrt(2)/4", "fact_spans": "[[[2, 59], [80, 81], [175, 177]], [[9, 59]], [[9, 59]], [[89, 92]], [[62, 70]], [[72, 79]], [[178, 182]], [[183, 186]], [[9, 59]], [[9, 59]], [[2, 59]], [[61, 83]], [[94, 121]], [[123, 148]], [[149, 186]], [[149, 186]], [[188, 205]]]", "query_spans": "[[[207, 238]]]", "process": "As shown in the figure, |PF₂| = |F₁F₂| = 2c, and PF₂ ⊥ F₁F₂, ∴ △PF₂F is an isosceles right triangle. Also, ∵ |PA| = |AF₁|, ∴ AF₁ = AF₂ = PA = ½ PF₁ = √2 c. Also |AF₁| + |AF₂| = 2a, ∴ 2a = 2√2 c, i.e., a = √2 c. Substitute into b² = a² − c², get b = c. From PF₂ ⊥ F₁F₂, we know x_B = c. Substitute into the ellipse equation to solve for |y_B| = b²/a = c²/b = c/√2, so |BF₂| = c/√2. Then |BF₂| / |PF₁| = (c/√2) / (2c) = √2 / 4." }, { "text": "The directrix of the parabola $y=-\\frac{1}{8} x^{2}$ intersects the axis of symmetry at point $A$. A line passing through point $A$ intersects the parabola at points $M$ and $N$. Point $B$ lies on the axis of symmetry of the parabola, and $(\\overrightarrow{B M}+\\frac{\\overrightarrow{M N}}{2}) \\cdot \\overrightarrow{M N}=0$. Then the range of $|\\overrightarrow{O B}|$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/8);A: Point;Intersection(Directrix(G), SymmetryAxis(G)) = A;H: Line;PointOnCurve(A, H) = True;Intersection(H, G) = {M, N};M: Point;N: Point;PointOnCurve(B, SymmetryAxis(G)) = True;B: Point;DotProduct((VectorOf(B, M) + VectorOf(M, N)/2), VectorOf(M, N)) = 0;O: Origin", "query_expressions": "Range(Abs(VectorOf(O, B)))", "answer_expressions": "(-oo,-6)", "fact_spans": "[[[0, 25], [48, 51], [67, 70]], [[0, 25]], [[34, 38], [40, 44]], [[0, 38]], [[45, 47]], [[39, 47]], [[45, 61]], [[52, 55]], [[56, 59]], [[62, 74]], [[62, 66]], [[76, 160]], [[162, 186]]]", "query_spans": "[[[162, 193]]]", "process": "The standard equation of the parabola is $x^{2}=-8y$, so the focus is $(0,-2)$, and the directrix equation is $y=2$. Let $A(0,2)$. Suppose the line $MN$ has equation $y=kx+2$, $M(x_{1},y_{1})$, $N(x_{2},y_{2})$. From $\\begin{cases}y=kx+2\\\\y=-\\frac{1}{8}x^{2}\\end{cases}$, we obtain $\\frac{1}{8}x^{2}+kx+2=0$, so $\\Delta=k^{2}-4\\times\\frac{1}{8}\\times2=k^{2}-1>0$, $k^{2}>1$. Thus $x_{1}+x_{2}=-8k$. Let the midpoint of $MN$ be $P(x_{0},y_{0})$, then $x_{0}=\\frac{x_{1}+x_{2}}{2}=-4k$, $y_{0}=kx_{0}+2=-4k^{2}+2$, $\\overrightarrow{BM}+\\frac{\\overrightarrow{MN}}{2}=\\overrightarrow{BM}+\\overrightarrow{MP}=\\overrightarrow{BP}$. From $(\\overrightarrow{BM}+\\frac{\\overrightarrow{MN}}{2})\\cdot\\overrightarrow{MN}=0$, we get $\\overrightarrow{BP}\\cdot\\overrightarrow{MN}=0$, i.e., $BP\\perp MN$. Let $B(0,t)$, then $\\frac{-4k^{2}+2-t}{-4k}\\times k=-1$, $t=-4k^{2}-2$. Since $k^{2}>1$, we have $t=-4k^{2}-2<-6$, so the range of the $y$-coordinate of point $B$ is $(-\\infty,-6)$." }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are the foci, if $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$, $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|=|\\overrightarrow{F_{1} F_{2}}|$, then the eccentricity of the 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);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(P, F1, F2) = pi/6;Abs(VectorOf(P, F1) + VectorOf(P, F2)) = Abs(VectorOf(F1, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[6, 58], [205, 207]], [[8, 58]], [[8, 58]], [[2, 5]], [[62, 69]], [[70, 77]], [[8, 58]], [[8, 58]], [[6, 58]], [[2, 61]], [[6, 80]], [[82, 118]], [[119, 203]]]", "query_spans": "[[[205, 213]]]", "process": "" }, { "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and $|F A|=2|F B|$. If the chord length intercepted by the circle $x^{2}+y^{2}=2 p$ on the line $A B$ is $4$, then $p=$?", "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);Intersection(L, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B));H: Circle;Expression(H) = (x^2 + y^2 = 2*p);Length(InterceptChord(LineOf(A, B), H)) = 4", "query_expressions": "p", "answer_expressions": "{3,6}", "fact_spans": "[[[4, 25], [36, 39]], [[4, 25]], [[107, 110]], [[7, 25]], [[28, 31]], [[4, 31]], [[32, 34]], [[2, 34]], [[32, 51]], [[42, 45]], [[46, 49]], [[53, 67]], [[77, 95]], [[77, 95]], [[69, 105]]]", "query_spans": "[[[107, 112]]]", "process": "The parabola $ y^{2} = 2px $ ($ p > 0 $) has focus $ F\\left(\\frac{p}{2}, 0\\right) $. Let the equation of line $ AB $ be $ x = my + \\frac{p}{2} $. Substituting into $ y^{2} = 2px $ gives $ y^{2} - 2pmx - p^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 2pm $ ①, $ y_{1}y_{2} = -p^{2} $ ②. From $ |FA| = 2|FB| $ we obtain $ y_{1} = -2y_{2} $ ③. Solving ①, ②, and ③ together yields $ m = \\frac{\\sqrt{2}}{4} $. Thus, the equation of line $ AB $ is $ x = \\frac{\\sqrt{2}}{4}y + \\frac{p}{2} $, or $ 4x - \\sqrt{2}y - 2p = 0 $. The distance from the center $ (0,0) $ to line $ AB $ is $ d = \\frac{2p}{\\sqrt{18}} $. The radius of the circle is $ r = \\sqrt{2p} $, and the chord length is 4. Hence, $ 2p - \\frac{4p^{2}}{18} = 4 $, solving gives $ p = 3 $ or $ 6 $." }, { "text": "There is a moving point $A$ on the parabola with vertex at the origin and focus $F(0,1)$, and a moving point $M$ on the circle $(x+1)^{2}+(y-4)^{2}=1$. Then the minimum value of $|AM|+|AF|$ is?", "fact_expressions": "G: Parabola;H: Circle;F: Point;A: Point;M: Point;O:Origin;Vertex(G)=O;Coordinate(F) = (0, 1);Focus(G)=F;Expression(H) = ((x + 1)^2 + (y - 4)^2 = 1);PointOnCurve(A,G);PointOnCurve(M,H)", "query_expressions": "Min(Abs(LineSegmentOf(A,M))+Abs(LineSegmentOf(A,F)))", "answer_expressions": "4", "fact_spans": "[[[20, 23]], [[32, 56]], [[11, 19]], [[28, 31]], [[61, 64]], [[3, 7]], [[0, 23]], [[11, 19]], [[8, 23]], [[32, 56]], [[20, 31]], [[32, 64]]]", "query_spans": "[[[66, 85]]]", "process": "Given the problem, the equation of the parabola is $ x^{2} = 4y $, and its directrix is $ y = -1 $. Let $ d = |AF| $ be the distance from point $ A $ to the directrix. Then the minimum value of $ |AM| + |AF| $ equals the distance from the center of the circle $ (-1, 4) $ to the directrix minus the radius, that is, $ 4 + 1 - 1 = 4 $." }, { "text": "The focal distance of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{4}=1$ is $2$, then what is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (y^2/4 + x^2/m = 1);FocalLength(G) = 2", "query_expressions": "m", "answer_expressions": "{5,3}", "fact_spans": "[[[0, 37]], [[46, 49]], [[0, 37]], [[0, 44]]]", "query_spans": "[[[46, 53]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$), the left focus is $F$. A line $l$ passing through $F$ and perpendicular to one asymptote of $C$ intersects the right branch of $C$ at point $P$. If $A$ is the midpoint of $PF$, and $|OA|=\\frac{3b}{2}-a$ ($O$ is the origin), then the eccentricity of $C$ is?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;P: Point;F: Point;O: Origin;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(F,l);IsPerpendicular(l,OneOf(Asymptote(C)));Intersection(l,RightPart(C))=P;MidPoint(LineSegmentOf(P, F)) = A;Abs(LineSegmentOf(O, A)) = -a + (3*b)/2", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(13)/2", "fact_spans": "[[[90, 95]], [[2, 63], [78, 81], [96, 99], [159, 162]], [[9, 63]], [[9, 63]], [[104, 108]], [[68, 71], [73, 76]], [[148, 151]], [[110, 113]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 71]], [[72, 95]], [[77, 95]], [[90, 108]], [[110, 122]], [[124, 147]]]", "query_spans": "[[[159, 168]]]", "process": "Let the right focus of the hyperbola be $ F_{1} $. Let the line $ l $ intersect the asymptote $ y = -\\frac{b}{a}x $ at point $ B $. It can be found that $ |BF| = b $, $ |OB| = a $, $ |PF_{1}| = 3b - 2a $. By the definition of ellipse, $ |PF| = 3b $, $ |AB| = \\frac{b}{2} $. In the right triangle $ ABO $, $ \\left( \\frac{3}{2}b - a \\right)^{2} = a^{2} + \\left( \\frac{b}{2} \\right)^{2} $, thus $ \\frac{b}{a} = \\frac{3}{2} $. As shown in the figure, let the right focus of the hyperbola be $ F_{1} $, and suppose without loss of generality that the line $ l $ intersects the asymptote $ y = -\\frac{b}{a}x $ at point $ B $. In the right triangle $ BOF $, by the distance from a point to a line, we have \n$$\n\\begin{matrix}\nBF & = \\frac{|-\\frac{b}{a} \\cdot (-c) - 0|}{\\sqrt{1 + \\frac{b^{2}}{a^{2}}}} = \\frac{bc}{a} = b,\n\\end{matrix}\n$$\nsince $ |OF| = c $, then $ |OB| = \\sqrt{|OF|^{2} - |BF|^{2}} = a $. Since $ OA $ is the midline of triangle $ \\triangle PFF_{1} $, $ |OA| = \\frac{3b}{2} - a $, so $ |PF_{1}| = 3b - 2a $. Because $ |PF| - |PF_{1}| = 2a $, it follows that $ |PF| = 3b $, hence $ |AF| = \\frac{3b}{2} $, $ |AB| = |AF| - |BF| = \\frac{b}{2} $. Then in the right triangle $ ABO $, $ \\left( \\frac{3}{2}b - a \\right)^{2} = a^{2} + \\left( \\frac{b}{2} \\right)^{2} $, simplifying yields $ \\frac{b}{a} = \\frac{3}{2} $." }, { "text": "It is known that point $P$ lies 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, respectively. Given that $\\angle F_{1} P F_{2}=60^{\\circ}$ and $|P F_{1}|=3|P F_{2}|$, find the eccentricity of the ellipse.", "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);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree);Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(7)/4", "fact_spans": "[[[7, 59], [82, 84], [153, 155]], [[9, 59]], [[9, 59]], [[64, 71]], [[2, 6]], [[72, 79]], [[9, 59]], [[9, 59]], [[7, 59]], [[2, 63]], [[64, 91]], [[64, 91]], [[94, 127]], [[129, 151]]]", "query_spans": "[[[153, 161]]]", "process": "Using the law of cosines and the definition of an ellipse, set up equations and simplify to find the eccentricity of the ellipse. Given |PF_{1}| = 3|PF_{2}|, combining the definition of the ellipse and the law of cosines yields:\n\\begin{cases}\n|PF_{1}| = 3|PF_{2}| \\\\\n|PF_{1}| + |PF_{2}| = 2a \\\\\n(2c)^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}| \\cdot |PF_{2}| \\cdot \\cos 60\n\\end{cases}\nSimplifying gives \\frac{c^{2}}{a^{2}} = \\frac{7}{16}, e = \\frac{c}{a} = \\frac{\\sqrt{7}}{4}." }, { "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{m}=1$ $(m>0)$, the distance from the right focus $F$ to one of the asymptotes is $2$. What is the focal length of the hyperbola?", "fact_expressions": "C: Hyperbola;m: Number;m>0;Expression(C) = (x^2 - y^2/m = 1);F: Point;RightFocus(C) = F;Distance(F,OneOf(Asymptote(C)))=2", "query_expressions": "FocalLength(C)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 40], [62, 65]], [[9, 40]], [[9, 40]], [[2, 40]], [[44, 47]], [[2, 47]], [[2, 60]]]", "query_spans": "[[[62, 70]]]", "process": "The hyperbola $ C: x^{2} - \\frac{y^{2}}{m} = 1 $ $ (m > 0) $, has its right focus at $ F(c, 0) $. The distance from point $ F $ to the asymptotes $ y = \\pm \\frac{b}{a}x $ is: $ \\frac{|bc|}{\\sqrt{a^{2} + b^{2}}} = b $, so $ b = 2 $. Thus, $ m = b^{2} = 4 $. Therefore, the focal length of the hyperbola is: $ 2\\sqrt{1 + 4} = 2\\sqrt{5} $." }, { "text": "If the length of the real axis of the hyperbola $y^{2}-\\frac{x^{2}}{m}=-1$ is $4$, then its focal distance is equal to?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (y^2 - x^2/m = -1);Length(RealAxis(G))=4", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[1, 30], [40, 41]], [[4, 30]], [[1, 30]], [[1, 38]]]", "query_spans": "[[[40, 46]]]", "process": "The hyperbola equation can be written as \\frac{x^{2}}{m}-y^{2}=1, so we must have m>0. Since the length of the real axis is 4, it follows that 2\\sqrt{m}=4, therefore m=4, so c^{2}=4+1=5, hence the focal distance is 2c=2\\sqrt{5}." }, { "text": "The asymptotes of hyperbola $H$ are $x+2y=0$ and $x-2y=0$. If $H$ passes through the point $P(2, 0)$, then the equation of hyperbola $H$ is?", "fact_expressions": "H: Hyperbola;P: Point;Coordinate(P) = (2, 0);Expression(Asymptote(H))=(pm*2*y+x=0);PointOnCurve(P,H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2/4-y^2=1", "fact_spans": "[[[0, 6], [33, 36], [51, 57]], [[38, 49]], [[38, 49]], [[0, 30]], [[33, 49]]]", "query_spans": "[[[51, 62]]]", "process": "According to the problem, the desired hyperbola equation can be set as (x+2y)(x-2y)=\\lambda(\\lambda\\neq0). Since H passes through the point P(2,0), we have (2+0)(2-0)=\\lambda, thus \\lambda=4. Therefore, the equation of the hyperbola H is x^{2}-4y^{2}=4, or equivalently \\frac{x^{2}}{4}-y^{2}=1." }, { "text": "Given a parabola $y^{2}=4x$ with a chord $AB$, where $A(x_{1},y_{1})$, $B(x_{2}, y_{2})$, and the line containing $AB$ intersects the $y$-axis at the point $(0,2)$, then $\\frac{1}{y_{1}} + \\frac{1}{y_{2}}$=?", "fact_expressions": "G: Parabola;I: Point;A: Point;B: Point;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y^2 = 4*x);Coordinate(I) = (0, 2);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);IsChordOf(LineSegmentOf(A, B), G);Intersection(OverlappingLine(LineSegmentOf(A, B)), yAxis) = I", "query_expressions": "1/y2 + 1/y1", "answer_expressions": "1/2", "fact_spans": "[[[2, 16]], [[82, 90]], [[28, 44]], [[45, 63]], [[28, 44]], [[28, 44]], [[45, 63]], [[45, 63]], [[2, 16]], [[82, 90]], [[28, 44]], [[45, 63]], [[2, 25]], [[65, 90]]]", "query_spans": "[[[92, 129]]]", "process": "" }, { "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{m+3}=1$ represents a hyperbola with foci on the $y$-axis, then the range of values for $m$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (y^2/(m + 3) + x^2/m = 1);PointOnCurve(Focus(G), yAxis);m:Number", "query_expressions": "Range(m)", "answer_expressions": "", "fact_spans": "[[[51, 54]], [[1, 54]], [[42, 54]], [[56, 59]]]", "query_spans": "[[[56, 66]]]", "process": "" }, { "text": "The coordinates of two vertices $A$ and $B$ of $\\triangle A B C$ are $(-5,0)$ and $(5,0)$, respectively. The product of the slopes of the lines containing sides $AC$ and $BC$ is equal to $-\\frac{3}{5}$. Then the equation of the locus of vertex $C$ is?", "fact_expressions": "A: Point;C: Point;B: Point;Coordinate(A) = (-5, 0);Coordinate(B) = (5, 0);Slope(OverlappingLine(LineSegmentOf(B, C)))*Slope(OverlappingLine(LineSegmentOf(A, C))) = -3/5", "query_expressions": "LocusEquation(C)", "answer_expressions": "(x^2/25 + y^2/15 = 1)&(Negation(y = 0))", "fact_spans": "[[[21, 24]], [[96, 99]], [[25, 28]], [[21, 52]], [[21, 52]], [[54, 92]]]", "query_spans": "[[[96, 106]]]", "process": "Let point C(x,y) express the product of the slopes in terms of the coordinates of point C(x,y); simplifying yields the trajectory of vertex C. Let point C(x,y), then k_{AC}\\cdotk_{BC}=\\frac{y}{x+5}\\times\\frac{y}{x-5}=\\frac{y^{2}}{x^{2}-25}(y\\neq0). Therefore, \\frac{y^{2}}{x^{2}-25}=-\\frac{3}{5}(y\\neq0). Simplifying gives \\frac{x^{2}}{25}+\\frac{y^{2}}{15}=1(y\\neq0)." }, { "text": "Given that a line passing through the point $M(p, 0)$ intersects the parabola $y^{2}=2 p x$ $(p>0)$ at two points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$, if $y_{1} y_{2}=-2$, then $x_{1} x_{2}$=?", "fact_expressions": "G: Parabola;p: Number;H: Line;M: Point;A: Point;p>0;x1:Number;y1:Number;x2:Number;y2:Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(M) = (p, 0);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H,G) = {A, B};y1*y2=-2;B:Point;PointOnCurve(M,H)", "query_expressions": "x1*x2", "answer_expressions": "1", "fact_spans": "[[[17, 40]], [[4, 13]], [[14, 16]], [[3, 13]], [[42, 59]], [[20, 40]], [[42, 59]], [[42, 59]], [[62, 79]], [[62, 79]], [[17, 40]], [[3, 13]], [[42, 59]], [[62, 79]], [[14, 81]], [[83, 100]], [[62, 79]], [2, 15]]", "query_spans": "[[[102, 117]]]", "process": "Let the equation of line AB be x = my + p. Combining equations, we have \n\\begin{cases} y^2 = 2px \\\\ x = my + p \\end{cases}, \nwhich simplifies to y^{2} - 2mpy - 2p^{2} = 0. \nThus, y_{1}y_{2} = -2p^{2}. Since y_{1}y_{2} = -2, it follows that -2p^{2} = -2, so p^{2} = 1. \nSince p > 0, we have p = 1. Therefore, \nx_{1}x_{2} = \\frac{y_{1}^{2}}{2} \\cdot \\frac{y_{2}^{2}}{2} = \\frac{1}{4}(y_{1}y_{2})^{2} = 1." }, { "text": "Given the equation of a hyperbola is $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, points $A$ and $B$ lie on the right branch of the hyperbola, the segment $AB$ passes through the right focus $F_2$ of the hyperbola, $|AB|=m$, and $F_{1}$ is the other focus. Then the perimeter of $\\triangle ABF_{1}$ is?", "fact_expressions": "A: Point;B: Point;G: Hyperbola;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(A, RightPart(G));PointOnCurve(B, RightPart(G));Focus(G)={F1,F2};RightFocus(G) = F2;PointOnCurve(F2, LineSegmentOf(A, B));Abs(LineSegmentOf(A, B)) = m;m:Number;b:Number;a:Number", "query_expressions": "Perimeter(TriangleOf(A, B, F))", "answer_expressions": "4*a+2*m", "fact_spans": "[[[53, 58]], [[60, 63]], [[2, 5], [64, 67], [80, 83]], [[104, 111]], [[87, 93]], [[2, 52]], [[53, 71]], [[53, 71]], [[80, 116]], [[80, 93]], [[72, 93]], [[94, 102]], [[94, 102]], [[2, 52]], [[2, 52]]]", "query_spans": "[[[118, 138]]]", "process": "Test analysis: ∵|AF₁||AF₂|=2a, |BF₁|+|BF₂|=2a, and |AF₂|+|BF₂|=|AB|=m, ∴|AF₁|+|BF₁|=4a+m, ∴ the perimeter of △ABF₁ = |AF₁|+|BF₁|+|AB|=4a+2|AB|=4a+2m" }, { "text": "Draw a line perpendicular to the $x$-axis from the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, intersecting the hyperbola at point $P$. Let $F_{2}$ be the right focus. If $\\angle F_{1} P F_{2}=45^{\\circ}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;a>0;b>0;L:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1,L);IsPerpendicular(L,xAxis);Intersection(L,G)=P;AngleOf(F1,P,F2)=ApplyUnit(45,degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "0.2", "fact_spans": "[[[1, 57], [77, 80], [77, 80]], [[4, 57]], [[4, 57]], [[61, 68]], [[81, 85]], [[86, 93]], [[4, 57]], [[4, 57]], [], [[1, 57]], [[1, 68]], [[77, 97]], [[0, 76]], [[0, 76]], [[0, 85]], [[99, 132]]]", "query_spans": "[[[134, 143]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4x$, a line $l$ passing through the point $(2,0)$ intersects $C$ at points $A$ and $B$. What is the product of the slopes of lines $OA$ and $OB$ ($O$ being the origin)?", "fact_expressions": "l: Line;C: Parabola;A: Point;O: Origin;H: Point;B: Point;Expression(C) = (y^2 = 4*x);Coordinate(H) = (2, 0);PointOnCurve(H, l);Intersection(l, C) = {A, B}", "query_expressions": "Slope(LineOf(O, A))*Slope(LineOf(O, B))", "answer_expressions": "-2", "fact_spans": "[[[33, 38]], [[2, 21], [39, 42]], [[43, 46]], [[69, 72]], [[24, 32]], [[47, 50]], [[2, 21]], [[24, 32]], [[23, 38]], [[33, 52]]]", "query_spans": "[[[54, 85]]]", "process": "Let points A(x_{A}, y_{A}), B(x_{B}, y_{B}). Let the equation of line l be x = ty + 2. Then solving \n\\begin{cases}x = ty + 2\\\\y^2 = 4x\\end{cases} \ngives y^{2} - 4ty - 8 = 0, so y_{A} \\cdot y_{B} = -8, thus x_{A} \\cdot x_{B} = \\frac{y_{1}^{2} \\cdot y_{2}^{2}}{16} = 4, and therefore k_{OA} \\cdot k_{OB} = \\frac{y_{A} \\cdot y_{B}}{x_{A} \\cdot x_{B}} = \\frac{-8}{4} = -2" }, { "text": "The distance from a focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ to one of its asymptotes is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)", "query_expressions": "Distance(OneOf(Focus(G)), OneOf(Asymptote(G)))", "answer_expressions": "4", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 55]]]", "process": "According to the problem, one focus of the hyperbola has coordinates (0,5), and the equation of one asymptote is 3x−4y=0. Using the point-to-line distance formula, we obtain $\\frac{|3\\times0-4\\times5|}{\\sqrt{3^{2}+4^{2}}}=4$, which means the distance from one focus of the hyperbola to one asymptote is 4." }, { "text": "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 point $ P $ lies on the right branch of the hyperbola such that $ |P F_{1}| = 3|P F_{2}| $. The lines $ y = \\pm x $ are the asymptotes of $ C $. If the area of $ \\Delta P F_{1} F_{2} $ is $ 4 \\sqrt{2} $, then what is the distance between the foci 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);Focus(C)={F1,F2};PointOnCurve(P,RightPart(C));Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2));Expression(Asymptote(C)) = (y = pm*x);Area(TriangleOf(P, F1, F2)) = 4*sqrt(2)", "query_expressions": "FocalLength(C)", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[0, 62], [86, 89], [129, 132], [180, 186]], [[7, 62]], [[7, 62]], [[82, 85]], [[66, 73]], [[74, 81]], [[7, 62]], [[7, 62]], [[0, 62]], [[0, 81]], [[82, 92]], [[94, 117]], [[118, 138]], [[140, 178]]]", "query_spans": "[[[180, 192]]]", "process": "" }, { "text": "Given that a hyperbola has common points with the line $y = -x$ and has no common points with the line $y = -2x$, find the range of values for the eccentricity of the hyperbola.", "fact_expressions": "G: Hyperbola;H: Line;L:Line;Expression(H) = (y = -2*x);Expression(L) = (y = -x);IsIntersect(G,L);NumIntersection(G, H) = 0", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(2),sqrt(5)]", "fact_spans": "[[[2, 5], [37, 40]], [[20, 30]], [[6, 14]], [[20, 30]], [[6, 14]], [[2, 18]], [[2, 35]]]", "query_spans": "[[[37, 49]]]", "process": "If the foci of the hyperbola lie on the x-axis, the standard equation of the hyperbola can be written as \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0). Since the hyperbola has common points with the line y=-x but no common points with the line y=-2x, it follows that 1<\\frac{b}{a}\\leqslant2. Therefore, e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}\\in(\\sqrt{2},\\sqrt{5}]. If the foci of the hyperbola lie on the y-axis, the standard equation of the hyperbola can be written as \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0,b>0). If the hyperbola has common points with the line y=-x but no common points with the line y=-2x, this does not satisfy the given conditions. In summary, the range of values for the eccentricity of the hyperbola is (\\sqrt{2},\\sqrt{5}]." }, { "text": "Through a point $P$ on the line $x - y - 4 = 0$ (where point $P$ is not on the $x$-axis), draw two tangent lines to the parabola $x^2 = 4y$. These two tangent lines intersect the $x$-axis at points $G$ and $H$, respectively. Then the minimum value of the area of the circumcircle of $\\Delta GHP$ is?", "fact_expressions": "L: Line;Expression(L) = (x - y - 4 = 0);P: Point;PointOnCurve(P,L) = True;Negation(PointOnCurve(P,xAxis));Z: Parabola;Expression(Z) = (x^2 = 4*y);L1: Line;L2: Line;TangentOfPoint(P,Z) = {L1,L2};Intersection(L1,xAxis) = G;G: Point;Intersection(L2,xAxis) = H;H: Point", "query_expressions": "Min(Area(CircumCircle(TriangleOf(G,H,P))))", "answer_expressions": "25*pi/8", "fact_spans": "[[[1, 12]], [[1, 12]], [[15, 18], [19, 23]], [[1, 18]], [[19, 30]], [[32, 46]], [[32, 46]], [], [], [[0, 51]], [[0, 72]], [[64, 68]], [[0, 72]], [[69, 72]]]", "query_spans": "[[[74, 99]]]", "process": "The focus of the parabola $ x^{2} = 4y $ is $ F(0,1) $. As shown in the figure, let the tangent points be $ A(4a, 4a^{2}) $ ($ a \\neq 0 $), $ B(4n, 4n^{2}) $ ($ n \\neq 0 $). The lines $ PA $ and $ PB $ intersect the $ x $-axis at points $ G $ and $ H $, respectively. Connect $ PF $, $ GF $, $ HF $. From $ x^{2} = 4y $, we get $ y = \\frac{1}{4}x^{2} $, then $ y' = \\frac{1}{2}x $. Thus, $ k_{PA} = \\frac{1}{2} \\times 4a = 2a $, $ k_{PB} = \\frac{1}{2} \\times 4n = 2n $. The equation of line $ PA $ is $ y - 4a^{2} = 2a(x - 4a) $, i.e., $ y = 2a(x - 2a) $, so $ G(2a, 0) $. The equation of line $ PB $ is $ y - 4n^{2} = 2n(x - 4n) $, i.e., $ y = 2n(x - 2n) $, so $ H(2n, 0) $. Since $ k_{PA} \\cdot k_{FG} = 2a \\cdot \\frac{1 - 0}{0 - 2a} = -1 $, we have $ GF \\perp PA $. Since $ k_{PB} \\cdot k_{FH} = 2n \\cdot \\frac{1 - 0}{0 - 2n} = -1 $, we have $ HF \\perp PB $. Therefore, points $ P $, $ G $, $ F $, $ H $ lie on a circle with diameter $ PF $, which is exactly the circumcircle of $ \\triangle GHP $. The distance from point $ F $ to the line $ x - y - 4 = 0 $ is $ d = \\frac{|0 - 1 - 4|}{\\sqrt{1^{2} + (-1)^{2}}} = \\frac{5\\sqrt{2}}{2} $. Then $ |PF| \\geqslant d = \\frac{5\\sqrt{2}}{2} $, so the circumradius $ r $ of $ \\triangle GHP $ satisfies $ r = \\frac{1}{2}|PF| \\geqslant \\frac{5\\sqrt{2}}{4} $. Hence, the area $ S $ of the circumcircle of $ \\triangle GHP $ is $ S = \\pi r^{2} \\geqslant \\pi \\times \\left( \\frac{5\\sqrt{2}}{4} \\right)^{2} = \\frac{25\\pi}{8} $. Therefore, the minimum area of the circumcircle of $ \\triangle GHP $ is $ \\frac{25\\pi}{8} $." }, { "text": "The focal distance of the hyperbola $x^{2}-y^{2}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 1)", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[0, 18]], [[0, 18]]]", "query_spans": "[[[0, 23]]]", "process": "From the equation of the hyperbola, we have: a² = b² = 1, then c² = a² + b² = 2. The focal distance of the hyperbola is 2c = 2√2." }, { "text": "The line $m$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $P_{1}$ and $P_{2}$, and the midpoint of segment $P_{1} P_{2}$ is $P$. Let the slope of line $m$ be $k_{1}$ $(k_{1} \\neq 0)$, and the slope of line $O P$ be $k_{2}$. Then the value of $k_{1} k_{2}$ is?", "fact_expressions": "m: Line;G: Ellipse;O: Origin;P: Point;P1: Point;P2: Point;Expression(G) = (x^2/2 + y^2 = 1);Intersection(m, G) = {P1, P2};MidPoint(LineSegmentOf(P1, P2)) = P;k1: Number;Slope(m) = k1;Negation(k1 = 0);k2: Number;Slope(LineOf(O, P)) = k2", "query_expressions": "k1*k2", "answer_expressions": "-1/2", "fact_spans": "[[[0, 5], [77, 82]], [[6, 33]], [[110, 115]], [[72, 75]], [[35, 42]], [[43, 50]], [[6, 33]], [[0, 52]], [[53, 75]], [[86, 107]], [[77, 107]], [[86, 107]], [[119, 126]], [[108, 126]]]", "query_spans": "[[[128, 145]]]", "process": "Using the point difference method, let P_{1}(x_{1},y_{1}), P_{2}(x_{2},y_{2}), and midpoint P(x_{0},y_{0}) satisfy \\begin{cases}\\frac{x_{1}}{2}+y_{1}=1\\\\\\frac{x_{2}}{2}+y_{2}=1,\\\\ rearranging yields \\frac{1}{2}+\\frac{y-y}{x_{1}}x_{2}.\\frac{y_{1}+y_{2}}{x}+x=0, i.e. \\frac{1}{2}+\\frac{y_{1}-y_{2}}{x}-x_{2}.\\frac{x_{0}}{x_{0}}=0, i.e. \\frac{1}{2}+k_{1}\\cdot k_{2}=0,\\end{cases}\\therefore k_{1}k_{2}=-\\frac{1}{2}" }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$ $(00\\\\m-1>0\\\\3-m≠m-1\\end{cases}, solving gives 10)$, find the value of $p$.", "fact_expressions": "G: Parabola;p: Number;H: Circle;p>0;Expression(G) = (x^2 = 2*(p*y));Expression(H) = (y^2 + (x - 1)^2 = 9);IsTangent(H, Directrix(G))", "query_expressions": "p", "answer_expressions": "6", "fact_spans": "[[[23, 44]], [[51, 54]], [[2, 22]], [[26, 44]], [[23, 44]], [[2, 22]], [[2, 49]]]", "query_spans": "[[[51, 58]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $P$ is a point on the ellipse $C$ such that $\\overrightarrow{P F_{1}} \\perp \\overrightarrow{P F_{2}}$, and the area of $\\Delta P F_{1} F_{2}$ is $9$, then $b=$?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: 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);IsPerpendicular(VectorOf(P, F2),VectorOf(P, F1));Area(TriangleOf(P, F1, F2)) = 9", "query_expressions": "b", "answer_expressions": "3", "fact_spans": "[[[18, 75], [85, 90]], [[187, 190]], [[24, 75]], [[81, 84]], [[2, 9]], [[10, 17]], [[24, 75]], [[24, 75]], [[18, 75]], [[2, 80]], [[81, 93]], [[95, 152]], [[155, 184]]]", "query_spans": "[[[187, 192]]]", "process": "" }, { "text": "Given the hyperbola $\\Gamma$: $\\frac{x^{2}}{m}-\\frac{y^{2}}{m+9}=1$ $(m>0)$, the length of its imaginary axis is twice the length of its real axis. Find the focal distance of $\\Gamma$.", "fact_expressions": "Gamma: Hyperbola;Expression(Gamma) = (x^2/m - y^2/(m + 9) = 1);m: Number;m>0;Length(ImageinaryAxis(Gamma)) = 2*Length(RealAxis(Gamma))", "query_expressions": "FocalLength(Gamma)", "answer_expressions": "2*sqrt(15)", "fact_spans": "[[[2, 57], [72, 80]], [[2, 57]], [[15, 57]], [[15, 57]], [[2, 70]]]", "query_spans": "[[[72, 85]]]", "process": "Given that the length of the imaginary axis is twice the length of the real axis, we obtain b = 2a, and further find m and [detailed solution]. From the problem, in the hyperbola, b = 2a, hence b^{2} = 4a^{2}, that is, m + 9 = 4m \\Rightarrow m = 3. Thus c^{2} = m + m9 = 15, so the focal distance 2c = 2\\sqrt{15}." }, { "text": "If a focus of the hyperbola $x^{2}+ky^{2}=1$ is $(3 , 0)$, then the real number $k$=?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (k*y^2 + x^2 = 1);Coordinate(OneOf(Focus(G))) = (3, 0)", "query_expressions": "k", "answer_expressions": "-1/8", "fact_spans": "[[[1, 20]], [[37, 42]], [[1, 20]], [[1, 35]]]", "query_spans": "[[[37, 44]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $y^{2}=-x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(-1/4,0)", "fact_spans": "[[[0, 13]], [[0, 13]]]", "query_spans": "[[[0, 20]]]", "process": "" }, { "text": "What is the equation of the parabola with vertex at the right focus of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ and focus at the left vertex?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1);H: Parabola;RightFocus(G) = Vertex(H);LeftVertex(G) = Focus(H)", "query_expressions": "Expression(H)", "answer_expressions": "y^2=-36*(x-5)", "fact_spans": "[[[1, 40]], [[1, 40]], [[55, 58]], [[0, 58]], [[0, 58]]]", "query_spans": "[[[55, 62]]]", "process": "" }, { "text": "The equation of one asymptote of a hyperbola centered at the origin is $y=\\sqrt{3} x$. What is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;O:Origin;Center(G)=O;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{2,2*sqrt(3)/3}", "fact_spans": "[[[6, 9], [6, 9]], [[3, 5]], [[0, 9]], [[6, 32]]]", "query_spans": "[[[35, 44]]]", "process": "If the hyperbola equation is \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1, since one of its asymptotes is y=\\sqrt{3}x, then \\frac{b}{a}=\\sqrt{3}, so e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=4, e=2. If the hyperbola equation is \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{\\frac{x}{12}}=1, since one of its asymptotes is y=\\sqrt{3}x, then \\frac{a}{b}=\\sqrt{3}, so e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=\\frac{4}{3}, e=\\frac{2\\sqrt{3}}{3}." }, { "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{64}+\\frac{y^{2}}{36}=1$ is at a distance of $4$ from one of its foci, then what is the distance from point $P$ to the other focus?", "fact_expressions": "G: Ellipse;P: Point;F1:Point;F2:Point;Expression(G) = (x^2/64 + y^2/36 = 1);PointOnCurve(P, G);OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 4", "query_expressions": "Distance(P,F2)", "answer_expressions": "12", "fact_spans": "[[[1, 40], [47, 48]], [[43, 46], [65, 69]], [], [], [[1, 40]], [[1, 46]], [[47, 53]], [[47, 75]], [[47, 75]], [[43, 61]]]", "query_spans": "[[[47, 81]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{6}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $AB$ passing through the right focus $F_{2}$ intersects the ellipse at points $A$ and $B$. Then, the perimeter of $\\Delta A B F_{1}$ is?", "fact_expressions": "G: Ellipse;B: Point;A: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 + y^2/6 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, LineOf(A, B));Intersection(LineOf(A, B), G) = {A, B}", "query_expressions": "Perimeter(TriangleOf(A, B, F1))", "answer_expressions": "16", "fact_spans": "[[[2, 40], [84, 86]], [[92, 95]], [[88, 91]], [[48, 55]], [[56, 63], [68, 75]], [[2, 40]], [[2, 63]], [[2, 63]], [[64, 83]], [[76, 97]]]", "query_spans": "[[[99, 122]]]", "process": "The ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{6}=1$ has $a=4$, the perimeter of triangle $ABF_{2}$ is $|AB|+|AF_{1}|+|BF_{1}|=|AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=2a+2a=4a=16$" }, { "text": "Given the line $l$: $a x + b y = 0$ intersects the ellipse $x^{2} + \\frac{y^{2}}{9} = 1$ at points $A$ and $B$. If the coordinates of point $C$ are $(1,1)$, then the range of values of $\\overrightarrow{C A} \\cdot \\overrightarrow{C B}$ is?", "fact_expressions": "l: Line;G: Ellipse;C: Point;A: Point;B: Point;Expression(G) = (x^2 + y^2/9 = 1);Coordinate(C) = (1, 1);a: Number;b: Number;Expression(l) = (a*x + b*y = 0);Intersection(l, G) = {A, B}", "query_expressions": "Range(DotProduct(VectorOf(C, A), VectorOf(C, B)))", "answer_expressions": "[-7, 1]", "fact_spans": "[[[2, 20]], [[21, 48]], [[61, 65]], [[50, 53]], [[54, 57]], [[21, 48]], [[61, 76]], [[9, 20]], [[9, 20]], [[2, 20]], [[2, 59]]]", "query_spans": "[[[78, 134]]]", "process": "Let A(x₀, y₀), B(-x₀, -y₀) be set according to the problem, then $\\overrightarrow{CA} \\cdot \\overrightarrow{CB} = (x₀ - 1, y₀ - 1) \\cdot (-x₀ - 1, -y₀ - 1) = 2 - (x₀² + y₀²)$, $|OA| = \\sqrt{x₀² + y₀²} \\in [1, 3]$, $\\therefore x₀² + y₀² \\in [1, 9]$, $\\therefore \\overrightarrow{CA} \\cdot \\overrightarrow{CB} \\in [-7, 1]$" }, { "text": "The line $y = kx + 1$ and the parabola $y^2 = 2x$ have at most one common point. Find the range of values for $k$.", "fact_expressions": "G: Parabola;H: Line;k: Number;Expression(G) = (y^2 = 2*x);Expression(H) = (y = k*x + 1);NumIntersection(H, G) <= 1", "query_expressions": "Range(k)", "answer_expressions": "{0}+[1/2, oo)", "fact_spans": "[[[12, 26]], [[0, 11]], [[36, 39]], [[12, 26]], [[0, 11]], [[0, 34]]]", "query_spans": "[[[36, 45]]]", "process": "Substitute y = kx + 1 into y^{2} = 2x to obtain k^{2}x^{2} + (2k - 2)x + 1 = 0. \n(1) If k = 0, then -2x + 1 = 0, the equation has only one solution, which satisfies the condition. \n(2) If k \\neq 0, \\Delta = (2k - 2)^{2} - 4k^{2} = 4 - 8k. \nThe line y = kx + 1 and the parabola y^{2} = 2x have at most one common point when \\Delta = 4 - 8k \\leqslant 0, solving gives k \\geqslant \\frac{1}{2}. \nTherefore, k \\geqslant \\frac{1}{2} or k = 0. \nHence, the answer is: 0 \\cup [\\frac{1}{2}, +\\infty)." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a}-\\frac{y^{2}}{a-2}=1$ has foci at $(0, \\pm \\sqrt{6})$, then the value of $a$ is?", "fact_expressions": "C: Hyperbola;a: Number;Expression(C) = (-y^2/(a - 2) + x^2/a = 1);Coordinate(Focus(C)) = (0,pm*sqrt(6))", "query_expressions": "a", "answer_expressions": "-2", "fact_spans": "[[[2, 47]], [[74, 77]], [[2, 47]], [[2, 72]]]", "query_spans": "[[[74, 81]]]", "process": "The coordinates of the foci of the hyperbola C: \\frac{x^{2}}{a}-\\frac{y^{2}}{a-2}=1 are (0,\\pm\\sqrt{6})." }, { "text": "If the eccentricity of the conic section $\\frac{x^{2}}{5-\\lambda}-\\frac{y^{2}}{\\lambda-1}=1$ is $\\sqrt{\\lambda}$, then $\\lambda$=?", "fact_expressions": "H: ConicSection;lambda: Number;Expression(H) = (x^2/(5 - lambda) - y^2/(lambda - 1) = 1);Eccentricity(H) = sqrt(lambda)", "query_expressions": "lambda", "answer_expressions": "4", "fact_spans": "[[[1, 56]], [[80, 89]], [[1, 56]], [[1, 77]]]", "query_spans": "[[[80, 91]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4 x$ with focus $F$, and the line $y=2 x-4$ intersecting $C$ at points $A$ and $B$. Then $\\cos \\angle A F B$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;Expression(G) = (y = 2*x - 4);A: Point;B: Point;Intersection(G, C) = {A, B}", "query_expressions": "Cos(AngleOf(A, F, B))", "answer_expressions": "-4/5", "fact_spans": "[[[2, 21], [41, 44]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 40]], [[29, 40]], [[46, 49]], [[50, 53]], [[29, 55]]]", "query_spans": "[[[58, 79]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the upper and lower vertices are $B_{1}$ and $B_{2}$, respectively. Point $M$ is an arbitrary point on $C$ not lying on the coordinate axes, and the product of the slopes of lines $M B_{1}$ and $M B_{2}$ is $-\\frac{b}{2 a}$. Then the eccentricity 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;B1: Point;B2: Point;UpperVertex(C) = B1;LowerVertex(C) = B2;M: Point;PointOnCurve(M, C);Negation(PointOnCurve(M, axis));Slope(LineOf(M, B1))*Slope(LineOf(M, B2)) = -b/(2*a)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[2, 59], [90, 93], [153, 156]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[68, 75]], [[76, 83]], [[2, 83]], [[2, 83]], [[85, 89]], [[85, 105]], [[85, 105]], [[107, 151]]]", "query_spans": "[[[153, 162]]]", "process": "Let M(x_{0},y_{0}), according to the problem we have \\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1, then x_{0}^{2}=\\frac{a^{2}}{b^{2}}(b^{2}-y_{0}^{2}) so k_{MB_{1}}\\cdot k_{MB_{2}}=\\frac{y_{0}-b}{x_{0}} \\cdot \\frac{y_{0}+b}{x_{0}}=\\frac{y_{0}^{2}-b^{2}}{x_{0}^{2}}=-\\frac{b^{2}}{a^{2}}=-\\frac{b}{2a}, then \\frac{b}{a}=\\frac{1}{2}, e=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{3}}{2}" }, { "text": "If the directrix of the parabola $y^{2}=8 x$ intersects the curve $\\frac{x^{2}}{a}+\\frac{y^{2}}{4}=1$ $(y \\geq 0)$ at exactly one point, then the condition satisfied by the real number $a$ is?", "fact_expressions": "G: Parabola;H: Curve;a: Real;Expression(G) = (y^2 = 8*x);Expression(H) = And(x^2/a+y^2/4=1,y>=0);NumIntersection(Directrix(G), H) = 1", "query_expressions": "a", "answer_expressions": "{(-oo,0)+[4,oo)}", "fact_spans": "[[[1, 15]], [[19, 66]], [[74, 79]], [[1, 15]], [[19, 66]], [[1, 72]]]", "query_spans": "[[[74, 86]]]", "process": "According to the problem, the directrix of the parabola is given by x = -2. Discuss the graph represented by the curve $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$) when $a > 0$ and $a < 0$, respectively, to solve the problem. The directrix of the parabola $y^{2} = 8x$ is $x = -2$. When $a > 0$, $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$) represents the upper part of an ellipse including the left and right vertices, so $-\\sqrt{a} \\leqslant x \\leqslant \\sqrt{a}$. If $x = -2$ intersects the curve $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$) at exactly one point, then $-\\sqrt{a} \\leqslant -2$, solving gives $a \\geqslant 4$. When $a < 0$, $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$) represents the upper branch of a hyperbola above the x-axis, in which case $x \\in (-\\infty, +\\infty)$, and $x = -2$ intersects the curve $\\frac{x^{2}}{a} + \\frac{y^{2}}{4} = 1$ ($y \\geqslant 0$) at exactly one point, so $a < 0$. In conclusion, the condition satisfied by the real number $a$ is $a < 0$ or $a \\geqslant 4$." }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and a line passing through the focus intersects the parabola at points $A$ and $B$, what is the sine of the inclination angle of the line $AB$ when $|AF|+4|BF|$ attains its minimum value?", "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};WhenMin(Abs(LineSegmentOf(A, F)) + 4*Abs(LineSegmentOf(B, F)))", "query_expressions": "Sin(Inclination(LineOf(A, B)))", "answer_expressions": "2*sqrt(2)/3", "fact_spans": "[[[2, 16], [31, 34]], [[2, 16]], [[20, 23]], [[2, 23]], [[28, 30]], [[2, 30]], [[36, 39]], [[40, 43]], [[28, 45]], [[47, 68]]]", "query_spans": "[[[69, 86]]]", "process": "① When the slope of a line passing through the focus does not exist, then |AF| = |BF| = 2, that is, \\frac{1}{|AF|} + \\frac{1}{|BF|} = 1. ② When the slope of a line passing through the focus exists, let the equation of the line be y = k(x - 1) (k \\neq 0). Solving the system \\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. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}}, x_{1}x_{2} = 1. \\frac{1}{|AF|} + \\frac{1}{|BF|} = \\frac{1}{x_{1} + 1} + \\frac{1}{x_{2} + 1} = \\frac{x_{1} + x_{2} + 2}{x_{1}x_{2} + x_{1} + x_{2} + 1} = \\frac{\\frac{2k^{2} + 4}{k^{2}} + 2}{1 + \\frac{2k^{2} + 4}{k^{2}} + 1} = 1. In conclusion, \\frac{1}{|AF|} + \\frac{1}{|BF|} = 1. Let |AF| = a, |BF| = b, then \\frac{1}{a} + \\frac{1}{b} = 1. \\therefore |AF| + 4|BF| = a + 4b = \\left( \\frac{1}{a} + \\frac{1}{b} \\right)(a + 4b) = 5 + \\frac{4b}{a} + \\frac{a}{b} \\geqslant 5 + 2\\sqrt{\\frac{4b}{a} \\cdot \\frac{a}{b}} = 9, equality holds if and only if a = 2b = 3. \\therefore when |AF| + 4|BF| attains its minimum value, x_{1} + 1 = 2(x_{2} + 1) = 3, that is, x_{1} = 2, x_{2} = \\frac{1}{2}, at this time the slope of line AB is k = \\pm 2\\sqrt{2}, that is, the sine of the inclination angle of line AB is \\frac{2\\sqrt{2}}{3}." }, { "text": "If on the line $y=x$ there exists a point $P$ such that the difference of the distances from $P$ to point $A(-m, 0)$ and to point $B(m, 0)$ $(m>0)$ is $2$, then the range of real values for $m$ is?", "fact_expressions": "G: Line;A: Point;B: Point;P:Point;m: Real;m>0;Expression(G) = (y = x);PointOnCurve(P, G);Coordinate(A) = (-m, 0);Coordinate(B) = (m, 0);Distance(P,A)-Distance(P,B)=2", "query_expressions": "Range(m)", "answer_expressions": "(sqrt(2), +oo)", "fact_spans": "[[[2, 9]], [[21, 32]], [[34, 49]], [[17, 20], [12, 16]], [[60, 65]], [[35, 49]], [[2, 9]], [[2, 16]], [[21, 32]], [[34, 49]], [[17, 58]]]", "query_spans": "[[[60, 72]]]", "process": "According to the definition of a hyperbola, transform the problem into determining whether the line y = x intersects the right branch of the hyperbola. From the definition of the hyperbola, we know that 2a = 2, so a = 1. For the right branch of the hyperbola x^{2} - \\frac{y^{2}}{b^{2}} = 1 to intersect the line y = x, the slope \\frac{b}{a} of its asymptote y = \\frac{b}{a}x must satisfy \\frac{b}{a} > 1, hence b > 1. Therefore, m = c = \\sqrt{a^{2} + b^{2}} = \\sqrt{1 + b^{2}} > \\sqrt{1 + 1} = \\sqrt{2}, so the range of m is (\\sqrt{2}, +\\infty)." }, { "text": "Given that point $A$ lies on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F$ is a focus of the ellipse, $A F \\perp x$-axis, and $|A F| =$ focal distance, 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;A: Point;PointOnCurve(A, G);F: Point;OneOf(Focus(G)) = F;IsPerpendicular(LineSegmentOf(A, F), xAxis);FocalLength(G) = Abs(LineSegmentOf(A, F))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)-1", "fact_spans": "[[[7, 59], [67, 69], [103, 105]], [[7, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[2, 6]], [[2, 62]], [[63, 66]], [[63, 74]], [[76, 90]], [[67, 101]]]", "query_spans": "[[[103, 111]]]", "process": "" }, { "text": "Let the focus of the parabola $x^{2}=4 y$ be $F$, and let points $A$ and $B$ lie on the parabola such that $\\overrightarrow{A F}=\\lambda \\overrightarrow{F B}$. If $|\\overrightarrow{A F}|=\\frac{3}{2}$, then the value of $\\lambda$ is?", "fact_expressions": "G: Parabola;A: Point;F: Point;B: Point;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;lambda:Number", "query_expressions": "lambda", "answer_expressions": "1/2", "fact_spans": "[[[1, 15], [33, 36]], [[23, 28]], [[19, 22]], [[29, 32]], [[1, 15]], [[1, 22]], [[24, 37]], [[29, 37]], [[42, 93]], [[96, 132]], [[134, 143]]]", "query_spans": "[[[134, 147]]]", "process": "From the given conditions, F(0,1) because AB is a focal chord of the parabola x^{2}=4y, so \\frac{1}{|AF|}+\\frac{1}{|BF|}=\\frac{2}{p}=1. Given |\\overrightarrow{AF}|=\\frac{3}{2}, solving gives |\\overrightarrow{BF}|=3. Also, since \\overrightarrow{AF}=\\lambda\\overrightarrow{FB}, it follows that \\lambda=\\frac{|\\overrightarrow{AF}|}{|\\overrightarrow{BF}|}=\\frac{\\frac{3}{2}}{3}=\\frac{1}{2}." }, { "text": "Given that the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) is $(1,0)$, a line passing through point $F$ intersects the directrix of the parabola at point $A$, and intersects the parabola at one point $B$, with $\\overrightarrow{F A}=-3 \\overrightarrow{F B}$. Then $|A B|$=?", "fact_expressions": "G: Parabola;p: Number;H: Line;F: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(F) = (1, 0);Focus(G) = F;PointOnCurve(F, H);Intersection(H, Directrix(G)) = A;OneOf(Intersection(H, G)) = B;VectorOf(F, A) = -3*VectorOf(F, B)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "32/3", "fact_spans": "[[[2, 23], [48, 51], [62, 65]], [[5, 23]], [[44, 46]], [[39, 43], [26, 29]], [[55, 59]], [[71, 74]], [[5, 23]], [[2, 23]], [[26, 37]], [[2, 29]], [[38, 46]], [[44, 59]], [[44, 74]], [[76, 122]]]", "query_spans": "[[[124, 133]]]", "process": "The parabola y^{2}=4x has focus F(1,0) and directrix x=-1. Draw BD perpendicular to the directrix at D, and let the directrix intersect the x-axis at point E. Then EF//DB, so \\triangleAEF\\sim\\triangleADB. Since \\overrightarrow{FA}=-3\\overrightarrow{FB}, it follows that |BD|=\\frac{4}{3}|EF|=\\frac{8}{3}. By the definition of a parabola, |BF|=|BD|=\\frac{8}{3}. Therefore, |AB|=4|BF|=\\frac{32}{3}." }, { "text": "Given that $F$ is the left focus of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, point $A(-3,-3)$ is a fixed point, and point $P$ is a moving point on the ellipse $C$, then the maximum value of $|P A|+|P F|$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F: Point;LeftFocus(C) = F;A: Point;Coordinate(A) = (-3, -3);P: Point;PointOnCurve(P, C)", "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "9", "fact_spans": "[[[6, 48], [71, 76]], [[6, 48]], [[2, 5]], [[2, 52]], [[55, 65]], [[55, 65]], [[66, 70]], [[66, 82]]]", "query_spans": "[[[84, 103]]]", "process": "Let the right focus of the ellipse be F_{1}(1,0), and then use a combination of algebraic and geometric methods to analyze and solve. Let the right focus of the ellipse be F_{1}(1,0), |PA|+|PF|=|PA|+2a-|PF|=4+|PA|-|PF|\\leqslant4+|AF|=4+\\sqrt{(-3-1)^{2}+3^{2}}=9" }, { "text": "Given that point $F$ is the focus of the parabola $y^{2}=-8x$, $O$ is the origin, point $P$ is a moving point on the directrix of the parabola, $A$ lies on the parabola and $|AF|=4$, then the minimum value of $|PA|+|PO|$ is?", "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;O: Origin;Expression(G) = (y^2 = -8*x);Focus(G) = F;PointOnCurve(P, Directrix(G));PointOnCurve(A, G);Abs(LineSegmentOf(A, F)) = 4", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, O)))", "answer_expressions": "2*sqrt(13)", "fact_spans": "[[[7, 22], [38, 41], [52, 55]], [[48, 51]], [[2, 6]], [[33, 37]], [[26, 29]], [[7, 22]], [[2, 25]], [[33, 47]], [[48, 56]], [[58, 67]]]", "query_spans": "[[[69, 88]]]", "process": "" }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$, with focus $F$. If points $A$ and $B$ lie on the parabola in the first quadrant, $|A F|=2$, $|B F|=4$, $|A B|=3$, find the slope of line $A B$?", "fact_expressions": "G: Parabola;p: Number;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(A, G);PointOnCurve(B, G);Quadrant(A) = 1;Quadrant(B) = 1;Abs(LineSegmentOf(A, F)) = 2;Abs(LineSegmentOf(B, F)) = 4;Abs(LineSegmentOf(A, B)) = 3", "query_expressions": "Slope(LineOf(A, B))", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[2, 23], [40, 43]], [[5, 23]], [[36, 39]], [[32, 35]], [[27, 30]], [[5, 23]], [[2, 23]], [[2, 30]], [[32, 44]], [[36, 44]], [[32, 50]], [[32, 50]], [[51, 60]], [[61, 70]], [[71, 80]]]", "query_spans": "[[[82, 94]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the slope of AB exists, denote the slope of AB as k. Points A and B are both above the x-axis, and from the given condition we know k > 0. By the definition of the parabola, AF = x_{1} + \\frac{p}{2}, BF = x_{2} + \\frac{p}{2}. Then \n\\begin{cases}x_{1} + \\frac{p}{2} = 2\\\\x_{2} + \\frac{p}{2} = 4\\end{cases} \n\\Rightarrow |x_{1} - x_{2}| = 2. By the chord length formula, AB = \\sqrt{1 + k^{2}}|x_{1} - x_{2}|, so \nAB = \\sqrt{1 + k^{2}}|x_{1} - x_{2}| = 3 \\Rightarrow \\sqrt{1 + k^{2}} = \\frac{3}{2} \\Rightarrow k = \\frac{\\sqrt{5}}{2}." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, a line passing through its right focus and perpendicular to the real axis intersects the hyperbola at points $M$ and $N$, and $O$ is the origin. If $OM \\perp ON$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;O: Origin;M: Point;N: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(G),H);IsPerpendicular(RealAxis(G),H);Intersection(H,G) = {M,N};IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(O, N))", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)-1)/2", "fact_spans": "[[[2, 59], [75, 78], [117, 120], [61, 62]], [[5, 59]], [[5, 59]], [[72, 74]], [[90, 93]], [[80, 83]], [[84, 87]], [[5, 59]], [[5, 59]], [[2, 59]], [[60, 74]], [[60, 74]], [[72, 89]], [[100, 115]]]", "query_spans": "[[[117, 126]]]", "process": "According to the problem, let the right focus be F. Since OM ⊥ ON, triangle OMN is an isosceles right triangle, so |MF| = |OF|, yielding \\frac{b^{2}}{a} = c. Also, from c^{2} = a^{2} + b^{2}, simplifying gives c^{2} - ac - a^{2} = 0, solving which yields \\frac{c}{a} = \\frac{1 \\pm \\sqrt{5}}{2}. Since e > 1, it follows that e = \\frac{1 + \\sqrt{5}}{2}." }, { "text": "Let $P$ be a point on the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, and let $M$, $N$ be points on the two 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;Z: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Expression(H) = (y^2 + (x - 5)^2 = 4);Expression(Z) = (y^2 + (x + 5)^2 = 1);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, Z)", "query_expressions": "Max(Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)))", "answer_expressions": "9", "fact_spans": "[[[5, 44]], [[63, 82]], [[83, 102]], [[1, 4]], [[48, 51]], [[54, 57]], [[5, 44]], [[63, 82]], [[83, 102]], [[1, 47]], [[48, 105]], [[48, 105]]]", "query_spans": "[[[107, 126]]]", "process": "Problem Analysis: Let the centers of the two circles $(x-5)^{2}+y^{2}=4$ and $(x+5)^{2}+y^{2}=1$ be $A$ and $B$ respectively. Then $A$ and $B$ are exactly the two foci of a hyperbola. $|PM|-|PN|\\leqslant|PA|+2-(|PB|-1)=|PA|-|PB|+3=2a+3=6+3=9$, so the maximum value is $9$." }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ has an asymptote with equation $y=\\frac{\\sqrt {2}}{2} x$. Then its eccentricity $e$=?", "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*(sqrt(2)/2));Eccentricity(G)=e;e:Number", "query_expressions": "e", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[0, 59], [95, 96]], [[3, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[0, 59]], [[0, 93]], [[95, 103]], [[100, 103]]]", "query_spans": "[[[100, 105]]]", "process": "" }, { "text": "Let $P$ be a moving point on the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, $O$ be the origin, and $M$ be the midpoint of the segment $OP$. Then the equation of the trajectory of point $M$ is?", "fact_expressions": "O: Origin;P: Point;G: Hyperbola;M: Point;Expression(G) = (x^2/4 - y^2 = 1);PointOnCurve(P, G);MidPoint(LineSegmentOf(O, P)) = M", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2-4*y^2=1", "fact_spans": "[[[38, 41]], [[1, 4]], [[5, 33]], [[47, 50], [62, 66]], [[5, 33]], [[1, 37]], [[47, 60]]]", "query_spans": "[[[62, 73]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{11}+\\frac{y^{2}}{7}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. Then, what is the perimeter of $\\Delta F_{1} A B$?", "fact_expressions": "l: Line;G: Ellipse;F1: Point;A: Point;B: Point;F2: Point;Expression(G) = (x^2/11 + y^2/7 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, l);Intersection(l, G) = {A, B}", "query_expressions": "Perimeter(TriangleOf(F1, A, B))", "answer_expressions": "4*sqrt(11)", "fact_spans": "[[[73, 78]], [[2, 40], [79, 81]], [[48, 55]], [[83, 86]], [[87, 90]], [[65, 72], [56, 63]], [[2, 40]], [[2, 63]], [[2, 63]], [[64, 78]], [[73, 92]]]", "query_spans": "[[[94, 117]]]", "process": "From the ellipse $\\frac{x^{2}}{11}+\\frac{y^{2}}{7}=1$, the semi-major axis length is $a=\\sqrt{11}$. By the definition of an ellipse, the perimeter of triangle $\\triangle F_{1}AB$ is $|AB|+|AF_{1}|+|BF_{1}|=|AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=2a+2a=4\\sqrt{11}$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with right focus $F(c, 0)$, point $P$ lies on the ellipse $C$, the segment $PF$ is tangent to the circle $(x-\\frac{c}{3})^{2}+y^{2}=\\frac{b^{2}}{9}$ at point $Q$, and $\\overrightarrow{P Q}=2 \\overrightarrow{Q F}$. 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;F: Point;c: Number;Coordinate(F) = (c, 0);RightFocus(C) = F;P: Point;PointOnCurve(P, C);G: Circle;Expression(G) = (y^2 + (-c/3 + x)^2 = b^2/9);Q: Point;TangentPoint(LineSegmentOf(P, F), G) = Q;VectorOf(P, Q) = 2*VectorOf(Q, F)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/3", "fact_spans": "[[[2, 59], [78, 83], [193, 198]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[63, 72]], [[63, 72]], [[63, 72]], [[2, 72]], [[73, 77]], [[73, 84]], [[93, 137]], [[93, 137]], [[140, 144]], [[85, 144]], [[146, 191]]]", "query_spans": "[[[193, 204]]]", "process": "According to geometric and algebraic analysis, we have PF\\bot PF', and by the Pythagorean theorem, b^{2}+(2a-b)^{2}=4c^{2}=4(a^{2}-b^{2}) to calculate the eccentricity. As shown in the figure, first draw the graph of the function, |EF|=|OF|-|OE|=c-\\frac{1}{3}c=\\frac{2}{3}c, \\therefore\\frac{|EF|}{|EF|}=\\frac{\\frac{2}{3}c}{c+\\frac{1}{3}c}=\\frac{1}{2}. Also, \\because\\overrightarrow{PQ}=2\\overrightarrow{QF}, \\therefore PF'//QE, and \\frac{|QE|}{|PF|}=\\frac{1}{3}, and PF\\bot PF'. \\because|QE|=\\frac{b}{3}, \\therefore|PF|=b. According to the definition of the ellipse, |PF|=2a-b. By the Pythagorean theorem, |PF|^{2}+|PF|^{2}=|F_{1}F_{2}|^{2}, i.e., b^{2}+(2a-b)^{2}=4c^{2}=4(a^{2}-b^{2}). Rearranging gives b^{2}+4a^{2}+b^{2}-4ab=4a^{2}-4b^{2}, i.e., \\frac{b}{a}=\\frac{2}{3}, \\therefore\\frac{c}{a}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{5}}{3}" }, { "text": "Through the focus $F$ of the parabola $C$: $x^{2}=4 y$, draw a line $l$ with slope $\\sqrt{3}$, intersecting the parabola at points $A$ and $B$. The two tangents to the parabola at $A$ and $B$ intersect at point $M$. Then $|M F|$=?", "fact_expressions": "l: Line;C: Parabola;M: Point;F: Point;A: Point;Expression(C) = (x^2 = 4*y);Focus(C)=F;PointOnCurve(F,l);Slope(l)=sqrt(3);Intersection(l,C)={A,B};Intersection(TangentOnPoint(A,C),TangentOnPoint(B,C))=M;B:Point", "query_expressions": "Abs(LineSegmentOf(M, F))", "answer_expressions": "4", "fact_spans": "[[[41, 46]], [[1, 20], [62, 65], [48, 51]], [[81, 85]], [[23, 26]], [[52, 55]], [[1, 20]], [[1, 26]], [[0, 46]], [[27, 46]], [[41, 61]], [[62, 85]], [[56, 59]]]", "query_spans": "[[[87, 96]]]", "process": "The parabola $ C: x^{2} = 4y $ has focus $ F(0,1) $, and the line $ l $ is $ y = \\sqrt{3}x + 1 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $ be points satisfying \n\\[\n\\begin{cases}\ny = \\sqrt{3}x + 1 \\\\\nx^{2} = 4y\n\\end{cases}\n\\]\nwhich gives $ x^{2} - 4\\sqrt{3}x - 4 = 0 $. Then $ x_{1} + x_{2} = 4\\sqrt{3} $, $ x_{1}x_{2} = -4 $. From $ y = \\frac{1}{4}x^{2} $, we get $ y' = \\frac{1}{2}x $, so the slope of the tangent line at point $ A(x_{1}, y_{1}) $ is $ \\frac{1}{2}x_{1} $. Thus, the equation of the tangent line at $ A(x_{1}, y_{1}) $ is $ y - \\frac{x_{1}^{2}}{4} = \\frac{1}{2}x_{1}(x - x_{1}) $, that is, $ y = \\frac{1}{2}x_{1}x - \\frac{x_{1}^{2}}{4} $. Similarly, the tangent line at $ B(x_{2}, y_{2}) $ is $ y = \\frac{1}{2}x_{2}x - \\frac{x_{2}^{2}}{4} $. Solving the two tangent equations simultaneously, we have $ \\frac{1}{2}x_{1}x - \\frac{x_{1}^{2}}{4} = \\frac{1}{2}x_{2}x - \\frac{x_{2}^{2}}{4} $, which gives $ x = \\frac{1}{2}(x_{1} + x_{2}) = 2\\sqrt{3} $. Then $ y = \\frac{1}{2}x_{1} \\cdot \\frac{1}{2}(x_{1} + x_{2}) - \\frac{x_{1}^{2}}{4} = \\frac{1}{4}x_{1}x_{2} = -1 $. Therefore, the coordinates of point $ M $ are $ (2\\sqrt{3}, -1) $. Hence, $ |MF| = \\sqrt{(2\\sqrt{3})^{2} + (-1 - 1)^{2}} = \\sqrt{12 + 4} = 4 $." }, { "text": "An ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$ has a focus $F_{1}$, and point $P$ lies on the ellipse. If the midpoint $M$ of segment $P F_{1}$ lies on the $y$-axis, then the distance $|M O|$ from point $M$ to the origin $O$ is $?$.", "fact_expressions": "G: Ellipse;P: Point;F1: Point;M: Point;O: Origin;Expression(G) = (x^2/12 + y^2/3 = 1);OneOf(Focus(G)) = F1;PointOnCurve(P, G);MidPoint(LineSegmentOf(P, F1)) = M;PointOnCurve(M, yAxis);Distance(M, O) = Abs(LineSegmentOf(M, O))", "query_expressions": "Abs(LineSegmentOf(M, O))", "answer_expressions": "sqrt(3)/4", "fact_spans": "[[[0, 38], [57, 59]], [[52, 56]], [[44, 51]], [[77, 80], [88, 92]], [[93, 100]], [[0, 38]], [[0, 51]], [[52, 60]], [[63, 80]], [[77, 86]], [[88, 110]]]", "query_spans": "[[[103, 112]]]", "process": "Let the coordinates of point M be (0, m). From the problem, we have F₁(-3, 0). Using the midpoint coordinate formula, the coordinates of point P are (3, 2m). Substituting into the equation of the ellipse gives the solution. Let the coordinates of point M be (0, m). From the problem, we have F₁(-3, 0). Using the midpoint coordinate formula, the coordinates of point P are (3, 2m). Substituting into the equation of the ellipse yields m = ±\\frac{\\sqrt{3}}{4}. Therefore, |MO| = \\frac{\\sqrt{3}}{4}." }, { "text": "If line $l$ passes through the focus of the parabola $x^{2}=-4 y$ and is tangent to the circle $(x-1)^{2}+(y-2)^{2}=1$, then the equation of line $l$ is?", "fact_expressions": "l: Line;G: Parabola;H: Circle;Expression(G) = (x^2 = -4*y);Expression(H) = ((x - 1)^2 + (y - 2)^2 = 1);PointOnCurve(Focus(G),l);IsTangent(l,H)", "query_expressions": "Expression(l)", "answer_expressions": "{x=0,4*x-3*y-3=0}", "fact_spans": "[[[1, 6], [56, 61]], [[8, 23]], [[28, 52]], [[8, 23]], [[28, 52]], [[1, 26]], [[1, 54]]]", "query_spans": "[[[56, 66]]]", "process": "Since the equation of the parabola is $x^{2} = -4y$, the coordinates of the focus are: $F(0, -1)$. When the slope of the line does not exist, let the equation of the line be: $x = 0$. The distance from the center of the circle to the line is $d = 1 = r$, which satisfies the condition. When the slope of the line exists, let the equation of the line be: $y = kx - 1$, that is, $kx - y - 1 = 0$. The distance from the center of the circle to the line is $d = \\frac{|k - 3|}{\\sqrt{1 + k^{2}}} = r = 1$, solving gives $k = \\frac{4}{3}$. Therefore, the equation of the line is $4x - 3y - 3 = 0$." }, { "text": "A point $P$ on the hyperbola $y^{2}-4x^{2}=64$ is at a distance of $7$ from one of its foci. Find the distance from $P$ to the other focus.", "fact_expressions": "G: Hyperbola;Expression(G) = (-4*x^2 + y^2 = 64);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G))=F1;Distance(P, F1) = 7;Negation(F1 = F2);OneOf(Focus(G))=F2", "query_expressions": "Distance(P, F2)", "answer_expressions": "23", "fact_spans": "[[[0, 21], [28, 29], [48, 49]], [[0, 21]], [[24, 27], [44, 47]], [[0, 27]], [], [], [[28, 34]], [[24, 42]], [[28, 55]], [[48, 55]]]", "query_spans": "[[[44, 60]]]", "process": "Transform the hyperbola $ y^{2} - 4x^{2} = 64 $ into standard form: $ \\frac{y^{2}}{64} - \\frac{x^{2}}{16} = 1 $, $ \\therefore a^{2} = 64 $, $ b^{2} = 16 $. The distance from point $ P $ to one of its foci is 7, let $ PF_{1} = 7 $. Since $ |PF_{1} - PF_{2}| = 2a = 16 $, $ \\therefore PF_{2} = PF_{1} \\pm 16 = 23 $ or $ -9 $ (discarded)." }, { "text": "The standard equation of an ellipse with eccentricity $e=\\frac{1}{2}$ and one focus at $F(0, -3)$ is?", "fact_expressions": "G: Ellipse;F: Point;Coordinate(F) = (0, -3);Eccentricity(G)=e;e: Number;e = 1/2;OneOf(Focus(G))=F", "query_expressions": "Expression(G)", "answer_expressions": "x^2/27+y^2/36=1", "fact_spans": "[[[37, 39]], [[26, 36]], [[26, 36]], [[0, 39]], [[3, 19]], [[3, 19]], [[21, 39]]]", "query_spans": "[[[37, 45]]]", "process": "Test analysis: First, assume the equation of the ellipse, set up equations about a, b, c according to the conditions, and solve for a, b, c to obtain the conclusion. According to the problem, the foci of the ellipse lie on the y-axis, so let the equation be: \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1. From the given conditions: \\begin{cases}c=3\\\\\\frac{c}{a}=\\frac{1}{2}\\\\a^{2}=b^{2}+c^{2}\\end{cases}, solving yields \\begin{cases}a=6\\\\b=3\\sqrt{3}\\end{cases}. Therefore, the standard equation of the ellipse is \\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ $(m>0)$ has asymptotes with equations $y=\\pm \\sqrt{2} x$, and $F_{1}$, $F_{2}$ are the left and right foci of $C$, respectively. Let $P$ be a point on the right branch of $C$. If $|P F_{1}|=m-1$, then the area of $\\Delta P F_{1} F_{2}$ is?", "fact_expressions": "C: Hyperbola;m: Number;P: Point;F1: Point;F2: Point;m>0;Expression(C) = (x^2/4 - y^2/m = 1);Expression(Asymptote(C)) = (y = pm*(sqrt(2)*x));LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, RightPart(C));Abs(LineSegmentOf(P, F1)) = m - 1", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "2*sqrt(26)", "fact_spans": "[[[2, 50], [98, 101], [112, 115]], [[9, 50]], [[108, 111]], [[77, 85]], [[88, 95]], [[10, 50]], [[2, 50]], [[2, 76]], [[77, 107]], [[77, 107]], [[108, 120]], [[123, 138]]]", "query_spans": "[[[140, 167]]]", "process": "Since the hyperbola $ C: \\frac{x^{2}}{4} - \\frac{y^{2}}{m} = 1 $ ($ m > 0 $) has asymptotes given by $ y = \\pm\\sqrt{2}x $, it follows that $ \\sqrt{\\frac{m}{4}} = \\sqrt{2} $, solving gives $ m = 8 $. Since $ |PF_{1}| - |PF_{2}| = 2a = 4 $, $ |PF_{1}| = m - 1 $, so $ |PF_{2}| = 3 $, and $ |F_{1}F_{2}| = 4\\sqrt{3} $, by the law of cosines we obtain $ \\cos\\angle F_{1}PF_{2} = \\frac{5}{21} $, then the area $ S $ of $ \\triangle PF_{1}F_{2} $ is $ \\frac{1}{2} \\times 3 \\times 7 \\times \\sin\\angle F_{1}PF_{2} = \\frac{21}{2} \\times \\frac{4\\sqrt{26}}{21} = 2\\sqrt{26} $." }, { "text": "$P$, $Q$ are two points on the parabola $y = x^{2}$ other than the vertex, and $O$ is the origin. $\\angle P O Q = \\frac{\\pi}{4}$. The lines $l_{1}$, $l_{2}$ are the tangent lines to the parabola at points $P$ and $Q$, respectively. (I) What is the trajectory equation of the intersection point $M$ of $l_{1}$ and $l_{2}$? (II) If $l_{1}$ and $l_{2}$ intersect the $x$-axis at points $A$ and $B$, respectively, what is the equation of the line passing through the orthocenter of $\\triangle A B M$ and the point $(0, -\\frac{1}{4})$?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);P: Point;Q: Point;PointOnCurve(P, G) = True;PointOnCurve(Q, G) = True;Negation(Vertex(G) = P);Negation(Vertex(G) = Q);O: Origin;AngleOf(P, O, Q) = pi/4;l1: Line;l2: Line;PointOnCurve(P, l1) = True;PointOnCurve(Q, l2) = True;IsTangent(l1, G) = True;IsTangent(l2, G) = True;M: Point;Intersection(l1, l2) = M;A: Point;B: Point;Intersection(l1, xAxis) = A;Intersection(l2, xAxis) = B;l3: Line;PointOnCurve(Orthocenter(TriangleOf(A, B, M)), l3) = True;I: Point;Coordinate(I) = (0, -1/4);PointOnCurve(I, l3) = True", "query_expressions": "LocusEquation(M);Expression(l3)", "answer_expressions": "(4*x^2 - y^2 - 6*y - 1 = 0)&Negation(y = 0)\ny = -1/4", "fact_spans": "[[[8, 20], [97, 100]], [[8, 20]], [[0, 3], [88, 91]], [[4, 7], [92, 95]], [[0, 28]], [[0, 28]], [[0, 28]], [[0, 28]], [[29, 32]], [[38, 66]], [[67, 76], [110, 117], [146, 153]], [[77, 84], [118, 125], [154, 161]], [[67, 103]], [[67, 103]], [[67, 103]], [[67, 103]], [[128, 132]], [[110, 132]], [[169, 172]], [[173, 176]], [[146, 178]], [[146, 178]], [[223, 225]], [[180, 225]], [[203, 222]], [[203, 222]], [[180, 225]]]", "query_spans": "[[[128, 139]], [[223, 229]]]", "process": "" }, { "text": "The line $ l $ intersects the parabola $ y = \\frac{x^{2}}{2} $ at points $ A $ and $ B $, and the tangents to the parabola at points $ A $ and $ B $ are perpendicular to each other, where point $ A $ has coordinates $ (2,\\ 2) $. Then the slope of line $ l $ is equal to?", "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y = x^2/2);Intersection(l, G) = {A, B};A: Point;B: Point;IsPerpendicular(TangentOnPoint(A,G),TangentOnPoint(B,G)) = True;Coordinate(A) = (2, 2)", "query_expressions": "Slope(l)", "answer_expressions": "3/4", "fact_spans": "[[[0, 5], [84, 89]], [[6, 28], [41, 44]], [[6, 28]], [[0, 39]], [[30, 33], [65, 68], [65, 69]], [[34, 37], [49, 52]], [[41, 62]], [[65, 81]]]", "query_spans": "[[[84, 95]]]", "process": "For the parabola y=\\frac{x^{2}}{2}, y=x; and A(2,2), we have: k_{A}=2. The tangents to the parabola at points A and B are perpendicular, \\therefore k_{B}=-\\frac{1}{2} \\therefore B(-\\frac{1}{2},\\frac{1}{8}) \\therefore k_{AB}=\\frac{2-\\frac{1}{8}}{2+\\frac{1}{2}}=\\frac{3}{4}. The correct result for this problem: \\frac{3}{3}" }, { "text": "Through the point $P(1,1)$, draw a line $l$ intersecting the hyperbola $x^{2}-\\frac{y^{2}}{2}=\\lambda$ at points $A$ and $B$. If point $P$ is exactly the midpoint of segment $AB$, then what is the range of real values of $\\lambda$?", "fact_expressions": "l: Line;G: Hyperbola;B: Point;A: Point;P: Point;Expression(G) = (x^2 - y^2/2 = lambda);Coordinate(P) = (1, 1);PointOnCurve(P,l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B))=P;lambda: Real", "query_expressions": "Range(lambda)", "answer_expressions": "(-oo,0)+(0,1/2)", "fact_spans": "[[[11, 16]], [[17, 51]], [[57, 60]], [[53, 56]], [[1, 10], [64, 68]], [[17, 51]], [[1, 10]], [[0, 16]], [[11, 62]], [[64, 80]], [[82, 93]]]", "query_spans": "[[[82, 100]]]", "process": "According to the midpoint coordinate formula and the point difference method, the equation of line $ l $ can be found. Combining the condition that the line intersects the hyperbola at two distinct points, we obtain $ \\Delta > 0 $, which allows us to find the range of $ \\lambda $. Since the hyperbola equation is $ x^{2} - \\frac{y^{2}}{2} = 2 $, then $ \\lambda \\neq 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Since point $ P $ is exactly the midpoint of segment $ AB $, then $ x_{1} + x_{2} = 2 $, $ y_{1} + y_{2} = 2 $. Then \n\\[\n\\begin{cases}\nx_{1}^{2} - \\frac{y_{1}^{2}}{2} = \\lambda \\\\\nx_{2}^{2} - \\frac{y_{2}^{2}}{2} = \\lambda\n\\end{cases}\n\\]\nSubtracting these two equations and simplifying yields \n\\[\n\\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = 2 \\times \\frac{x_{1} + x_{2}}{y_{1} + y_{2}} = 2\n\\]\nThus, the slope of line $ l $ is 2. Therefore, the equation of line $ l $ is $ y = 2x - 1 $, and $ x^{2} - \\frac{y^{2}}{2} = \\lambda $. Simplifying gives $ 2x^{2} - 4x + 2\\lambda + 1 = 0 $. Since line $ l $ intersects the hyperbola at two distinct points, we have $ \\Delta = 16 - 4 \\times 2 \\times (2\\lambda + 1) > 0 $. Solving yields $ \\lambda < \\frac{1}{2} $ and $ \\lambda \\neq 0 $. Therefore, the range of $ \\lambda $ is $ (-\\infty, 0) \\cup (0, \\frac{1}{2}) $." }, { "text": "Given that the parabolas $\\Gamma_{1}$ and $\\Gamma_{2}$ both have focus at point $F(2,1)$, and their directrix equations are $x=0$ and $5x+12y=0$, respectively. Suppose the two parabolas intersect at points $A$ and $B$. Then the equation of line $AB$ is?", "fact_expressions": "Gamma_1: Parabola;A: Point;B: Point;F: Point;Gamma_2: Parabola;Focus(Gamma_1)=F;Focus(Gamma_2)=F;Coordinate(F)=(2,1);Expression(Directrix(Gamma_2)) = (5*x + 12*y = 0);Expression(Directrix(Gamma_1)) = (x = 0);Intersection(Gamma_1,Gamma_2)={A,B}", "query_expressions": "Expression(LineOf(A,B))", "answer_expressions": "y=(2/3)*x", "fact_spans": "[[[2, 17]], [[77, 80]], [[81, 84]], [[35, 44]], [[18, 30]], [[2, 44]], [[2, 44]], [[35, 44]], [[18, 68]], [[2, 55]], [[71, 86]]]", "query_spans": "[[[88, 100]]]", "process": "According to the definition of a parabola, write the equations of the two parabolas (squared), subtract them to obtain the equation satisfied by the coordinates of points A and B, and simplify this equation (determine the sign based on the positions of points A and B relative to the two directrices) to get the equation of line AB. Detailed solution: By the definition of a parabola, we have \n_{T_{1}}:(x-2)^{2}+(y-1)^{2}=x^{2}; \nI_{2}:(x-2)^{2}+(y-1)^{2}=\\frac{(5x+12y)^{2}}{13^{2}}. \nSubtracting these two equations gives \nx^{2}=\\frac{(5x+12y)^{2}}{13^{2}}. \nSince points A and B lie to the right of x=0 and above the line 5x+12y=0, it follows that \nx=\\frac{5x+12y}{13}, \nwhich simplifies to \ny=\\frac{2}{3}x." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ has two foci $F_{1}(-\\frac{\\sqrt{3}}{2}, 0)$ , $F_{2}(\\frac{\\sqrt{3}}{2}, 0)$, and point $P$ lies on the hyperbola in the first quadrant such that $\\tan \\angle P F_{1} F_{2}=\\frac{1}{2}$, $\\tan \\angle P F_{2} F_{1}=-2$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-sqrt(3)/2, 0);Coordinate(F2) = (sqrt(3)/2, 0);Focus(G) = {F1, F2};Quadrant(P) = 1;PointOnCurve(P, G);Tan(AngleOf(P, F1, F2)) = 1/2;Tan(AngleOf(P, F2, F1)) = -2", "query_expressions": "Eccentricity(G)", "answer_expressions": "3*sqrt(5)/5", "fact_spans": "[[[2, 61], [143, 146], [223, 226]], [[5, 61]], [[5, 61]], [[67, 98]], [[100, 131]], [[133, 137]], [[5, 61]], [[5, 61]], [[2, 61]], [[67, 98]], [[100, 131]], [[2, 131]], [[133, 149]], [[133, 149]], [[151, 190]], [[191, 221]]]", "query_spans": "[[[223, 232]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the circle $O$: $x^{2}+y^{2}=b^{2}$, if there exists a point $P$ on $C$ such that two tangents drawn from $P$ to circle $O$ touch the circle at points $A$ and $B$ respectively, satisfying $\\angle A P B=60^{\\circ}$, then the range of the eccentricity of ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;O: Circle;A: Point;P: Point;B: Point;l1: Line;l2: Line;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = b^2);TangentOfPoint(P, O) = {l1, l2};TangentPoint(l1,O)=A;TangentPoint(l2,O)=B;AngleOf(A, P, B) = ApplyUnit(60, degree);PointOnCurve(P,C)", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[sqrt(3)/2,1)", "fact_spans": "[[[2, 58], [86, 89], [159, 164]], [[8, 58]], [[8, 58]], [[59, 83], [106, 110]], [[121, 124]], [[101, 105], [101, 105]], [[125, 128]], [], [], [[8, 58]], [[8, 58]], [[2, 58]], [[59, 83]], [[100, 115]], [[100, 128]], [[100, 128]], [[131, 156]], [[86, 96]]]", "query_spans": "[[[159, 175]]]", "process": "Connect OP, ∴∠OPB=30°, ∵|OB|=b, ∴|OP|=2b, ∴2b⩽a, ∴4a²−4c²⩽a², ∴e⩾√3⁄2." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$, point $M$ lies on $E$, and $\\angle F_{1} M F_{2}=\\frac{2 \\pi}{3}$, then the area of $\\Delta F_{1} M F_{2}$ is?", "fact_expressions": "E: Ellipse;a: Number;F1: Point;M: Point;F2: Point;Expression(E) = (y^2/3 + x^2/a^2 = 1);RightFocus(E) = F2;LeftFocus(E)=F1;PointOnCurve(M, E);AngleOf(F1, M, F2) = (2*pi)/3", "query_expressions": "Area(TriangleOf(F1, M, F2))", "answer_expressions": "3*sqrt(3)", "fact_spans": "[[[18, 64], [76, 79]], [[25, 64]], [[2, 9]], [[71, 75]], [[10, 17]], [[18, 64]], [[2, 70]], [[2, 70]], [[71, 80]], [[82, 120]]]", "query_spans": "[[[122, 149]]]", "process": "" }, { "text": "Let the vertex of the parabola be at the origin, and the equation of the directrix be $x = -2$. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;Expression(Directrix(G)) = (x = -2)", "query_expressions": "Expression(G)", "answer_expressions": "y^2 = 8*x", "fact_spans": "[[[1, 4], [24, 27]], [[8, 10]], [[1, 10]], [[1, 22]]]", "query_spans": "[[[24, 32]]]", "process": "" }, { "text": "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 line $y=\\frac{4}{3} x$ intersects the hyperbola at points $A$ and $B$. If $A F \\perp B F$, then the equation of the asymptotes of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;F: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = (4/3)*x);RightFocus(G) = F;Intersection(H, G) = {A, B};IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[0, 59], [88, 91], [122, 125]], [[3, 59]], [[3, 59]], [[68, 87]], [[94, 97]], [[64, 67]], [[98, 101]], [[3, 59]], [[3, 59]], [[0, 59]], [[68, 87]], [[0, 67]], [[68, 103]], [[105, 120]]]", "query_spans": "[[[122, 133]]]", "process": "According to the problem: the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ has its foci on the $x$-axis, and the right focus is $F(c,0)$. Then\n$$\n\\begin{cases}\ny=\\frac{4}{3}x \\\\\n\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\n\\end{cases}\n$$\nSimplifying yields: $(9b^{2}-16a^{2})x^{2}=9a^{2}b^{2}$, so $x^{2}=\\frac{9a^{2}b^{2}}{9b^{2}-16a^{2}}$. Therefore, points $A$ and $B$ are symmetric about the origin. Let $A(x,\\frac{4}{3}x)$, $B(-x,-\\frac{4}{3}x)$, then $\\overrightarrow{FA}=(x-c,\\frac{4}{3}x)$, $\\overrightarrow{FB}=(-x-c,-\\frac{4}{3}x)$. Since $AF\\bot BF$, it follows that $\\overrightarrow{FA}\\cdot\\overrightarrow{FB}=0$, so $(x-c)(-x-c)+\\frac{4}{3}x\\times(-\\frac{4}{3}x)=0$. Simplifying gives: $c^{2}=\\frac{25}{9}x^{2}$. Hence,\n$$\na^{2}+b^{2}=\\frac{25}{9}\\times\\frac{9a^{2}b^{2}}{9b^{2}-16a^{2}},\n$$\nwhich leads to $9b^{4}-32a^{2}b^{2}-16a^{4}=0$. Therefore, $(b^{2}-4a^{2})(9b^{2}+4a^{2})=0$. Since $a>0$, $b>0$, we have $9b^{2}+4a^{2}\\neq0$, so $b^{2}-4a^{2}=0$, thus $b=2a$. The asymptotes of the hyperbola are $y=\\pm\\frac{b}{a}x=\\pm2x$." }, { "text": "Let the parabola $C$: $x^{2}=2 p y(p>0)$ have focus $F$ and directrix $l$, with $A \\in C$. A circle centered at $F$ with radius $FA$, denoted as circle $F$, intersects $l$ at points $B$ and $D$. Given that $\\angle B F D=90^{\\circ}$ and the area of $\\triangle A B D$ is $4 \\sqrt{2}$, find the value of $p$.", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*(p*y));p: Number;p>0;F: Point;Focus(C) = F;l: Line;Directrix(C) = l;A: Point;In(A,C);F1: Circle;Center(F1) = F;Radius(F1) = LineSegmentOf(F,A);B: Point;D: Point;Intersection(F1, l) = {B, D};AngleOf(B, F, D) = ApplyUnit(90, degree);Area(TriangleOf(A, B, D)) = 4*sqrt(2)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[1, 27]], [[1, 27]], [[155, 158]], [[9, 27]], [[31, 34], [57, 60]], [[1, 34]], [[38, 41], [78, 81]], [[1, 41]], [[44, 53]], [[44, 53]], [[73, 77]], [[56, 77]], [[64, 77]], [[82, 85]], [[86, 89]], [[73, 91]], [[93, 118]], [[120, 153]]]", "query_spans": "[[[155, 162]]]", "process": "By the given condition, $\\triangle BFD$ is an isosceles right triangle with hypotenuse $|BD|=2p$. Combining with $S_{\\triangle ABD}=4\\sqrt{2}$, the solution is obtained. By symmetry: $\\triangle BFD$ is an isosceles right triangle, hypotenuse $|BD|=2p$. The distance from point $A$ to the directrix $l$ is $d=|FA|=|FB|=\\sqrt{2}p$. $S_{\\Delta ABD}=4\\sqrt{2} \\Leftrightarrow \\frac{1}{2}\\times|BD|\\times d=4\\sqrt{2} \\Leftrightarrow p=2$" }, { "text": "Given that point $F_{1}$ is the left focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, point $A(1,1)$, and a moving point $P$ lies on the ellipse, find the minimum value of $|P A + P F_{1}|$.", "fact_expressions": "G: Ellipse;A: Point;P: Point;F1: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Coordinate(A) = (1, 1);LeftFocus(G) = F1;PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, A) + LineSegmentOf(P, F1)))", "answer_expressions": "6-sqrt(2)", "fact_spans": "[[[11, 48], [69, 71]], [[53, 62]], [[65, 68]], [[2, 10]], [[11, 48]], [[53, 62]], [[2, 52]], [[65, 72]]]", "query_spans": "[[[74, 96]]]", "process": "" }, { "text": "Through a vertex of the major axis of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, draw two tangent lines to the circle $x^{2}+y^{2}=b^{2}$, with points of tangency $A$ and $B$. If $\\angle A O B=90^{\\circ}$ ($O$ is the origin), then the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;A: Point;O: Origin;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = b^2);l1: Line;l2: Line;TangentOfPoint(OneOf(Endpoint(MajorAxis(C))),G) = {l1,l2};TangentPoint(l1,G)=A;TangentPoint(l2,G)=B;AngleOf(A, O, B) = ApplyUnit(90, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[1, 58], [145, 150]], [[7, 58]], [[7, 58]], [[66, 86]], [[97, 100]], [[133, 136]], [[101, 104]], [[7, 58]], [[7, 58]], [[1, 58]], [[66, 86]], [], [], [[0, 91]], [[0, 104]], [[0, 104]], [[107, 132]]]", "query_spans": "[[[145, 156]]]", "process": "" }, { "text": "The radius of a circle passing through the focus and vertex of the parabola $y^{2}=8 x$ and tangent to the directrix is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);H: Circle;PointOnCurve(Focus(G),H) = True;PointOnCurve(Vertex(G),H) = True;IsTangent(Directrix(G),H) = True", "query_expressions": "Radius(H)", "answer_expressions": "3", "fact_spans": "[[[2, 16]], [[2, 16]], [[29, 30]], [[0, 30]], [[0, 30]], [[0, 30]]]", "query_spans": "[[[29, 35]]]", "process": "Test analysis: The focus is at (2,0), so the center of the circle lies on the line x=1. Let the coordinates of the center be (1,a). Therefore, the distance to the directrix, which is the radius, is 1+2=3." }, { "text": "Given that the center of an ellipse is at the origin and its axes of symmetry are the coordinate axes, the perimeter of the triangle formed by one endpoint $B$ of the minor axis and the two foci $F_{1}$, $F_{2}$ of the ellipse is $4+2 \\sqrt{3}$, and $\\angle F_{1} B F_{2}=\\frac{2 \\pi}{3}$. Find the equation of the ellipse?", "fact_expressions": "G: Ellipse;Center(G) = O;O: Origin;SymmetryAxis(G) = axis;OneOf(Endpoint(MinorAxis(G))) = B;B: Point;F1: Point;F2: Point;Focus(G) = {F1, F2};Perimeter(TriangleOf(B, F1, F2)) = 4 + 2*sqrt(3);AngleOf(F1, B, F2) = 2*pi/3", "query_expressions": "Expression(G)", "answer_expressions": "{(x^2/4+y^2=1),(y^2/4+x^2=1)}", "fact_spans": "[[[2, 4], [30, 32], [120, 122]], [[2, 10]], [[8, 10]], [[2, 18]], [[2, 29]], [[26, 29]], [[37, 44]], [[45, 52]], [[30, 52]], [[26, 76]], [[79, 117]]]", "query_spans": "[[[120, 127]]]", "process": "Let the foci of the ellipse lie on the x-axis, the length of the major axis be 2a, and the focal distance be 2c. As shown in the figure, in \\triangle F_{2}OB, from \\angle F_{2}BO = \\frac{\\pi}{3} we get: c = \\frac{\\sqrt{3}}{2}a. Therefore, the perimeter of \\triangle F_{2}BF_{1} is 2a + 2c = 2a + \\sqrt{3}a = 4 + 2\\sqrt{3}, \\therefore a = 2, c = \\sqrt{3}. \\therefore b^{2} = 1; hence, the standard equation of the required ellipse is \\frac{x^{2}}{4} + y^{2} = 1. When the foci of the ellipse lie on the y-axis, similarly, the equation is: \\frac{y^{2}}{4} + x^{2} = 1." }, { "text": "Given the curve $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$, when $m \\in[-2 ,-1]$, what is the range of values for the eccentricity of this curve?", "fact_expressions": "G: Curve;Expression(G) = (x^2/4 + y^2/m = 1);m: Number;In(m, [-2, -1])", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(5)/2, sqrt(6)/2]", "fact_spans": "[[[2, 39], [59, 61]], [[2, 39]], [[41, 56]], [[41, 56]]]", "query_spans": "[[[59, 72]]]", "process": "" }, { "text": "Draw a line through point $M(1,1)$ with slope $-\\frac{1}{2}$, intersecting the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$. If $M$ is the midpoint of segment $AB$, then the eccentricity of ellipse $C$ is?", "fact_expressions": "M: Point;Coordinate(M) = (1, 1);G: Line;Slope(G) = -1/2;PointOnCurve(M, G);C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[1, 10], [101, 104]], [[1, 10]], [[29, 31]], [[11, 31]], [[0, 31]], [[32, 89], [117, 122]], [[32, 89]], [[39, 89]], [[39, 89]], [[39, 89]], [[39, 89]], [[92, 95]], [[96, 99]], [[29, 99]], [[101, 115]]]", "query_spans": "[[[117, 128]]]", "process": "Test Analysis: Let A(x₁, y₁), B(x₂, y₂), 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 passing through point M(1,1) with slope -\\frac{1}{2} intersects the ellipse C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 (a > b > 0) at points A, 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, that is, \\frac{2}{a^{2}} + (-\\frac{1}{2}) \\cdot \\frac{2}{b^{2}} = 0. Therefore, a = \\sqrt{2}b, so c = b, thus e = \\frac{c}{a} = \\frac{\\sqrt{2}}{2}." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{m^{2}} - \\frac{y^{2}}{m^{2}+5} = 1$ $(m>0)$, respectively. If there exists a point $A$ on the hyperbola such that $\\angle F_{1} A F_{2} = \\frac{2\\pi}{3}$ and $|A F_{1}| = 3|A F_{2}|$, then $m = $?", "fact_expressions": "G: Hyperbola;m: Number;F1: Point;A: Point;F2: Point;m>0;Expression(G) = (-y^2/(m^2 + 5) + x^2/m^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A, G);AngleOf(F1, A, F2) = (2*pi)/3;Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(A, F2))", "query_expressions": "m", "answer_expressions": "2", "fact_spans": "[[[19, 72], [80, 83]], [[156, 159]], [[1, 8]], [[86, 90]], [[9, 16]], [[22, 72]], [[19, 72]], [[1, 78]], [[1, 78]], [[80, 90]], [[92, 130]], [[132, 154]]]", "query_spans": "[[[156, 161]]]", "process": "From the equation of the hyperbola, we obtain: the real semi-axis length $ a = m $. Let the semi-focal distance be $ c $, then $ c^{2} = 2m^{2} + 5 $. By the definition of the hyperbola, we have $ |AF_{1}| - |AF_{2}| = 2|AF_{2}| = 2a $, so $ |AF_{2}| = a = m $, $ |AF_{1}| = 3m $. In $ \\triangle F_{1}AF_{2} $, by the cosine law, $ |F_{1}F_{2}|^{2} = |F_{1}A|^{2} + |F_{2}A|^{2} - 2|F_{1}A||F_{2}A|\\cos\\angle F_{1}AF_{2} $, that is, $ 8m^{2} + 20 = 9m^{2} + m^{2} - 2 \\times 3m \\times m \\times (-\\frac{1}{2}) $. Solving gives: $ m = 2 $." }, { "text": "Given that the asymptotes of a hyperbola are $3x \\pm 4y = 0$, and the coordinates of the foci are $(-5, 0)$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Coordinate(OneOf(Focus(G))) = (-5, 0);Expression(Asymptote(G))=(pm*4*y+3*x=0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16-y^2/9=1", "fact_spans": "[[[2, 5], [43, 46]], [[2, 41]], [[2, 27]]]", "query_spans": "[[[43, 51]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, then $|F_{1} F_{2}|$=?", "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2", "query_expressions": "Abs(LineSegmentOf(F1, F2))", "answer_expressions": "10", "fact_spans": "[[[18, 57]], [[2, 9]], [[10, 17]], [[18, 57]], [[2, 63]], [[2, 63]]]", "query_spans": "[[[65, 82]]]", "process": "From \\frac{x^2}{16}-\\frac{y^{2}}{9}=1, we have a^{2}=16, b^{2}=9, so c=\\sqrt{a^{2}+b^{2}}=5, thus |F_{1}F_{2}|=2c=10." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Point $P$ lies on the left branch of the hyperbola, and the minimum value of $\\frac{|\\overrightarrow{P F_{2}}|^{2}}{|P F_{1}|}$ is $8 a$. Then the range of 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;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, LeftPart(G)) = True;Min(Abs(VectorOf(P, F2))^2/Abs(LineSegmentOf(P, F1))) = 8*a", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, 3]", "fact_spans": "[[[1, 8]], [[9, 16]], [[17, 63], [74, 77], [145, 148]], [[17, 63]], [[20, 63]], [[138, 143]], [[1, 68]], [[1, 68]], [[69, 73]], [[69, 81]], [[83, 143]]]", "query_spans": "[[[145, 159]]]", "process": "Problem Analysis: \nSince the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has left and right foci $F_{1}$, $F_{2}$ respectively, and $P$ is an arbitrary point on the left branch of the hyperbola, \n$\\therefore |PF_{2}|-|PF_{1}|=2a$, $|PF_{2}|=2a+|PF_{1}|$, \n$\\therefore \\frac{|\\overrightarrow{PF_{2}}|^{2}}{|PF_{1}|}=\\frac{(|PF_{1}|+2a)^{2}}{|PF_{1}|}=|PF_{1}|+\\frac{4a^{2}}{|PF_{1}|}+4a\\geqslant8a$ (equality holds if and only if $|PF_{1}|=2a$), \nso $|PF_{2}|=2a+|PF_{1}|=4a$, \n$\\because |PF_{2}|-|PF_{1}|=2a<2c$, $|PF_{1}|+|PF_{2}|=6a\\geqslant2c$, \n$\\therefore e\\in(1,3]$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the left focus is $F$, and $M$ is a point on one of the asymptotes of the hyperbola such that $OM \\perp MF$, where $O$ is the origin. If the area of triangle $OMF$ is $4$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/16 - y^2/b^2 = 1);b: Number;b>0;F: Point;LeftFocus(C) = F;M: Point;PointOnCurve(M, OneOf(Asymptote(C)));O: Origin;IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(M, F));Area(TriangleOf(O, M, F)) = 4", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[2, 55], [136, 142], [69, 72]], [[2, 55]], [[10, 55]], [[10, 55]], [[60, 63]], [[2, 63]], [[64, 67]], [[64, 80]], [[100, 103]], [[82, 97]], [[110, 134]]]", "query_spans": "[[[136, 148]]]", "process": "As shown in the figure, by the symmetry of the hyperbola, assume without loss of generality that point M is in the second quadrant, then the corresponding asymptote equation is $ y = -\\frac{b}{a}x \\Rightarrow bx + ay = 0 $. Since $ OM \\perp MF $, it follows that $ MF = \\frac{|bc|}{\\sqrt{a^{2}+b^{2}}} = \\frac{bc}{c} = b $. Given that the area of $ \\triangle OMF $ is 4, then $ \\frac{1}{2} \\times 4 \\times b = 4 \\Rightarrow b = 2 $. Thus, the eccentricity $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\frac{\\sqrt{20}}{4} = \\frac{\\sqrt{5}}{2} $." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ with right focus $F$, a line passing through the origin $O$ with slope $\\frac{4}{3}$ intersects the right branch of $C$ at point $A$. If $|O A|=|O F|$, 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;RightFocus(C) = F;O: Origin;PointOnCurve(O, G) = True;Slope(G) = 4/3;G: Line;Intersection(G, RightPart(C)) = A;A: Point;Abs(LineSegmentOf(O, A)) = Abs(LineSegmentOf(O, F))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 66], [102, 105], [130, 133]], [[2, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[10, 66]], [[71, 74]], [[2, 74]], [[76, 81]], [[75, 101]], [[82, 101]], [[99, 101]], [[99, 113]], [[109, 113]], [[115, 128]]]", "query_spans": "[[[130, 139]]]", "process": "Suppose the inclination angle of the given line is $\\theta$, from the problem we know: $\\tan\\theta=\\frac{4}{3}$, so $\\sin\\theta=\\frac{4}{5}$, $\\cos\\theta=\\frac{3}{5}$. Then, according to $|OA|=|OF|$, find the coordinates of point $A$, and finally substitute into the hyperbola equation, simplify, and combine with $e=\\frac{c}{a}$ ($e>1$) to obtain the result. Let the inclination angle of the given line be $\\theta$, from the problem we know: $\\tan\\theta=\\frac{4}{3}$, so $\\sin\\theta=\\frac{4}{5}$, $\\cos\\theta=\\frac{3}{5}$. Since $|OA|=|OF|=c$, therefore, $A\\left(\\frac{3}{5}c,\\frac{4}{5}c\\right)$. Also, since $A$ lies on hyperbola $C$, substituting into $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, we get $\\frac{9c^{2}}{25a^{2}}-\\frac{16c^{2}}{25b^{2}}=1$, so $9e^{2}-\\frac{16e^{2}}{e^{2}-1}=25$, that is, $e^{4}-9e^{2}-16e^{2}=25e^{2}-25$. Simplifying yields: $9e^{4}-50e^{2}+25=0$, which is $(9e^{2}-5)(e^{2}-5)=0$. Solving gives: $e=+\\frac{\\sqrt{5}}{5}$ or $e=\\pm\\sqrt{5}$. Since $e>1$, therefore $e=\\sqrt{5}$." }, { "text": "Given that a line passing through the focus of the parabola $C$: $y^{2}=8x$ intersects the parabola at points $A$ and $B$. A perpendicular is drawn from point $A$ to the directrix of the parabola, with foot of the perpendicular at $M$. If $|AB|=|BM|$, then the horizontal coordinate of point $A$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);G: Line;PointOnCurve(Focus(C), G);A: Point;B: Point;Intersection(G, C) = {A, B};L: Line;PointOnCurve(A, L);IsPerpendicular(L, Directrix(C));M: Point;FootPoint(L, Directrix(C)) = M;Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(B, M))", "query_expressions": "XCoordinate(A)", "answer_expressions": "4", "fact_spans": "[[[3, 22], [28, 31], [48, 51]], [[3, 22]], [[25, 27]], [[2, 27]], [[32, 35], [43, 47], [81, 85]], [[36, 39]], [[25, 41]], [], [[42, 56]], [[42, 56]], [[60, 63]], [[42, 63]], [[65, 78]]]", "query_spans": "[[[81, 91]]]", "process": "The parabola C: y^{2}=8x has focus F(2,0) and directrix x=-2. Let A(x_{1},y_{1}), B(x_{2},y_{2}), M(-2,y_{1}). Take N as the midpoint of AM, connect BN intersecting the x-axis at T. Given |AB|=|BM|, then BN\\botAM. From \\frac{|TF|}{AN}=\\frac{|BF|}{|AB|}, we obtain \\frac{2-x_{2}}{2+x_{1}}=\\frac{2+x_{2}}{x_{1}+x_{2}+4}. Also, x_{1}=2x_{2}+2, then (3-\\frac{x_{1}}{2})(\\frac{3x_{1}}{2}+3)=\\frac{2}{2}(1+\\frac{x_{1}}{2})(2+x_{1}). Solving gives x_{1}=4 or x_{1}=-2 (discarded). Ultimately, A" }, { "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, $A$ is the top vertex of the ellipse, $D$ is the midpoint of $A F_{2}$, if $\\overrightarrow{A F_{2}} \\perp \\overrightarrow{D F_{1}}$, then the eccentricity of ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;F2: Point;D: Point;F1: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;UpperVertex(C)=A;MidPoint(LineSegmentOf(A, F2)) = D;IsPerpendicular(VectorOf(A,F2),VectorOf(D,F1))", "query_expressions": "Eccentricity(C)", "answer_expressions": "1/2", "fact_spans": "[[[2, 59], [88, 90], [172, 177]], [[8, 59]], [[8, 59]], [[84, 87]], [[76, 83]], [[94, 97]], [[68, 75]], [[8, 59]], [[8, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 93]], [[94, 110]], [[112, 169]]]", "query_spans": "[[[172, 183]]]", "process": "" }, { "text": "If the minimum distance from a point on the ellipse to the focus is $5$ and the maximum distance is $15$, then what is the length of the minor axis of the ellipse?", "fact_expressions": "G: Ellipse;P: Point;PointOnCurve(P, G);Min(Distance(P, Focus(G))) = 5;Max(Distance(P, Focus(G))) = 15", "query_expressions": "Length(MinorAxis(G))", "answer_expressions": "10*sqrt(3)", "fact_spans": "[[[1, 3], [31, 33]], [[5, 6]], [[1, 6]], [[1, 20]], [[1, 29]]]", "query_spans": "[[[31, 38]]]", "process": "Assume the equation of the ellipse is: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). From the given conditions, we have \\begin{cases}a+c=15\\\\a-c=5\\end{cases}, solving -c=5 gives \\begin{cases}a=10\\\\c=5\\end{cases}, then the length of the minor axis of the ellipse is: 2b=2\\sqrt{10^{2}-5^{2}}=10\\sqrt{3}." }, { "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $G$ lies on the branch of the hyperbola in the first quadrant. If point $H$ satisfies $\\overrightarrow{O H}=\\overrightarrow{O G}+\\lambda\\left(\\frac{\\overrightarrow{G F_{1}}}{|\\overrightarrow{G F_{1}}|}+\\frac{\\overrightarrow{G F_{2}}}{|\\overrightarrow{G F_{2}}|}\\right)$ $(\\lambda \\neq 0)$, and the intersection point of line $G H$ with the $x$-axis is $P\\left(\\frac{\\sqrt{3}}{3}, 0\\right)$, then what are the coordinates of point $G$?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Point;PointOnCurve(G, C) ;Quadrant(G) = 1;H: Point;VectorOf(O, H) = VectorOf(O, G) + lambda*(VectorOf(G, F1)/Abs(VectorOf(G, F1)) + VectorOf(G, F2)/Abs(VectorOf(G, F2)));lambda: Number;Negation(lambda = 0);O: Origin;Intersection(LineOf(G,H), xAxis) = P;P : Point;Coordinate(P) = (sqrt(3)/3, 0)", "query_expressions": "Coordinate(G)", "answer_expressions": "(sqrt(3),2)", "fact_spans": "[[[2, 35], [71, 74]], [[2, 35]], [[44, 51]], [[52, 59]], [[2, 59]], [[2, 59]], [[60, 64], [319, 323]], [[60, 75]], [[60, 75]], [[77, 81]], [[83, 273]], [[83, 273]], [[83, 273]], [[83, 273]], [[275, 317]], [[291, 317]], [[291, 317]]]", "query_spans": "[[[319, 328]]]", "process": "From $\\overrightarrow{OH}=\\overrightarrow{OG}+\\lambda\\left(\\frac{\\overrightarrow{GF_1}}{|\\overrightarrow{GF_1}|}+\\frac{\\overrightarrow{GF_2}}{|\\overrightarrow{GF_2}|}\\right)$ it follows that $GH$ is the angle bisector of $\\angle F_1GF_2$. According to the angle bisector property and the definition of a hyperbola, the distance from $G$ to the foci can be found, and using the fact that $G$ lies on the hyperbola, the coordinates of $G$ can be determined. Since $\\overrightarrow{OH}=\\overrightarrow{OG}+\\lambda\\left(\\frac{\\overrightarrow{GF_1}}{|\\overrightarrow{GF_1}|}+\\frac{\\overrightarrow{GF_2}}{|\\overrightarrow{GF_2}|}\\right)$, we have $\\overrightarrow{GH}=\\overrightarrow{OH}-\\overrightarrow{OG}=\\lambda\\left(\\frac{\\overrightarrow{GF_1}}{|\\overrightarrow{GF_1}|}+\\frac{\\overrightarrow{GF_2}}{|\\overrightarrow{GF_2}|}\\right)$, so $GH$ is the angle bisector of $\\angle F_1GF_2$. From the hyperbola equation $x^2-\\frac{y^2}{2}=1$, we get $a=1$, $b=\\sqrt{2}$, thus $c=\\sqrt{a^2+b^2}=\\sqrt{3}$, so $F_1(-\\sqrt{3},0)$, $F_2(\\sqrt{3},0)$. Also $P\\left(\\frac{\\sqrt{3}}{3},0\\right)$, hence $|PF_1|=\\frac{4\\sqrt{3}}{3}$, $|PF_2|=\\frac{2\\sqrt{3}}{3}$, $\\frac{|GF_1|}{|GF_2|}=\\frac{|PF_1|}{|PF_2|}=\\frac{\\frac{4\\sqrt{3}}{3}}{\\frac{2\\sqrt{3}}{3}}=2$. By the definition of the hyperbola, $|GF_1|-|GF_2|=2a=2$, thus $|GF_2|=2$, $|GF_1|=4$. Let $G(x_0,y_0)$ ($y_0>0$), then\n$$\n\\begin{cases}\nx_0^2-\\frac{y_0^2}{2}=1 \\\\\n(x_0-\\sqrt{3})^2+y_0^2=4\n\\end{cases}\n$$\nSolving gives\n$$\n\\begin{cases}\nx_0=\\sqrt{3} \\\\\ny_0=2\n\\end{cases}\n$$\nTherefore, $G(\\sqrt{3},2)$." }, { "text": "If one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ is given by $3x-2y=0$, then what is the eccentricity of the ellipse for which the vertices and foci of the hyperbola are respectively the foci and vertices?", "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G)))=(3*x-2*y=0);Vertex(G)=Focus(H);Focus(G)=Vertex(H)", "query_expressions": "Eccentricity(H)", "answer_expressions": "2*sqrt(13)/13", "fact_spans": "[[[1, 48], [71, 74]], [[4, 48]], [[89, 91]], [[4, 48]], [[1, 48]], [[1, 68]], [[70, 91]], [[70, 91]]]", "query_spans": "[[[89, 97]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with two vertices $A(a, 0)$, $B(0, b)$. Perpendiculars to $AB$ are drawn through $A$ and $B$, intersecting the ellipse at distinct points $C$ and $D$, respectively. If $2|BD|=3|AC|$, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;D: Point;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (a, 0);Coordinate(B) = (0, b);In(A,Vertex(G));In(B,Vertex(G));L1:Line;L2:Line;PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(L1,LineSegmentOf(A,B));IsPerpendicular(L2,LineSegmentOf(A,B));Intersection(L1, G) = C;Intersection(L2, G) = D;Negation(C=D);2*Abs(LineSegmentOf(B, D)) = 3*Abs(LineSegmentOf(A, C))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 54], [102, 104], [136, 138]], [[4, 54]], [[59, 68]], [[59, 68], [82, 85]], [[71, 80], [86, 89]], [[112, 115]], [[108, 111]], [[4, 54]], [[4, 54]], [[2, 54]], [[59, 68]], [[71, 80]], [[2, 80]], [[2, 80]], [], [], [[81, 100]], [[81, 100]], [[81, 100]], [[81, 100]], [[81, 117]], [[81, 117]], [[105, 117]], [[119, 134]]]", "query_spans": "[[[136, 144]]]", "process": "The equation of the line perpendicular to AB passing through A is y=\\frac{a}{b}(x-a), and together with \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1, solve the system of equations. The equation of the line perpendicular to AB passing through B is y=\\frac{a}{b}x+b, and together with \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1. Since 2|BD|=3|AC|, it follows that 2|x_{D}-0|=3|x_{C}-a|. \\frac{4a3b^{2}}{a^{4}+b^{4}}=\\frac{3a\\times2b^{4}}{a^{4}+b^{4}} \\therefore 2a^{2}=3b^{2}=3a^{2}-3c^{2}, a^{2}=3c^{2} \\therefore e^{2}=\\frac{1}{3}, e=\\frac{\\sqrt{3}}{3}" }, { "text": "If the curve represented by the equation $x^{2}-3 m y^{2}=1$ is an ellipse with foci on the $x$-axis, then the range of real values for $m$ is?", "fact_expressions": "G: Ellipse;H: Curve;Expression(H) = (x^2 - 3*m*y^2 = 1);m: Real;PointOnCurve(Focus(G), xAxis) ;H = G", "query_expressions": "Range(m)", "answer_expressions": "(-oo, -1/3)", "fact_spans": "[[[37, 39]], [[25, 27]], [[1, 27]], [[41, 46]], [[28, 39]], [[25, 39]]]", "query_spans": "[[[41, 53]]]", "process": "Rewrite the equation as \\frac{x^{2}}{1}+\\frac{y^{2}}{-\\frac{1}{3m}}=1. According to the problem, we have \\begin{cases}-\\frac{1}{3m}>0,\\\\1>-\\frac{1}{3m},\\end{cases} solving which yields m<-\\frac{1}{3}, that is, the range of real number m is {m|m<-\\frac{1}{3}}" }, { "text": "The length of the real axis of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1)", "query_expressions": "Length(RealAxis(G))", "answer_expressions": "4", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 44]]]", "process": "From the given condition, a=2, so the length of the real axis is 2a=4." }, { "text": "If a hyperbola has the same foci as the ellipse $x^{2}+4 y^{2}=64$, and one of its asymptotes is given by the equation $x+\\sqrt{3} y=0$, then what is the standard equation of this hyperbola?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2 + 4*y^2 = 64);Focus(G) = Focus(H);Expression(OneOf(Asymptote(G))) = (x+sqrt(3)*y=0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/36-y^2/12=1", "fact_spans": "[[[1, 4], [32, 33], [61, 64]], [[5, 25]], [[5, 25]], [[1, 31]], [[32, 58]]]", "query_spans": "[[[61, 71]]]", "process": "\\because x^{2}+4y^{2}=64 \\Leftrightarrow \\frac{x^{2}}{64}+\\frac{y^{2}}{16}=1 \\therefore the foci of this ellipse lie on the x-axis, and the coordinates of the foci are: (\\pm 4\\sqrt{3},0) \\because one asymptote equation of the hyperbola is x+\\sqrt{3}y=0, \\therefore assume the equation of the hyperbola is x^{2}-3y^{2}=\\lambda, namely \\frac{x^{2}}{\\lambda}-\\frac{y^{2}}{\\frac{\\lambda}{3}}=1 \\because the hyperbola shares the same foci with x^{2}+4y^{2}=64, \\therefore \\lambda+\\frac{1}{3}\\lambda=48, \\therefore \\lambda=36, \\therefore the equation of the hyperbola is \\frac{x^{2}}{36}-\\frac{y^{2}}{12}=1" }, { "text": "If the eccentricity of the ellipse $W$: $\\frac{x^{2}}{2}+\\frac{y^{2}}{m}=1$ is $\\frac{\\sqrt{6}}{3}$, then $m=$?", "fact_expressions": "W: Ellipse;m: Number;Expression(W) = (x^2/2 + y^2/m = 1);Eccentricity(W) = sqrt(6)/3", "query_expressions": "m", "answer_expressions": "{2/3,6}", "fact_spans": "[[[1, 42]], [[69, 72]], [[1, 42]], [[1, 67]]]", "query_spans": "[[[69, 74]]]", "process": "\\textcircled{1} When the foci of the ellipse lie on the x-axis, then $ a = \\sqrt{2} $, $ c = \\sqrt{2 - m} $. From the given condition, $ \\frac{\\sqrt{2 - m}}{\\sqrt{2}} = \\frac{\\sqrt{6}}{3} $, solving gives $ m = \\frac{2}{3} $. \\textcircled{2} When the foci of the ellipse lie on the y-axis, then $ a = \\sqrt{m} $, $ c = \\sqrt{m - 2} $. From the given condition, $ \\frac{\\sqrt{m - 2}}{\\sqrt{m}} = \\frac{\\sqrt{6}}{3} $, solving gives $ m = 6 $. In conclusion, $ m = \\frac{2}{3} $ or $ m = 6 $." }, { "text": "The eccentricity of the hyperbola $9 x^{2}-4 y^{2}=36$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (9*x^2 - 4*y^2 = 36)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(13)/2", "fact_spans": "[[[0, 23]], [[0, 23]]]", "query_spans": "[[[0, 29]]]", "process": "From 9x^{2}-4y^{2}=36, we have \\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1, so a=2, b=3. Also, c^{2}=a^{2}+b^{2}, so c=\\sqrt{13}, then e=\\frac{c}{a}=\\frac{\\sqrt{13}}{2}" }, { "text": "The line $ l $ intersects the parabola $ y = \\frac{x^{2}}{2} $ at points $ A $ and $ B $. When $ |AB| = 4 $, what is the minimum value of the distance from the midpoint $ M $ of chord $ AB $ to the $ x $-axis?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;M: Point;Expression(G)=(y=x^2/2);Intersection(l,G)={A,B};Abs(LineSegmentOf(A,B))=4;IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M", "query_expressions": "Min(Distance(M, xAxis))", "answer_expressions": "3/2", "fact_spans": "[[[0, 5]], [[6, 28]], [[31, 34]], [[35, 38]], [[62, 65]], [[6, 28]], [[0, 40]], [[42, 51]], [[6, 60]], [[55, 65]]]", "query_spans": "[[[62, 78]]]", "process": "According to the problem, the focus of the parabola $ y = \\frac{x^2}{2} $ has coordinates $ (0, \\frac{1}{2}) $. By the definition of a parabola, as shown in the figure, the required distance is $ d = MM_1 = \\frac{AA_1 + BB_1}{2} = \\frac{AF + BF - 1}{2} \\geqslant \\frac{AB - 1}{2} = \\frac{3}{2} $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, respectively, point $P$ lies on $C$, and $\\angle F_{1} P F_{2}=60^{\\circ}$, then what is the value of $|P F_{1}| \\cdot |P F_{2}|$?", "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, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))", "answer_expressions": "4", "fact_spans": "[[[20, 43], [55, 58]], [[20, 43]], [[2, 9]], [[10, 17]], [[2, 49]], [[2, 49]], [[50, 54]], [[50, 59]], [[60, 93]]]", "query_spans": "[[[95, 124]]]", "process": "From the hyperbola equation, we know: |F_{1}F_{2}| = 2c = 2\\sqrt{2}. In \\triangle PF_{1}F_{2}, by the law of cosines: |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF_{2} = (|PF_{1}| - |PF_{2}|)^{2} + |PF_{1}|\\cdot|PF_{2}|, \\therefore |PF_{1}|\\cdot|PF_{2}| = 8 - (|PF_{1}| - |PF_{2}|)^{2}. Since ||PF_{1}| - |PF_{2}|| = 2, \\therefore |PF_{1}|\\cdot|PF_{2}| = 4." }, { "text": "The equation of an ellipse centered at the origin, with one focus at $(0 , 5)$ and minor axis of length $4$, is?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;Coordinate(OneOf(Focus(G))) = (0, 5);Length(MinorAxis(G)) = 4", "query_expressions": "Expression(G)", "answer_expressions": "y^2/29+x^2/4=1", "fact_spans": "[[[31, 33]], [[3, 5]], [[0, 33]], [[6, 33]], [[23, 33]]]", "query_spans": "[[[31, 37]]]", "process": "" }, { "text": "If the equation $\\frac{x^{2}}{5-m}+\\frac{y^{2}}{m^{2}-2 m-3}=1$ represents a hyperbola with foci on the $y$-axis, then the range of real values for $m$ is?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/(5 - m) + y^2/(m^2 - 2*m - 3) = 1);PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(m)", "answer_expressions": "(5,oo)", "fact_spans": "[[[61, 64]], [[66, 71]], [[3, 64]], [[52, 64]]]", "query_spans": "[[[66, 78]]]", "process": "" }, { "text": "The parabola $C$: $x^{2}=2 p y$ has focus $F$, and the point $P(x_{0}, \\frac{p}{8})$ is a point on the parabola $C$ satisfying $|P F|=\\frac{5}{8}$. Then the equation of the parabola $C$ is?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;F: Point;Focus(C) = F;P: Point;Coordinate(P) = (x0, p/8);x0: Number;PointOnCurve(P,C) = True;Abs(LineSegmentOf(P,F)) = 5/8", "query_expressions": "Expression(C)", "answer_expressions": "x^2=pm*2*y", "fact_spans": "[[[0, 21], [54, 60], [87, 93]], [[0, 21]], [[8, 21]], [[25, 28]], [[0, 28]], [[29, 53]], [[29, 53]], [[30, 53]], [[29, 64]], [[66, 85]]]", "query_spans": "[[[87, 98]]]", "process": "When p>0, PF=\\frac{p}{8}+\\frac{p}{2}=\\frac{5}{8}, solving gives p=1, then the equation of parabola C is: x^{2}=2y. When p<0, |PF|=-\\frac{p}{2}-\\frac{p}{8}=\\frac{5}{8}, solving gives p=-1, then the equation of parabola C is: x^{2}=-2y;" }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2$, and the length of the chord intercepted on the hyperbola by a line passing through the right focus of the hyperbola and perpendicular to the $x$-axis is $m$, then $\\frac{m}{a}=$?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;m: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 2;PointOnCurve(RightFocus(G), H);IsPerpendicular(xAxis, H);Length(InterceptChord(H, G)) = m", "query_expressions": "m/a", "answer_expressions": "6", "fact_spans": "[[[2, 58], [68, 71], [86, 89]], [[5, 58]], [[5, 58]], [[83, 85]], [[95, 98]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[67, 85]], [[75, 85]], [[83, 98]]]", "query_spans": "[[[100, 115]]]", "process": "The focal distance of the hyperbola is 2c, then \\frac{c}{a}=2, that is, c=2a, then b=\\sqrt{a}. Substitute x=c=2a into the hyperbola to obtain y=\\pm\\frac{b^{2}}{a}, hence m=\\frac{2b^{2}}{a}, so \\frac{m}{a}=\\frac{2b^{2}}{a^{2}}=6^{-}" }, { "text": "Given the parabola $M$: $y^{2}=3 x$, a line $l$ passing through the point $(3,0)$ intersects the parabola $M$ at points $A$ and $B$. Then $\\angle A O B$=?", "fact_expressions": "M: Parabola;Expression(M) = (y^2 = 3*x);l: Line;H: Point;Coordinate(H) = (3, 0);PointOnCurve(H, l);A: Point;B: Point;Intersection(l, M) = {A, B};O: Origin", "query_expressions": "AngleOf(A, O, B)", "answer_expressions": "ApplyUnit(90, degree)", "fact_spans": "[[[2, 21], [39, 45]], [[2, 21]], [[33, 38]], [[24, 32]], [[24, 32]], [[23, 38]], [[46, 49]], [[50, 53]], [[33, 55]], [[57, 71]]]", "query_spans": "[[[57, 73]]]", "process": "Since the slope of line $ l $ is non-zero by the given condition, we can set the equation of line $ l $ as $ x = ty + 3 $. Substituting into $ y^2 = 3x $ gives $ y^2 - 3ty - 9 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ y_1 + y_2 = 3t $, $ y_1 y_2 = -9 $. Because $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_1 x_2 + y_1 y_2 = (ty_1 + 3)(ty_2 + 3) + y_1 y_2 = (1 + t^2)y_1 y_2 + 3t(y_1 + y_2) + 9 = (1 + t^2)(-9) + 3t \\cdot 3t + 9 = 0 $, so $ \\overrightarrow{OA} \\perp \\overrightarrow{OB} $, that is, $ \\angle AOB = 90^{\\circ} $. Hence, fill in $ 90^{\\circ} $." }, { "text": "If points $O$ and $F$ are the center and the left focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, respectively, and point $P$ is any 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;O: Origin;P: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);PointOnCurve(P, RightPart(G));Center(G)=O;LeftFocus(G)=F", "query_expressions": "Range(DotProduct(VectorOf(O, P), VectorOf(F, P)))", "answer_expressions": "[3+2*sqrt(3),+oo)", "fact_spans": "[[[13, 41], [54, 57]], [[1, 5]], [[49, 53]], [[6, 10]], [[13, 41]], [[49, 65]], [[1, 48]], [[1, 48]]]", "query_spans": "[[[67, 123]]]", "process": "Problem Analysis: From the given, let P(x, y), and F(-2, 0), we have: \\overrightarrow{OP}=(x,y), \\overrightarrow{FP}=(x+2,y), \\overrightarrow{OP}\\cdot\\overrightarrow{FP}=x(x+2)+y^{2}. Since P lies on \\frac{x^{2}}{3}-y^{2}=1, it follows that \\overrightarrow{OP}\\cdot\\overrightarrow{FP}=\\frac{4x^{2}}{3}+2x-1=\\frac{4}{3}(x+\\frac{3}{4})^{2}-\\frac{7}{4}, x\\geqslant\\sqrt{3}. When x=\\sqrt{3}, the minimum value is 3+2\\sqrt{3}. Thus, the range is [3+2\\sqrt{3}, +\\infty)." }, { "text": "Given that the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{m^{2}}=1$ $(m>0)$ have no common points with the circle $x^{2}+(y+2)^{2}=1$, find the range of values for the focal distance of this hyperbola.", "fact_expressions": "G: Hyperbola;m: Number;H: Circle;m>0;Expression(G) = (x^2 - y^2/m^2 = 1);Expression(H) = (x^2 + (y + 2)^2 = 1);NumIntersection(Asymptote(G),H)=0", "query_expressions": "Range(FocalLength(G))", "answer_expressions": "(2,4)", "fact_spans": "[[[2, 39], [72, 75]], [[5, 39]], [[44, 64]], [[5, 39]], [[2, 39]], [[44, 64]], [[2, 69]]]", "query_spans": "[[[72, 85]]]", "process": "" }, { "text": "Given that points $A$ and $B$ lie on the parabola $y^{2}=2 p x (p>0)$ with focus $F$, if $|A F|+|B F|=4$ and the distance from the midpoint of segment $A B$ to the line $x=\\frac{p}{2}$ is $1$, then what is the value of $p$?", "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x = p/2);PointOnCurve(A,G);PointOnCurve(B,G);Focus(G)=F;Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 4;Distance(MidPoint(LineSegmentOf(A,B)),H)=1", "query_expressions": "p", "answer_expressions": "{1,3}", "fact_spans": "[[[20, 43]], [[99, 102]], [[73, 90]], [[2, 5]], [[6, 9]], [[16, 19]], [[23, 43]], [[20, 43]], [[73, 90]], [[2, 44]], [[2, 44]], [[13, 43]], [[46, 61]], [[62, 97]]]", "query_spans": "[[[99, 106]]]", "process": "Draw perpendiculars from A and B to the directrix $ l: x = -\\frac{p}{2} $, with feet of perpendiculars C and D respectively. Let M be the midpoint of AB, and N be the projection of M onto the directrix. Connect MN. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $, $ M(x_{0},y_{0}) $. According to the definition of a parabola, $ |AF| + |BF| = |AC| + |BD| = 4 $. Thus, in trapezoid ACDB, the median line $ MN = \\frac{1}{2}(AC + BD) = 2 $, giving $ x_{0} = 2 - \\frac{p}{2} $. Since the distance from the midpoint of segment AB to the line $ x = \\frac{p}{2} $ is 1, we have $ \\left|x_{0} - \\frac{p}{2}\\right| = 1 $, so $ |2 - p| = 1 $. Solving gives $ p = 1 $ or $ p = 3 $." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}$, $F_{2}$, upper vertex $A$, and point $P$ a point on the ellipse in the first quadrant such that $|P F_{1}|+|P F_{2}|=4|F_{1} F_{2}|$, and $S_{\\Delta P F_{1} A}=2 S_{\\Delta P F_{1} F_{2}}$, find the slope of the line $P F_{1}$.", "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;A: Point;UpperVertex(G) = A;P: Point;Quadrant(P) = 1;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2)) = 4*Abs(LineSegmentOf(F1, F2));Area(TriangleOf(P, F1, A))=2*Area(TriangleOf(P, F1, F2))", "query_expressions": "Slope(LineOf(P, F1))", "answer_expressions": "sqrt(15)/5", "fact_spans": "[[[2, 54], [95, 97]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[61, 68]], [[69, 76]], [[2, 76]], [[2, 76]], [[81, 84]], [[2, 84]], [[85, 89]], [[85, 101]], [[85, 101]], [[102, 138]], [[139, 188]]]", "query_spans": "[[[190, 206]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x (p>0)$ has focus $F$, directrix $l$: $x=-\\frac{5}{4}$, point $M$ lies on the parabola $C$, point $A$ lies on the directrix $l$, if $M A \\perp l$, and the inclination angle of line $A F$ is $\\frac{\\pi}{3}$, then $|M F|$=?", "fact_expressions": "C: Parabola;p: Number;l: Line;A: Point;F: Point;M: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Expression(l)=(x=-5/4);PointOnCurve(M, C);Directrix(C)=l;PointOnCurve(A, l);IsPerpendicular(LineSegmentOf(M, A),l);Inclination(LineOf(A,F)) = pi/3", "query_expressions": "Abs(LineSegmentOf(M, F))", "answer_expressions": "5", "fact_spans": "[[[2, 30], [67, 73]], [[10, 30]], [[40, 61], [82, 85]], [[75, 79]], [[34, 37]], [[62, 66]], [[10, 30]], [[2, 30]], [[2, 37]], [[40, 61]], [[62, 74]], [[67, 85]], [[67, 86]], [[88, 101]], [[102, 129]]]", "query_spans": "[[[131, 140]]]", "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 $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the hyperbola at points $P$ and $Q$, such that $P Q \\perp P F_{1}$ and $|P Q|=\\frac{5}{12}|P F_{1}|$. Find the eccentricity of the hyperbola.", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;Q: 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(F2, H);Intersection(H, G) = {P, Q};IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(P, F1));Abs(LineSegmentOf(P, Q)) = (5/12)*Abs(LineSegmentOf(P, F1))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(37)/5", "fact_spans": "[[[2, 58], [94, 97], [161, 164]], [[5, 58]], [[5, 58]], [[91, 93]], [[98, 101]], [[102, 105]], [[66, 73]], [[74, 81], [83, 90]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 81]], [[2, 81]], [[82, 93]], [[91, 107]], [[109, 129]], [[130, 159]]]", "query_spans": "[[[161, 170]]]", "process": "As shown in the figure, let P and Q be points on the right branch of a hyperbola, with PQ ⊥ PF₁, |PQ| = (5/12)|PF₁|. In the right triangle PF₁Q, |QF₁| = √(|PF₁|² + |PQ|²) = (13/12)|PF₁|. By the definition of the hyperbola, we have: 2a = |PF₁| - |PF₂| = |QF₁| - |QF₂|. From |PQ| = (5/12)|PF₁|, it follows that |PF₂| + |QF₂| = (5/12)|PF₁|, which gives |PF₁| - 2a + (13/12)|PF₁| - 2a = (5/12)|PF₁|. Therefore, (1 - 5/12 + 13/12)|PF₁| = 4a, solving yields |PF₁| = (12a)/5, |PF₂| = |PF₁| - 2a = (2a)/5. By the Pythagorean theorem, we get: 2c = |F₁F₂| = √((12a/5)² + (2a/5)²) = (2√37)/5 a, thus e = √37/5." }, { "text": "Given the parabola $E$: $y^{2}=x$, two mutually perpendicular lines $l_{1}$ and $l_{2}$ are drawn through the focus $F$ of the parabola, intersecting the parabola $E$ at points $A$, $B$ and $C$, $D$ respectively. What is the minimum value of $|A B|+4|C D|$?", "fact_expressions": "E: Parabola;l1:Line;l2:Line;A: Point;B: Point;C: Point;D: Point;F:Point;Expression(E) = (y^2 = x);Focus(E)=F;PointOnCurve(F,l1);PointOnCurve(F,l2);IsPerpendicular(l1,l2);Intersection(l1,E)={A,B};Intersection(l2,E)={C,D}", "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + 4*Abs(LineSegmentOf(C, D)))", "answer_expressions": "9", "fact_spans": "[[[1, 18], [56, 62]], [[33, 42]], [[45, 52]], [[64, 67]], [[68, 71]], [[74, 77]], [[78, 81]], [[21, 24]], [[1, 18]], [[1, 24]], [[0, 52]], [[0, 52]], [[26, 52]], [[33, 83]], [[33, 83]]]", "query_spans": "[[[85, 105]]]", "process": "" }, { "text": "The equation of the line containing the chord of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ for which $P(8,1)$ is the midpoint is?", "fact_expressions": "G: Hyperbola;H: LineSegment;P: Point;Expression(G) = (x^2/16 - y^2/4 = 1);Coordinate(P) = (8, 1);IsChordOf(H, G);MidPoint(H) = P", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "2*x - y - 15 = 0", "fact_spans": "[[[0, 39]], [], [[41, 49]], [[0, 39]], [[41, 49]], [[0, 54]], [[0, 54]]]", "query_spans": "[[[0, 63]]]", "process": "Let the line containing the chord with midpoint P(8,1) intersect the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ at points $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Since point $P(8,1)$ is the midpoint of segment $AB$, we have\n$$\n\\begin{cases}\nx_{1}+x_{2}=16 \\\\\ny_{1}+y_{2}=2\n\\end{cases}\n$$\nSubstituting $A(x_{1},y_{1})$, $B(x_{2},y_{2})$ into the hyperbola $x^{2}-4y^{2}=16$, we get\n$$\n\\begin{cases}\nx_{1}^{2}-4y_{1}^{2}=16 \\quad\\textcircled{1} \\\\\nx_{2}^{2}-4y_{2}^{2}=16 \\quad\\textcircled{2}\n\\end{cases}\n$$\n$\\textcircled{1}-\\textcircled{2}$ yields $(x_{1}+x_{2})(x_{1}-x_{2})-4(y_{1}+y_{2})(y_{1}-y_{2})=0$ \n$\\therefore 16(x_{1}-x_{2})-8(y_{1}-y_{2})=0$ \n$k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=2$ \n$\\therefore$ The equation of the line containing the chord with midpoint $P(8,1)$ is $y-1=2(x-8)$ \nSimplifying, we get $2x-y-15=0$." }, { "text": "A focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ coincides with the focus of the parabola $y^{2}=8x$, and the distance from the focus to its asymptote is $1$. Find the length of the real axis of this hyperbola?", "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);OneOf(Focus(G))=Focus(H);Distance(Focus(G),Asymptote(G))=1", "query_expressions": "Length(RealAxis(G))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[0, 57], [89, 90], [103, 106]], [[3, 57]], [[3, 57]], [[63, 79]], [[3, 57]], [[3, 57]], [[0, 57]], [[63, 79]], [[0, 84]], [[0, 100]]]", "query_spans": "[[[103, 111]]]", "process": "Problem Analysis: According to the problem, draw the figure. Then, using the parabola equation, find the focus coordinates, which give the focus coordinates of the hyperbola. Use the condition that the distance from the focus to one asymptote of the hyperbola is 1 to set up an equation, and solve it together with the implicit condition. As shown in the figure, from the parabola equation $ y^{2} = 8x $, we obtain the focus coordinates of the parabola $ F(2,0) $, which is also the right focus coordinates of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $). The asymptotes of the hyperbola are $ y = \\pm\\frac{b}{a}x $. Take $ y = \\frac{b}{a}x $, rewritten in general form: $ bx - ay = 0 $. Then $ \\frac{|2b|}{\\sqrt{a^{2} + b^{2}}} = 1 $, i.e., $ 4b^{2} = a^{2} + b^{2} $. Also, $ a^{2} = 4b^{2} $. Solving these equations simultaneously gives $ a^{2} = 3 $, $ \\therefore a = \\sqrt{3} $. Thus, the length of the real axis of the hyperbola is $ 2\\sqrt{3} $." }, { "text": "Given that the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $\\frac{\\sqrt{5}}{2}$, then the equations of the asymptotes of $C$ are?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(5)/2", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(1/2)*x", "fact_spans": "[[[2, 53], [80, 83]], [[9, 53]], [[9, 53]], [[2, 53]], [[2, 78]]]", "query_spans": "[[[80, 91]]]", "process": "From the given condition, $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\frac{\\sqrt{5}}{2} $, it follows that $ a^{2} = 4b^{2} $. Also, the asymptotes of $ C $ are $ y = \\pm\\frac{b}{a}x $, hence $ y = \\pm\\frac{1}{2}x $." }, { "text": "Given that point $P(6 , y)$ lies on the parabola $y^{2}=2 px(p>0)$, $F$ is the focus of the parabola, and if $|PF|=8$, then the distance from point $F$ to the directrix of the parabola equals?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;P: Point;y0: Number;Coordinate(P) = (6, y0);PointOnCurve(P, G);F: Point;Focus(G) = F;Abs(LineSegmentOf(P, F)) = 8", "query_expressions": "Distance(F, Directrix(G))", "answer_expressions": "4", "fact_spans": "[[[14, 34], [40, 43], [62, 65]], [[14, 34]], [[17, 34]], [[17, 34]], [[2, 13]], [[3, 13]], [[2, 13]], [[2, 35]], [[36, 39], [57, 61]], [[36, 45]], [[47, 55]]]", "query_spans": "[[[57, 73]]]", "process": "" }, { "text": "Given that a focus of the hyperbola $x^{2}-k y^{2}=1$ is $(\\sqrt{5}, 0)$, then $k$=?", "fact_expressions": "G: Hyperbola;k: Number;H: Point;Expression(G) = (-k*y^2 + x^2 = 1);Coordinate(H) = (sqrt(5), 0);OneOf(Focus(G))= H", "query_expressions": "k", "answer_expressions": "1/4", "fact_spans": "[[[2, 22]], [[45, 48]], [[28, 43]], [[2, 22]], [[28, 43]], [[2, 43]]]", "query_spans": "[[[45, 50]]]", "process": "Test analysis: Rewrite the hyperbola equation in standard form to get \\( x^{2} - \\frac{y^{2}}{1} = 1 \\), then \\( 1 + \\frac{1}{k} = 5 \\), solving gives \\( k = \\frac{1}{4} \\)." }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, the line $y=x-3$ intersects the parabola $C$ at points $A$ and $B$, and $|A F|+|B F|=|A B|$, then $p$=?", "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Expression(G) = (y = x - 3);Focus(C) = F;Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = Abs(LineSegmentOf(A, B))", "query_expressions": "p", "answer_expressions": "6", "fact_spans": "[[[2, 28], [46, 52]], [[85, 88]], [[36, 45]], [[53, 56]], [[32, 35]], [[57, 60]], [[10, 28]], [[2, 28]], [[36, 45]], [[2, 35]], [[36, 62]], [[64, 83]]]", "query_spans": "[[[85, 90]]]", "process": "Since |AF| + |BF| = |AB|, the points A, B, F are collinear. Given that the focus of the parabola C: y^{2} = 2px (p > 0) is F(\\frac{p}{2}, 0), substituting the focus F(\\frac{p}{2}, 0) into the line y = x - 3 gives \\frac{p}{2} - 3 = 0, solving yields p = 6." }, { "text": "If the directrix of the parabola $y^{2}=2 p x (p>0)$ passes through a focus of the hyperbola $x^{2}-y^{2}=2$, then $p$=?", "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2 = 2);p>0;Expression(H) = (y^2 = 2*(p*x));PointOnCurve(OneOf(Focus(G)), Directrix(H))", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[29, 47]], [[1, 24]], [[54, 57]], [[29, 47]], [[4, 24]], [[1, 24]], [[1, 52]]]", "query_spans": "[[[54, 59]]]", "process": "The left focus of the hyperbola \\( x^{2} - y^{2} = 2 \\) is \\( (-2, 0) \\), so the directrix of the parabola \\( v^2 = 2px \\) is \\( x = -2 \\), \\( \\therefore \\frac{p}{4} = 2 \\), \\( \\therefore p = 4 \\), the answer is." }, { "text": "If a point $A(a , b)$ on the right branch of the hyperbola $x^{2}-y^{2}=1$ is at a distance of $\\sqrt{2}$ from the line $y=x$, then $a+b=$?", "fact_expressions": "G: Hyperbola;H: Line;A: Point;a:Number;b:Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = x);Coordinate(A) = (a, b);PointOnCurve(A,RightPart(G));Distance(A, H) = sqrt(2)", "query_expressions": "a + b", "answer_expressions": "1/2", "fact_spans": "[[[1, 19]], [[35, 42]], [[24, 34]], [[24, 34]], [[24, 34]], [[1, 19]], [[35, 42]], [[24, 34]], [[1, 34]], [[24, 57]]]", "query_spans": "[[[59, 66]]]", "process": "" }, { "text": "A circle centered at a vertex of a hyperbola passes through one focus of the hyperbola and is tangent to one directrix of the hyperbola; then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Center(H) = OneOf(Vertex(G));PointOnCurve(OneOf(Focus(G)), H) = True;H: Circle;IsTangent(H, OneOf(Directrix(G))) = True", "query_expressions": "Eccentricity(G)", "answer_expressions": "1+sqrt(2)", "fact_spans": "[[[1, 4], [17, 20], [29, 32], [43, 46]], [[0, 14]], [[13, 25]], [[13, 14]], [[13, 39]]]", "query_spans": "[[[43, 52]]]", "process": "" }, { "text": "If the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$ is $e$, then $e=$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/15 = 1);e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "4", "fact_spans": "[[[1, 30]], [[1, 30]], [[40, 43], [35, 38]], [[1, 38]]]", "query_spans": "[[[40, 45]]]", "process": "" }, { "text": "A point $P$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ is at a distance of $2$ from one of its foci. What is the distance from point $P$ to the other focus?", "fact_expressions": "G: Ellipse;P: Point;F1:Point;F2:Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);Distance(P,F1)=2;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)", "query_expressions": "Distance(P,F2)", "answer_expressions": "2", "fact_spans": "[[[0, 37], [44, 45]], [[40, 43], [61, 65]], [], [], [[0, 37]], [[0, 43]], [[40, 58]], [[44, 50]], [[44, 71]], [[44, 71]]]", "query_spans": "[[[44, 77]]]", "process": "From the equation of the ellipse, we know that a=13, 2a=26. By the definition of the ellipse, the distance from point x to the other focus is equal to 2a-2=4-2=2" }, { "text": "The ellipse $3 x^{2}+k y^{2}=3$ has a focal distance of $2\\sqrt{2}$. Then $k=$?", "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (k*y^2 + 3*x^2 = 3);FocalLength(G) = 2*sqrt(2)", "query_expressions": "k", "answer_expressions": "1", "fact_spans": "[[[0, 21]], [[37, 40]], [[0, 21]], [[0, 35]]]", "query_spans": "[[[37, 42]]]", "process": "" }, { "text": "If the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}(a>b>0)$ and the circle $x^{2}+y^{2}=\\left(\\frac{b}{2}+c\\right)^{2}$ have four intersection points, where $c$ is the semi-focal length of the ellipse, then what is the range of values for the eccentricity $e$ of the ellipse?", "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Circle;c: Number;e: Number;a > b;b > 0;Expression(G) = (a^2*y^2 + b^2*x^2 = a^2*b^2);Expression(H) = (x^2 + y^2 = (b/2 + c)^2);NumIntersection(G, H) = 4;HalfFocalLength(G)=c;Eccentricity(G) = e", "query_expressions": "Range(e)", "answer_expressions": "(sqrt(5)/3,3/5)", "fact_spans": "[[[1, 47], [94, 96], [102, 104]], [[3, 47]], [[3, 47]], [[48, 82]], [[90, 93]], [[108, 111]], [[3, 47]], [[3, 47]], [[1, 47]], [[48, 82]], [[1, 87]], [[90, 100]], [[102, 111]]]", "query_spans": "[[[108, 118]]]", "process": "According to the problem, the upper and lower vertices of the ellipse lie inside the circle, while the left and right vertices lie outside the circle. Then \n\\begin{cases}a>\\frac{b}{2}+c\\\\b<\\frac{b}{2}+c\\end{cases}, \nrearranging yields \n\\begin{cases}(a-c)^{2}>\\frac{1}{4}(a^{2}-c^{2})\\\\\\sqrt{a^{2}-c^{2}}<2c\\end{cases}, \nsolving gives \n\\frac{\\sqrt{5}}{3}0 , b>0)$ and hyperbola $D$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{6}=1$ have the same asymptotes, and $C$ passes through the point $(2,6)$, then the length of the real axis of $C$ is?", "fact_expressions": "C: Hyperbola;D:Hyperbola;b: Number;a: Number;G: Point;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);Coordinate(G) = (2, 6);Expression(D)=(x^2/4-y^2/6=1);Asymptote(C)=Asymptote(D);PointOnCurve(G, C)", "query_expressions": "Length(RealAxis(C))", "answer_expressions": "2*sqrt(30)", "fact_spans": "[[[1, 65], [118, 121], [133, 136]], [[66, 109]], [[9, 65]], [[9, 65]], [[123, 131]], [[9, 65]], [[9, 65]], [[1, 65]], [[123, 131]], [[66, 109]], [[1, 116]], [[118, 131]]]", "query_spans": "[[[133, 142]]]", "process": "Find the relationship between a and b according to the given conditions, then use the point through which hyperbola C passes to compute the answer. [Detailed solution] The asymptotes of hyperbola C: \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 are y=\\pm\\frac{a}{b}x, while the asymptotes of hyperbola D: \\frac{x^{2}}{4}-\\frac{y^{2}}{6}=1 are y=\\pm\\frac{\\sqrt{6}}{2}x. According to the problem, \\frac{a}{b}=\\frac{\\sqrt{6}}{2}. Since hyperbola C passes through the point (2,6), we have \\frac{36}{a^{2}}-\\frac{4}{b^{2}}=1. Solving gives: a=\\sqrt{30}, b=2\\sqrt{5}. Therefore, the length of the real axis of hyperbola C is 2\\sqrt{30}." }, { "text": "Draw tangents from point $P(2,0)$ to the circle $O$: $x^{2}+y^{2}=1$, with points of tangency $A$ and $B$. If $A$ and $B$ lie exactly on the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "O: Circle;Expression(O) = (x^2 + y^2 = 1);P: Point;Coordinate(P) = (2, 0);L1: Line;L2: Line;TangentOfPoint(P, O) = {L1, L2};TangentPoint(L1, O) = A;TangentPoint(L2, O) = B;A: Point;B: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Z1: Line;Z2: Line;Asymptote(C) = {Z1, Z2};PointOnCurve(A, Z1);PointOnCurve(B, Z2)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[11, 32]], [[11, 32]], [[1, 10]], [[1, 10]], [], [], [[0, 35]], [[0, 48]], [[0, 48]], [[41, 44], [51, 54]], [[55, 58], [55, 58]], [[61, 122], [131, 137]], [[61, 122]], [[68, 122]], [[68, 122]], [[68, 122]], [[68, 122]], [], [], [[61, 128]], [[51, 129]], [[51, 129]]]", "query_spans": "[[[131, 143]]]", "process": "As shown in the figure, the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has asymptotes given by $ y = \\pm \\frac{b}{a}x $. A tangent is drawn from point $ P(2,0) $ to the circle $ O: x^{2} + y^{2} = 1 $, where $ A $ and $ B $ are the points of tangency. By the property of tangents to a circle, we have $ OA \\perp AP $. Since $ OA = 1 $, $ OP = 2 $, it follows that $ \\angle AOP = 60^{\\circ} $, so the tangent of the inclination angle of the asymptote $ y = \\frac{b}{a}x $ is $ \\sqrt{3} $. Therefore, $ \\frac{b}{a} = \\sqrt{3} $, which implies $ b^{2} = 3a^{2} $. From $ b^{2} = c^{2} - a^{2} $, it follows that $ c^{2} = 4a^{2} $, so $ e = \\frac{c}{a} = 2 $, meaning the eccentricity of hyperbola $ C $ is 2." }, { "text": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{4}=1(a>0)$ and the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1$ have the same foci, find the value of $a$.", "fact_expressions": "G: Hyperbola;H: Ellipse;a: Number;Expression(G) = (x^2/9 - y^2/3 = 1);a>0;Expression(H) = (y^2/4 + x^2/a^2 = 1);Focus(H) = Focus(G)", "query_expressions": "a", "answer_expressions": "4", "fact_spans": "[[[49, 87]], [[2, 48]], [[95, 98]], [[49, 87]], [[4, 48]], [[2, 48]], [[2, 93]]]", "query_spans": "[[[95, 102]]]", "process": "From the given condition, we have $ a^{2} - 4 = 9 + 3 $, thus the value of $ a $ can be found. Solving $ a^{2} - 4 = 9 + 3 $, we get $ a = 4 $ or $ a = -4 $ (discarded)." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{7}=1$ $(a>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, the left vertex is $A$, and a circle centered at $F_{2}$ with radius $|F_{2} A|$ intersects the right branch of the hyperbola at points $M$ and $N$. If the perpendicular bisector of segment $AM$ passes through point $N$, then $a=$?", "fact_expressions": "G: Hyperbola;a: Number;H: Circle;A: Point;M: Point;F2: Point;F1: Point;N: Point;a>0;Expression(G) = (-y^2/7 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;LeftVertex(G) = A;Center(H) = F2;Radius(H) = Abs(LineSegmentOf(F2, A));Intersection(H, RightPart(G)) = {M, N};PointOnCurve(N, PerpendicularBisector(LineSegmentOf(A, M)))", "query_expressions": "a", "answer_expressions": "3", "fact_spans": "[[[2, 49], [114, 117]], [[151, 154]], [[112, 113]], [[80, 84]], [[120, 123]], [[67, 74], [86, 93]], [[57, 64]], [[124, 127], [145, 149]], [[5, 49]], [[2, 49]], [[2, 75]], [[2, 75]], [[2, 84]], [[85, 113]], [[97, 113]], [[112, 129]], [[131, 149]]]", "query_spans": "[[[151, 156]]]", "process": "From the given, |AN| = |MN|. Combining the symmetry of the hyperbola, △AMN is an equilateral triangle, and the circumradius is a + c. Find the coordinates of point M. By the definition of the hyperbola, |MF₁| = 3a + c. Then use the distance formula between two points to find |MF₁|, thereby obtaining the relationship between a and c. Given b = √7, solve accordingly. The perpendicular bisector of segment AM passes through point N, |AN| = |MN|. With F₂ as the center and |F₂A| as the radius, the circle intersects the right branch of the hyperbola at points M and N. Thus, the radius of the circle is a + c, and |AM| = |AN|, so △AMN is an equilateral triangle, and F₂ is the circumcenter of △AMN. Therefore, the coordinates of M are ( (a + 3c)/2 , ±√3(a + c)/2 ). |MF₂| = a + c, ∴ |MF₁| = 3a + c. F₁(-c, 0), ∴ MF₁² = (3a + c)² = ( (a + 3c)/2 + c )² + [ √3(a + c)/2 ]². Simplifying yields 4a² - ac - 3c² = 0, (4a - 3c)(a + c) = 0, c = (4/3)a, b² = (16/9)a² - a² = 7, ∴ a = 3." }, { "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the ellipse $\\frac{x^{2}}{5}+y^{2}=1$, then $p$=?", "fact_expressions": "G: Parabola;p: Number;H: Ellipse;Expression(G) = (y^2 = 2*(p*x));Expression(H) = (x^2/5 + y^2 = 1);Focus(G) = RightFocus(H)", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[1, 17]], [[56, 59]], [[21, 48]], [[1, 17]], [[21, 48]], [[1, 54]]]", "query_spans": "[[[56, 61]]]", "process": "The equation of the ellipse is $\\frac{x^{2}}{5}+y^{2}=1$, from which we can obtain that the coordinates of the right focus of the ellipse are $(2,0)$. From the equation of the parabola $y^{2}=2px$, we can obtain its focus at $(\\frac{p}{2},0)$. According to the problem, $\\frac{p}{2}=2$, and we solve for $p$. From the ellipse equation $\\frac{x^{2}}{5}+y^{2}=1$, we have $c=\\sqrt{5-1}=2$, so the coordinates of the right focus of the ellipse are $(2,0)$. The focus of the parabola $y^{2}=2px$ is $(\\frac{p}{2},0)$. Since the focus of the parabola $y^{2}=2px$ coincides with the right focus of the ellipse $\\frac{x^{2}}{5}+y^{2}=1$, we have $\\frac{p}{2}=2$, solving gives $p=4$." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$. Then, what is the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(H) = sqrt(3)/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[80, 126]], [[2, 52]], [[2, 52]], [[0, 52]], [[80, 126]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 78]]]", "query_spans": "[[[80, 132]]]", "process": "" }, { "text": "The minor axis length of an ellipse is $\\sqrt{5}$, and the eccentricity is $e = \\frac{2}{3}$. The foci of the ellipse are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Then, what is the perimeter of $\\triangle A B F_{2}$?", "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;Length(MinorAxis(G))=sqrt(5);Eccentricity(G)=e;e:Number;e=2/3;Focus(G) = {F1, F2};PointOnCurve(F1, H);Intersection(H, G) = {A, B}", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "6", "fact_spans": "[[[34, 36], [69, 71]], [[66, 68]], [[72, 75]], [[76, 79]], [[49, 56]], [[41, 48], [58, 65]], [[0, 36]], [[15, 36]], [[18, 33]], [[18, 33]], [[34, 56]], [[57, 68]], [[66, 81]]]", "query_spans": "[[[83, 109]]]", "process": "Let the major axis of the ellipse be 2a, the minor axis be 2b, and the focal distance be 2c. Since the minor axis is \\sqrt{5} and the eccentricity is e=\\frac{2}{3}, we have \\begin{cases}\\frac{c}{a}=\\frac{2}{3}\\\\2b=\\sqrt{5}\\\\a^{2}=b^{2}+c^{2}\\end{cases}. Thus, a^{2}=\\frac{9}{4}, so a=\\frac{3}{2}. Moreover, the perimeter of \\triangle ABF_{2} is |AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|. By the definition of an ellipse, the perimeter of \\triangle ABF_{2} is 4a=6." }, { "text": "Given that a line $l$ passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, where point $A$ has coordinates $(4,4)$, then the distance from the midpoint of segment $AB$ to the directrix is equal to?", "fact_expressions": "l: Line;G: Parabola;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 4);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B}", "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), Directrix(G))", "answer_expressions": "25/8", "fact_spans": "[[[23, 28]], [[3, 17], [29, 32]], [[37, 40]], [[33, 36], [45, 49]], [[19, 22]], [[3, 17]], [[45, 59]], [[3, 22]], [[2, 28]], [[23, 42]]]", "query_spans": "[[[29, 80]]]", "process": "From $ y^{2} = 4x $, we have $ 2p = 4 $, $ p = 2 $. Therefore, the focus is $ F(1,0) $. Given $ A(4,4) $, the slope of line $ AB $ is $ k = \\frac{4}{3} $. Thus, the equation of line $ AB $ is $ y = \\frac{4}{3}(x - 1) $. Solving the system of equations consisting of the line and the parabola:\n\\[\n\\begin{cases}\ny = \\frac{4}{3}(x - 1)\n\\end{cases}\n\\]\nwe obtain $ 4x^{2} - 17x + 4 = 0 $. Then, $ x_{1} + x_{2} = \\frac{17}{4} $. Therefore, the distance from the midpoint of segment $ AB $ to the directrix is $ \\frac{x_{1} + x_{2}}{2} + \\frac{p}{2} = \\frac{25}{8} $." }, { "text": "Given that on the parabola $y^{2}=2 p x(p>0)$, the distance from the point with horizontal coordinate $1$ to the focus is $\\frac{5}{2}$, then $p=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;XCoordinate(A) = 1;PointOnCurve(A, G);Distance(A, Focus(G)) = 5/2", "query_expressions": "p", "answer_expressions": "3", "fact_spans": "[[[2, 23]], [[2, 23]], [[55, 58]], [[5, 23]], [[32, 33]], [[24, 33]], [[2, 33]], [[2, 53]]]", "query_spans": "[[[55, 60]]]", "process": "According to the definition of a parabola, the solution can be obtained. From the definition of a parabola, we have $1 - \\left(-\\frac{p}{2}\\right) = \\frac{5}{2}$, solving for $p$ gives $p = 3$. Therefore, the answer is: $3$." }, { "text": "Given that the directrix of the parabola $y^{2}=8 x$ is $l$, and $C(-3,-4)$ is a fixed point. Let $P$ be any point on the parabola, and let $d$ be the distance from $P$ to $l$. What is the minimum value of $d+|P C|$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);l: Line;Directrix(G) = l;C: Point;Coordinate(C) = (-3, -4);P: Point;PointOnCurve(P, G);d: Number;Distance(P, l) = d", "query_expressions": "Min(d + Abs(LineSegmentOf(P, C)))", "answer_expressions": "sqrt(41)", "fact_spans": "[[[2, 16], [43, 46]], [[2, 16]], [[20, 23], [54, 57]], [[2, 23]], [[26, 36]], [[26, 36]], [[50, 53]], [[43, 53]], [[61, 64]], [[50, 64]]]", "query_spans": "[[[66, 82]]]", "process": "The parabola y^{2}=8x has the directrix l: x=-2, and focus F(2,0). From point P, draw PM\\botl, intersecting at point M. When points F, P, and M are collinear, d+|PC| attains its minimum value. \\frac{1}{2} examines the application of the shortest distance between two points, tests computational ability, and is a basic problem." }, { "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 $F_{1}$ and $F_{2}$ respectively. The line passing through $F_{1}$ and perpendicular to the $x$-axis intersects the left branch of the hyperbola at points $A$ and $B$. Lines $AF_{2}$ and $BF_{2}$ intersect the $y$-axis at points $P$ and $Q$ respectively. If the perimeter of $\\triangle P Q F_{2}$ is $4$, then the range of values of $\\frac{5-b^{2}}{2-a}$ is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;F2: Point;B: Point;P: Point;Q: Point;F1: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F2;LeftFocus(G)=F1;PointOnCurve(F1, H);IsPerpendicular(H,xAxis);Intersection(H, LeftPart(G)) = {A, B};Intersection(LineSegmentOf(A,F2),yAxis)=P;Intersection(LineSegmentOf(B, F2), yAxis) = Q;Perimeter(TriangleOf(P, Q, F2)) = 4", "query_expressions": "Range((5 - b^2)/(2 - a))", "answer_expressions": "(5/2,+oo)", "fact_spans": "[[[2, 58], [107, 110]], [[5, 58]], [[5, 58]], [[103, 105]], [[115, 118]], [[77, 84]], [[119, 122]], [[153, 156]], [[157, 160]], [[67, 74], [67, 74]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 84]], [[2, 84]], [[86, 105]], [[95, 105]], [[103, 124]], [[125, 162]], [[125, 162]], [[164, 192]]]", "query_spans": "[[[194, 222]]]", "process": "" }, { "text": "In the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, what is the equation of the line containing the chord for which $M(-1 , 2)$ is the midpoint?", "fact_expressions": "G: Ellipse;H: LineSegment;M: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Coordinate(M) = (-1, 2);IsChordOf(H, G);MidPoint(H) = M", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "9*x-32*y+73=0", "fact_spans": "[[[0, 38]], [], [[42, 53]], [[0, 38]], [[42, 53]], [[0, 58]], [[0, 58]]]", "query_spans": "[[[0, 67]]]", "process": "Use the point-difference method to find the slope of the line, then find the equation of the line containing the midpoint chord. Let the line intersect the ellipse at points A(x_{1},y_{1}), B(x_{2},y_{2}), x_{1}\\neq x_{2}. \n\\begin{cases}\\frac{x_{1}^{2}}{16}+\\frac{y_{1}^{2}}{9}=1\\\\\\frac{x_{2}^{2}}{16}+\\frac{y_{2}^{2}}{9}=1\\end{cases}, \nsubtracting the two equations gives \\frac{x_{1}^{2}-x_{2}^{2}}{16}+\\frac{y_{1}^{2}-y_{2}^{2}}{9}=0, \nrearranged as: \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{16}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{9}=0. \nFrom the given conditions, x_{1}+x_{2}=-2, y_{1}+y_{2}=4, substituting yields: \n-\\frac{2(x_{1}-x_{2})}{16}+\\frac{4(y_{1}-y_{2})}{9}=0, \nthat is, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{9}{32}, so the slope of the line k=\\frac{9}{32}. \nHence, the equation of the line is y-2=\\frac{9}{32}(x+1), i.e., 9x-32y+73=0." }, { "text": "The equation of an ellipse that has the same foci as the ellipse $9 x^{2} + 4 y^{2} = 36$ and has a minor axis length of $4 \\sqrt{5}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 4*y^2 = 36);H:Ellipse;Focus(G)=Focus(H);Length(MinorAxis(H))=4*sqrt(5)", "query_expressions": "Expression(H)", "answer_expressions": "y^2/25+x^2/20=1", "fact_spans": "[[[1, 23]], [[1, 23]], [[47, 49]], [[0, 49]], [[30, 49]]]", "query_spans": "[[[47, 53]]]", "process": "Ellipse $ 9x^{2} + 4y^{2} = 36 $, $ \\frac{x^{2}}{4} + \\frac{y^{2}}{9} = 1 $, $ \\therefore c = \\sqrt{5} $, $ \\because $ the foci of the ellipse are the same as those of the ellipse $ 9x^{2} + 4y^{2} = 36 $, $ \\therefore $ the semi-focal length of the ellipse is $ c = \\sqrt{5} $, i.e., $ a^{2} - b^{2} = 5 $, $ \\because $ the minor axis length is $ 4\\sqrt{5} $, $ \\therefore b = 2\\sqrt{5} $, $ a = 5 $, $ \\therefore $ the standard equation of the ellipse is $ \\frac{y^{2}}{25} + \\frac{x^{2}}{20} = 1 $." }, { "text": "$P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$, and $M$, $N$ are points on the circles $(x+4)^{2}+y^{2}=4$ and $(x-4)^{2}+y^{2}=1$, respectively. Then the maximum value of $PM - PN$ is?", "fact_expressions": "G: Hyperbola;H: Circle;U: Circle;P: Point;M: Point;N: Point;Expression(G) = (x^2 - y^2/15 = 1);Expression(H) = (y^2 + (x + 4)^2 = 4);Expression(U) = (y^2 + (x - 4)^2 = 1);PointOnCurve(P, RightPart(G));PointOnCurve(M, H);PointOnCurve(N, U)", "query_expressions": "Max(LineSegmentOf(P, M) - LineSegmentOf(P, N))", "answer_expressions": "5", "fact_spans": "[[[4, 33]], [[49, 69]], [[70, 89]], [[0, 3]], [[39, 42]], [[43, 46]], [[4, 33]], [[49, 69]], [[70, 89]], [[0, 38]], [[39, 92]], [[39, 92]]]", "query_spans": "[[[94, 109]]]", "process": "From the given conditions, we know that the two foci of the hyperbola are the centers of the two circles. Then, using plane geometry knowledge, |PM| - |PN| can be transformed into the difference of distances from a point on the hyperbola to the two foci, thus obtaining the maximum value of |PM| - |PN|. As shown in the figure, the two foci of the hyperbola are: F_{1}(-4,0), F_{2}(4,0), which are the centers of the two circles, with radii r_{1}=2, r_{2}=1 respectively. |PM|_{\\max}=|PF_{1}|+2, |PN|_{\\min}=|PF_{2}|-1. Hence, the maximum value of |PM| - |PN| is: (|PF_{1}|+2-|PF_{2}|+1)=|PF_{1}|-|PF_{2}|+3=5" }, { "text": "Given the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and a fixed point $A(1,3)$. The point $P$ moves on the right branch of the hyperbola. Then the minimum value of $|P F_{1}|+|P A|$ is equal to?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;Coordinate(A) = (1, 3);P: Point;PointOnCurve(P, RightPart(G))", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1)))", "answer_expressions": "11", "fact_spans": "[[[2, 41], [81, 84]], [[2, 41]], [[49, 56]], [[57, 64]], [[2, 64]], [[2, 64]], [[67, 75]], [[67, 75]], [[76, 80]], [[76, 90]]]", "query_spans": "[[[92, 116]]]", "process": "Since point P lies on the right branch of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, then $|PF_{1}|-|PF_{2}|=6$. Thus, $|PF_{1}|=|PF_{2}|+6$. Given $A(1,3)$ and the right focus of the hyperbola $F_{2}(5,0)$, we have $|PF_{1}|+|PA|=|PF_{2}|+|PA|+6$. When points $A$, $P$, and $F_{2}$ are collinear, $|PF_{2}|+|PA|=|AF_{2}|$; when points $A$, $P$, and $F_{2}$ are not collinear, $|PF_{2}|+|PA|>|AF_{2}|$. Therefore, $|PF_{1}|+|PA|=|PF_{2}|+|PA|+6\\geqslant|AF_{2}|+6=\\sqrt{(5-1)^{2}+(0-3)^{2}}+6=5+6=11$ (equality holds if and only if points $A$, $P$, and $F_{2}$ are collinear)." }, { "text": "The coordinates of the focus of the parabola $y=x^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,-1/4)", "fact_spans": "[[[0, 12]], [[0, 12]]]", "query_spans": "[[[0, 19]]]", "process": "" }, { "text": "Given that the directrix of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$ passes through the focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1(b>0)$, then $b$=?", "fact_expressions": "G: Hyperbola;H: Ellipse;b: Number;Expression(G) = (x^2/2 - y^2/2 = 1);b>0;Expression(H) = (x^2/4 + y^2/b^2 = 1);PointOnCurve(Focus(H),Directrix(G))", "query_expressions": "b", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 40]], [[45, 91]], [[96, 99]], [[2, 40]], [[47, 91]], [[45, 91]], [[2, 94]]]", "query_spans": "[[[96, 101]]]", "process": "Problem Analysis: For the hyperbola \\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1, we have a=b=\\sqrt{2}, c=\\sqrt{a^{2}+b^{2}}=2, and its directrix is x=\\frac{a^{2}}{c}=1, so 4-b^{2}=1, b=\\sqrt{3}" }, { "text": "Given that the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4}=1$ $(a>0)$ with eccentricity $\\frac{3 \\sqrt{5}}{5}$ has its left focus coinciding with the focus of the parabola $y^{2}=2 m x$, find the real number $m=?$", "fact_expressions": "Eccentricity(C) = 3*sqrt(5)/5;C: Hyperbola;Expression(C) = (-y^2/4 + x^2/a^2 = 1);a: Number;a>0;G: Parabola;m: Real;Expression(G) = (y^2 = 2*(m*x));LeftFocus(C) = Focus(G)", "query_expressions": "m", "answer_expressions": "-6", "fact_spans": "[[[2, 83]], [[29, 83]], [[29, 83]], [[37, 83]], [[37, 83]], [[88, 103]], [[110, 115]], [[88, 103]], [[29, 108]]]", "query_spans": "[[[110, 117]]]", "process": "" }, { "text": "If the parabola is $x^{2}=12 y$, then what are the coordinates of the focus $F$?", "fact_expressions": "G: Parabola;F: Point;Expression(G) = (x^2 = 12*y);Focus(G)=F", "query_expressions": "Coordinate(F)", "answer_expressions": "(0,3)", "fact_spans": "[[[1, 16]], [[21, 24]], [[1, 16]], [[1, 24]]]", "query_spans": "[[[21, 29]]]", "process": "Since 2p = 12, we get p = 6, \\frac{p}{2} = 3, so the focus F of the parabola x^{2} = 12y is (0, 3)." }, { "text": "Given that point $P(6 , y)$ lies on the parabola $y^{2}=2 p x(p>0)$, $F$ is the focus of the parabola, and $|PF|=8$, then the distance from point $F$ to the directrix of the parabola equals?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;P: Point;y0: Number;Coordinate(P) = (6, y0);PointOnCurve(P, G);F: Point;Focus(G) = F;Abs(LineSegmentOf(P, F)) = 8", "query_expressions": "Distance(F, Directrix(G))", "answer_expressions": "4", "fact_spans": "[[[14, 35], [41, 44], [64, 67]], [[14, 35]], [[17, 35]], [[17, 35]], [[2, 13]], [[3, 13]], [[2, 13]], [[2, 36]], [[37, 40], [59, 63]], [[37, 47]], [[49, 57]]]", "query_spans": "[[[59, 75]]]", "process": "Test Analysis: Let point P(6, y) have its projection on the parabola y^{2} = 2px (p > 0) as M; then M(-\\frac{p}{2}, y). According to the given condition, |PM| = |PF| = 8, so 6 - (-\\frac{p}{2}) = 8, \\therefore p = 4. Thus, the distance from point F to the directrix of the parabola equals 4. The answer is 4." }, { "text": "Given that $F_{1}(-1,0)$, $F_{2}(1,0)$ are the two foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and a point $P$ on the ellipse satisfies $|\\overrightarrow{P F_{1}}|+|\\overrightarrow{P F_{2}}|=4$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;F2: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(VectorOf(P,F1))+Abs(VectorOf(P,F2))=4", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[29, 81], [88, 90], [157, 159]], [[31, 81]], [[31, 81]], [[2, 15]], [[16, 28]], [[93, 96]], [[31, 81]], [[31, 81]], [[29, 81]], [[2, 15]], [[16, 28]], [[2, 86]], [[88, 96]], [[98, 155]]]", "query_spans": "[[[157, 165]]]", "process": "By the given condition, 2a = |\\overrightarrow{PF}| + |\\overrightarrow{PF_{2}}| = 4. Then a = 2, and c = 1, so the eccentricity e = \\frac{c}{a} = \\frac{1}{2}." }, { "text": "The focus of the parabola $y^{2}=6 x$ is $F$, and $A(x , y)$ is any point on it. Given point $P(2 , 2)$, then the minimum value of $|A F|+|A P|$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 6*x);Coordinate(A) = (x1, y1);Coordinate(P) = (2, 2);Focus(G) = F;x1: Number;y1: Number;PointOnCurve(A, G)", "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(A, P)))", "answer_expressions": "7/2", "fact_spans": "[[[0, 14], [22, 23]], [[28, 38]], [[39, 50]], [[18, 21]], [[0, 14]], [[28, 38]], [[39, 50]], [[0, 21]], [[28, 38]], [[28, 38]], [[22, 38]]]", "query_spans": "[[[52, 71]]]", "process": "" }, { "text": "The moving circle $M$ passes through the point $(3 , 2)$ and is tangent to the line $y=1$. What is the trajectory equation of the center $M$ of the moving circle?", "fact_expressions": "M: Circle;G: Line;H: Point;Expression(G) = (y = 1);Coordinate(H) = (3, 2);PointOnCurve(H,M);IsTangent(M,G);M1:Point;Center(M)=M1", "query_expressions": "LocusEquation(M1)", "answer_expressions": "x^2-6*x-2*y+12=0", "fact_spans": "[[[2, 5], [29, 31]], [[18, 25]], [[6, 16]], [[18, 25]], [[6, 16]], [[0, 16]], [[0, 27]], [[33, 36]], [[29, 36]]]", "query_spans": "[[[33, 43]]]", "process": "Let the center of the moving circle be M(x, y). The moving circle M passes through the point (3, 2) and is tangent to the line y = 1. We obtain \\sqrt{(x-3)^{2}+(y-2)^{2}}=|1-y|. Simplifying this yields x^{2}-6x-2y+12=0. Thus, the trajectory equation of the center M of the moving circle is x^{2}-6x-2y+12=0." }, { "text": "Given the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Find the area of $\\Delta F_{1} P F_{2}$.", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/36 + y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;F1: Point;F2: Point;P: Point;PointOnCurve(P,G) = True;AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "3*sqrt(3)", "fact_spans": "[[[2, 40], [70, 72]], [[2, 40]], [[2, 65]], [[2, 65]], [[50, 57]], [[58, 65]], [[66, 69]], [[66, 75]], [[77, 110]]]", "query_spans": "[[[112, 139]]]", "process": "Let |PF_{1}| = m, |PF_{2}| = n. According to the definition of an ellipse, |PF_{1}| + |PF_{2}| = m + n = 2a = 12. From the ellipse \\frac{x^{2}}{36} + \\frac{y^{2}}{9} = 1, we obtain the focal distance |F_{1}F_{2}| = 2\\sqrt{36 - 9} = 6\\sqrt{3}. Given \\angle F_{1}PF_{2} = 60^{\\circ}, it follows that \\cos \\angle F_{1}PF_{2} = \\cos 60^{\\circ} = \\frac{|PF_{1}|^{2} + |PF_{2}|^{2} - |F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|} = \\frac{m^{2} + n^{2} - 108}{2mn} = \\frac{(m+n)^{2} - 2mn - 108}{2mn} = \\frac{18}{mn} - 1, so mn = 12. Therefore, the area of \\triangle F_{1}PF_{2} is S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2}|PF_{1}||PF_{2}|\\sin \\angle F_{1}PF_{2} = \\frac{1}{2}mn \\times \\frac{\\sqrt{3}}{2} = 3\\sqrt{3}." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has its left focus at $F_{1}(-3 \\sqrt{2}, 0)$ and eccentricity $3$, then the standard equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-3*sqrt(2), 0);LeftFocus(C) = F1;Eccentricity(C)=3", "query_expressions": "Expression(C)", "answer_expressions": "x^2/2-y^2/16=1", "fact_spans": "[[[2, 63], [102, 108]], [[10, 63]], [[10, 63]], [[68, 91]], [[10, 63]], [[10, 63]], [[2, 63]], [[68, 91]], [[2, 91]], [[2, 100]]]", "query_spans": "[[[102, 115]]]", "process": "Problem Analysis: From the given conditions, \n\\begin{cases}\\frac{c}{a}=3,\\\\c=3\\sqrt{2}\\end{cases} \nsolving gives \n\\begin{cases}a=\\sqrt{2},\\\\c=3\\sqrt{2}\\end{cases} \nthen \nb^{2}=c^{2}-a^{2}=16. \nTherefore, the standard equation of hyperbola C is \n\\frac{x^{2}}{2}-\\frac{y^{2}}{16}=1" }, { "text": "The distance from the focus to the directrix of the parabola $y^{2}=a x(a \\neq 0)$ is?", "fact_expressions": "G: Parabola;a: Number;Negation(a=0);Expression(G) = (y^2 = a*x)", "query_expressions": "Distance(Focus(G), Directrix(G))", "answer_expressions": "Abs(a)/2", "fact_spans": "[[[0, 24]], [[3, 24]], [[3, 24]], [[0, 24]]]", "query_spans": "[[[0, 35]]]", "process": "Since 2p = |a|, it follows that p = \\frac{|a|}{|}|, that is, the distance from the focus to the directrix is \\frac{|a|}{|}" }, { "text": "The standard equation of an ellipse centered at the origin, with foci on the coordinate axes, major axis length twice the minor axis length, and passing through the point $ P(4,0) $ is?", "fact_expressions": "O: Origin;Center(G) = O;G: Ellipse;PointOnCurve(Focus(G), axis);Length(MajorAxis(G)) = 2*Length(MinorAxis(G));P: Point;Coordinate(P) = (4, 0);PointOnCurve(P, G)", "query_expressions": "Expression(G)", "answer_expressions": "{x^2/16+y^2/4=1,y^2/64+x^2/16=1}", "fact_spans": "[[[3, 5]], [[0, 42]], [[40, 42]], [[6, 42]], [[14, 42]], [[29, 39]], [[29, 39]], [[28, 42]]]", "query_spans": "[[[40, 49]]]", "process": "Analyze: Discuss the cases when the foci of the ellipse lie on the x-axis or on the y-axis. Using the fact that the major axis length is twice the minor axis length and the ellipse passes through point P(4,0), determine the values of a and b respectively, to obtain the standard equation of the ellipse. If the foci of the ellipse lie on the x-axis, then a=4, and since the major axis length is twice the minor axis length, we have a=4=2b ⇒ b=2, and the equation of the ellipse is \\frac{x}{16}^{2}+\\frac{y^{2}}{4}=1; if the foci of the ellipse lie on the y-axis, then b=4, and since the major axis length is twice the minor axis length, we have a=2b=8, and the equation of the ellipse is \\frac{y^{2}}{64}+\\frac{x}{16}^{2}=1." }, { "text": "Given circle $C$: $(x+3)^{2}+y^{2}=4$, a moving circle $M$ passes through point $A(3,0)$ and is externally tangent to circle $C$. What is the equation of the locus of the center $M$ of the moving circle $M$?", "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x + 3)^2 = 4);A: Point;M: Circle;Coordinate(A) = (3, 0);PointOnCurve(A, M);IsOutTangent(C, M);M1: Point;Center(M) = M1", "query_expressions": "LocusEquation(M1)", "answer_expressions": "(x^2-y^2/8=1)&(x>=1)", "fact_spans": "[[[2, 26], [44, 45]], [[2, 26]], [[33, 42]], [[29, 32], [46, 50], [56, 59]], [[33, 42]], [[29, 42]], [[44, 52]], [[62, 65]], [[56, 65]]]", "query_spans": "[[[62, 72]]]", "process": "Let the center of circle M be M(x, y). According to the problem, the center of circle C is C(-3, 0) and its radius r = 2. Since the moving circle M passes through the point A(3, 0) and is externally tangent to circle C, then |MC| - |MA| = r = 2 < 6. Thus, the locus of point M is the right branch of a hyperbola. From the properties of the hyperbola, the length of the transverse axis is 2a = 2 and the focal distance is 2c = 6. Therefore, a^{2} = 1, b^{2} = 8. Hence, the equation of this hyperbola is x^{2} - \\frac{y^{2}}{8} = 1, that is, the trajectory equation of the center M of the moving circle M is x^{2} - \\frac{y^{2}}{8} = 1 (x \\geqslant 1)." }, { "text": "Given the parabola $C$: $y^{2}=4x$, with focus $F$. A line $l$ with slope $k$ ($k>0$) passes through the point $P(-1,0)$ and intersects the parabola $C$ at points $A$ and $B$. Connect $AF$ and $BF$ ($AF > BF$). If $|AF| = 2|BF|$, then $k = $?", "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;P: Point;Expression(C) = (y^2 = 4*x);Coordinate(P) = (-1, 0);Focus(C) = F;PointOnCurve(P, l);Slope(l) = k;k:Number;k>0;Intersection(l, C) = {A, B};LineSegmentOf(A, F)>LineSegmentOf(B, F);Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(B, F))", "query_expressions": "k", "answer_expressions": "2*sqrt(2)/3", "fact_spans": "[[[54, 59]], [[2, 21], [60, 66]], [[68, 71]], [[25, 28]], [[72, 75]], [[30, 40]], [[2, 21]], [[30, 40]], [[2, 28]], [[29, 59]], [[41, 59]], [[120, 123]], [[44, 53]], [[54, 77]], [[92, 101]], [[104, 118]]]", "query_spans": "[[[120, 125]]]", "process": "The focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, and the equation of line $ AB $ is $ y - 0 = k(x + 1) $, $ k > 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Substituting into the parabola $ y^{2} = 4x $, we obtain $ k^{2}x^{2} + (2k^{2} - 4)x + k^{2} = 0 $, so $ x_{1} + x_{2} = \\frac{4}{k^{2}} - 2 $, $\\textcircled{1}$ $ x_{1} \\cdot x_{2} = 1 $, $\\textcircled{2}$ From the focal radius formula of the parabola, we have: $ |AF| = x_{1} + \\frac{p}{2} = x_{1} + 1 $, $ |BF| = x_{2} + \\frac{p}{2} = x_{2} + 1 $. Given $ |AF| = 2|BF| $, then $ x_{2} - 2x_{1} = 1 $, $\\textcircled{3}$ From $\\textcircled{1}$ and $\\textcircled{2}$, solving gives: $ x_{1} = \\frac{4}{3k^{2}} - 1 $, $ x_{2} = \\frac{8}{3k^{2}} $, $ x_{1} \\cdot x_{2} = \\left( \\frac{4}{3k^{2}} - 1 \\right) \\times \\left( \\frac{8}{3k^{2}} \\right) = 1 $, simplifying yields: $ k^{2} = \\frac{8}{9} $, solving gives: $ k = \\pm \\frac{2\\sqrt{2}}{3} $. Since $ k > 0 $, then $ k = \\frac{2\\sqrt{2}}{3}\\sqrt{2} $" }, { "text": "Given point $Q(2 \\sqrt{2}, 0)$ and a moving point $P(x_{0}, y_{0})$ on the parabola $y = \\frac{x^{2}}{4}$, find the minimum value of $y_{0} + |PQ|$.", "fact_expressions": "Q: Point;Coordinate(Q) = (2*sqrt(2), 0);G: Parabola;Expression(G) = (y = x^2/4);P: Point;Coordinate(P) = (x0, y0);x0: Number;y0: Number;PointOnCurve(P, G)", "query_expressions": "Min(y0 + Abs(LineSegmentOf(P, Q)))", "answer_expressions": "2", "fact_spans": "[[[2, 22]], [[2, 22]], [[23, 45]], [[23, 45]], [[49, 67]], [[49, 67]], [[49, 67]], [[49, 67]], [[23, 67]]]", "query_spans": "[[[69, 89]]]", "process": "" }, { "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{k+4}+\\frac{y^{2}}{9}=1$ is $2$, then what is the value of $k$?", "fact_expressions": "G: Hyperbola;k: Number;Expression(G) = (x^2/(k + 4) + y^2/9 = 1);Eccentricity(G) = 2", "query_expressions": "k", "answer_expressions": "-31", "fact_spans": "[[[1, 41]], [[51, 54]], [[1, 41]], [[1, 49]]]", "query_spans": "[[[51, 58]]]", "process": "" }, { "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ has eccentricity $\\frac{\\sqrt{6}}{2}$, then what is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(6)/2", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(2)/2)*x", "fact_spans": "[[[1, 38], [65, 66]], [[4, 38]], [[4, 38]], [[1, 38]], [[1, 63]]]", "query_spans": "[[[65, 73]]]", "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ $(a>0)$ has eccentricity $\\frac{\\sqrt{6}}{2}$, then ${\\frac{c}{a}=\\frac{\\sqrt{6}}{2}\\begin{matrix}_{c^{2}-a^{2}}=1& so $a=\\sqrt{2}$, $b=1$, then the asymptotes are $y=\\pm\\frac{\\sqrt{2}}{2}x$" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=2 p x$ ($p>0$), and points $A$, $B$ on the parabola $C$ satisfy $\\overrightarrow{A F}=4 \\overrightarrow{F B}$. If the projections of $A$, $B$ onto the directrix are $M$, $N$ respectively, and the area of $\\Delta M F N$ is $5$, then $|A B|=$?", "fact_expressions": "F: Point;C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;Focus(C) = F;A: Point;B: Point;PointOnCurve(A, C) = True;PointOnCurve(B, C) = True;VectorOf(A, F) = 4*VectorOf(F, B);Projection(A, Directrix(C)) = M;Projection(B, Directrix(C)) = N;M: Point;N: Point;Area(TriangleOf(M, F, N)) = 5", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "25/4", "fact_spans": "[[[2, 5]], [[6, 32], [36, 42]], [[6, 32]], [[13, 32]], [[13, 32]], [[2, 35]], [[44, 48], [101, 104]], [[49, 52], [105, 108]], [[36, 48]], [[36, 52]], [[54, 99]], [[36, 125]], [[36, 125]], [[118, 121]], [[122, 125]], [[127, 148]]]", "query_spans": "[[[150, 159]]]", "process": "Let line AB be $ x = ky + \\frac{p}{2} $. Combining with the parabola gives: $ y^2 - 2pky - p^2 = 0 $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $, then $ y_1 + y_2 = 2pk $, $ y_1 y_2 = -p^2 $, where $ M(-\\frac{p}{2}, y_1) $, $ N(-\\frac{p}{2}, y_2) $, then $ S_{\\triangle MFN} = \\frac{1}{2} \\cdot |y_1 - y_2| = 5 $. From $ \\overrightarrow{AF} = 4\\overrightarrow{FB} $ we get: $ y_1 = -4y_2 $, then $ -4y_2^2 = -p^2 $, solving gives $ y_2 = \\pm\\frac{F}{2}\\frac{2}{2} $, at this time $ y_1 = -4y_2 = \\frac{2}{2}p $, so $ |y_1 - y_2| = \\frac{5}{2}p $, hence $ \\frac{5}{4}p^2 = 5 $, solving gives: $ p = 2 $. When $ y_1 = 2_1 $, $ y_2 = -\\frac{p}{2} $, $ k = \\frac{3}{4} $, at this time $ x_1 + x_2 = k(y_1 + y_2) + p = \\frac{3}{4} \\times 3 + 2 = \\frac{17}{4} $. When $ y_1 = -2p $, $ y_2 = \\frac{p}{2} $, $ |AB| = x_1 + x_2 + p = \\frac{17}{4} + 2 = \\frac{25}{4} $." }, { "text": "Given that the asymptotes of a hyperbola are $y = \\pm \\frac{3}{4} x$, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3/4)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{5/3,5/4}", "fact_spans": "[[[2, 5], [36, 39]], [[2, 33]]]", "query_spans": "[[[36, 45]]]", "process": "This problem examines the eccentricity of a hyperbola. Since the asymptotes of the hyperbola are given by y=\\pm\\frac{3}{4}x, it follows that a=4, b=3 or a=3, b=4, hence c=5. Therefore, the eccentricity e=\\frac{5}{3} or \\frac{5}{4}. Answer: \\frac{5}{3} or \\frac{5}{4}" }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$, a line passing through its focus intersects the parabola at points $A$ and $B$. If $|A B|=6$ and the x-coordinate of the midpoint of $A B$ is $2$, then the equation of this parabola is?", "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);PointOnCurve(Focus(G),H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 6;XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2", "query_expressions": "Expression(G)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[2, 23], [25, 26], [32, 35], [78, 81]], [[5, 23]], [[29, 31]], [[36, 39]], [[40, 43]], [[5, 23]], [[2, 23]], [[24, 31]], [[29, 45]], [[47, 56]], [[59, 75]]]", "query_spans": "[[[78, 86]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since AB passes through the focus of the parabola, ∴ AB = AF + BF = x_{1} + \\frac{p}{2} + x_{2} + \\frac{p}{2} = x_{1} + x_{2} + p = 2 \\times 2 + p = 6. Solving gives p = 2, so the equation of the parabola is y^{2} = 4x." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}(-c, 0)$, $F_{2}(c, 0)$, respectively. Point $P$ lies on the ellipse $C$, $P F_{1} \\perp x$-axis, and $\\angle P F_{2} F_{1}=45^{\\circ}$. 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;F1: Point;F2: Point;Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);c: Number;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), xAxis) ;AngleOf(P, F2, F1) = ApplyUnit(45, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "", "fact_spans": "[[[0, 56], [100, 105], [163, 168]], [[0, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[6, 56]], [[64, 78]], [[81, 94]], [[64, 78]], [[81, 94]], [[81, 94]], [[0, 94]], [[0, 94]], [[95, 99]], [[95, 107]], [[108, 126]], [[128, 161]]]", "query_spans": "[[[163, 174]]]", "process": "" }, { "text": "Given that an ellipse passes through the points $(\\sqrt{2}, 1)$ and $(\\sqrt{3}, \\frac{\\sqrt{2}}{2})$, what is the standard equation of the ellipse?", "fact_expressions": "G: Ellipse;H: Point;I: Point;Coordinate(H) = (sqrt(2), 1);Coordinate(I) = (sqrt(3), sqrt(2)/2);PointOnCurve(H, G);PointOnCurve(I, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4 + y^2/2 = 1", "fact_spans": "[[[3, 5], [58, 60]], [[6, 22]], [[24, 56]], [[6, 22]], [[24, 56]], [[3, 22]], [[3, 56]]]", "query_spans": "[[[58, 67]]]", "process": "Let the equation of the ellipse be $mx^{2}+ny^{2}=1$ ($m>0$, $n>0$ and $m\\neq n$). $\\therefore\\begin{cases}2m+n=1\\\\3m+\\frac{1}{2}n=1\\end{cases}$, solving gives: $\\begin{cases}m=\\frac{1}{4}\\\\n=\\frac{1}{2}\\end{cases}$, $\\therefore$ the standard equation of the ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the right branch of the hyperbola. The midpoint $M$ of $P F_{2}$ lies on the circle centered at $O$ with radius $O F_{1}$. Then $|P F_{2}|$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/20 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G)) = True;MidPoint(LineSegmentOf(P, F2)) = M;M: Point;O: Origin;Center(H) = O;Radius(H) = LineSegmentOf(O, F1);H: Circle;PointOnCurve(M, H) = True", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "4", "fact_spans": "[[[2, 42], [71, 74]], [[2, 42]], [[51, 58]], [[59, 66]], [[2, 66]], [[2, 66]], [[67, 70]], [[67, 79]], [[81, 96]], [[93, 96]], [[98, 101]], [[97, 119]], [[105, 119]], [[118, 119]], [[93, 120]]]", "query_spans": "[[[122, 135]]]", "process": "According to the problem, draw the figure and use the given information together with the triangle midline theorem and the definition of a hyperbola to solve. As shown in the figure, from the hyperbola \\frac{x^{2}}{16}-\\frac{y^{2}}{20}=1, we obtain a^{2}=16, b^{2}=20, then c=\\sqrt{a^{2}+b^{2}}=6. Thus |OM|=|OF_{1}|=6, |PF_{1}|=2|OM|=12, \\therefore |PF_{2}|=|PF_{1}|-8=4," }, { "text": "Given that the distance from the focus $F$ to the directrix of the parabola $y^{2}=2 p x$ ($p>0$) is $2$, a line passing through the focus $F$ intersects the parabola at points $A$ and $B$, and $|A F|=3|F B|$. Then, the distance from the midpoint of segment $A B$ to the $y$-axis is?", "fact_expressions": "G: Parabola;p: Number;H: Line;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Distance(F,Directrix(G)) = 2;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(F, B))", "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)), yAxis)", "answer_expressions": "5/3", "fact_spans": "[[[2, 23], [50, 53]], [[5, 23]], [[47, 49]], [[59, 62]], [[55, 58]], [[26, 29], [43, 46]], [[5, 23]], [[2, 23]], [[2, 29]], [[2, 39]], [[40, 49]], [[47, 64]], [[66, 80]]]", "query_spans": "[[[82, 102]]]", "process": "The distance from the focus F to the directrix is p=2. Draw AD perpendicular to the directrix l from point A, meeting l at D. Draw BE perpendicular to l from point B, meeting l at E. Extend AB to intersect l at C. Then ABCEAACD, so \\frac{BC}{AC}=\\frac{BE}{AD}=\\frac{BF}{AF}=\\frac{1}{3}. Let BC=x, then AC=3x. Since |AF|=3|FB|, it follows that BF=\\frac{1}{4}AB=\\frac{1}{2}x, and AF=3BF=\\frac{3}{2}x. Because CF=BC+BF=\\frac{3}{2}x, and F is the midpoint of AC, AD=2FG=4. Thus AF=\\frac{3}{2}x=4, so x=\\frac{8}{3}, and |BE|=\\frac{4}{3}. Therefore, the distance from the midpoint of segment AB to the y-axis is \\frac{|AD|-1+|BE|-1}{2}=\\frac{5}{3}." }, { "text": "Given that the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ $(m>0)$ is $2$, then the value of $m$ is?", "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2 - y^2/m = 1);Eccentricity(G) = 2", "query_expressions": "m", "answer_expressions": "3", "fact_spans": "[[[2, 35]], [[45, 48]], [[5, 35]], [[2, 35]], [[2, 43]]]", "query_spans": "[[[45, 52]]]", "process": "Problem Analysis: From the given conditions, a^{2}=1, b^{2}=m, c=\\sqrt{1+m}. According to the hyperbola's eccentricity \\frac{c}{a}=\\frac{\\sqrt{1+m}}{1}=2, we get m=3." }, { "text": "Given point $A(0,4)$, the parabola $C$: $x^{2}=2 p y (0 0, b > 0) $, intersecting the left and right branches of the hyperbola at points $ A $ and $ B $, respectively. If the perpendicular bisector of segment $ F_1 B $ passes through the right focus $ F_2 $, then the eccentricity of the hyperbola is?", "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;F1: Point;F2:Point;B: Point;A: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Slope(l)=sqrt(5)/2;LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1,l);Intersection(l,LeftPart(G))=A;Intersection(l,RightPart(G))=B;PointOnCurve(F2,PerpendicularBisector(LineSegmentOf(F1,B)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "3", "fact_spans": "[[[24, 29]], [[31, 87], [100, 103], [155, 158]], [[34, 87]], [[34, 87]], [[91, 98]], [[146, 153]], [[117, 120]], [[113, 116]], [[34, 87]], [[34, 87]], [[31, 87]], [[0, 29]], [[31, 98]], [[100, 153]], [[24, 98]], [[24, 122]], [[24, 122]], [[124, 153]]]", "query_spans": "[[[155, 164]]]", "process": "According to the property of the perpendicular bisector, we have |BF₂| = |F₁F₂| = 2c. Then by the definition of the hyperbola, we get: |BF₁| = 2c + 2a. Let the inclination angle of line l be θ, then tanθ = √5⁄2, from which the value of cosθ can be obtained. In triangle F₁BF₂, using the law of cosines, the relationship between a and c can be found, and thus the eccentricity can be determined. \nSolution: Let the focal distance of the hyperbola be 2c. Since the perpendicular bisector of segment F₁B passes through the right focus F₂, we have |BF₂| = |F₁F₂| = 2c. By the definition of the hyperbola, we obtain: |BF₁| = |BF₂| + 2a = 2c + 2a. Let the inclination angle of line l be θ, then tanθ = √5⁄2, so θ is an acute angle. Therefore, from \n{ sinθ = √5⁄2 , \n sin²θ + cos²θ = 1 } \nwe get: \n{ sinθ = √5⁄3 , \n cosθ = 2⁄3 }. \nIn triangle F₁BF₂, by the law of cosines, we have: \ncosθ = cos∠BF₁F₂ = (|BF₁|² + |F₁F₂|² − |BF₂|²) / (2|BF₁|·|F₁F₂|) = (4(a+c)² + 4c² − 4c²) / (2·2(a+c)·2c) = (a+c)/(2c) = 2⁄3. \nSolving gives: c = 3a, so the eccentricity e = c/a = 3." }, { "text": "Given that the equation of a hyperbola is $\\frac{x^{2}}{m-3}-\\frac{y^{2}}{m+2}=1$, what is the range of values for $m$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m - 3) - y^2/(m + 2) = 1);m: Number", "query_expressions": "Range(m)", "answer_expressions": "{(3, +oo), (-oo, -2)}", "fact_spans": "[[[4, 7]], [[4, 50]], [[52, 55]]]", "query_spans": "[[[52, 62]]]", "process": "From the condition satisfied by the hyperbola equation, it follows that (m-3) and (m+2) have the same sign, so the range of m can be obtained. From the equation of the hyperbola, we get (m-3)(m+2)>0, solving yields m>3 or m<-2." }, { "text": "The line $y=\\frac{\\sqrt{3}}{3} x$ intersects the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$, $F$ is the right focus of the ellipse. If $A F \\perp B F$, then the eccentricity of the ellipse is?", "fact_expressions": "H: Line;Expression(H) = (y = (sqrt(3)/3)*x);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;B: Point;Intersection(H, G) = {A, B};F: Point;RightFocus(G) = F;IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)/3", "fact_spans": "[[[0, 26]], [[0, 26]], [[27, 79], [95, 97], [120, 122]], [[27, 79]], [[29, 79]], [[29, 79]], [[29, 79]], [[29, 79]], [[81, 84]], [[85, 88]], [[0, 90]], [[91, 94]], [[91, 101]], [[103, 118]]]", "query_spans": "[[[120, 128]]]", "process": "According to the problem, convert the condition to point A(\\frac{\\sqrt{3}c}{2},\\frac{c}{2}), substituting into the ellipse equation gives \\frac{3c^{2}}{4a^{2}}+\\frac{c^{2}}{4b^{2}}=1, simplifying yields the solution. Let point F(c,0), and the ellipse is shown in the figure: \n\\because line_{AB}: y=\\frac{\\sqrt{3}}{3}x' \n\\therefore_{\\tan}\\angle AOF = \\frac{\\sqrt{3}}{3}, thus \\angle AOF = 30^{\\circ}. \nAlso, AF \\bot BF, O is the midpoint of AB, \n\\therefore OA = OF = c, \n\\therefore point A(c\\cos\\angle AOF, c\\sin\\angle AOF), i.e., A(\\frac{\\sqrt{3}c}{2},\\frac{c}{2}). \n\\because point A(\\frac{\\sqrt{3}c}{2},\\frac{c}{2}) lies on the ellipse, \n\\therefore \\frac{3c^{2}}{4a^{2}} + \\frac{c^{2}}{4b^{2}} = 1. \nCombining with b^{2} = a^{2} - c^{2}, simplifying gives 3c^{4} - 8a^{2}c^{2} + 4a^{4} = 0. \nFrom e = \\frac{c}{a}, we get 3e^{4} - 8e^{2} + 4 = 0, solving yields e^{2} = \\frac{2}{3} or e^{2} = 2 (discarded). \n\\therefore e = \\sqrt{\\frac{2}{3}} = \\frac{\\sqrt{6}}{3}." }, { "text": "Given that a line passing through the point $P(2,0)$ intersects the parabola $C$: $y^{2}=4x$ at points $A$ and $B$, and the lines $OA$, $OB$ ($O$ being the origin) intersect the line $x=-2$ at points $M$ and $N$ respectively, then what is the length of the chord intercepted on the $x$-axis by the circle with $MN$ as diameter?", "fact_expressions": "P: Point;Coordinate(P) = (2, 0);H: Line;PointOnCurve(P, H) = True;C: Parabola;Expression(C) = (y^2 = 4*x);Intersection(H, C) = {A, B};A: Point;O: Origin;B: Point;H1: Line;Expression(H1) = (x=-2);Intersection(LineOf(O,A),H1) = M;Intersection(LineOf(O,B),H1) = N;M: Point;N: Point;G: Circle;IsDiameter(LineSegmentOf(M,N),G) = True", "query_expressions": "Length(InterceptChord(xAxis,G))", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[3, 12]], [[3, 12]], [[13, 15]], [[2, 15]], [[16, 35]], [[16, 35]], [[13, 45]], [[36, 39]], [[62, 65]], [[40, 43]], [[75, 83]], [[75, 83]], [[46, 92]], [[46, 92]], [[84, 88]], [[89, 92]], [[104, 105]], [[94, 105]]]", "query_spans": "[[[104, 117]]]", "process": "" }, { "text": "Through the point $M(1,1)$, draw a moving line intersecting the parabola $y^{2}=8x$ at points $A$ and $B$. Then the equation of the trajectory of the midpoint of segment $AB$ is?", "fact_expressions": "M: Point;Coordinate(M) = (1, 1);H: Line;PointOnCurve(M, H);G: Parabola;Expression(G) = (y^2 = 8*x);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "y^2-4*x-y+4=0", "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 14]], [[0, 14]], [[15, 29]], [[15, 29]], [[30, 33]], [[34, 37]], [[11, 39]]]", "query_spans": "[[[41, 57]]]", "process": "Let the midpoint coordinates be (x_{0}, y_{0}), A(x_{1}, y_{1}), B(x_{1}, y_{1}), y_{1}^{2}=8x_{1}, y_{2}^{2}=8x_{2}. Subtracting the two equations yields =\\frac{4}{y_{0}}=\\frac{y_{0}-1}{x_{0}-1}\\Rightarrow y^{2}-4x-y+4=0." }, { "text": "If the line segment $AB$ has length $4$, and its endpoints $A$ and $B$ move on the $x$-axis and $y$-axis respectively, then what is the equation of the trajectory of the midpoint $M$ of $AB$?", "fact_expressions": "A: Point;B: Point;M:Point;Length(LineSegmentOf(A, B)) = 4;Endpoint(LineSegmentOf(A,B))={A,B};PointOnCurve(A, xAxis);PointOnCurve(B, yAxis);MidPoint(LineSegmentOf(A,B))=M", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2+y^2=4", "fact_spans": "[[[16, 19]], [[20, 23]], [[47, 50]], [[1, 12]], [[1, 23]], [[16, 38]], [[16, 38]], [[40, 50]]]", "query_spans": "[[[47, 57]]]", "process": "Let the coordinates of the midpoint P of AB be (x, y), A(m, 0), B(0, n), ∴\\begin{cases}x=\\frac{m}{2}\\\\y=\\frac{n}{2}\\end{cases}, ∴\\begin{cases}m=2x\\\\n=2y\\end{cases}, ∵ the length of segment AB is 4, ∴\\sqrt{m^{2}+n^{2}}=4, ∴m^{2}+n^{2}=16, ∴4x^{2}+4y^{2}=16, ∴x^{2}+y^{2}=4, the trajectory equation of the midpoint M of AB is x^{2}+y^{2}=4." }, { "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{8}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a moving point on the ellipse, and suppose a moving point $Q$ satisfies $\\overrightarrow{F_{1} P}=\\lambda \\overrightarrow{P Q}(\\lambda \\in R, \\lambda>0)$ and $|\\overrightarrow{P Q}|=|\\overrightarrow{P F_{2}}|$. Then the maximum distance from point $Q$ to an asymptote of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/9 + y^2/8 = 1);F1: Point;F2: Point;LeftFocus(H) = F1;RightFocus(H) = F2;P: Point;PointOnCurve(P, H);Q: Point;lambda: Real;VectorOf(F1, P) = lambda*VectorOf(P, Q);Abs(VectorOf(P, Q)) = Abs(VectorOf(P, F2));G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1);lambda>0", "query_expressions": "Max(Distance(Q, OneOf(Asymptote(G))))", "answer_expressions": "(sqrt(21)/7)+6", "fact_spans": "[[[2, 39], [68, 70]], [[2, 39]], [[48, 55]], [[56, 63]], [[2, 63]], [[2, 63]], [[64, 67]], [[64, 74]], [[78, 81], [218, 222]], [[83, 164]], [[83, 164]], [[165, 216]], [[223, 261]], [[223, 261]], [[83, 164]]]", "query_spans": "[[[218, 274]]]", "process": "The ellipse has $ a=3, b=2\\sqrt{2} $, then $ c=1 $, $ F(-1,0) $. If a moving point $ Q $ satisfies $ \\overrightarrow{F_{1}P} = \\lambda \\overrightarrow{PQ} $ ($ \\lambda \\in \\mathbb{R}, \\lambda > 0 $) and $ |\\overrightarrow{PQ}| = |\\overrightarrow{PF}_{2}| $, then $ F_{1}, P, Q $ are collinear and in the same direction. By $ |QF| = |PQ| + |PF| = |PF_{2}| + |PF_{1}| = 2a = 6 $, the trajectory of $ Q $ is a circle with center $ F_{1} $ and radius 6. The hyperbola $ \\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1 $ has an asymptote equation given as $ \\sqrt{3}x - 2y = 0 $. The distance from the center of the circle to the asymptote is $ \\frac{\\sqrt{3}}{\\sqrt{3+4}} = \\frac{\\sqrt{21}}{7} $. Therefore, the maximum distance from point $ Q $ to one asymptote of the hyperbola $ \\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1 $ is $ \\frac{\\sqrt{21}}{7} + 6 $." }, { "text": "It is known that a line passing through the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with an inclination angle of $45^{\\circ}$ intersects the right branch of the hyperbola at two points. What is the range of values for the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Line;PointOnCurve(RightFocus(G), H) = True;Inclination(H) = ApplyUnit(45, degree);NumIntersection(H, RightPart(G)) = 2", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,sqrt(2))", "fact_spans": "[[[3, 49], [73, 76], [85, 88]], [[3, 49]], [[6, 49]], [[6, 49]], [[70, 72]], [[2, 72]], [[53, 72]], [[70, 83]]]", "query_spans": "[[[85, 99]]]", "process": "" }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, respectively. If $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = pi/3", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[6, 43], [65, 67]], [[47, 54]], [[2, 5]], [[55, 62]], [[6, 43]], [[2, 46]], [[47, 73]], [[47, 73]], [[75, 111]]]", "query_spans": "[[[114, 141]]]", "process": "Test analysis: By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 4. In \\Delta F_{1}PF_{2}, |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos\\frac{\\pi}{3} = (|PF_{1}| + |PF_{2}|)^{2} - 3|PF_{1}||PF_{2}|. Since |F_{1}F_{2}| = 2\\sqrt{4-3} = 2, \\therefore |PF_{1}||PF_{2}| = 4, \\therefore S_{\\Delta PF_{1}PF_{2}} = \\frac{1}{2}|PF_{1}||PF_{2}|\\sin\\frac{\\pi}{3} = \\sqrt{3}" }, { "text": "A line $l$ passing through the point $M(2,1)$ intersects the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$ at points $A$ and $B$, such that $M$ is the midpoint of $AB$. What is the slope of the line $l$?", "fact_expressions": "M: Point;Coordinate(M) = (2, 1);PointOnCurve(M, l) ;l: Line;G: Ellipse;Expression(G) = (x^2/12 + y^2/4 = 1);Intersection(l, G) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Slope(l)", "answer_expressions": "-2/3", "fact_spans": "[[[2, 11], [68, 71]], [[2, 11]], [[0, 17]], [[12, 17], [82, 87]], [[18, 56]], [[18, 56]], [[12, 66]], [[57, 60]], [[61, 64]], [[68, 80]]]", "query_spans": "[[[82, 92]]]", "process": "Let the coordinates of points A and B be (x_{1}, y_{1}) and (x_{2}, y_{2}), respectively. Substituting into the ellipse equation and subtracting gives \\frac{1}{12} + \\frac{1}{4} \\cdot \\frac{y_{2}-y_{1}}{x_{1}-x_{2}} = 0. Substituting the slope and midpoint coordinates yields the solution. [Detailed Solution] Let the coordinates of points A and B be (x_{1}, y_{1}) and (x_{2}, y_{2}). Substituting into the ellipse equation gives: \\frac{x_{1}^{2}}{12} + \\frac{y_{1}^{2}}{4} = 1, \\frac{x_{2}^{2}}{12} + \\frac{y_{2}^{2}}{4} = 1. Subtracting these two equations gives: \\frac{1}{12}(x_{1}^{2} - x_{2}^{2}) + \\frac{1}{4}(y_{1}^{2} - y_{2}^{2}) = 0, which simplifies to \\frac{1}{12} + \\frac{1}{4} \\cdot \\frac{y_{1}^{2} - y_{2}^{2}}{x_{1}^{2} - x_{2}^{2}} = 0. Let the slope be k, then k = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}}. Given the midpoint M(2, 1), we have x_{1} + x_{2} = 4, y_{1} + y_{2} = 2. Substituting these values gives \\frac{1}{12} + \\frac{1}{8}k = 0, so k = -\\frac{2}{3}." }, { "text": "If an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ passes through the point $(3,-4)$, 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;H: Point;Coordinate(H) = (3, -4);PointOnCurve(H, OneOf(Asymptote(G)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/3", "fact_spans": "[[[1, 47], [67, 70]], [[1, 47]], [[4, 47]], [[4, 47]], [[55, 64]], [[55, 64]], [[1, 64]]]", "query_spans": "[[[67, 76]]]", "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are given by $y=\\pm\\frac{b}{a}x$. Since the asymptote passes through the point $(3,-4)$, we have $-4=-\\frac{3b}{a}$, thus $b=\\frac{4}{3}a$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{a^{2}+\\frac{16}{9}a^{2}}=\\frac{5}{3}a$, and therefore $e=\\frac{c}{a}=\\frac{5}{3}$." }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$ passes through the point $M(1,2)$, and a line $l$ intersects the parabola at two distinct points $A$, $B$. If the incenter of $\\Delta M A B$ is $(1, t)$, then what is the slope of line $l$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;Coordinate(M) = (1, 2);PointOnCurve(M, G);l: Line;A: Point;B: Point;Negation(A=B);Intersection(l, G) = {A, B};t: Number;Coordinate(Center(InscribedCircle(TriangleOf(M, A, B)))) = (1, t)", "query_expressions": "Slope(l)", "answer_expressions": "-1", "fact_spans": "[[[2, 23], [41, 44]], [[2, 23]], [[5, 23]], [[5, 23]], [[25, 34]], [[25, 34]], [[2, 34]], [[35, 40], [90, 95]], [[50, 53]], [[54, 57]], [[46, 57]], [[35, 57]], [[80, 88]], [[59, 88]]]", "query_spans": "[[[90, 100]]]", "process": "Substitute point M(1,2) into y²=2px to obtain p=2, so the equation of the parabola is y²=4x. According to the problem, the slope of line l exists and is not zero; let the equation of line l be x=my+n (m≠0). Substituting into y²=4x gives y²−4my−4n=0. Let A(x₁,y₁), B(x₂,y₂); then y₁+y₂=4m, y₁y₂=−4n. Given that the incenter of triangle AMB is (1,t), it follows that k_{MA}+k_{MB}=\\frac{y₁−2}{x₁−1}. Rearranging yields y₁+y₂+4=4m+4=0, solving gives m=−1, thus the equation of line l is y=−x+n. Therefore, the slope of line l is −1. This problem examines the relationship between a line and a parabola, and uses the method of setting variables without solving to express relationships between intersection points, belonging to medium-difficulty problems." }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and a point $P$ on $C$ such that $|PF|=8$, what is the distance from the midpoint $M$ of $PF$ to the $y$-axis?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C) = True;Abs(LineSegmentOf(P, F)) = 8;M: Point;MidPoint(LineSegmentOf(P, F)) = M", "query_expressions": "Distance(M, yAxis)", "answer_expressions": "4", "fact_spans": "[[[2, 21], [30, 33]], [[2, 21]], [[25, 28]], [[2, 28]], [[37, 40]], [[29, 40]], [[42, 51]], [[61, 64]], [[53, 64]]]", "query_spans": "[[[61, 74]]]", "process": "Let P(x₀, y₀), p = 2, so |PF| = x₀ + 1 = 8, solving gives x₀ = 7, therefore the x-coordinate of the midpoint M of PF is (1 + 7)/2 = 4, that is, the distance from the midpoint M of PF to the y-axis is 4." }, { "text": "Given the hyperbola $\\Gamma$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the line $l$: $x=\\sqrt{a^{2}+b^{2}}$ intersects the hyperbola $\\Gamma$ at points $A$ and $B$, $O$ is the origin. If $(\\overrightarrow{A O}+\\overrightarrow{A B}) \\cdot \\overrightarrow{B O}=0$, then the eccentricity of the hyperbola $\\Gamma$ is?", "fact_expressions": "l: Line;Gamma: Hyperbola;A: Point;O: Origin;B: Point;Expression(Gamma) = (-y^2/b^2 + x^2/a^2 = 1);a:Number;b:Number;a>0;b>0;Expression(l)=(x = sqrt(a^2 + b^2));Intersection(l, Gamma) = {A, B};DotProduct((VectorOf(A, B) + VectorOf(A, O)),VectorOf(B, O)) = 0", "query_expressions": "Eccentricity(Gamma)", "answer_expressions": "(sqrt(3)+sqrt(39))/6", "fact_spans": "[[[69, 98]], [[2, 68], [99, 110], [208, 219]], [[112, 115]], [[122, 125]], [[116, 119]], [[2, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[69, 98]], [[69, 121]], [[132, 206]]]", "query_spans": "[[[208, 225]]]", "process": "Problem Analysis: Since $ a^{2} + b^{2} = c^{2} $, the line $ l: x = c $ ($ c > 0 $), then $ A(c, \\frac{b^{2}}{a}) $, $ B(c, -\\frac{b^{2}}{a}) = (-c, -\\frac{b^{2}}{a}) $, $ \\overrightarrow{AB} = (0, -\\frac{2b^{2}}{a}) $, $ \\overrightarrow{BO} = (-c, \\frac{b^{2}}{a}) $, from the dot product of planar vectors we get: $ c^{2} - \\frac{3b^{4}}{a^{2}} = 0 $, solving yields: $ e = \\frac{\\sqrt{3} + \\sqrt{39}}{6} $." }, { "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, if the line $2x - y = 0$ is an asymptote of $C$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Line;Expression(G) = (2*x - y = 0);OneOf(Asymptote(C)) = G", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[2, 63], [77, 80], [88, 91]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[65, 76]], [[65, 76]], [[65, 86]]]", "query_spans": "[[[88, 97]]]", "process": "From the given condition, the asymptotes of the hyperbola are $ y = \\pm\\frac{a}{b}x $. Since one asymptote of the hyperbola is $ 2x - y = 0 $, we have $ y = 2x $, so $ \\frac{a}{b} = 2 $. Therefore, $ e = \\sqrt{1+\\left(\\frac{b}{a}\\right)^{2}} = \\sqrt{1+\\frac{1}{4}} = \\frac{\\sqrt{5}}{2} $." }, { "text": "The moving point $P(x, y)$ has the sum of distances to two fixed points $F_{1}(0,-3)$, $F_{2}(0, 3)$ equal to $10$. Then the equation of the trajectory of point $P$ is?", "fact_expressions": "P: Point;F1: Point;F2: Point;x1:Number;y1: Number;Coordinate(P) = (x1, y1);Coordinate(F1) = (0,-3);Coordinate(F2) = (0, 3);Distance(P,F1)+Distance(P,F2)=10", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/16+y^2/25=1", "fact_spans": "[[[2, 12], [56, 60]], [[16, 29]], [[32, 46]], [[2, 12]], [[2, 12]], [[2, 12]], [[16, 29]], [[32, 46]], [[2, 54]]]", "query_spans": "[[[56, 67]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$ and directrix $l$. A line passing through $F$ with an inclination angle of $60^{\\circ}$ intersects $C$ at points $A$ and $B$. $A M \\perp l$, $B N \\perp l$, where $M$ and $N$ are the feet of the perpendiculars. Point $Q$ is the midpoint of $M N$, and $|Q F|=2$. Then $p=$?", "fact_expressions": "C: Parabola;p: Number;F: Point;l: Line;H: Line;A: Point;B: Point;M: Point;N: Point;Q: Point;p > 0;Expression(C) = (y^2 = 2*p*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(F, H);Inclination(H) = ApplyUnit(60, degree);Intersection(H, C) = {A, B};IsPerpendicular(LineSegmentOf(A, M), l);IsPerpendicular(LineSegmentOf(B, N), l);FootPoint(LineSegmentOf(A, M), l) = M;FootPoint(LineSegmentOf(B, N), l) = N;MidPoint(LineSegmentOf(M, N)) = Q;Abs(LineSegmentOf(Q, F)) = 2", "query_expressions": "p", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 28], [68, 71]], [[150, 153]], [[32, 35], [45, 48]], [[39, 42]], [[65, 67]], [[72, 75]], [[76, 79]], [[113, 116]], [[118, 121]], [[125, 129]], [[9, 28]], [[2, 28]], [[2, 35]], [[2, 42]], [[44, 67]], [[48, 67]], [[65, 81]], [[82, 95]], [[97, 110]], [[82, 124]], [[82, 124]], [[125, 138]], [[139, 148]]]", "query_spans": "[[[150, 155]]]", "process": "As shown in the figure, by the geometric properties of the parabola, the circle with AB as diameter is tangent to the directrix l, and the point of tangency is Q. Triangle MFN is a right triangle with ∠MFN being the right angle, so |MN| = 2|QF| = 4, i.e., |BD| = 4, ∴ |AB| = \\frac{|BD|}{\\sin60^{\\circ}} = \\frac{4}{\\sqrt{3}} = \\frac{8\\sqrt{3}}{3}. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system \\begin{cases}y^{2}=2px\\\\y-0=\\sqrt{3}(x-\\frac{p}{2})\\end{cases}, we obtain 12x^{2}-20px+3p=0. Then x_{1}+x_{2}=\\frac{20}{12}p=\\frac{5}{3}p. Then x_{1}+x_{2}=\\frac{20}{12}p=\\frac{2}{3}p, ∴ |AB|=x_{1}+x_{2}+p=\\frac{5}{2}p+p=\\frac{8}{3}p. ∴ \\frac{8}{3}p=\\frac{8\\sqrt{3}}{3}, then p=\\sqrt{3}." }, { "text": "Let the ellipse and hyperbola share a common focus, and the sum of their eccentricities be $2$. If the equation of the ellipse is $25 x^{2}+9 y^{2}=225$, then what is the equation of the hyperbola?", "fact_expressions": "G: Hyperbola;H: Ellipse;Focus(H) = Focus(G);Eccentricity(G)+Eccentricity(H)=2;Expression(H) = (25*x^2 + 9*y^2 = 225)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/(100/9)-x^2/(44/9)=1", "fact_spans": "[[[4, 7], [56, 59]], [[1, 3], [27, 29]], [[1, 12]], [[13, 25]], [[27, 54]]]", "query_spans": "[[[56, 64]]]", "process": "" }, { "text": "Given that point $M(2,1)$ lies on the parabola $x^{2}=2 p y$, then the equation of the tangent line to the parabola at point $M$ is?", "fact_expressions": "G: Parabola;p: Number;M: Point;Expression(G) = (x^2 = 2*(p*y));Coordinate(M) = (2, 1);PointOnCurve(M, G);l:Line;IsTangent(l,G);TangentPoint(l,G)=M", "query_expressions": "Expression(l)", "answer_expressions": "x-y-1=0", "fact_spans": "[[[12, 28], [42, 45]], [[15, 28]], [[2, 11], [34, 38]], [[12, 28]], [[2, 11]], [[2, 31]], [], [[42, 48]], [[33, 48]]]", "query_spans": "[[[33, 52]]]", "process": "" }, { "text": "The eccentricities of a confocal ellipse and hyperbola are $e_{1}$ and $e_{2}$, respectively. If the minor axis length of the ellipse is 3 times the imaginary axis length of the hyperbola, then the maximum value of $\\frac{1}{e_{1}}+\\frac{1}{e_{2}}$ is?", "fact_expressions": "H: Ellipse;G: Hyperbola;Focus(H) = Focus(G);e1: Number;Eccentricity(H) = e1;e2: Number;Eccentricity(G) = e2;Length(MinorAxis(H)) = Length(ImageinaryAxis(G))*3", "query_expressions": "Max(1/e2 + 1/e1)", "answer_expressions": "10/3", "fact_spans": "[[[4, 6], [35, 37]], [[7, 10], [42, 45]], [[0, 10]], [[17, 24]], [[4, 33]], [[26, 33]], [[4, 33]], [[35, 53]]]", "query_spans": "[[[55, 94]]]", "process": "Analysis: Let the semi-minor axis of the ellipse and the semi-imaginary axis of the hyperbola be $ b_{1} $, $ b_{2} $ respectively, and the semi-major axis of the ellipse and the semi-real axis of the hyperbola be $ a_{1} $, $ a_{2} $ respectively. According to the given condition, $ a_{1}^{2} + 9a_{2}^{2} = 10c^{2} $. Let $ a_{1} = \\sqrt{10}c\\sin\\theta $, $ a_{2} = \\frac{\\sqrt{10}}{3}c\\cos\\theta $. Use trigonometric functions to find the maximum value. Detailed: Let the semi-minor axis of the ellipse and the semi-imaginary axis of the hyperbola be $ b_{1} $, $ b_{2} $ respectively, and the semi-major axis of the ellipse and the semi-real axis of the hyperbola be $ a_{1} $, $ a_{2} $ respectively. Then $ b_{1} = 3b_{2} \\Rightarrow a_{1}^{2} + 9a_{2}^{2} = 10c^{2} $. Let $ a_{1} = \\sqrt{10}c\\sin\\theta $, $ a_{2} = \\frac{\\sqrt{10}}{3}c\\cos\\theta $, $ \\frac{1}{e_{1}} + \\frac{1}{e_{2}} = \\frac{\\sqrt{10}}{3}(3\\sin\\theta + \\cos\\theta) = \\frac{10}{3}\\sin(\\theta + \\varphi) \\leqslant \\frac{10}{3} $." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. If there exists a point $P$ on $C$ such that $P F_{1} \\perp P F_{2}$ and $\\angle P F_{1} F_{2}=45^{\\circ}$, then the eccentricity of $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F1: 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);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));AngleOf(P, F1, F2) = ApplyUnit(45, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[17, 74], [82, 85], [155, 158]], [[24, 74]], [[24, 74]], [[90, 93]], [[1, 8]], [[9, 16]], [[24, 74]], [[24, 74]], [[17, 74]], [[1, 79]], [[82, 93]], [[95, 118]], [[120, 153]]]", "query_spans": "[[[155, 164]]]", "process": "From the given information, triangle $ PF_{1}F_{2} $ is an isosceles right triangle, with $ \\angle F_{1}PF_{2} = 90^{\\circ} $, $ |PF_{1}| = |PF_{2}| $. By the definition of the ellipse, $ |PF_{1}| + |PF_{2}| = 2a $, so $ |PF_{1}| = |PF_{2}| = a $. Also, $ |F_{1}F_{2}| = 2c $. Therefore, in $ \\triangle PF_{1}F_{2} $, by the Pythagorean theorem, we have: $ 2|PF_{1}|^{2} = |F_{1}F_{2}|^{2} $, that is, $ 2a^{2} = 4c^{2} $, so $ e = \\frac{c}{a} = \\frac{\\sqrt{2}}{2} $." }, { "text": "Given $M(0, -5)$, $N(0, 5)$, a moving point $P$ satisfies $|PM| - |PN| = 6$, then the equation of the trajectory of point $P$ is?", "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (0,-5);Coordinate(N) = (0, 5);Abs(LineSegmentOf(P, M)) - Abs(LineSegmentOf(P, N)) = 6", "query_expressions": "LocusEquation(P)", "answer_expressions": "(y^2/9 - x^2/16 = 1)&(y > 0)", "fact_spans": "[[[2, 12]], [[14, 25]], [[28, 31], [48, 52]], [[2, 12]], [[14, 25]], [[33, 46]]]", "query_spans": "[[[48, 59]]]", "process": "" }, { "text": "The eccentricity $e$ of the hyperbola $\\frac{x^{2}}{4}+\\frac{y^{2}}{k}=1$ satisfies $e \\in(1,2)$. What is the range of values for $k$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 + y^2/k = 1);k: Number;e: Number;Eccentricity(G) = e;In(e, (1, 2))", "query_expressions": "Range(k)", "answer_expressions": "(-12, 0)", "fact_spans": "[[[0, 38]], [[0, 38]], [[56, 59]], [[42, 54]], [[0, 54]], [[42, 54]]]", "query_spans": "[[[56, 65]]]", "process": "" }, { "text": "Given the parabola $y^{2}=4x$, a line passing through the focus $F$ intersects the parabola at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the $x$-axis and $y$-axis, with feet of perpendiculars at $C$ and $D$, respectively. Then the minimum value of $|AC|+|BD|$ is?", "fact_expressions": "G: Parabola;H:Line;A: Point;C: Point;B: Point;D: Point;F:Point;l1:Line;l2:Line;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(F, H);Intersection(H,G)={A,B};PointOnCurve(A,l1);PointOnCurve(B,l2);IsPerpendicular(l1,xAxis);IsPerpendicular(l2,yAxis);FootPoint(l1,xAxis)=C;FootPoint(l2,yAxis)=D", "query_expressions": "Min(Abs(LineSegmentOf(A, C)) + Abs(LineSegmentOf(B, D)))", "answer_expressions": "3", "fact_spans": "[[[2, 16], [27, 30]], [[24, 26]], [[43, 46], [43, 46]], [[71, 74]], [[47, 50], [47, 50]], [[75, 78]], [[20, 23]], [], [], [[2, 16]], [[2, 23]], [[17, 26]], [[24, 41]], [[42, 65]], [[42, 65]], [[42, 65]], [[42, 65]], [[42, 78]], [[42, 78]]]", "query_spans": "[[[80, 99]]]", "process": "" }, { "text": "Given that the focal distance of the curve $\\frac{x^{2}}{a}+\\frac{y^{2}}{16}=1$ is $10$, and the distance from a point $P$ on the curve to one focus is $2$, then the distance from point $P$ to the other focus is?", "fact_expressions": "G: Curve;a: Number;F1: Point;F2: Point;P: Point;Expression(G) = (y^2/16 + x^2/a = 1);FocalLength(G) = 10;PointOnCurve(P, G);OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 2", "query_expressions": "Distance(P, F2)", "answer_expressions": "{2*sqrt(41) - 2, 10}", "fact_spans": "[[[2, 40], [49, 51]], [[4, 40]], [], [], [[53, 57]], [[2, 40]], [[2, 48]], [[49, 57]], [[49, 62]], [[49, 81]], [[49, 81]], [[49, 69]]]", "query_spans": "[[[49, 86]]]", "process": "According to the problem, the semi-focal length of the curve is 5. If the curve is an ellipse with foci on the x-axis, then a > 16, so a - 16 = 25 ⇒ a = 41, and since the distance from a point P on the ellipse to one focus is 2, the distance from point P to the other focus is 2\\sqrt{41} - 2; if the curve is an ellipse with foci on the y-axis, then 0 < a < 16, so 16 - a = 25 ⇒ a = -9, which is discarded; if the curve is a hyperbola, then a < 0, and it is easy to determine that the foci of the hyperbola lie on the y-axis, so 16 + (-a) = 25 ⇒ a = -9. Without loss of generality, assume point P lies on the upper branch of the hyperbola, with upper and lower foci F_{2}(0,5), F_{1}(0,-5) respectively. Since the real semi-axis length is 4, it is easy to determine that the minimum distance from point P to the lower focus is 4 + 5 = 9 > 2, which does not satisfy the condition. Therefore, the distance from point P to the upper focus is 2, so its distance to the lower focus PF_{1} = |PF_{2}| + 8 = 10." }, { "text": "The left focus of ellipse $C$ is $F_{1}(-6 , 0)$, and it passes through the point $P(5 , 2)$. What is the standard equation of ellipse $C$?", "fact_expressions": "C: Ellipse;F1: Point;P: Point;Coordinate(F1) = (-6, 0);Coordinate(P) = (5, 2);LeftFocus(C) = F1;PointOnCurve(P,C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/45+ y^2/9=1", "fact_spans": "[[[0, 5], [42, 47]], [[10, 25]], [[29, 40]], [[10, 25]], [[29, 40]], [[0, 25]], [[0, 40]]]", "query_spans": "[[[42, 54]]]", "process": "According to the problem, the foci of the ellipse lie on the x-axis and are given by F_{1}(-6,0), F_{2}(6,0). By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a, which leads to the solution. The left focus of ellipse C is F_{1}(-6,0), so the foci of the ellipse lie on the x-axis and the right focus is F_{2}(6,0). By the definition of the ellipse, |PF_{1}| + |PF_{2}| = \\sqrt{(5+6)^{2}+2^{2}} + \\sqrt{(5-6)^{2}+2^{2}} = 6\\sqrt{5} = 2a. Therefore, a = 3\\sqrt{5}, c = 6, and b^{2} = 45 - 36 = 9. Thus, the standard equation of ellipse C is: \\frac{x^{2}}{45} + \\frac{y^{2}}{9} = 1" }, { "text": "Given that the major axis of the ellipse $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m-2}=1$ lies on the $y$-axis, and the focal distance is $4$, then $m$=?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(10 - m) + y^2/(m - 2) = 1);OverlappingLine(MajorAxis(G),yAxis);FocalLength(G)=4", "query_expressions": "m", "answer_expressions": "8", "fact_spans": "[[[2, 44]], [[63, 66]], [[2, 44]], [[2, 53]], [[2, 61]]]", "query_spans": "[[[63, 68]]]", "process": "Since the major axis of the ellipse $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m-2}=1$ lies on the $y$-axis and the focal distance is $4$, we have $m-2-(10-m)=(\\frac{4}{2})^{2}$, therefore $m=8$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with foci $F_{1}$, $F_{2}$, and a point $P$ on the hyperbola $C$ satisfying $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, $|\\overrightarrow{P F_{1}}|=3$, $|\\overrightarrow{P F_{2}}|=4$, then the eccentricity of the hyperbola $C$ 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);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Abs(VectorOf(P, F1)) = 3;Abs(VectorOf(P, F2)) = 4", "query_expressions": "Eccentricity(C)", "answer_expressions": "5", "fact_spans": "[[[2, 63], [84, 90], [225, 231]], [[10, 63]], [[10, 63]], [[92, 96]], [[67, 74]], [[75, 82]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 82]], [[84, 96]], [[98, 157]], [[159, 189]], [[191, 223]]]", "query_spans": "[[[225, 237]]]", "process": "By the definition of a hyperbola, we have 2a = |\\overrightarrow{PF_{2}}| - |\\overrightarrow{PF_{1}}| = 1, so a = \\frac{1}{2}. Since \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = 0, it follows that \\overrightarrow{PF_{1}} \\bot \\overrightarrow{PF_{2}}. Therefore, (2c)^{2} = |\\overrightarrow{PF_{2}}|^{2} + |\\overrightarrow{PF}|^{2} = 25, solving gives c = \\frac{5}{2}. Thus, the eccentricity of hyperbola C is e = \\frac{c}{a} = 5." }, { "text": "The coordinates of the focus $F$ of the parabola $y^{2}=4x$ are? The distance from point $F$ to the asymptotes of the hyperbola $x^{2}-y^{2}=1$ is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y^2 = 4*x);F: Point;Focus(H) = F", "query_expressions": "Coordinate(F);Distance(F, Asymptote(G))", "answer_expressions": "(1,0);sqrt(2)/2", "fact_spans": "[[[30, 48]], [[0, 14]], [[30, 48]], [[0, 14]], [[17, 20], [25, 29]], [[0, 20]]]", "query_spans": "[[[17, 25]], [[25, 57]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ with left and right foci $F_{1}$, $F_{2}$, and $|F_{1} F_{2}|=6$, then the distance from $F_{1}$ to an asymptote is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/b^2 = 1);b: Number;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Abs(LineSegmentOf(F1, F2)) = 6", "query_expressions": "Distance(F1, OneOf(Asymptote(C)))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 49]], [[2, 49]], [[10, 49]], [[55, 62], [91, 98]], [[63, 70]], [[2, 70]], [[2, 70]], [[72, 89]]]", "query_spans": "[[[2, 108]]]", "process": "According to the problem, find the equation of the hyperbola $ C: \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $, obtain one of its asymptotes $ \\sqrt{5}x - 2y = 0 $, then use the point-to-line distance formula to solve. \nSolution: From the given conditions, the left and right foci of the hyperbola $ C: \\frac{x^{2}}{4} - \\frac{y^{2}}{b^{2}} = 1 $ are $ F_{1}, F_{2} $. Since $ |F_{1}F_{2}| = 6 $, i.e., $ 2c = 6 $, solving gives $ c = 3 $, so $ 4 + b^{2} = 3^{2} $, solving gives $ b^{2} = 5 $, thus $ b = \\sqrt{5} $. Therefore, the equation of the hyperbola is $ C: \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $, then one of its asymptotes is $ y = \\frac{\\sqrt{5}}{2}x $, i.e., $ \\sqrt{5}x - 2y = 0 $. Hence, the distance from focus $ F_{1}(-3,0) $ to the line $ \\sqrt{5}x - 2y = 0 $ is $ d = \\frac{|\\sqrt{5} \\times (-3) - 2 \\times 0|}{\\sqrt{(\\sqrt{5})^{2} + (-2)^{2}}} = \\sqrt{5} $." }, { "text": "The focus of the parabola $y^{2}=6 x$ is $F$. A line passing through point $F$ intersects the parabola at two points $A$ and $B$. If the horizontal coordinate of the midpoint of $A B$ is $2$, then $|A B|$=?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 6*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "7", "fact_spans": "[[[0, 14], [31, 34]], [[28, 30]], [[40, 43]], [[44, 47]], [[23, 27], [18, 21]], [[0, 14]], [[0, 21]], [[22, 30]], [[28, 47]], [[49, 63]]]", "query_spans": "[[[65, 74]]]", "process": "From the given condition, p=3. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since the horizontal coordinate of the midpoint of segment AB is 2, we have x_{1}+x_{2}=4. By the definition of the parabola, |AB|=x_{1}+x_{2}+p=4+3=" }, { "text": "The coordinates of the foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*sqrt(5),0)", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 35]]]", "process": "" }, { "text": "Given that the asymptotes of a hyperbola are $y = \\pm x$, and it passes through the point $(1, \\sqrt{2})$, then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*x);H: Point;Coordinate(H) = (1, sqrt(2));PointOnCurve(H,G) = True", "query_expressions": "Expression(G)", "answer_expressions": "y^2-x^2=1", "fact_spans": "[[[2, 5], [42, 45]], [[2, 21]], [[24, 40]], [[24, 40]], [[2, 40]]]", "query_spans": "[[[42, 52]]]", "process": "Let the equation with asymptotes y = ±x be x^{2} - y^{2} = \\lambda. Since this hyperbola passes through the point (1, \\sqrt{2}), it follows that \\lambda = 1 - 2 = -1, so the standard equation of the hyperbola is y^{2} - x^{2} = 1." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ has a focal distance of?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/2 = 1)", "query_expressions": "FocalLength(C)", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[0, 42]], [[0, 42]]]", "query_spans": "[[[0, 47]]]", "process": "c^{2}=a^{2}-b^{2}=4-2=2, so c=\\sqrt{2}, therefore the focal distance of the ellipse C: \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1 is 2\\sqrt{2}." }, { "text": "The coordinates of the focus of the parabola $x^{2}=2 y$ are?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*y)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, 1/2)", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "From the given condition, 2p = 2, solving gives p = 1. Since the parabola x^{2} = 2y opens upwards, the focus of the parabola x^{2} = 2y has coordinates (0, \\frac{1}{2})." }, { "text": "If the focus of the parabola $y^{2}=2 p x(p>0)$ coincides with the right focus of the hyperbola $x^{2}-y^{2}=2$, then what is the value of $p$?", "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2 = 2);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[26, 44]], [[1, 22]], [[52, 55]], [[26, 44]], [[4, 22]], [[1, 22]], [[1, 50]]]", "query_spans": "[[[52, 59]]]", "process": "From $x^{2}-y^{2}=2$ we obtain the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$, then $c=\\sqrt{2+2}=2$, therefore, the focus of the parabola is $(2,0)$, $\\frac{p}{2}=2$, $p=4$." }, { "text": "Given that the eccentricity of the hyperbola $\\frac{y^{2}}{n-6}-\\frac{x^{2}}{n}=1$ is $\\sqrt{3}$, then $n=$?", "fact_expressions": "G: Hyperbola;n: Number;Expression(G) = (y^2/(n - 6) - x^2/n = 1);Eccentricity(G) = sqrt(3)", "query_expressions": "n", "answer_expressions": "{-6,12}", "fact_spans": "[[[2, 42]], [[59, 62]], [[2, 42]], [[2, 57]]]", "query_spans": "[[[59, 64]]]", "process": "When the foci are on the y-axis, \n\\begin{cases}n-6>0\\\\n>0\\\\\\frac{n-6+n}{n-6}=3\\end{cases}, \nsolving gives n=12. When the foci are on the x-axis, the standard equation of the hyperbola is \\frac{x^{2}}{n}-\\frac{y^{2}}{6-n}=1,\\begin{cases}6-n>0\\\\-n>0\\\\\\frac{-n+6-n}{a}=3\\end{cases}, solving gives n=-6. Combining the results, n=12 or n=-6." }, { "text": "Given that $P$ and $Q$ are two points on the parabola $x^{2}=2 y$, the horizontal coordinates of points $P$ and $Q$ are $4$ and $-2$, respectively. Tangents to the parabola at $P$ and $Q$ are drawn, and these two tangents intersect at point $A$. What is the vertical coordinate of point $A$?", "fact_expressions": "G: Parabola;P: Point;Q: Point;A: Point;L1:Line;L2:Line;Expression(G) = (x^2 = 2*y);PointOnCurve(P,G);PointOnCurve(Q,G);XCoordinate(P)=4;XCoordinate(Q)=-2;TangentOfPoint(P,G)=L1;TangentOfPoint(Q,G)=L2;Intersection(L1,L2)=A", "query_expressions": "YCoordinate(A)", "answer_expressions": "-4", "fact_spans": "[[[10, 24], [63, 66]], [[2, 5], [28, 32], [53, 56]], [[6, 9], [33, 36], [57, 60]], [[75, 79], [81, 85]], [], [], [[10, 24]], [[2, 27]], [[2, 27]], [[28, 50]], [[28, 50]], [[52, 69]], [[52, 69]], [[52, 79]]]", "query_spans": "[[[81, 91]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with foci $F_{1}$ and $F_{2}$, 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;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/2 + y^2 = 1);Coordinate(P) = (x0, y0);Focus(C)={F1,F2};x0:Number;y0:Number;01\\Leftrightarrow P is outside the ellipse. Therefore, in this problem, P is inside the ellipse, |PF_{1}|+|PF_{2}|<2a=2\\sqrt{2}, and |PF_{1}|+|PF_{2}|\\geqslant|F_{1}F_{2}|=2c=2, so the required range is [2,2\\sqrt{2})." }, { "text": "A line passing through the focus $F$ of the parabola $y^{2}=8x$ intersects the parabola at points $A$ and $B$. Then $\\frac{1}{|A F|}+\\frac{1}{|B F|}$=?", "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 8*x);Focus(G)=F;PointOnCurve(F, H);Intersection(H, G) = {A, B}", "query_expressions": "1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F))", "answer_expressions": "1/2", "fact_spans": "[[[1, 15], [25, 28]], [[22, 24]], [[29, 32]], [[18, 21]], [[33, 36]], [[1, 15]], [[1, 21]], [[0, 24]], [[22, 38]]]", "query_spans": "[[[40, 75]]]", "process": "" }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 34]]]", "process": "" }, { "text": "A line passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If point $A$ lies in the first quadrant and $|AF|=3|FB|$, then what is the inclination angle of the line $AB$?", "fact_expressions": "C: Parabola;G: Line;B: Point;A: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Quadrant(A)=1;Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(F, B))", "query_expressions": "Inclination(LineOf(A,B))", "answer_expressions": "pi/3", "fact_spans": "[[[1, 20], [29, 32]], [[26, 28]], [[37, 40]], [[33, 36], [44, 48]], [[22, 25]], [[1, 20]], [[1, 25]], [[0, 28]], [[26, 42]], [[44, 53]], [[55, 69]]]", "query_spans": "[[[71, 84]]]", "process": "Since |AF| = 3|FB|, the line AB has a slope, denoted as k (k ≠ 0). The focus F of the parabola C: y² = 4x has coordinates (1, 0), so the equation of line AB is y = k(x - 1). Solving the system of equations: \n\\begin{cases} y^2 = 4x \\\\ y = k(x - 1) \\end{cases} \\Rightarrow ky^2 - 4y - 4k = 0. \nLet A(x₁, y₁), B(x₂, y₂), with x₁ > 0, y₁ > 0, x₂ > 0, y₂ < 0. Thus, we have y₁ + y₂ = \\frac{4}{k}, y₁y₂ = -4. Since point A lies in the first quadrant and |AF| = 3|FB|, it follows that \\overrightarrow{AF} = 3\\overrightarrow{FB} \\Rightarrow (1 - x₁, -y₁) = 3(x₂ - 1, y₂) \\Rightarrow y₁ = -3y₂. Therefore, -3y₂ + y₂ = \\frac{4}{k}, -3y₂ \\cdot y₂ = -4 \\Rightarrow -2y₂ = \\frac{4}{k} > 0, -3y₂^2 = -4 \\Rightarrow k > 0, k^2 = 3 \\Rightarrow k = \\sqrt{3}. Let the inclination angle of line AB be α, so \\tan\\alpha = \\sqrt{3}, solving gives: α = \\frac{\\pi}{3}." }, { "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, with focal distance $2c$. If a point of intersection $M$ between the line $y=\\sqrt{3}(x+c)$ and ellipse $C$ satisfies $\\angle M F_{1} F_{2}=2 \\angle M F_{2} F_{1}$, then the eccentricity of this ellipse is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;c: Number;M: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = sqrt(3)*(c + x));LeftFocus(C) = F1;RightFocus(C) = F2;FocalLength(C) = 2*c;OneOf(Intersection(G, C)) = M;AngleOf(M, F1, F2) = 2*AngleOf(M, F2, F1)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3) - 1", "fact_spans": "[[[0, 57], [112, 117], [175, 177]], [[7, 57]], [[7, 57]], [[92, 111]], [[85, 90]], [[122, 125]], [[66, 73]], [[74, 81]], [[7, 57]], [[7, 57]], [[0, 57]], [[92, 111]], [[0, 81]], [[0, 81]], [[0, 90]], [[92, 125]], [[127, 172]]]", "query_spans": "[[[175, 183]]]", "process": "" }, { "text": "The parabola $C$: $x^{2}=4 y$ has focus $F$. The directrix of the parabola $C$ intersects the coordinate axes at point $A$. Point $P$ lies on the parabola $C$. Then the maximum value of $\\frac{|P A|}{|P F|}$ is?", "fact_expressions": "C: Parabola;P: Point;A: Point;F: Point;Expression(C) = (x^2 = 4*y);Intersection(Directrix(C), axis) = A;PointOnCurve(P, C);Focus(C)=F", "query_expressions": "Max(Abs(LineSegmentOf(P, A))/Abs(LineSegmentOf(P, F)))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[7, 26], [45, 51]], [[40, 44]], [[35, 39]], [[3, 6]], [[7, 26]], [[7, 39]], [[40, 52]], [[0, 26]]]", "query_spans": "[[[54, 81]]]", "process": "According to the problem, draw PM perpendicular from P to the directrix, with foot of perpendicular at M. Let ∠PAM = θ, then |PA|/|PF| = |PA|/|MP| = 1/sinθ. If |PA|/|PF| attains its maximum value, sinθ must attain its minimum value, so θ attains its minimum value, and at this moment AP is tangent to the parabola. Let the equation of line AP be y = kx − 1. Solve the system: \n\\begin{cases} x^{2} = 4y \\\\ y = kx - \\end{cases} \nEliminating y gives: x² − 4kx + 4 = 0. From Δ = 16k² − 16 = 0, solve to get: k = 1 or k = −1. Take k = tanθ = 1 for calculation. Since 0 ≤ θ < π, we have θ = π/4. Therefore, the maximum value of |PA|/|PF| is (1/2)/(√2/2) = √2." }, { "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. Let $P$ be an arbitrary point on the left branch, and let line $l$ be an asymptote of the hyperbola. The projection of point $P$ onto line $l$ is $Q$. When $|P F_{2}|+|P Q|$ attains its minimum value of $5$, what is the maximum value of $S_{\\Delta F_{1} Q F_{2}}$?", "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, LeftPart(G));l: Line;OneOf(Asymptote(G)) = l;Q: Point;Projection(P, l) = Q;Min(Abs(LineSegmentOf(P, F2)) + Abs(LineSegmentOf(P, Q))) = 5", "query_expressions": "Max(Area(TriangleOf(F1, Q, F2)))", "answer_expressions": "25/8", "fact_spans": "[[[2, 58], [101, 104]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[66, 73]], [[74, 81]], [[2, 81]], [[2, 81]], [[82, 86], [111, 115]], [[2, 94]], [[95, 100], [116, 121]], [[95, 110]], [[126, 129]], [[111, 129]], [[131, 157]]]", "query_spans": "[[[158, 190]]]", "process": "" }, { "text": "If a line $l$ passing through the point $A(0, 2)$ intersects the curve $x^{2}-y^{2}=1$, then what is the range of possible values for the slope of line $l$?", "fact_expressions": "A: Point;Coordinate(A) = (0, 2);l: Line;PointOnCurve(A, l);G: Curve;Expression(G) = (x^2 - y^2 = 1);IsIntersect(l, G)", "query_expressions": "Range(Slope(l))", "answer_expressions": "[-sqrt(5), -1) + (-1, 1) + (1, sqrt(5)]", "fact_spans": "[[[2, 13]], [[2, 13]], [[14, 19], [43, 48]], [[1, 19]], [[20, 37]], [[20, 37]], [[14, 41]]]", "query_spans": "[[[43, 58]]]", "process": "" }, { "text": "Given that one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has the equation $y=3x$, find the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Number;b>0;Expression(OneOf(Asymptote(G))) = (y = 3*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(10)", "fact_spans": "[[[2, 39], [58, 61]], [[2, 39]], [[5, 39]], [[5, 39]], [[2, 56]]]", "query_spans": "[[[58, 67]]]", "process": "Test analysis: Given that one asymptote of the hyperbola is y = bx, it follows that b = 3. Since a = 1, the eccentricity of the hyperbola is e = \\frac{c}{a} = \\frac{\\sqrt{10}}{1} = \\sqrt{10}." }, { "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 any point on the ellipse, then the minimum value of $|OP|^{2}+|PF|^{2}$ is?", "fact_expressions": "G: Ellipse;O: Point;P: Point;F: Point;Expression(G) = (x^2/2 + y^2 = 1);PointOnCurve(P, G);Center(G) = O;LeftFocus(G) = F", "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]], [[48, 61]], [[1, 47]], [[1, 47]]]", "query_spans": "[[[63, 91]]]", "process": "" }, { "text": "Let the right vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $A(a, 0)$, and suppose there exists a point $P$ on it such that $\\angle A P O=90^{\\circ}$, then the range of 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;A: Point;Coordinate(A) = (a, 0);RightVertex(G) = A;P: Point;PointOnCurve(P, G) = True;O: Origin;AngleOf(A, P, O) = ApplyUnit(90, degree)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(2)/2,1)", "fact_spans": "[[[1, 53], [68, 69], [106, 108]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[58, 67]], [[58, 67]], [[1, 67]], [[74, 77]], [[68, 77]], [[79, 104]], [[79, 104]]]", "query_spans": "[[[106, 119]]]", "process": "Let P(x,y). Since \\angle APO = 90^{\\circ}, the point P lies on a circle with OA as diameter. The center of the circle is (\\frac{a}{2}, 0) and the radius is \\frac{|OA|}{2} = \\frac{a}{2}. Thus, the equation of the circle is (x - \\frac{a}{2})^{2} + y^{2} = (\\frac{a}{2})^{2}. Therefore, y^{2} = ax - x^{2} \\textcircled{1}. Since point P also lies on the ellipse, \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\textcircled{2}. Substituting \\textcircled{1} into \\textcircled{2} gives \\frac{x^{2}}{a^{2}} + \\frac{ax - x^{2}}{b^{2}} = 1, so (a^{2} - b^{2})x^{2} - a^{3}x + a^{2}b^{2} = 0. Hence, (x - a) \\cdot [(a^{2} - b^{2})x - ab^{2}] = 0. Since x \\neq a and x \\neq 0, it follows that x = \\frac{ab^{2}}{a^{2} - b^{2}}. Given 0 < x < a, we have 0 < \\frac{ab^{2}}{a^{2} - b^{2}} < a, so 2b^{2} < a^{2}, which implies a^{2} < 2c^{2}, thus e > \\frac{\\sqrt{2}}{2}. Since 0 < e < 1, the required range for the eccentricity of the ellipse is (\\frac{\\sqrt{2}}{2}, 1)." }, { "text": "Given that the right focus of a hyperbola is $(5, 0)$ and one asymptote has the equation $2x - y = 0$, then the standard equation of this hyperbola is?", "fact_expressions": "E: Hyperbola;Coordinate(RightFocus(E)) = (5, 0);Expression(OneOf(Asymptote(E))) = (2*x - y = 0)", "query_expressions": "Expression(E)", "answer_expressions": "x^2/5 - y^2/20 = 1", "fact_spans": "[[[2, 5], [40, 43]], [[2, 19]], [[2, 37]]]", "query_spans": "[[[40, 50]]]", "process": "According to the problem, one asymptote of the hyperbola is 2x - y = 0, so the equation of the hyperbola can be written as x^{2} - \\frac{y^{2}}{4} = \\lambda, \\lambda \\neq 0; also, the right focus of the hyperbola is (5, 0), meaning the foci lie on the x-axis and c = 5, thus \\lambda > 0; then the equation of the hyperbola can be rewritten as \\frac{x^{2}}{\\lambda} - \\frac{y^{2}}{4\\lambda} = 1; given c = 5, we have 5\\lambda = 25, solving gives \\lambda = 5; therefore, the standard equation of this hyperbola is \\frac{x^{2}}{5} - \\frac{y^{2}}{20} = 1" }, { "text": "The coordinates of the focus of the parabola $y = a x^{2}$ passing through the point $(2,4)$ are?", "fact_expressions": "G: Parabola;a: Number;H: Point;Expression(G) = (y = a*x^2);Coordinate(H) = (2, 4);PointOnCurve(H, G)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,1/4)", "fact_spans": "[[[11, 25]], [[14, 25]], [[2, 10]], [[11, 25]], [[2, 10]], [[0, 25]]]", "query_spans": "[[[11, 31]]]", "process": "\\because the parabola y=ax^2 passes through the point (2,4) \\therefore a=1, \\therefore the standard equation of the parabola is x^{2}=y, \\therefore the coordinates of the focus of the parabola are (0,\\frac{1}{4})" }, { "text": "If the ellipse $C$: $\\frac{x^{2}}{2 m+1}+\\frac{y^{2}}{2 m}=1$ has eccentricity $\\frac{1}{2}$, then what is the length of the minor axis of $C$?", "fact_expressions": "C: Ellipse;m: Number;Expression(C) = (x^2/(2*m + 1) + y^2/(2*m) = 1);Eccentricity(C) = 1/2", "query_expressions": "Length(MinorAxis(C))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[1, 49], [69, 72]], [[8, 49]], [[1, 49]], [[1, 67]]]", "query_spans": "[[[69, 78]]]", "process": "Determine that the foci of the ellipse lie on the x-axis, find the values of a, b, c, and according to the concept of eccentricity, we have e = \\frac{1}{\\sqrt{2m+1}} = \\frac{1}{2}; solve for the value of m to find the length of the minor axis. From the ellipse C: \\frac{x^2}{2m+1} + \\frac{y^2}{2m} = 1, we get that the foci lie on the x-axis, a^{2} = 2m+1, b^{2} = 2m, c^{2} = 1, and e = \\frac{1}{\\sqrt{2m+1}} = \\frac{1}{2}; solving gives m = \\frac{3}{2}, so b^{2} = 2m = 3, b = \\sqrt{3}, therefore the length of the minor axis of C is 2\\sqrt{3}." }, { "text": "Let a line passing through the origin intersect the hyperbola $ C $: $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) at two distinct points $ P $ and $ Q $, and let $ F $ be a focus of $ C $. If $ \\tan \\angle P F Q = \\frac{4}{3} $, $ |Q F| = 5|P F| $, then the eccentricity of the hyperbola $ C $ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);G: Line;O:Origin;PointOnCurve(O,G);P: Point;F: Point;Q: Point;a>0;b>0;Intersection(G, C) = {P, Q};OneOf(Focus(C)) = F;Tan(AngleOf(P, F, Q)) = 4/3;Abs(LineSegmentOf(Q, F)) = 5*Abs(LineSegmentOf(P, F));Negation(P=Q)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[8, 69], [88, 91], [146, 152]], [[15, 69]], [[15, 69]], [[8, 69]], [[5, 7]], [[2, 4]], [[1, 7]], [[71, 74]], [[84, 87]], [[75, 78]], [[15, 69]], [[15, 69]], [[5, 83]], [[84, 96]], [[98, 129]], [[130, 144]], [69, 80]]", "query_spans": "[[[146, 158]]]", "process": "As shown in the figure: connect PF₁, QF₁, by symmetry PF₁QF is a parallelogram, calculate to get |PF₁| = \\frac{5a}{2}, |PF| = \\frac{a}{2}, use the law of cosines to compute the answer. As shown in the figure: connect PF₁, QF₁, by symmetry PF₁QF is a parallelogram, \\tan\\angle PFQ = \\frac{4}{3}, then \\cos\\angle PFQ = \\frac{3}{5}, \\cos\\angle F₁PF = -\\frac{3}{5}, |QF| = |PF₁| = 5|PF|, |PF₁| - |PF| = 2a, hence |PF₁| = \\frac{5a}{2}, |PF| = \\frac{a}{2}. By the law of cosines: 4c^{2} = (\\frac{5a}{2})^{2} + (\\frac{a}{2})^{2} - 2 \\cdot \\frac{5a}{2} \\cdot \\frac{a}{2} \\cdot (-\\frac{3}{5}), simplify to get c^{2} = 2a^{2}, hence e = \\sqrt{2}. Comprehensive application ability and computational ability." }, { "text": "Given that the circle $O$: $x^{2}+y^{2}=1$ passes through the right focus $F_{2}$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and the chord intercepted by the ellipse $C$ on the tangent line to the circle $O$ at point $F_{2}$ has length $\\sqrt{2}$. Then the equation of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;O: Circle;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = 1);RightFocus(C) = F2;L:Line;TangentOnPoint(F2, O)=L;Length(InterceptChord(L,C))=sqrt(2)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/2 + y^2 = 1", "fact_spans": "[[[24, 81], [113, 118], [137, 142]], [[31, 81]], [[31, 81]], [[2, 22], [105, 109]], [[85, 92], [96, 104]], [[31, 81]], [[31, 81]], [[24, 81]], [[2, 22]], [[24, 92]], [], [[94, 112]], [[94, 134]]]", "query_spans": "[[[137, 147]]]", "process": "Since the circle O: $x^{2}+y^{2}=1$ passes through the right focus $F_{2}$ of the ellipse $C$, we have $c=1$, then $a^{2}=b^{2}+1$, and the chord intercepted on the ellipse $C$ by the tangent line drawn from point $F_{2}$ to the circle $O$ has length $\\sqrt{2}$, so the point $(1,\\frac{\\sqrt{2}}{2})$ lies on the ellipse, that is, $\\frac{1}{1+b^{2}}+\\frac{1}{2b^{2}}=1$, thus $b^{2}=1$, $a^{2}=2$, therefore the equation of the ellipse $C$ is $\\frac{x^{2}}{2}+y^{2}=1$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $e$. If both points $(e, 1)$ and $(-\\sqrt{3}, \\sqrt{2})$ lie on the hyperbola $C$, then what is the distance from a focus of the hyperbola to one of its asymptotes?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Point;H: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(G) = (e, 1);Coordinate(H) = (-sqrt(3), sqrt(2));Eccentricity(C) = e;PointOnCurve(G, C);PointOnCurve(H, C)", "query_expressions": "Distance(Focus(C), OneOf(Asymptote(C)))", "answer_expressions": "1", "fact_spans": "[[[2, 63], [109, 115], [119, 122], [126, 127]], [[10, 63]], [[10, 63]], [[73, 82]], [[83, 107]], [[68, 71]], [[10, 63]], [[10, 63]], [[2, 63]], [[73, 82]], [[83, 107]], [[2, 71]], [[73, 116]], [[73, 116]]]", "query_spans": "[[[119, 137]]]", "process": "Since the point (e,1) and the point (-\\sqrt{3},\\sqrt{2}) both lie on the hyperbola C, we have \\begin{cases}\\frac{e^{2}}{a^{2}}-\\frac{1}{b^{2}}=1\\\\\\frac{3}{a^{2}}-\\frac{2}{b^{2}}=1\\end{cases}. Eliminating e^{2} yields a^{2}=b^{2}, so a^{2}=b^{2}=1. Therefore, the foci of the hyperbola are (\\pm\\sqrt{2},0), and the asymptotes are y=\\pm x. Taking the focus (\\sqrt{2},0) and the asymptote y=x, the distance from the focus to the asymptote is d=\\frac{\\sqrt{2}}{\\sqrt{1+1}}=1." }, { "text": "The standard equation of a circle centered at the right focus of the hyperbola $\\frac{x^{2}}{8}-y^{2}=1$ and tangent to its asymptotes is?", "fact_expressions": "G: Hyperbola;H: Circle;Expression(G) = (x^2/8 - y^2 = 1);Center(H) = RightFocus(G);IsTangent(H, Asymptote(G))", "query_expressions": "Expression(H)", "answer_expressions": "(x - 3)^2 + y^2 = 1", "fact_spans": "[[[1, 29], [39, 40]], [[46, 47]], [[1, 29]], [[0, 47]], [[38, 47]]]", "query_spans": "[[[46, 54]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $8 y^{2}-x=0$ are?", "fact_expressions": "G: Parabola;Expression(G) = (-x + 8*y^2 = 0)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(1/32,0)", "fact_spans": "[[[0, 16]], [[0, 16]]]", "query_spans": "[[[0, 23]]]", "process": "Converting the parabola into standard form gives y^{2}=\\frac{1}{8}x. According to the basic concept of a parabola, the coordinates of the focus can be calculated. \\because the equation of the parabola is x=8y^{2}, \\therefore converting it into standard form yields y^{2}=\\frac{1}{8}x, from which we obtain that the parabola's 2p=\\frac{1}{8}, so \\frac{p}{2}=\\frac{1}{32}. \\therefore the coordinates of the focus of the parabola are (\\frac{1}{32},0)." }, { "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, with two foci $F_{1}$ and $F_{2}$. If a circle centered at the origin $O$ with radius $b$ intersects the hyperbola $C$ at point $P$ (where $P$ lies in the first quadrant), and $O P \\perp P F_{2}$, then what is the equation of the asymptotes of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;F1: Point;F2: Point;Focus(C) = {F1, F2};O: Origin;G: Circle;Center(G) = O;Radius(G) = b;P: Point;Intersection(G, C) = P;Quadrant(P) = 1;IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(P, F2))", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(sqrt(2)/2)*x", "fact_spans": "[[[2, 63], [109, 115], [155, 161]], [[2, 63]], [[100, 103]], [[9, 63]], [[9, 63]], [[9, 63]], [[71, 78]], [[79, 86]], [[2, 86]], [[89, 96]], [[107, 108]], [[88, 108]], [[100, 108]], [[117, 121], [122, 126]], [[107, 121]], [[122, 131]], [[134, 153]]]", "query_spans": "[[[155, 169]]]", "process": "From the given information, the foci of hyperbola C lie on the y-axis. As shown in the figure, a circle centered at the origin O with radius b intersects hyperbola C at point P in the first quadrant. It follows that |OP| = b, and since |OF₂| = c, we have |PF₂| = a. By the definition of a hyperbola, |PF₁| = 3a. In right triangle OF₂P, cos∠OF₂P = a/c. In triangle F₁F₂P, by the law of cosines, cos∠F₁F₂P = (a² + (2c)² - (3a)²)/(2a·2c) = cos∠OF₂P = a/c. Simplifying yields c = √3a, so b = √2a. Therefore, the asymptotes of hyperbola C are y = ±(√2/2)x." }, { "text": "It is known that one focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is the center of the circle $x^{2}+y^{2}-6x+8=0$, and the length of the minor axis is $8$. Then, what is the length of the major axis of the ellipse?", "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Circle;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a = 1);Expression(H) = (-6*x + x^2 + y^2 + 8 = 0);OneOf(Focus(G)) = Center(H);Length(MinorAxis(G))=8", "query_expressions": "Length(MajorAxis(G))", "answer_expressions": "10", "fact_spans": "[[[2, 50], [92, 94]], [[4, 50]], [[4, 50]], [[56, 78]], [[4, 50]], [[4, 50]], [[2, 50]], [[56, 78]], [[2, 81]], [[2, 90]]]", "query_spans": "[[[92, 100]]]", "process": "From $x^{2}+y^{2}-6x+8=0$ we get $(x-3)^{2}+y^{2}=1$, whose center is $(3,0)$, which is a focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$. Therefore, $a^{2}-b^{2}=9$. Also, $2b=8$, so $a^{2}=25$, hence $a=5$, and $2a=10$. The major axis length of the ellipse is 10." }, { "text": "The left focus of the hyperbola $\\frac{x^{2}}{3}-\\frac{16 y^{2}}{p^{2}}=1(p>0)$ lies on the directrix of the parabola $y^{2}=2 p x$. Find the eccentricity of the hyperbola.", "fact_expressions": "G: Hyperbola;p: Number;H: Parabola;p>0;Expression(G) = (x^2/3 - 16*y^2/p^2 = 1);Expression(H) = (y^2 = 2*(p*x));PointOnCurve(LeftFocus(G), Directrix(H))", "query_expressions": "Eccentricity(G)", "answer_expressions": "(2/3)*sqrt(3)", "fact_spans": "[[[0, 50], [77, 80]], [[3, 50]], [[55, 71]], [[3, 50]], [[0, 50]], [[55, 71]], [[0, 75]]]", "query_spans": "[[[77, 86]]]", "process": "" }, { "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and a line passing through $F_{1}$ intersects the ellipse at points $A$, $B$. If $|F_{2} A|+|F_{2} B|=12$, then $|A B|$=?", "fact_expressions": "E: Ellipse;Expression(E) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(E) = {F1, F2};l: Line;PointOnCurve(F1, l);A: Point;B: Point;Intersection(l, E) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[20, 58], [76, 78]], [[20, 58]], [[2, 11], [65, 72]], [[12, 19]], [[2, 63]], [[73, 75]], [[64, 75]], [[79, 82]], [[83, 86]], [[73, 88]], [[91, 115]]]", "query_spans": "[[[117, 126]]]", "process": "" }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are tangent to the circle $(x-2)^{2}+y^{2}=3$, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 + (x - 2)^2 = 3);IsTangent(H, Asymptote(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 47], [77, 80]], [[4, 47]], [[4, 47]], [[52, 72]], [[1, 47]], [[52, 72]], [[1, 74]]]", "query_spans": "[[[77, 86]]]", "process": "" }, { "text": "The equation of the hyperbola with foci at the vertices of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{5}=1$ and vertices at the foci of the ellipse is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/8 + y^2/5 = 1);Focus(H) = Vertex(C);C: Hyperbola;Focus(C) = Vertex(H)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/3-y^2/5=1", "fact_spans": "[[[1, 38], [46, 48]], [[1, 38]], [[0, 58]], [[55, 58]], [[45, 58]]]", "query_spans": "[[[55, 62]]]", "process": "The ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{5}=1$ has vertices at $(-2\\sqrt{2},0)$ and $(2\\sqrt{2},0)$, and foci at $(-\\sqrt{3},0)$ and $(\\sqrt{3},0)$. Therefore, the hyperbola has foci at $(-2\\sqrt{2},0)$ and $(2\\sqrt{2},0)$, and vertices at $(-\\sqrt{3},0)$ and $(\\sqrt{3},0)$. Therefore, for the hyperbola, $a=\\sqrt{3}$, $c=2\\sqrt{2}$ $\\Rightarrow b=\\sqrt{5}$. Therefore, the equation of the hyperbola is $\\frac{x^{2}}{3}-\\frac{y^{2}}{5}=1$. (This question examines the properties and applications of hyperbolas and ellipses; when solving, pay attention to distinguishing the basic properties of hyperbolas and ellipses.)" }, { "text": "The point $P(8,1)$ bisects a chord of the hyperbola $x^{2}-4 y^{2}=4$. Then the general form equation of the line containing this chord is?", "fact_expressions": "P: Point;Coordinate(P) = (8, 1);G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);H: LineSegment;IsChordOf(H,G) = True;MidPoint(H) = P", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "2*x-y-15=0", "fact_spans": "[[[0, 9]], [[0, 9]], [[11, 31]], [[11, 31]], [], [[11, 35]], [[0, 35]]]", "query_spans": "[[[11, 52]]]", "process": "Let the endpoints of the chord be A(x_{1},y_{1}) and B(x_{2},y_{2}). Since the midpoint of AB is P(8,1), we have x_{1}+x_{2}=16, y_{1}+y_{2}=2. Using the point difference method, we can find the equation of the line containing this chord. Let the two endpoints of the chord be A(x_{1},y_{1}) and B(x_{2},y_{2}), then x_{1}^{2}-4y_{1}^{2}=4, x_{2}^{2}-4y_{2}^{2}=4. Subtracting these two equations gives (x_{1}+x_{2})(x_{1}-x_{2})-4(y_{1}+y_{2})(y_{1}-y_{2})=0. Since the midpoint of segment AB is P(8,1), it follows that x_{1}+x_{2}=16, y_{1}+y_{2}=2. Therefore, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{2}{2}=\\overline{4}\\frac{x_{1}+x_{2}}{(y_{1}+y_{2})}=2. Hence, the equation of line AB is y-1=2(x-8). Substituting into x^{2}-4y2=4 satisfies A>0, so the equation of the line is 2x-y-15=0" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left vertex is $A$, the right focus is $F$. A circle centered at $F$ is tangent to one asymptote of the hyperbola $C$ at a point $B$ in the first quadrant. If the slope of line $AB$ is $\\frac{1}{2}$, 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;A: Point;LeftVertex(C) = A;F: Point;RightFocus(C) = F;G: Circle;Center(G) = F;B: Point;Quadrant(B) = 1;TangentPoint(G, OneOf(Asymptote(C))) = B;Slope(LineOf(A, B)) = 1/2", "query_expressions": "Eccentricity(C)", "answer_expressions": "5/3", "fact_spans": "[[[2, 63], [90, 96], [145, 151]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[76, 79], [81, 84]], [[2, 79]], [[88, 89]], [[80, 89]], [[113, 116]], [[105, 116]], [[88, 116]], [[119, 143]]]", "query_spans": "[[[145, 157]]]", "process": "F(c,0), A(-a,0), according to the problem set B(x,\\frac{b}{a}x), then \\frac{b}{x-c}\\times\\frac{b}{a}=-1, solving gives x=\\frac{a^{2}}{c}, that is, B(\\frac{a^{2}}{c},\\frac{ab}{c}). Therefore, k_{AB}=\\frac{\\frac{ab}{c}}{\\frac{a^{2}}{c}+a}=\\frac{1}{2}, 2b=a+c, 3c^{2}-2ac-5a^{2}=0, 3e^{2}-2e-5=0, solving gives e=\\frac{5}{2} or e=-1 (discarded)" }, { "text": "Given that point $P(2,6)$ lies on the parabola $C$: $y^{2}=2 p x$ ($p>0$), and the focus of parabola $C$ is $F$, then $|P F|$=?", "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(P) = (2, 6);PointOnCurve(P, C);Focus(C) = F", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "13/2", "fact_spans": "[[[11, 37], [41, 47]], [[19, 37]], [[2, 10]], [[51, 54]], [[19, 37]], [[11, 37]], [[2, 10]], [[2, 40]], [[41, 54]]]", "query_spans": "[[[56, 65]]]", "process": "Substituting P(2,6) into the parabola equation gives p=9. Thus, from P(2,6) being a point on the parabola C: y^{2}=2px (p>0), we obtain 6^{2}=4p, so p=9. Then |PF|=2+\\frac{9}{2}=\\frac{13}{2}" }, { "text": "Let the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$ be $F_{1}$ and $F_{2}$, respectively. The line $x=1$ intersects one of the asymptotes of the hyperbola at point $P$. Then the area of $\\triangle P F_{1} F_{2}$ is?", "fact_expressions": "G: Hyperbola;H: Line;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/9 = 1);Expression(H) = (x = 1);LeftFocus(G)=F1;RightFocus(G)=F2;Intersection(H,OneOf(Asymptote(G)))=P", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "3*sqrt(10)", "fact_spans": "[[[1, 29], [62, 65]], [[54, 61]], [[75, 79]], [[38, 45]], [[46, 53]], [[1, 29]], [[54, 61]], [[1, 53]], [[1, 53]], [[54, 79]]]", "query_spans": "[[[81, 111]]]", "process": "Find the asymptotes and foci of the hyperbola, then determine the points where the line x=1 intersects the asymptotes of the hyperbola, and use the triangle area formula to solve. From the hyperbola equation, its asymptotes are: y=\\pm3x, and the foci are F_{1}(-\\sqrt{10},0), F_{2}(\\sqrt{10},0). Thus, the line x=1 intersects the asymptotes of the hyperbola at points (1,3), (1,-3). Without loss of generality, let P(1,3). Then S_{\\trianglePF_{1}F_{2}}=\\frac{1}{2}\\times2\\sqrt{10}\\times3=3\\sqrt{10}." }, { "text": "The line $l$ passes through the focus $F$ of the parabola $y^{2}=2x$, and intersects the parabola at two distinct points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If $x_{1}+x_{2}=3$, then what is the length of the chord $AB$?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;F: Point;x1: Number;x2: Number;y1: Number;y2: Number;Expression(G) = (y^2 = 2*x);Coordinate(A) = (x1, y1) ;Coordinate(B) = (x2, y2);Focus(G) = F;PointOnCurve(F, l);x1 + x2 = 3;Intersection(l, G) = {A, B};Negation(A = B);IsChordOf(LineSegmentOf(A, B), G)", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "4", "fact_spans": "[[[0, 5], [28, 31]], [[6, 20], [33, 36]], [[43, 60]], [[62, 80]], [[23, 26]], [[43, 60]], [[62, 80]], [[43, 60]], [[62, 80]], [[6, 20]], [[43, 60]], [[62, 80]], [[6, 26]], [[0, 26]], [[83, 98]], [[28, 80]], [[38, 80]], [[33, 106]]]", "query_spans": "[[[101, 110]]]", "process": "From the given condition, we have p=1. By the definition of the parabola: AB=AF+BF=x_{1}+\\frac{p}{2}+x_{2}+\\frac{p}{2}=x_{1}+x_{2}+p=3+1=4," }, { "text": "The coordinates of the focus of the parabola $y=\\frac{1}{4} x^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,1)", "fact_spans": "[[[0, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "Test analysis: The parabola equation is x^{2}=4y, thus p=2, \\frac{p}{2}=1, from which the focus coordinates of the parabola can be found. The parabola y-\\frac{1}{4}x^{2} is equivalent to x^{2}=4y, \\therefore p=2, \\frac{p}{2}=1, hence the focus coordinates are (0,1)." }, { "text": "Given a point $P$ on the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{144}=1$ such that the distance from $P$ to point $A(13,0)$ is $20$, what is the distance from point $P$ to point $B(-13,0)$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/25 - y^2/144 = 1);P: Point;PointOnCurve(P, G);A: Point;Coordinate(A) = (13, 0);Distance(P, A) = 20;B: Point;Coordinate(B) = (-13, 0)", "query_expressions": "Distance(P, B)", "answer_expressions": "{30,10}", "fact_spans": "[[[2, 43]], [[2, 43]], [[46, 49], [70, 74]], [[2, 49]], [[50, 60]], [[50, 60]], [[46, 68]], [[75, 86]], [[75, 86]]]", "query_spans": "[[[70, 91]]]", "process": "" }, { "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ is $y=\\sqrt{3} x$, and one of its foci coincides with the focus of the parabola $y^{2}=16 x$, find the equation 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;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);H: Parabola;Expression(H) = (y^2 = 16*x);OneOf(Focus(G)) = Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^/12=1", "fact_spans": "[[[2, 61], [85, 86], [115, 118]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[2, 84]], [[92, 108]], [[92, 108]], [[85, 113]]]", "query_spans": "[[[115, 123]]]", "process": "" }, { "text": "Given the ellipse $C$: $x^{2}+2 y^{2}=4$, a line passing through the point $P(1,1)$ intersects the ellipse $C$ at points $A$ and $B$. If point $P$ is exactly the midpoint of segment $A B$, then the equation of line $A B$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2 + 2*y^2 = 4);P: Point;Coordinate(P) = (1, 1);L: Line;PointOnCurve(P, L) = True;Intersection(L, C) = {A, B};A: Point;B: Point;MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(LineOf(A, B))", "answer_expressions": "x+2y-3=0", "fact_spans": "[[[2, 26], [41, 46]], [[2, 26]], [[28, 37], [59, 63]], [[28, 37]], [[38, 40]], [[27, 40]], [[38, 57]], [[48, 51]], [[52, 55]], [[59, 75]]]", "query_spans": "[[[77, 89]]]", "process": "" }, { "text": "Given that the circle $x^{2}+y^{2}-6 x-7=0$ is tangent to the directrix of the parabola $y=a x^{2}$ ($a>0$), then $a$=?", "fact_expressions": "H: Circle;Expression(H) = (-6*x + x^2 + y^2 - 7 = 0);G: Parabola;Expression(G) = (y = a*x^2);a: Number;a>0;IsTangent(H, Directrix(G))", "query_expressions": "a", "answer_expressions": "1/16", "fact_spans": "[[[2, 24]], [[2, 24]], [[25, 44]], [[25, 44]], [[51, 54]], [[28, 44]], [[2, 49]]]", "query_spans": "[[[51, 56]]]", "process": "" }, { "text": "Given that the focus $F$ of the parabola $y^{2}=2 p x (p>0)$ is exactly the right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, and the line connecting the intersection points of the two curves passes through the point $F$, then the eccentricity of the ellipse is?", "fact_expressions": "E: Parabola;p: Number;C: Ellipse;a: Number;b: Number;F: Point;p>0;Expression(E) = (y^2 = 2*(p*x));Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Focus(E) = F;RightFocus(C) = F;A: Point;B: Point;Intersection(C, E) = {A, B};PointOnCurve(F, LineOf(A, B))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2) - 1", "fact_spans": "[[[2, 25]], [[5, 25]], [[34, 79], [102, 104]], [[36, 79]], [[36, 79]], [[28, 31], [95, 99]], [[5, 25]], [[2, 25]], [[34, 79]], [[2, 31]], [[28, 83]], [[89, 91]], [[89, 91]], [[85, 91]], [[89, 99]]]", "query_spans": "[[[102, 110]]]", "process": "Let the intersection points of two curves be A and B. According to the symmetry of the ellipse and parabola, AB \\bot x-axis. Without loss of generality, let point A lie in the first quadrant. Since A lies on the parabola, A(\\frac{p}{2}, p). Since A lies on the ellipse, A(c, \\frac{b^{2}}{a}). e^{2}-2e-1=0. Solving gives e=1+\\sqrt{2}, e=1-\\sqrt{2} (discarded)." }, { "text": "If the coordinates of two vertices of $\\triangle A B C$ are $A(-4,0)$, $B(4,0)$, and the perimeter of $\\triangle A B C$ is $18$, then the equation of the trajectory of vertex $C$ is?", "fact_expressions": "A: Point;B: Point;C: Point;Coordinate(A) = (-4, 0);Coordinate(B) = (4, 0);Perimeter(TriangleOf(A, B, C)) = 18", "query_expressions": "LocusEquation(C)", "answer_expressions": "{(x^2/25+y^2/9=1)&(Negation(y=0))}", "fact_spans": "[[[25, 34]], [[37, 45]], [[77, 80]], [[25, 34]], [[37, 45]], [[47, 73]]]", "query_spans": "[[[77, 86]]]", "process": "\\because AABC has two vertices with coordinates A(-4,0), B(4,0), and a perimeter of 18 \\therefore AB=8, BC+AC=10 \\because 10>8, \\therefore the sum of distances from point C to two fixed points is constant. \\therefore the locus of point C is an ellipse with foci at A and B \\because 2a=10, 2c=8, \\therefore b=3 \\because the standard equation of the ellipse is \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1 (y\\neq0)" }, { "text": "If the two asymptotes of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ are exactly the two tangent lines to the curve $y=a x^{2}+\\frac{1}{3}$, then the value of $a$ is?", "fact_expressions": "G: Hyperbola;H: Curve;a: Number;Expression(G) = (x^2/9 - y^2/4 = 1);Expression(H) = (y = a*x^2 + 1/3);L1:Line;L2:Line;Asymptote(G)={L1,L2};IsTangent(L1,H);IsTangent(L2,H)", "query_expressions": "a", "answer_expressions": "", "fact_spans": "[[[1, 39]], [[48, 73]], [[80, 83]], [[1, 39]], [[48, 73]], [], [], [[1, 45]], [[1, 78]], [[1, 78]]]", "query_spans": "[[[80, 87]]]", "process": "Since y = 2ax, and the two asymptotes of the hyperbola \\frac{x^{2}}{9} - \\frac{y^{2}}{4} = 1 are y = \\pm\\frac{2}{3}x, and when x = 0, y = ax^{2} + \\frac{1}{3} = \\frac{1}{3} > 0, therefore a > 0, 2a = \\frac{2}{3} \\Rightarrow a = \\frac{1}{3}." }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$, a line passing through its focus with slope $-1$ intersects the parabola at points $A$ and $B$. If the horizontal coordinate of the midpoint of segment $AB$ is $3$, then what is the equation of the directrix of this parabola?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Line;PointOnCurve(Focus(G), H);Slope(H) = -1;A: Point;B: Point;Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 3", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-1", "fact_spans": "[[[2, 23], [25, 26], [40, 43], [76, 79]], [[2, 23]], [[5, 23]], [[5, 23]], [[37, 39]], [[24, 39]], [[29, 39]], [[44, 47]], [[48, 51]], [[37, 53]], [[55, 73]]]", "query_spans": "[[[76, 86]]]", "process": "Given that the line passes through the focus and has a slope of -1, the equation is y = -x + \\frac{p}{2}. Substituting into the parabola's equation and transforming it into a quadratic equation in terms of x, the value of p can be found using the relationship between roots and coefficients and the midpoint coordinate formula, which leads to the directrix equation of the parabola. Let the line equation be y - 0 = -1 \\cdot (x - \\frac{p}{2}), yielding y = -x + \\frac{p}{2}. Substituting into the parabola equation gives (\\frac{p}{2} - x)^{2} = 2px, simplifying to x^{2} - 3px + \\frac{p^{2}}{4} = 0. Then x_{1} + x_{2} = 3p = 2 \\times 3, \\therefore p = 2, the directrix equation is x = -\\frac{p}{2} = -1" }, { "text": "If the distance from point $A(x_{0},-2)$ on the parabola $y^{2}=2 p x(p>0)$ to its focus is 3 times the distance from point $A$ to the $y$-axis, then what is the value of $p$?", "fact_expressions": "G: Parabola;p: Number;A: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (x0, -2);x0:Number;PointOnCurve(A,G);Distance(A,Focus(G))=3*Distance(A,yAxis)", "query_expressions": "p", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[1, 22], [39, 40]], [[64, 67]], [[24, 38], [46, 50]], [[4, 22]], [[1, 22]], [[24, 38]], [[25, 38]], [[1, 38]], [[24, 62]]]", "query_spans": "[[[64, 70]]]", "process": "The parabola y^{2}=2px (p>0) opens to the right, with directrix x=-\\frac{p}{2}. Substituting the coordinates of point A into the parabola equation gives 4=2px_{0}, so x_{0}=\\frac{2}{p}. Since the distance from point A(x_{0},-2) on the parabola y^{2}=2px (p>0) to its focus is three times the distance from point A to the y-axis, by the definition of a parabola we have x_{0}+\\frac{p}{2}=3x_{0}. Therefore, \\frac{2}{p}+\\frac{p}{2}=3\\times\\frac{2}{p}, \\frac{p}{2}=\\frac{4}{p}, p^{2}=8, p=2\\sqrt{2}." }, { "text": "Given that the equation $\\frac{x^{2}}{2-k}+\\frac{y^{2}}{k-1}=1$ represents a hyperbola, what is the range of real values for $k$?", "fact_expressions": "G: Hyperbola;k: Real;Expression(G)=(x^2/(2 - k) + y^2/(k - 1) = 1)", "query_expressions": "Range(k)", "answer_expressions": "{(-oo,1),(2,+oo)}", "fact_spans": "[[[45, 48]], [[50, 55]], [[2, 48]]]", "query_spans": "[[[50, 62]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, $F_{1}$, $F_{2}$ are the two foci of the ellipse, point $P$ lies on the ellipse, and $|P F_{2}|=3$, then the area of $\\triangle P F_{1} F_{2}$ is?", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/16 + y^2/12 = 1);Focus(C) = {F1,F2};PointOnCurve(P, C);Abs(LineSegmentOf(P, F2)) = 3", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "6", "fact_spans": "[[[2, 46], [68, 70], [80, 82]], [[75, 79]], [[48, 56]], [[58, 65]], [[2, 46]], [[49, 74]], [[75, 83]], [[85, 98]]]", "query_spans": "[[[100, 130]]]", "process": "\\because|PF|+|PF_{2}|=2a=8\\therefore|PF_{1}|=8-3=5,|F_{1}F_{2}|=2c=4\\therefore PF_{2}\\bot F_{1}F_{2}\\therefore S=\\frac{1}{2}|PF_{2}||PF_{1}|=\\frac{1}{2}\\times3\\times4=6" }, { "text": "A line $ l $ passing through the focus $ F $ of the parabola $ C: y^2 = 4x $ intersects the parabola $ C $ at points $ P $ and $ Q $, and intersects its directrix at point $ M $, with $ \\overrightarrow{F M} = 3 \\overrightarrow{F P} $. Then $ |\\overrightarrow{F P}| = $?", "fact_expressions": "l: Line;C: Parabola;F: Point;M: Point;P: Point;Q: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {P, Q};Intersection(l, Directrix(C)) = M;VectorOf(F, M) = 3*VectorOf(F, P)", "query_expressions": "Abs(VectorOf(F, P))", "answer_expressions": "4/3", "fact_spans": "[[[27, 32]], [[1, 20], [33, 39], [52, 53]], [[23, 26]], [[57, 61]], [[41, 44]], [[45, 48]], [[1, 20]], [[1, 26]], [[0, 32]], [[27, 50]], [[27, 61]], [[63, 108]]]", "query_spans": "[[[110, 136]]]", "process": "" }, { "text": "Given that the vertex of a parabola is at the origin, the axis of symmetry is the $x$-axis, and the distance from point $M(-3 , m)$ on the parabola to the focus is equal to $5$, then what is the equation of the directrix of the parabola?", "fact_expressions": "G: Parabola;M: Point;O: Origin;m:Number;Coordinate(M) = (-3, m);Vertex(G) = O;SymmetryAxis(G)=xAxis;PointOnCurve(M, G);Distance(M, Focus(G)) = 5", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=2", "fact_spans": "[[[2, 5], [23, 26], [53, 56]], [[28, 40]], [[9, 13]], [[29, 40]], [[28, 40]], [[2, 13]], [[2, 22]], [[23, 40]], [[23, 51]]]", "query_spans": "[[[53, 63]]]", "process": "Let the equation of the parabola be \\( y^{2} = -2px \\) (\\( p > 0 \\)), then the focus is \\( F\\left(-\\frac{p}{2}, 0\\right) \\). From the given conditions, we have \n\\[\n\\begin{cases}\nm^{2} \\\\\n\\sqrt{m^{2} + (3}\n\\end{cases}\n\\]\nSolving gives \n\\[\n\\begin{cases}\np = 4, \\\\\nm = 2\\sqrt{6}\n\\end{cases}\n\\quad \\text{or} \\quad\n\\begin{cases}\n\\frac{\\sqrt{3}}{2}, \\\\\nm = -2\\sqrt{6}\n\\end{cases}\n\\]\nThus, the required parabola equation is \\( y^{2} = -8x \\). Hence, the directrix equation is \\( x = 2b \\) woman safety is." }, { "text": "The equation of the ellipse that shares foci with the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ and passes through the point $(1, \\frac{\\sqrt{3}}{2})$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;I: Point;Expression(G) = (x^2/2 - y^2 = 1);Coordinate(I) = (1, sqrt(3)/2);PointOnCurve(I, H);Focus(G)=Focus(H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2/4+y^2=1", "fact_spans": "[[[1, 29]], [[62, 64]], [[35, 61]], [[1, 29]], [[35, 61]], [[34, 64]], [[0, 64]]]", "query_spans": "[[[62, 68]]]", "process": "From the given condition: ∵\\frac{x^{2}}{2}-y^{2}=1, so its foci are (\\pm\\sqrt{3},0), and the foci lie on the x-axis. Since the ellipse shares the same foci with the hyperbola, assume the equation of the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1. ∴a^{2}-b^{2}=3. Also, the ellipse passes through the point (1,\\frac{\\sqrt{3}}{2}), so \\frac{1}{a^{2}}+\\frac{(\\frac{\\sqrt{3}}{2})^{2}}{b^{2}}=1. Solving gives a=2, b=1, ∴\\frac{x^{2}}{4}+y^{2}=1. The equation of the ellipse sharing foci with the hyperbola \\frac{x^{2}}{2}-y^{2}=1 and passing through the point (1,\\frac{\\sqrt{3}}{2}) is \\frac{x^{2}}{4}+y^{2}=1." }, { "text": "Given point $P(5,0)$, if there exist two moving points $M$, $N$ on the right branch of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$ such that $\\overrightarrow{P M} \\perp \\overrightarrow{P N}$, then the minimum value of $\\overrightarrow{M P} \\cdot \\overrightarrow{M N}$ is?", "fact_expressions": "P: Point;Coordinate(P) = (5, 0);C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);M: Point;N: Point;PointOnCurve(M,RightPart(C)) ;PointOnCurve(N,RightPart(C)) ;IsPerpendicular(VectorOf(P, M), VectorOf(P, N))", "query_expressions": "Min(DotProduct(VectorOf(M, P), VectorOf(M, N)))", "answer_expressions": "63/4", "fact_spans": "[[[2, 11]], [[2, 11]], [[13, 46]], [[13, 46]], [[55, 58]], [[59, 62]], [[13, 62]], [[13, 62]], [[65, 114]]]", "query_spans": "[[[116, 171]]]", "process": "Let M(x_{1},y_{1}) with x_{1}\\geqslant1, then x_{1}^{2}-\\frac{y_{1}^{2}}{3}=1, that is, y_{1}^{2}=3(x_{1}^{2}-1). Since \\overrightarrow{PM}\\bot\\overrightarrow{PN}, it follows that \\overrightarrow{PM}\\cdot\\overrightarrow{PN}=0. Then \\overrightarrow{MP}\\cdot\\overrightarrow{MN}=\\overrightarrow{MP}\\cdot(\\overrightarrow{MP}+\\overrightarrow{PN})=\\overrightarrow{MP}^{2}+\\overrightarrow{MP}\\cdot\\overrightarrow{PN}=\\overrightarrow{MP}^{2}=(x_{1}-5)^{2}+y_{1}^{2}=x_{1}^{2}-10x_{1}+25+3(x_{1}^{2}-1)=4(x_{1}-\\frac{5}{4})^{2}+\\frac{63}{4}. Since x_{1}\\geqslant1, we have 4(x_{1}-\\frac{5}{4})^{2}+\\frac{63}{4}\\geqslant\\frac{63}{4}, and equality holds if and only if x_{1}=\\frac{5}{4}. Thus, the minimum value of \\overrightarrow{MP}\\cdot\\overrightarrow{MN} is \\frac{63}{4}." }, { "text": "Let the hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$ have left and right foci $F_{1}$ and $F_{2}$, respectively. Let $P$ be a point on the right branch of the hyperbola $C$. If $|P F_{2}|=5$, then the area of $\\Delta P F_{1} F_{2}$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;PointOnCurve(P, RightPart(C)) = True;Abs(LineSegmentOf(P, F2)) = 5", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "4*sqrt(6)", "fact_spans": "[[[1, 34], [63, 69]], [[1, 34]], [[43, 50]], [[51, 58]], [[1, 58]], [[1, 58]], [[59, 62]], [[59, 74]], [[76, 89]]]", "query_spans": "[[[91, 118]]]", "process": "Hyperbola C: x^{2}-\\frac{y^{2}}{3}=1, then |PF_{1}|-|PF_{2}|=2a=2, |PF_{2}|=5, hence |PF_{1}|=7, |F_{1}F_{2}|=2c=2\\sqrt{1+3}=4. According to the law of cosines: \\cos\\angle PF_{2}F_{1}=\\frac{5^{2}+4^{2}-7^{2}}{2\\times5\\times4}=-\\frac{1}{5}, hence \\sin\\angle PF_{2}F_{1}=\\frac{2\\sqrt{6}}{5}. Then S_{\\Delta PF_{1}F_{2}}=\\frac{1}{2}|F_{1}F_{2}|\\cdot|F_{2}P|\\sin\\angle PF_{2}F_{1}=4\\sqrt{6}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, the left vertex is $M$, the right focus is $F$. A line $l$ passing through point $F$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$, satisfying $M A \\perp M B$. Then the eccentricity of this hyperbola is?", "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;M: Point;A: Point;B: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = M;RightFocus(G) = F;PointOnCurve(F, l);IsPerpendicular(xAxis, l);Intersection(l, G) = {A,B};IsPerpendicular(LineSegmentOf(M, A), LineSegmentOf(M, B))", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[90, 95]], [[2, 59], [96, 99], [132, 135]], [[5, 59]], [[5, 59]], [[64, 67]], [[101, 104]], [[105, 108]], [[72, 75], [77, 81]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 67]], [[2, 75]], [[76, 95]], [[82, 95]], [[90, 110]], [[114, 129]]]", "query_spans": "[[[132, 141]]]", "process": "From MA⊥MB and symmetry, we know that △MAB is an isosceles right triangle, thus |MF| = |FA|. Based on this, we can set up an equation to find the eccentricity. [Detailed solution] From the given conditions, AF = MF, and AF = \\frac{b^{2}}{a}, MF = a + c, then \\frac{b^{2}}{a} = a + c, that is, b^{2} = a^{2} + ac = c^{2} - a^{2}, hence e^{2} - e\\cdot2 = 0 (e > 1), solving gives e = 2." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $e=2$, and one of its foci coincides with the focus of the parabola $y^{2}=8x$, then the equation of the hyperbola is?", "fact_expressions": "G:Hyperbola;a:Number;b:Number;a>0;b>0;C:Parabola;e:Number;Expression(G)=(x^2/a^2-y^2/b^2=1);Expression(C)=(y^2=8*x);Eccentricity(G)=e;e=2;OneOf(Focus(G))=Focus(C)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[2, 58], [68, 69], [96, 99]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[75, 89]], [[61, 66]], [[2, 58]], [[75, 89]], [[2, 66]], [[61, 66]], [[68, 94]]]", "query_spans": "[[[96, 104]]]", "process": "From the equation of the parabola, its focus is obtained; from the given conditions, the values of a, b, and c for the hyperbola are found, thus deriving the equation of the hyperbola. Let the semi-focal length of the hyperbola be c (c>0). Since the focus of the parabola y^{2}=8x is (2,0), we have c=2. Given that the eccentricity of the hyperbola is e=2, it follows that a=1. Using c^{2}=a^{2}+b^{2}, we obtain b=\\sqrt{3}. Therefore, the equation of the hyperbola is x^{2}-\\frac{y^{2}}{3}=1." }, { "text": "Given the parabola $C_{1}$: $y^{2}=8x$, the circle $C_{2}$: $x^{2}+y^{2}-4x+3=0$, and the point $M(1,1)$. If points $A$ and $B$ are moving points on $C_{1}$ and $C_{2}$ respectively, then the minimum value of $|AM|+|AB|$ is?", "fact_expressions": "C1: Parabola;C2: Circle;M: Point;A: Point;B: Point;Coordinate(M) = (1, 1);Expression(C1) = (y^2 = 8*x);Expression(C2) = (x^2+y^2-4*x+3=0);PointOnCurve(A, C1);PointOnCurve(B, C2)", "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(A, M)))", "answer_expressions": "2", "fact_spans": "[[[2, 24], [77, 84]], [[25, 55], [85, 92]], [[56, 65]], [[67, 70]], [[71, 74]], [[56, 65]], [[2, 24]], [[25, 55]], [[67, 96]], [[67, 96]]]", "query_spans": "[[[98, 117]]]", "process": "" }, { "text": "Given points $M(-3 , 0)$, $N(3 , 0)$, $B(1 , 0)$, $\\odot O$ is tangent to $MN$ at point $B$, two lines passing through $M$, $N$ respectively and tangent to $\\odot O$ intersect at point $P$, then the equation of the trajectory of point $P$ is?", "fact_expressions": "M: Point;Coordinate(M) = (-3, 0);N: Point;Coordinate(N) = (3, 0);B: Point;Coordinate(B) = (1, 0);O: Circle;TangentPoint(O, LineSegmentOf(M, N)) = B;L1: Line;L2: Line;TangentOfPoint(M, O) = L1;TangentOfPoint(N, O) = L2;P: Point;Intersection(L1, L2) = P", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x^2-y^2/8=1)&(x>1)", "fact_spans": "[[[2, 14], [63, 66]], [[2, 14]], [[16, 26], [68, 71]], [[16, 26]], [[28, 38], [57, 61]], [[28, 38]], [[40, 49], [72, 81]], [[40, 61]], [], [], [[62, 87]], [[62, 87]], [[90, 94], [96, 100]], [[62, 94]]]", "query_spans": "[[[96, 107]]]", "process": "" }, { "text": "The line with slope $k$ passing through the left vertex $A$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the ellipse at another point $B$, and the projection of point $B$ onto the $x$-axis is exactly the right focus $F$. If $\\frac{1}{3} b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;PointOnCurve(A, G);Slope(G) = k;Intersection(G, C) = {A, B};Projection(B, xAxis) = F;RightFocus(C) = F;1/3 < k;k < 1/2", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(1/2, 2/3)", "fact_spans": "[[[1, 58], [76, 78], [141, 143]], [[7, 58]], [[7, 58]], [[73, 75]], [[62, 65]], [[82, 86], [88, 92]], [[107, 110]], [[69, 72]], [[7, 58]], [[7, 58]], [[1, 58]], [[1, 65]], [[0, 75]], [[66, 75]], [[1, 86]], [[88, 110]], [[76, 110]], [[112, 139]], [[112, 139]]]", "query_spans": "[[[141, 153]]]", "process": "" }, { "text": "The standard equation of an ellipse with foci at $(-4,0)$ and $(4,0)$, and for which the sum of the distances from any point $P$ on the ellipse to the two foci is equal to $10$, is?", "fact_expressions": "G: Ellipse;P: Point;PointOnCurve(P, G);F1:Point;F2:Point;Focus(G)={F1,F2};Coordinate(F1)=(-4,0);Coordinate(F2)=(4,0);Distance(P, F1)+Distance(P, F2)=10", "query_expressions": "Expression(G)", "answer_expressions": "x^2/25+y^2/9=1", "fact_spans": "[[[28, 30], [52, 54]], [[33, 36]], [[28, 36]], [], [], [[0, 30]], [[0, 30]], [[0, 30]], [[28, 51]]]", "query_spans": "[[[52, 60]]]", "process": "\\because the coordinates of the two foci are (-4,0) and (4,0), \\therefore the foci of the ellipse lie on the horizontal axis, and c=4. \\therefore from the definition of the ellipse we have: 2a=10, that is, a=5, \\therefore using the relationship among a, b, c: a^2=b^{2}+c^2, we solve to get b=3, \\therefore the equation of the ellipse is \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=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, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "Problem Analysis: According to the parabola equation, find p, then use the properties of a parabola to determine the coordinates of its focus. In the parabola equation, p = 2, ∴ the coordinates of the focus of the parabola are (-1, 0). Hence, fill in (-1, 0)." }, { "text": "Given point $P(2 , 1)$, if a chord $AB$ of the parabola $y^{2}=4x$ has $P$ as its midpoint, then what is the equation of the line on which chord $AB$ lies?", "fact_expressions": "P: Point;Coordinate(P) = (2, 1);G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A, B)))", "answer_expressions": "", "fact_spans": "[[[2, 13], [42, 45]], [[2, 13]], [[15, 29]], [[15, 29]], [[33, 38]], [[33, 38]], [[15, 38]], [[33, 48]]]", "query_spans": "[[[51, 64]]]", "process": "" }, { "text": "Given that the line $y=(a+1) x-1$ and the curve $y^{2}=ax$ have exactly one common point, find the value of the real number $a$.", "fact_expressions": "G: Line;Expression(G) = (y = x*(a + 1) - 1);H: Curve;Expression(H) = (y^2 = a*x);a: Real;NumIntersection(G, H)=1", "query_expressions": "a", "answer_expressions": "{0,-1,-4/5}", "fact_spans": "[[[2, 17]], [[2, 17]], [[18, 30]], [[18, 30]], [[39, 44]], [[2, 37]]]", "query_spans": "[[[39, 48]]]", "process": "" }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b}=1$ $(b>0)$ are given by $y=\\pm x$, then what is the value of $b$?", "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b = 1);Expression(Asymptote(G)) = (y = pm*x)", "query_expressions": "b", "answer_expressions": "14", "fact_spans": "[[[1, 44]], [[62, 65]], [[4, 44]], [[1, 44]], [[1, 60]]]", "query_spans": "[[[62, 68]]]", "process": "" }, { "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an eccentricity of $2$, and its foci coincide with the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, what is the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Eccentricity(G) = 2;H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);Focus(G) = Focus(H)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(3)*x", "fact_spans": "[[[2, 48], [106, 109]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 56]], [[60, 98]], [[60, 98]], [[2, 103]]]", "query_spans": "[[[106, 116]]]", "process": "" }, { "text": "Given fixed points $A$ and $B$, with $|A B|=4$, and a moving point $P$ satisfying $|P A|-|P B|=3$, then the minimum value of $|P A|$ is?", "fact_expressions": "A: Point;B: Point;P: Point;Abs(LineSegmentOf(A, B)) = 4;Abs(LineSegmentOf(P, A)) - Abs(LineSegmentOf(P, B)) = 3", "query_expressions": "Min(Abs(LineSegmentOf(P, A)))", "answer_expressions": "7/2", "fact_spans": "[[[4, 7]], [[8, 11]], [[25, 28]], [[13, 22]], [[30, 45]]]", "query_spans": "[[[47, 60]]]", "process": "" }, { "text": "What is the distance from point $A(1,2)$ on the parabola $y=2 x^{2}$ to the focus $F$?", "fact_expressions": "G: Parabola;A: Point;Expression(G) = (y = 2*x^2);Coordinate(A) = (1, 2);PointOnCurve(A, G);Focus(G) = F;F: Point", "query_expressions": "Distance(A, F)", "answer_expressions": "17/8", "fact_spans": "[[[0, 14]], [[16, 25]], [[0, 14]], [[16, 25]], [[0, 25]], [[0, 31]], [[28, 31]]]", "query_spans": "[[[16, 36]]]", "process": "\\because the parabola is y=2x^{2}, \\therefore the standard equation is: x^{2}=\\frac{1}{2}y. The directrix equation is: y=-\\frac{1}{8}. The distance from point A(1,2) to the focus F is the distance from A to the directrix: 2+\\frac{1}{8}=\\frac{17}{8}" }, { "text": "Given that the semi-focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $c$, a line passing through the right focus with slope $1$ intersects the right branch of the hyperbola at two points. If the length of the chord intercepted by the hyperbola on the directrix of the parabola $y^{2}=4 c x$ is $\\frac{2 \\sqrt{2}}{3} b e^{2}$ ($e$ being the eccentricity of the hyperbola), then the value of $e$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;HalfFocalLength(G) = c;c: Number;PointOnCurve(RightFocus(G), L);Slope(L) = 1;L: Line;NumIntersection(L, RightPart(G)) = 2;H: Parabola;Expression(H) = (y^2 = 4*c*x);Length(InterceptChord(Directrix(H), G)) = (2*sqrt(2)/3)*b*e^2;e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[2, 58], [82, 85], [114, 117], [158, 161]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 66]], [[63, 66]], [[2, 81]], [[72, 81]], [[79, 81]], [[79, 92]], [[94, 110]], [[94, 110]], [[94, 153]], [[154, 157], [168, 171]], [[154, 165]]]", "query_spans": "[[[168, 175]]]", "process": "By the given conditions, the directrix of the parabola is x = -c, which passes exactly through the left focus of the hyperbola. Therefore, the chord length intercepted by the hyperbola on the directrix is $\\frac{2b^{2}}{a}$, so $\\frac{2b^{2}}{a} = \\frac{2\\sqrt{2}}{3}be^{2}$, that is, $\\frac{b}{a} = \\frac{\\sqrt{2}}{3}e^{2}$. Hence, $e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^{2}} = \\sqrt{1 + \\left(\\frac{\\sqrt{2}e^{2}}{3}\\right)^{2}}$. Simplifying yields $2e^{4} - 9e^{2} + 1 = 0$, solving gives $e = \\frac{\\sqrt{6}}{2}$ or $e = \\sqrt{3}$. Since the line passing through the focus with slope 1 intersects the right branch of the hyperbola, $e = \\frac{\\sqrt{6}}{2}$." }, { "text": "If a hyperbola has the same foci as the ellipse $\\frac{x^{2}}{27}+\\frac{y^{2}}{36}=1$ and passes through the point $(\\sqrt{15}, 4)$, then the equation of the hyperbola is?", "fact_expressions": "C: Hyperbola;F:Ellipse;P:Point;Expression(F) = (x^2/27 + y^2/36 = 1);Coordinate(P) = (sqrt(15), 4);Focus(C) = Focus(F);PointOnCurve(P,C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/4-x^2/5=1", "fact_spans": "[[[2, 5], [75, 78]], [[6, 45]], [[54, 72]], [[6, 45]], [[54, 72]], [[2, 50]], [[2, 72]]]", "query_spans": "[[[75, 83]]]", "process": "" }, { "text": "Through the point $M(1,1)$, draw a line intersecting the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ at points $A$ and $B$. If point $M$ is exactly the midpoint of chord $AB$, then the equation of the line $AB$ is?", "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;M: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Coordinate(M) = (1, 1);Intersection(G, H) = {A,B};IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A, B)) = M;PointOnCurve(M,H)", "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A,B)))", "answer_expressions": "4*x+9*y-13=0", "fact_spans": "[[[15, 52]], [[12, 14]], [[55, 58]], [[60, 63]], [[1, 10], [67, 71]], [[15, 52]], [[1, 10]], [[12, 65]], [[15, 80]], [[67, 83]], [0, 13]]", "query_spans": "[[[85, 99]]]", "process": "" }, { "text": "Let $F$ be the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$. The line $l$ passing through $F$ with slope $\\frac{a}{b}$ intersects the two asymptotes of the hyperbola $C$ at points $A$ and $B$, respectively, and $\\overrightarrow{A F}=2 \\overrightarrow{F B}$. Find the eccentricity of the hyperbola $C$.", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F: Point;B: Point;a>0;b>0;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(L1,l)=A;Intersection(L2,l)=B;VectorOf(A, F) = 2*VectorOf(F, B)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[96, 101]], [[5, 69], [102, 108], [176, 182]], [[12, 69]], [[12, 69]], [[118, 121]], [[1, 4], [75, 78]], [[122, 125]], [[12, 69]], [[12, 69]], [[5, 69]], [[1, 73]], [[74, 101]], [[79, 101]], [], [], [[102, 114]], [[96, 127]], [[96, 127]], [[129, 174]]]", "query_spans": "[[[176, 188]]]", "process": "\\because the slope of asymptote OB is -\\frac{b}{a}. \\therefore l\\bot OB, and \\overrightarrow{AF}=2\\overrightarrow{FB}. In Rt\\triangle OBA, by the angle bisector theorem we have \\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}" }, { "text": "Given that a line with slope $2$ passes through the right focus $F_{2}$ of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ and intersects the ellipse at points $A$ and $B$, what is the length of chord $AB$?", "fact_expressions": "L: Line;Slope(L) = 2;G: Ellipse;Expression(G) = (x^2/5 + y^2/4 = 1);RightFocus(G) = F2;F2: Point;PointOnCurve(F2, L) = True;Intersection(L, G) = {A, B};A: Point;B: Point;IsChordOf(LineSegmentOf(A,B), G) = True", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "5*sqrt(5)/3", "fact_spans": "[[[9, 11]], [[2, 11]], [[13, 50], [63, 65]], [[13, 50]], [[13, 61]], [[54, 61]], [[9, 61]], [[9, 77]], [[68, 71]], [[72, 75]], [[63, 85]]]", "query_spans": "[[[80, 89]]]", "process": "From the given conditions, the right focus $ F_{2} $ is at $ (1,0) $, so the equation of line $ AB $ is $ y = 2(x - 1) $. Solving the system of the line and the ellipse equation yields $ 3x^{2} - 5x = 0 $, then $ x_{1} = \\frac{5}{3}, x_{2} = 0 $. Substituting into $ |AB| = \\sqrt{1 + k^{2}} |x_{1} - x_{2}| $ gives the solution. According to the problem, the coordinates of the right focus $ F_{2} $ of the ellipse are $ (1,0) $. The equation of line $ AB $ is $ y = 2(x - 1) $. Solving the system \n\\[\n\\begin{cases}\ny = 2(x - 1) \\\\\n\\frac{x^{2}}{6} + \\frac{y^{2}}{4} = 1\n\\end{cases}\n\\]\neliminating $ y $, we get $ 3x^{2} - 5x = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} = \\frac{5}{3}, x_{2} = 0 $, so $ |AB| = \\sqrt{1 + k^{2}} |x_{1} - x_{2}| = \\sqrt{1 + 2^{2}} \\times \\frac{5}{3} = \\frac{5\\sqrt{5}}{3} $." }, { "text": "The line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Then the minimum value of $|AB|$ 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}", "query_expressions": "Min(Abs(LineSegmentOf(A, B)))", "answer_expressions": "4", "fact_spans": "[[[2, 16], [25, 28]], [[2, 16]], [[18, 21]], [[2, 21]], [[22, 24]], [[0, 24]], [[29, 32]], [[33, 36]], [[22, 38]]]", "query_spans": "[[[40, 53]]]", "process": "The focus of the parabola has coordinates (1,0). When the slope of the line does not exist, let x=1, we get: y=\\pm2, so |AB|=4. When the slope of the line exists, let the equation of the line be y=k(x-1). Solving simultaneously \n\\begin{cases}y=k(x-1)\\\\y^2=4x\\end{cases} \nwe obtain: k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0, k\\neq0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=\\frac{2k^{2}+4}{k^{2}}=2+\\frac{4}{k^{2}}, |AB|=x_{1}+x_{2}+p=2+\\frac{4}{k^{2}}+2=4+\\frac{4}{k^{2}}>4. Therefore, the minimum value of |AB| is 4. Hence the answer is: 4" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$. Draw a perpendicular line from point $M(-1,4)$ to the $y$-axis, intersecting the parabola $C$ at point $A$, such that $|A F|=|A M|$. Let the line $A F$ intersect the parabola $C$ at another point $B$. Then the ordinate of point $B$ is?", "fact_expressions": "C: Parabola;p: Number;Z: Line;A: Point;F: Point;M: Point;B: Point;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(M) = (-1, 4);Focus(C) = F;PointOnCurve(M, Z);IsPerpendicular(Z, yAxis);Intersection(Z, C) = A;Abs(LineSegmentOf(A, F)) = Abs(LineSegmentOf(A, M));Intersection(LineOf(A, F), C) = {A, B}", "query_expressions": "YCoordinate(B)", "answer_expressions": "-1", "fact_spans": "[[[2, 28], [56, 62], [94, 100]], [[10, 28]], [], [[63, 67]], [[32, 35]], [[37, 47]], [[104, 107], [109, 113]], [[10, 28]], [[2, 28]], [[37, 47]], [[2, 35]], [[36, 55]], [[36, 55]], [[36, 67]], [[71, 84]], [[63, 107]]]", "query_spans": "[[[109, 119]]]", "process": "From |AF| = |AM|, it follows that point M lies on the directrix, so -\\frac{p}{2} = -1, and thus the equation of the parabola C is y^{2} = 4x. Therefore, A(4,4), and the equation of line AB is y = \\frac{4}{3}x - \\frac{4}{3}. Then, solving the system of equations formed by the line and the parabola gives the y-coordinate of point B. From the given conditions, since |AF| = |AM|, point M lies on the directrix. Since the equation of the directrix is x = -\\frac{p}{2}, we have -\\frac{p}{2} = -1, so p = 2, and thus the equation of parabola C is y^{2} = 4x. Since the coordinates of point M are (-1,4), A(4,4). Hence, the equation of line AB is y = \\frac{4}{3}x - \\frac{4}{3}. Solving the system \\begin{cases} y = \\frac{4}{3}x - \\frac{4}{3}, \\\\ y^{2} = 4x \\end{cases} yields y^{2} - 3y - 4 = 0, whose solutions are y = 4 (discarded) or y = -1. Therefore, the y-coordinate of point B is -1." }, { "text": "Let point $P$ be a moving point on the ellipse $C$. Two tangents are drawn from point $P$ to a circle having the minor axis of the ellipse as its diameter, with points of tangency $M$ and $N$. If $\\angle M P N = 60^{\\circ}$, then the range of values for the eccentricity of ellipse $C$ is?", "fact_expressions": "C: Ellipse;G: Circle;M: Point;P: Point;N: Point;L1:Line;L2:Line;PointOnCurve(P, C);IsDiameter(MinorAxis(C),G);TangentOfPoint(P,G)={L1,L2};TangentPoint(L1,G)=M;TangentPoint(L2,G)=N;AngleOf(M, P, N) = ApplyUnit(60, degree)", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[sqrt(3)/2,1)", "fact_spans": "[[[5, 10], [22, 24], [78, 83]], [[30, 31]], [[42, 45]], [[0, 4], [16, 20]], [[46, 49]], [], [], [[0, 14]], [[21, 31]], [[15, 36]], [[15, 49]], [[15, 49]], [[51, 76]]]", "query_spans": "[[[78, 94]]]", "process": "According to the problem, find the relationship among a, b, c, and determine the range of the eccentricity. Let the center of the ellipse be O. Since \\angle MPN = 60^{\\circ}, it follows that \\angle POM = 60^{\\circ}, so \\frac{|OP|}{|OM|} = 2, hence |OP| = 2b. The farthest distance from a point on the ellipse to the origin is at the endpoint of the major axis, so a \\geqslant 2b, i.e., \\frac{b}{a} \\leqslant \\frac{1}{2}. Therefore, \\frac{b^{2}}{a^{2}} \\leqslant \\frac{1}{4}, thus \\frac{a^{2}-c^{2}}{a^{2}} \\leqslant \\frac{1}{4}. Therefore, the eccentricity e = \\sqrt{1-\\frac{b^{2}}{a^{2}}} \\geqslant \\sqrt{1-(\\frac{1}{2})^{2}} = \\frac{\\sqrt{3}}{2}, so e \\in [\\frac{\\sqrt{3}}{2}," }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the left branch of the hyperbola $C$, and $|A F_{1}|=12$. Find $|A F_{2}|$.", "fact_expressions": "C: Hyperbola;A: Point;F1: Point;F2: Point;Expression(C) = (x^2/16 - y^2/4 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(A, LeftPart(C));Abs(LineSegmentOf(A, F1)) = 12", "query_expressions": "Abs(LineSegmentOf(A, F2))", "answer_expressions": "20", "fact_spans": "[[[2, 46], [76, 82]], [[71, 75]], [[55, 62]], [[63, 70]], [[2, 46]], [[2, 70]], [[2, 70]], [[71, 86]], [[88, 102]]]", "query_spans": "[[[104, 117]]]", "process": "Using the definition of a hyperbola, |AF₂| can be found. Since A lies on the left branch of the hyperbola C, by the definition of a hyperbola, |AF₂| - |AF₁| = 2√16 = 8. Therefore, |AF₂| = |AF₁| + 8 = 20" }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{k+8}+\\frac{y^{2}}{9}=1$ is $e=\\frac{1}{2}$. Then the value of $k$ is?", "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/(k + 8) + y^2/9 = 1);Eccentricity(G) = e ;e:Number;e= 1/2", "query_expressions": "k", "answer_expressions": "{4,-5/4}", "fact_spans": "[[[0, 39]], [[61, 64]], [[0, 39]], [[0, 58]], [[43, 58]], [[43, 58]]]", "query_spans": "[[[61, 68]]]", "process": "" }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F(-1,0)$, and the line $l$: $y=-x+m$ intersects the parabola at two distinct points $A$, $B$. If $0 \\leq m<1$, then the maximum area of $\\Delta F A B$ is?", "fact_expressions": "l: Line;G: Parabola;p: Number;F: Point;A: Point;B: Point;m:Number;p>0;Expression(G) = (y^2 = 2*p*x);Expression(l)=(y=-x+m);m>=0;m<1;Coordinate(F) = (-1, 0);Focus(G) = F;Intersection(l,G) = {A, B};Negation(A = B)", "query_expressions": "Max(Area(TriangleOf(F, A, B)))", "answer_expressions": "8*sqrt(6)/9", "fact_spans": "[[[37, 52]], [[2, 23], [53, 56]], [[5, 23]], [[27, 36]], [[63, 66]], [[67, 70]], [[72, 84]], [[5, 23]], [[2, 23]], [[37, 52]], [[72, 84]], [[72, 84]], [[27, 36]], [[2, 36]], [[37, 70]], [[58, 70]]]", "query_spans": "[[[86, 109]]]", "process": "\\because the parabola y^{2}=2px (p>0) has focus F(-1,0), \\therefore the equation of the parabola is y^{2}=-4x. \nSolving the parabola and the line equation together yields x^{2}+(4-2m)x+m^{2}=0. \nSince there are two intersection points, \\triangle=(4-2m)^{2}-4m^{2}>0, \nthat is, m<1. Thus 0\\leqslant m<1 always holds. \nLet points A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=2m-4, x_{1}x_{2}=m^{2}. \n|AB|=\\sqrt{2}|x_{1}-x_{2}|=4\\sqrt{2}\\sqrt{1-m}. \n\\because the distance from F(-1,0) to the line y=-x+m is d=\\frac{|1+m|}{\\sqrt{2}}, \n\\therefore S_{\\Delta ABF}=\\frac{1}{2}d|AB|=2\\sqrt{(1-m)(1+m)^{2}}=\\sqrt{2}\\sqrt{(2-2m)(1+m)(1+m)} \n\\leqslant \\sqrt{2}\\sqrt{\\left(\\frac{2-2m+1+m+1+m}{3}\\right)^{3}}=\\frac{8\\sqrt{6}}{9}. \nEquality holds if and only if 2-2m=1+m, that is, m=\\frac{1}{3}, so the maximum area of \\triangle ABF is \\frac{8\\sqrt{6}}{9}." }, { "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, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "" }, { "text": "The equation of a parabola with vertex at the origin, focus on the coordinate axis, and passing through the point $(1,2)$ is?", "fact_expressions": "O: Origin;Vertex(G) = O;G: Parabola;PointOnCurve(Focus(G), axis) = True;H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, G) = True", "query_expressions": "Expression(G)", "answer_expressions": "{(y^2=4*x), (x^2=y/2)}", "fact_spans": "[[[1, 5]], [[0, 31]], [[28, 31]], [[9, 31]], [[19, 27]], [[19, 27]], [[17, 31]]]", "query_spans": "[[[28, 36]]]", "process": "Let the equation of the parabola be y^{2}=mx or x^{2}=ny, then 4=m or 1=2n, n=\\frac{1}{2}, so the equation of the parabola is y^{2}=8x or x^{2}=\\frac{1}{2}y" }, { "text": "The ellipse $\\frac{x^{2}}{4}+y^{2}=1$ has two foci $F_{1}$, $F_{2}$. A line passing through point $F_{1}$ and perpendicular to the $x$-axis intersects the ellipse, with one of the intersection points being $P$. Then $|P F_{2}|=$?", "fact_expressions": "G: Ellipse;H: Line;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/4 + y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(F1, H);IsPerpendicular(H, xAxis);IsIntersect(H, G);OneOf(Intersection(H, G)) = P", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "7/2", "fact_spans": "[[[0, 27], [69, 71]], [[66, 68]], [[81, 84]], [[40, 47]], [[32, 39], [49, 57]], [[0, 27]], [[0, 47]], [[48, 68]], [[58, 68]], [[66, 73]], [[66, 84]]]", "query_spans": "[[[86, 99]]]", "process": "By the given condition, the left focus of the ellipse is $ F_{1}(-\\sqrt{3},0) $, $ a=2 $, then the perpendicular segment $ |PF_{1}| = \\sqrt{1 - \\frac{3}{4}} = \\frac{1}{2} $, so $ |PF_{2}| = 2a - |PF_{1}| = 4 - \\frac{1}{2} = \\frac{7}{2} $." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$. A line passing through $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at point $P$. Then $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ = ?", "fact_expressions": "G: Hyperbola;H: Line;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);IsPerpendicular(xAxis, H);Intersection(H, G) = P", "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))", "answer_expressions": "4", "fact_spans": "[[[17, 45], [72, 75]], [[69, 71]], [[76, 80]], [[1, 8]], [[9, 16], [53, 60]], [[17, 45]], [[1, 51]], [[1, 51]], [[52, 71]], [[61, 71]], [[69, 80]]]", "query_spans": "[[[82, 141]]]", "process": "" }, { "text": "Given that the distance from the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ to one of its asymptotes is $2a$, what is the value of its eccentricity?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Distance(RightFocus(G), OneOf(Asymptote(G))) = 2*a", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 58], [79, 80]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 77]]]", "query_spans": "[[[79, 87]]]", "process": "The right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ is $(c,0)$, and one of its asymptotes is given by $bx-ay=0$. The distance from the focus to the asymptote is $d=\\frac{|bc|}{\\sqrt{2+b^{2}}}=b=2a$, so $\\frac{b}{a}=2$. Then $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{1+4}=\\sqrt{5}$. The answer is: $\\sqrt{5}$" }, { "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$ to its asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = 1)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 50]]]", "process": "Test analysis: From the equation, we know a^{2}=4, b^{2}=3, \\therefore c^{2}=7, \\therefore c=\\sqrt{7}, \\therefore the foci are (+\\sqrt{7},0), and the distance to the asymptotes y=\\pm\\frac{\\sqrt{3}}{2}x is \\sqrt{3}." }, { "text": "Given two points $A$ and $B$ on the parabola $y^{2}=4x$ with focus $F$ such that $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then the distance from the midpoint of $AB$ to the $y$-axis is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;B: Point;PointOnCurve(A,G) = True;PointOnCurve(B,G) = True;VectorOf(A,F)= 3*VectorOf(F,B)", "query_expressions": "Distance(MidPoint( LineSegmentOf(A,B)),yAxis)", "answer_expressions": "5/3", "fact_spans": "[[[10, 24]], [[10, 24]], [[3, 6]], [[3, 24]], [[28, 31]], [[32, 35]], [[10, 35]], [[10, 35]], [[37, 82]]]", "query_spans": "[[[84, 102]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) be two points A, B satisfying \\overrightarrow{AF}=3\\overrightarrow{FB}. y_{1}, \\overrightarrow{FB}=(x_{2}-1,y_{2}), that is, \\begin{cases}y_{1}=4x_{1}\\\\y_{2}=4x_{2}\\\\y_{1}=-3x_{2}\\end{cases} Solving gives: \\begin{cases}-y_{1}=3y_{2}\\\\x_{2}=\\frac{1}{3}\\end{cases} Hence \\begin{cases}x_{1}=3\\\\y_{1}=12\\end{cases} The distance from the midpoint of B to the y-axis is \\frac{x_{1}+x_{2}}{2}=\\frac{5}{3}. (Note) This problem examines coordinate operations of vectors, the definition of parabola, and other knowledge. When solving, set variables, list equations, find variable values, and solve according to the graph." }, { "text": "A point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has distance $2$ to the left focus $F_{1}$. $N$ is the midpoint of $M F_{1}$, and $O$ is the coordinate origin. Then $|O N|=$?", "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(M,G);LeftFocus(G)=F1;Distance(M,F1)=2;MidPoint(LineSegmentOf(M,F1))=N", "query_expressions": "Abs(LineSegmentOf(O, N))", "answer_expressions": "4", "fact_spans": "[[[0, 38]], [[41, 44]], [[48, 55]], [[80, 83]], [[63, 66]], [[0, 38]], [[0, 44]], [[0, 55]], [[41, 62]], [[63, 79]]]", "query_spans": "[[[90, 99]]]", "process": "" }, { "text": "The maximum distance from points on the curve $\\frac{x|x|}{2}-\\frac{y|y|}{4}=1$ to the line $y=\\sqrt{2} x$ is?", "fact_expressions": "G: Curve;Expression(G) = ((-y*Abs(y))/4 + (x*Abs(x))/2 = 1);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (y = sqrt(2)*x)", "query_expressions": "Max(Distance(P, H))", "answer_expressions": "(2*sqrt(6))/3", "fact_spans": "[[[0, 35]], [[0, 35]], [[37, 38]], [[0, 38]], [[39, 55]], [[39, 55]]]", "query_spans": "[[[37, 64]]]", "process": "The curve $\\frac{x|x|}{2}-\\frac{y|y|}{4}=1$ is equivalent to the following equations:\n$$\n\\begin{matrix}\n\\frac{x^{2}}{2}-\\frac{y^{2}}{4}=1 & (x\\geqslant0,y\\geqslant0) \\\\\n\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1 & (x\\geqslant0,y<0)\n\\end{matrix}\n\\quad \\text{and the graph also includes:} \\quad\n\\frac{y^{2}}{4}-\\frac{x^{2}}{2}=1 \\quad (x<0,y<0)\n$$\nThus, $y=\\sqrt{2}x$ is the asymptote of the hyperbolas $\\frac{x^{2}}{2}-\\frac{y^{2}}{4}=1$ $(x\\geqslant0,y\\geqslant0)$ and $\\frac{y^{2}}{4}-\\frac{x^{2}}{2}=1$ $(x<0,y<0)$. Therefore, the maximum distance from points on the curve $\\frac{x|x|}{2}-\\frac{y|y|}{4}=1$ to the line $y=\\sqrt{2}x$ equals the maximum distance from points on the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$ $(x\\geqslant0,y<0)$ to the line $y=\\sqrt{2}x$. Let the line $y=\\sqrt{2}x+m$ $(m<0)$ be tangent to the curve $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$ $(x\\geqslant0,y<0)$. Solving the system and simplifying yields $4x^{2}+2\\sqrt{2}mx+m^{2}-4=0$. Setting discriminant $\\Delta = 8m^{2}-16(m^{2}-4)=0$, we solve and get $m=-2\\sqrt{2}$. Thus, the tangent line is $y=\\sqrt{2}x-2\\sqrt{2}$. The distance between the two parallel lines $y=\\sqrt{2}x-2\\sqrt{2}$ and $y=\\sqrt{2}x$ is $d=\\frac{|0+2\\sqrt{2}|}{\\sqrt{3}}=\\frac{2\\sqrt{6}}{3}$. Therefore, the maximum distance from points on the curve $\\frac{x|x|}{2}-\\frac{y|y|}{4}=1$ to the line $y=\\sqrt{2}x$ is $\\frac{2\\sqrt{6}}{3}$." }, { "text": "Given that the slope of an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\sqrt{3}$, 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);Slope(OneOf(Asymptote(G)))=sqrt(3)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[2, 58], [81, 84]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 78]]]", "query_spans": "[[[81, 90]]]", "process": "Given the slope of the asymptote $ k = \\frac{b}{a} = \\sqrt{3} $, then the eccentricity $ e = \\frac{c}{a} = \\sqrt{1 + 3} = 2 $." }, { "text": "Given a moving point $M$ such that the product of the slopes of the lines connecting $M$ to two fixed points $A(1,0)$ and $B(-1,0)$ is a constant $m$ $(m \\neq 0)$, if the locus of point $M$ is an ellipse with foci on the $x$-axis (excluding points $A$ and $B$), then what is the range of values for $m$?", "fact_expressions": "M:Point;A:Point;B:Point;G:Ellipse;m:Number;Negation(m=0);Coordinate(A)=(1,0);Coordinate(B)=(-1,0);Slope(LineOf(M,A))*Slope(LineOf(M,B))=m;Locus(M)=G;PointOnCurve(Focus(G),xAxis);Negation(PointOnCurve(A,G));Negation(PointOnCurve(B,G))", "query_expressions": "Range(m)", "answer_expressions": "(-1, 0)", "fact_spans": "[[[4, 7], [59, 63]], [[13, 21], [81, 86]], [[24, 33], [87, 91]], [[76, 78]], [[94, 97], [44, 57]], [[44, 57]], [[13, 21]], [[24, 33]], [[4, 57]], [[59, 78]], [[67, 78]], [[76, 92]], [[76, 92]]]", "query_spans": "[[[94, 104]]]", "process": "Simplify to obtain \\( k_{AM} \\cdot k_{BM} = \\frac{y^{2}}{x^{2}-1} = m \\), that is, \\( x^{2} - \\frac{y^{2}}{m} = 1 \\). According to the condition \\( 0 < -m < 1 \\), solving gives the answer representing an ellipse with foci on the x-axis (excluding points A and B), hence \\( 0 < -m < 1 \\), so \\( m \\in (-1, 0) \\)." }, { "text": "Let $O$ be the origin, and let a moving point $M$ lie on the circle $C$: $x^{2}+y^{2}=4$. Draw a perpendicular from $M$ to the $x$-axis, with foot $N$. Point $P$ satisfies $\\overrightarrow{N P}=\\overrightarrow{PM}$. What is the equation of the trajectory of point $P$?", "fact_expressions": "C: Circle;N: Point;P: Point;M: Point;O: Origin;l: Line;Expression(C) = (x^2 + y^2 = 4);PointOnCurve(M, C);PointOnCurve(M, l);IsPerpendicular(l, xAxis);FootPoint(l, xAxis) = N;VectorOf(N, P) = VectorOf(P, M)", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/4 + y^2 = 1", "fact_spans": "[[[16, 37]], [[55, 58]], [[109, 113], [59, 63]], [[12, 15], [40, 43]], [[1, 4]], [], [[16, 37]], [[12, 38]], [[39, 51]], [[39, 51]], [[39, 58]], [[65, 107]]]", "query_spans": "[[[109, 120]]]", "process": "Let M(x_{0},y_{0}), then by the given condition we have N(x_{0},0). Let P(x,y). Using coordinate operations of vectors, and combining with the fact that M satisfies the ellipse equation, we can simplify and obtain the trajectory equation of P. [Detailed solution] Let M(x_{0},y_{0}), then by the given condition we have N(x_{0},0). Let P(x,y). Since point P satisfies \\overrightarrow{NP}=\\overrightarrow{PM}, it follows that P is the midpoint of MN, so we get x=x_{0}, y=\\frac{1}{2}y_{0}. Thus x_{0}=x, y_{0}=2y. Substituting into the circle C: x^{2}+y^{2}=4, we obtain x^{2}+4y^{2}=4, i.e., \\frac{x^{2}}{4}+y^{2}=1." }, { "text": "Given that $c$ is the semi-focal length of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, $(a>0, b>0)$, and the eccentricity is $e$, then the maximum value of $\\frac{1}{e}+\\frac{b}{c}$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;c: Number;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);HalfFocalLength(C) = c;Eccentricity(C) = e", "query_expressions": "Max(b/c + 1/e)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[6, 67]], [[13, 67]], [[13, 67]], [[2, 5]], [[76, 79]], [[13, 67]], [[13, 67]], [[6, 67]], [[2, 71]], [[6, 79]]]", "query_spans": "[[[81, 112]]]", "process": "From the given condition, $ c = \\sqrt{a^{2} + b^{2}} $, $ \\therefore \\frac{1}{e} + \\frac{b}{c} = \\frac{a + b}{c} = \\frac{a + b}{\\sqrt{a^{2} + b}} - \\frac{1}{2} $ holds. $ \\therefore $ the maximum value of $ \\frac{1}{e} + \\frac{b}{c} $ is $ \\sqrt{2} $." }, { "text": "Given that point $P$ moves on the parabola $x^{2}=4 y$, $F$ is the focus of the parabola, and the coordinates of point $A$ are $(2,3)$, find the coordinates of point $P$ when $P A+P F$ is minimized.", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (2, 3);WhenMin(LineSegmentOf(P, A) + LineSegmentOf(P, F))", "query_expressions": "Coordinate(P)", "answer_expressions": "(2, 1)", "fact_spans": "[[[7, 21], [29, 32]], [[7, 21]], [[2, 6], [68, 72]], [[2, 24]], [[25, 28]], [[25, 35]], [[36, 40]], [[36, 51]], [[53, 68]]]", "query_spans": "[[[68, 77]]]", "process": "" }, { "text": "Given that $M$ is a point on the hyperbola $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, and $|M F_{1}|=17$, then $|M F_{2}|=$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/64 - y^2/36 = 1);M: Point;PointOnCurve(M, G) = True;LeftFocus(G) = F1;RightFocus(G) = F2;F1: Point;F2: Point;Abs(LineSegmentOf(M, F1)) = 17", "query_expressions": "Abs(LineSegmentOf(M, F2))", "answer_expressions": "33", "fact_spans": "[[[6, 46], [66, 69]], [[6, 46]], [[2, 5]], [[2, 49]], [[50, 75]], [[50, 75]], [[50, 57]], [[58, 65]], [[77, 91]]]", "query_spans": "[[[93, 106]]]", "process": "From the hyperbola equation $\\frac{x^{2}}{64}-\\frac{y^{2}}{36}=1$, we have $a=8$, $b=6$, $c=10$. From the graph of the hyperbola, the distance $d$ from point $M$ to the right focus $F_{2}$ satisfies $d \\geqslant c-a = 2$. Since $|MF_{1}| - |MF_{2}| = 2a = 16$ and $|MF_{1}| = 17$, it follows that $|MF_{2}| = 1$ (discarded) or $|MF_{2}| = 33$." }, { "text": "Given that the eccentricity of ellipse $C$ is $e=\\frac{1}{3}$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and $P$ is a moving point on ellipse $C$, then the range of values of $\\frac{|P F_{1}|}{|P F_{2}|}$ is?", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;e:Number;Eccentricity(C) = e ;e= 1/3;LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P,C)", "query_expressions": "Range(Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2)))", "answer_expressions": "[1/2,2]", "fact_spans": "[[[2, 7], [54, 59]], [[50, 53]], [[34, 41]], [[42, 49]], [[11, 26]], [[2, 26]], [[11, 26]], [[2, 49]], [[2, 49]], [[50, 63]]]", "query_spans": "[[[65, 101]]]", "process": "Let P(x₀,y₀), \\frac{|PF₁|}{|PF₂|} = \\frac{a+ex₀}{a−ex₀} = \\frac{a+\\frac{1}{3}x₀}{a−\\frac{1}{3}x₀} = -1 + \\frac{2a}{a−\\frac{1}{3}x₀}, and −a \\leqslant x₀ \\leqslant a, then: \\frac{|PF₁|}{|PF₂|} \\in \\left[\\frac{1}{2}, 2\\right]" }, { "text": "Point $P$ lies on the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, 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 point $P$ is in the first quadrant, what is its $y$-coordinate?", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/16 + y^2/7 = 1);PointOnCurve(P, C);Focus(C) = {F1, F2};Radius(InscribedCircle(TriangleOf(P,F1,F2)))=1;Quadrant(P)=1", "query_expressions": "YCoordinate(P)", "answer_expressions": "7/3", "fact_spans": "[[[5, 48], [69, 71]], [[113, 117], [0, 4], [124, 125]], [[53, 60]], [[61, 68]], [[5, 48]], [[0, 52]], [[53, 76]], [[78, 110]], [[113, 122]]]", "query_spans": "[[[124, 131]]]", "process": "The problem of the focal triangle of an ellipse, making full use of the definition of the ellipse, express S_{\\trianglePF_{1}F_{2}} from two perspectives, and establish an equation in terms of y_{p} to solve. Since |PF_{1}|+|PF_{2}|=8, |F_{1}F_{2}|=6, it follows that S_{\\trianglePFF}F_{2}=\\frac{1}{2}(|PF_{1}|+|PF_{2}|+|F_{1}F_{2}|)\\times1=7; also because S_{\\DeltaPF_{1}F_{2}}=\\frac{1}{2}|F_{1}F_{2}|\\cdoty_{p}=3y_{p}=7, so y_{p}=\\frac{7}{3}" }, { "text": "Given that point $F$ is the focus of the parabola $C$: $x^{2}=2 y$. Two mutually perpendicular lines $l_{1}$ and $l_{2}$ are drawn through point $F$. Line $l_{1}$ intersects $C$ at points $A$ and $B$, and line $l_{2}$ intersects $C$ at points $D$ and $E$. Then the minimum value of $|A B|+4|D E|$ is?", "fact_expressions": "C: Parabola;l1: Line;l2: Line;A: Point;B: Point;D: Point;E: Point;F: Point;Expression(C) = (x^2 = 2*y);Focus(C) = F;PointOnCurve(F,l1);PointOnCurve(F,l2);IsPerpendicular(l1,l2);Intersection(l1, C) = {A, B};Intersection(l2, C) = {D,E}", "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + 4*Abs(LineSegmentOf(D, E)))", "answer_expressions": "18", "fact_spans": "[[[7, 26], [73, 76]], [[43, 53], [63, 72]], [[55, 62], [88, 97]], [[78, 81]], [[82, 85]], [[103, 106]], [[107, 110]], [[2, 6], [31, 35]], [[7, 26]], [[2, 29]], [[30, 62]], [[30, 62]], [[36, 62]], [[63, 87]], [[88, 112]]]", "query_spans": "[[[114, 134]]]", "process": "Let the equation of line $ l_{1} $ be $ y = kx + \\frac{1}{2} $. By solving the system of equations, we obtain $ |AB| = 2k^{2} + 2 $ and $ |DE| = \\frac{2}{k^{2}} + 2 $. Combining with the basic inequality, we can find the minimum value of $ |AB| + 4|DE| $, leading to the answer. According to the problem, the parabola $ C: x^{2} = 2y $ has focus $ F(0, \\frac{1}{2}) $ and directrix $ y = -\\frac{1}{2} $. Let the equation of line $ l_{1} $ be $ y = kx + \\frac{1}{2} $, $ k \\neq 0 $. Solving the system\n$$\n\\begin{cases}\nx^{2} = 2y \\\\\ny = kx + \\frac{1}{2}\n\\end{cases}\n$$\nwe get $ x^{2} - 2kx - 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Then $ x_{1} + x_{2} = 2k $, $ y_{1} + y_{2} = kx_{1} + \\frac{1}{2} + kx_{2} + \\frac{1}{2} = k(x_{1} + x_{2}) + 1 = 2k^{2} + 1 $. By the definition of the parabola, $ |AB| = y_{1} + y_{2} + 1 = 2k^{2} + 2 $. Since $ l_{1} \\perp l_{2} $, replacing $ k $ by $ -\\frac{1}{k} $ in the above expression gives $ |DE| = \\frac{2}{k^{2}} + 2 $. Then $ |AB| + 4|DE| = 10 + 2(k^{2} + \\frac{4}{k^{2}}) \\geqslant 10 + 2 \\times 2\\sqrt{k^{2} \\cdot \\frac{4}{k^{2}}} = 18 $. The equality holds if and only if $ k = \\pm \\sqrt{2} $. Therefore, the minimum value of $ |AB| + 4|DE| $ is 18." }, { "text": "The left focus of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is $F_{1}$, $P$ is a moving point on the ellipse, and $M$ is a moving point on the circle $(x+1)^{2}+(y-3)^{2}=1$. Then the maximum value of $|P M|+|P F_{1}|$ is?", "fact_expressions": "G: Ellipse;H: Circle;P: Point;M: Point;F1: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H) = ((x + 1)^2 + (y - 3)^2 = 1);LeftFocus(G)=F1;PointOnCurve(P,G);PointOnCurve(M, H)", "query_expressions": "Max(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "16", "fact_spans": "[[[0, 39], [56, 58]], [[67, 91]], [[52, 55]], [[63, 66]], [[44, 51]], [[0, 39]], [[67, 91]], [[0, 51]], [[52, 62]], [[63, 95]]]", "query_spans": "[[[97, 120]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$. It is known that point $P$ lies on this hyperbola and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. If the eccentricity of this hyperbola equals $\\frac{\\sqrt{6}}{2}$, then the distance from point $P$ to the $y$-axis equals?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/2 + x^2/a^2 = 1);a: Number;F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = pi/3;Eccentricity(G) = sqrt(6)/2", "query_expressions": "Distance(P, yAxis)", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[17, 59], [73, 76], [118, 121]], [[17, 59]], [[20, 59]], [[1, 8]], [[9, 16]], [[1, 64]], [[67, 71], [149, 153]], [[67, 77]], [[79, 115]], [[118, 147]]]", "query_spans": "[[[149, 164]]]", "process": "According to the problem, from \\begin{matrix}b=\\sqrt{2}\\\\c&\\sqrt{6}\\\\a&\\frac{\\sqrt{2}}{2}\\\\c^{2}=a^{2}+b^{2}\\end{matrix}, solving gives a=2, c=\\sqrt{6}. Using the area formula for a hyperbola focal triangle, S=b^{2}\\cot\\frac{\\angle F_{1}PF_{2}}{2}=\\frac{2}{\\tan\\frac{\\pi}{6}}=\\frac{1}{2}\\cdot2\\sqrt{6}\\cdot, solving gives |y|=\\sqrt{2}, substituting into the hyperbola equation yields |x|=2\\sqrt{2}" }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$, and its directrix intersects the $x$-axis at point $C$. If a circle centered at $F$ with radius $p$ intersects the parabola at points $A$ and $B$, then $\\sin \\angle A C F$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;C: Point;Intersection(Directrix(G), xAxis) = C;H: Circle;Center(H) = F;Radius(H) = p;A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Sin(AngleOf(A, C, F))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[2, 23], [65, 68]], [[2, 23]], [[56, 59]], [[5, 23]], [[27, 30], [49, 52]], [[2, 30]], [[41, 45]], [[2, 45]], [[63, 64]], [[48, 64]], [[56, 64]], [[71, 75]], [[76, 79]], [[63, 79]]]", "query_spans": "[[[81, 102]]]", "process": "" }, { "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A variable chord $AB$ passes through the left focus $F_{1}$. If $|\\overrightarrow {F_{2} A}-\\overrightarrow{F_{2} B}| \\geq|\\overrightarrow {F_{2} A}+\\overrightarrow{F_{2} B}|$ always holds, then the range of the eccentricity of the ellipse $E$ is?", "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;F2: Point;F1: Point;LeftFocus(E) = F1;RightFocus(E) = F2;A: Point;B: Point;IsChordOf(LineSegmentOf(A,B),E);PointOnCurve(F1,LineSegmentOf(A,B));Abs(VectorOf(F2,A)-VectorOf(F2,B)) >= Abs(VectorOf(F2,A)+VectorOf(F2,B))", "query_expressions": "Range(Eccentricity(E))", "answer_expressions": "(0,sqrt(2)-1]", "fact_spans": "[[[2, 52], [213, 218]], [[2, 52]], [[9, 52]], [[9, 52]], [[68, 75]], [[60, 67], [87, 94]], [[2, 75]], [[2, 75]], [[78, 83]], [[78, 83]], [[2, 83]], [[78, 94]], [[97, 208]]]", "query_spans": "[[[213, 229]]]", "process": "From the condition we get $\\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}\\leqslant0$, which translates to $\\frac{2b^{2}}{a}\\geqslant4c'$, thereby obtaining the range of eccentricity of ellipse $E$. From $|\\overrightarrow{F_{2}A}-\\overrightarrow{F_{2}B}|\\geqslant|\\overrightarrow{F_{2}A}+\\overrightarrow{F_{2}B}|$ we obtain $\\overrightarrow{F_{2}A}\\cdot\\overrightarrow{F_{2}B}\\leqslant0$, $\\frac{2b^{2}}{a}\\geqslant4c'$, that is, $2ac\\leqslanta^{2}-c^{2}$, $\\therefore e^{2}+2e-1\\leqslant0$, $\\therefore 00)$, a line $l$ passing through point $M(1,0)$ with positive slope intersects the parabola $E$ at points $A$ and $B$. If $\\overrightarrow{A M}=3 \\overrightarrow{M B}$, then what is the inclination angle of line $l$?", "fact_expressions": "P: Point;Coordinate(P) = (1, 2);E: Parabola;Expression(E) = (y^2 = 2*p*x);p: Number;p>0;PointOnCurve(P, E);M: Point;Coordinate(M) = (1, 0);PointOnCurve(M, l);Slope(l) > 0;l: Line;Intersection(l, E) = {A, B};A: Point;B: Point;VectorOf(A, M) = 3*VectorOf(M, B)", "query_expressions": "Inclination(l)", "answer_expressions": "pi/3", "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 38], [63, 69]], [[12, 38]], [[20, 38]], [[20, 38]], [[2, 39]], [[41, 50]], [[41, 50]], [[40, 62]], [[51, 62]], [[57, 62], [128, 133]], [[57, 79]], [[70, 73]], [[74, 77]], [[81, 126]]]", "query_spans": "[[[128, 139]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, let $A$ and $B$ be the left vertex and the top vertex of the ellipse, respectively, $F$ be the right focus, and $AB \\perp BF$. Find the eccentricity of the ellipse.", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;F: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G)=A;UpperVertex(G)=B;RightFocus(G) = F;IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(B, F));a > b;b > 0", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)-1)/2", "fact_spans": "[[[2, 54], [68, 70], [105, 107]], [[4, 54]], [[4, 54]], [[56, 59]], [[61, 65]], [[79, 82]], [[2, 54]], [[56, 78]], [[56, 78]], [[68, 86]], [[88, 103]], [[4, 54]], [[4, 54]]]", "query_spans": "[[[105, 113]]]", "process": "Given that A(-a,0), B(0,b), F(c,0), and since AB ⊥ BF, we have k_{AB} · k_{BF} = -1, so \\frac{b}{a} · (-\\frac{b}{c}) = -1, which implies b^{2} = ac. Thus a^{2} - c^{2} = ac. Dividing both sides by a^{2} gives 1 - \\frac{c^{2}}{a^{2}} = \\frac{c}{a}, or e^{2} + e - 1 = 0. Solving yields e = \\frac{\\sqrt{5}-1}{2}. Therefore, the eccentricity of the ellipse is \\underline{\\sqrt{5}-1}." }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, $O$ is the origin, and $M$ is a point on the parabola such that $|MF|=3$, then the area of $\\Delta OMF$ is?", "fact_expressions": "G: Parabola;O: Origin;M: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(M, G);Abs(LineSegmentOf(M, F)) = 3", "query_expressions": "Area(TriangleOf(O, M, F))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 16], [37, 40]], [[24, 27]], [[33, 36]], [[20, 23]], [[2, 16]], [[2, 23]], [[33, 43]], [[44, 53]]]", "query_spans": "[[[55, 74]]]", "process": "Let M(m, n). From the parabola equation y^{2} = 4x, we get: 2p = 4, so p = 2. By the definition of the parabola, |MF| = m + \\frac{p}{2} = 3, solving gives: m = 2. Also, n^{2} = 4m, solving gives: n = \\pm2\\sqrt{2}. Therefore, the area of triangle OMF is: S = \\frac{1}{2} \\times |OF| \\times |n| = \\frac{1}{2} \\times 1 \\times 2\\sqrt{2} = \\sqrt{2}. This problem mainly examines the definition of a parabola, equation thinking, and triangle area calculation, and is a basic problem." }, { "text": "The equation of the hyperbola passing through the point $M(4 , 3)$ with asymptotes given by $y=\\pm 2 x$ is?", "fact_expressions": "G: Hyperbola;M: Point;Coordinate(M) = (4, 3);PointOnCurve(M, G);Expression(Asymptote(G)) = (y = pm*(2*x))", "query_expressions": "Expression(G)", "answer_expressions": "4*x^2 - y^2 = 55", "fact_spans": "[[[32, 35]], [[2, 13]], [[2, 13]], [[0, 35]], [[14, 35]]]", "query_spans": "[[[32, 40]]]", "process": "" }, { "text": "Let the left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ be $F$, and the upper vertex be $B$. Given that the minor axis length of the ellipse is $4$ and the eccentricity is $\\frac{\\sqrt{5}}{5}$, then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;UpperVertex(G) = B;Length(MinorAxis(G)) = 4;Eccentricity(G)=sqrt(5)/5", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5+y^2/4=1", "fact_spans": "[[[1, 53], [72, 74], [109, 111]], [[3, 53]], [[3, 53]], [[58, 61]], [[66, 69]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 61]], [[1, 69]], [[72, 82]], [[72, 107]]]", "query_spans": "[[[109, 116]]]", "process": "From the given conditions, we can derive a system of equations in terms of $a$, $b$, and $c$. Solving for the values of $a$, $b$, and $c$, we can then determine the equation of the ellipse. According to the problem, we have\n\\[\n\\begin{cases}\n2b=4 \\\\\nc=\\frac{\\sqrt{5}}{5} \\\\\nb=\\sqrt{a^2-c^{2}}\n\\end{cases}\n\\]\nSolving this system yields\n\\[\n\\begin{cases}\na=\\sqrt{5} \\\\\nb=2 \\\\\nc=1\n\\end{cases}\n\\]\nTherefore, the equation of the ellipse is $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$." }, { "text": "$A$ is a point on the parabola $y^{2}=2px$ $(p>0)$, $F$ is the focus of the parabola, and $O$ is the origin. When $|AF|=4$, $\\angle OFA=120^{\\circ}$. Then the standard equation of the parabola is?", "fact_expressions": "G: Parabola;p: Number;A: Point;F: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(A, G);Focus(G) = F;Abs(LineSegmentOf(A,F))=4;AngleOf(O, F, A) = ApplyUnit(120, degree)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[4, 24], [32, 35], [85, 88]], [[7, 24]], [[0, 3]], [[28, 31]], [[39, 42]], [[7, 24]], [[4, 24]], [[0, 27]], [[28, 38]], [[49, 57]], [[59, 83]]]", "query_spans": "[[[85, 95]]]", "process": "By the given condition, ∠BFA = ∠OFA - 90° = 30°. Draw a perpendicular line AC from A to the directrix, and draw a perpendicular line from F to AC, with the foot of the perpendiculars being C and B respectively. The distance from point A to the directrix is: d = |AB| + |BC| = p + 2 = 4. Solving gives p = 2, then the standard equation of the parabola is: y² = 4x." }, { "text": "Given that the distance from a point $M(1, m)$ ($m>0$) on the parabola $y^{2}=2 p x$ ($p>0$) to its focus $F$ is $5$, and the distance from the point $F$ to one of the asymptotes of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{b^{2}}=1$ is $2 \\sqrt{2}$, find the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;H: Parabola;p: Number;M: Point;F: Point;m:Number;m>0;Expression(G) = (x^2/2 - y^2/b^2 = 1);p>0;Expression(H) = (y^2 = 2*p*x);Coordinate(M) = (1, m);PointOnCurve(M, H);Focus(H) = F;Distance(M, F) = 5;Distance(F,OneOf(Asymptote(G))) = 2*sqrt(2)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[60, 102], [127, 130]], [[63, 102]], [[2, 23], [41, 42]], [[5, 23]], [[26, 40]], [[55, 59], [44, 47]], [[26, 40]], [[26, 40]], [[60, 102]], [[5, 23]], [[2, 23]], [[26, 40]], [[2, 40]], [[41, 47]], [[26, 54]], [[55, 124]]]", "query_spans": "[[[127, 136]]]", "process": "(Analysis) From the given conditions and using the definition of a parabola, find p, obtain the coordinates of F, then use the point-to-line distance formula to solve for b, further find c, and thus obtain the answer. By the definition of the parabola, the distance from point M(1, m) to the focus equals the distance to the directrix x = -\\frac{p}{2}. That is, 1 + \\frac{p}{2} = 5, solving gives p = 8, so the standard equation of the parabola is y^{2} = 16x, and the focus is F(4, 0). Take one asymptote of the hyperbola \\frac{x^{2}}{2} - \\frac{y^{2}}{b^{2}} = 1 as y = \\frac{b}{\\sqrt{2}}x, or equivalently bx - \\sqrt{2}y = 0. Then we have \\frac{|4b|}{\\sqrt{b^{2} + 2}} = 2\\sqrt{2}, solving gives b = \\sqrt{2}. Therefore, c = \\sqrt{a^{2} + b^{2}} = 2, so e = \\frac{c}{a} = \\sqrt{2}. The answer is:,,," }, { "text": "A focus of the hyperbola $8 kx^{2}-ky^{2}=8$ is $(0 , 3)$. Find the value of $k$ and the equations of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;k: Number;H: Point;Expression(G) = (8*(k*x^2) - k*y^2 = 8);Coordinate(H) = (0, 3);OneOf(Focus(G)) = H", "query_expressions": "k;Expression(Asymptote(G))", "answer_expressions": "-1\ny=pm*2*sqrt(2)*x", "fact_spans": "[[[0, 22], [46, 49]], [[39, 42]], [[28, 37]], [[0, 22]], [[28, 37]], [[0, 37]]]", "query_spans": "[[[39, 46]], [[46, 57]]]", "process": "" }, { "text": "Given that point $P$ is an arbitrary point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, if the product of the distances from point $P$ to the two asymptotes of the hyperbola is equal to $\\frac{b^{2}}{3}$, then the eccentricity of the hyperbola is?", "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);L1: Line;L2: Line;Asymptote(G) = {L1, L2};Distance(P, L1)*Distance(P, L2) = b^2/3", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[7, 63], [76, 79], [110, 113]], [[91, 108]], [[10, 63]], [[2, 6], [71, 75]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 69]], [], [], [[76, 84]], [[71, 108]]]", "query_spans": "[[[110, 119]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the right vertex of $C$ is $A$. A circle $A1$ is drawn with center $A$ and radius $b$. The circle $A1$ intersects one asymptote of the hyperbola $C$ at points $M$ and $N$. If $\\angle M A N=60^{\\circ}$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;A1: Circle;M: Point;A: Point;N: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;Center(A1)=A;Length(Radius(A1))=b;Intersection(A1, OneOf(Asymptote(C))) = {M, N};AngleOf(M, A, N) = ApplyUnit(60, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 63], [100, 106], [152, 155]], [[10, 63]], [[81, 84]], [[94, 99], [88, 93]], [[114, 117]], [[68, 71]], [[118, 121]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[73, 93]], [[81, 93]], [[94, 123]], [[125, 150]]]", "query_spans": "[[[152, 161]]]", "process": "As shown in the figure, from the given conditions we have |OA| = a, |AN| = |AM| = b. Since ∠MAN = 60°, ∴ |AP| = (√3)/2 b, ∴ |OP| = √(|OA|² - |PA|²) = √(a² - (3/4)b²). Let θ be the inclination angle of one asymptote y = (b/a)x of hyperbola C, then tanθ = |AP| / |OP| = ((√3)/2 b) / √(a² - (3/4)b²). ∴ (√3) / (√2 * b/3) = b/a, solving gives a² = 3b², e = √(1 + b²/a²) = √(1 + 1/3) = (2√3)/3. Answer: √(1 + a²) / (√3/3)" }, { "text": "Point $A(x_{0}, y_{0})$ lies on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{32}=1$. If the distance from point $A$ to the right focus is $2 x_{0}$, then $x_{0}=$?", "fact_expressions": "G: Hyperbola;A: Point;x0: Number;y0:Number;Expression(G) = (x^2/4 - y^2/32 = 1);Coordinate(A) = (x0, y0);PointOnCurve(A, RightPart(G));Distance(A,RightFocus(G)) = 2*x0", "query_expressions": "x0", "answer_expressions": "2", "fact_spans": "[[[19, 58]], [[0, 18], [64, 68]], [[87, 94]], [[1, 18]], [[19, 58]], [[0, 18]], [[0, 62]], [[19, 85]]]", "query_spans": "[[[87, 96]]]", "process": "" }, { "text": "If the line $y = kx + 1$ and the parabola $y^2 = 4x$ have exactly one common point, then the value of $k$ is?", "fact_expressions": "H: Line;Expression(H) = (y = k*x + 1);k: Number;G: Parabola;Expression(G) = (y^2 = 4*x);NumIntersection(H, G) = 1", "query_expressions": "k", "answer_expressions": "{0,1}", "fact_spans": "[[[1, 12]], [[1, 12]], [[38, 41]], [[13, 27]], [[13, 27]], [[1, 36]]]", "query_spans": "[[[38, 45]]]", "process": "If the line y = kx + 1 and the parabola y^{2} = 4x have exactly one common point, discuss by cases: when the line is parallel to the x-axis, or when the line is tangent to the parabola, solve for the value of k, thereby obtaining the answer. ① When the line y = kx + 1 is parallel to the x-axis, the equation is y = 1, k = 0, and it has only one common point with the parabola y^{2} = 4x, with coordinates (\\frac{1}{4}, 1). ② When k ≠ 0, combining the equation y = kx + 1 with the parabola equation, eliminating y gives k^{2}x^{2} + (2k - 4)x + 1 = 0, Δ = (2k - 4)^{2} - 4k^{2} = 0, solving yields k = 1, the tangent line equation is y = x + 1. In conclusion, k = 0 or 1." }, { "text": "The parabola $y^{2}=6x$, a chord is drawn through the point $P(4,1)$ such that it is exactly bisected by the point $P$. Then the equation of the line containing this chord is?", "fact_expressions": "G: Parabola;H: LineSegment;P: Point;Expression(G) = (y^2 = 6*x);Coordinate(P) = (4, 1);PointOnCurve(P,H);MidPoint(H)=P;IsChordOf(H,G)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "3*x-y-11=0", "fact_spans": "[[[0, 14]], [[31, 32]], [[16, 25]], [[0, 14]], [[16, 25]], [[0, 29]], [[31, 41]], [[0, 29]]]", "query_spans": "[[[31, 54]]]", "process": "Problem Analysis: Let the two endpoints of the chord passing through point P(4,1) be A(x_{1},y_{1}), B(x_{2},y_{2}), then: \n\\begin{cases}y_{2}=6x_{1}\\\\y_{2}=6x_{2},x_{1}=\\frac{2}{x_{2}}-x_{1}=y_{2}+y,\\end{cases} \nobtain: y_{2}^{2}-y_{1}=6(x_{2}-x_{1}), and since point P(4,1) is exactly the midpoint of segment AB. \n\\therefore\\frac{y_{2}+y_{1}}{2}=1\\Rightarrow y_{2}+y_{1}=2. Hence, the slope of the line containing the chord is k=\\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\\frac{6}{y_{2}+y_{1}}=\\frac{6}{2}=3, so the equation of the line containing the chord is: y-1=3(x-4), i.e., 3x-y-11=0. Therefore, the answer should be filled in as: 3x-y-11=0" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line passing through the right focus $F$ with slope $\\sqrt{3}$ intersects the ellipse $C$ at points $A$ and $B$. If $\\overrightarrow{A F}=\\frac{1}{2} \\overrightarrow{F B}$, then the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C)=F;PointOnCurve(F, G);Slope(G) = sqrt(3);Intersection(G, C) = {A, B};VectorOf(A, F) = VectorOf(F, B)/2", "query_expressions": "Eccentricity(C)", "answer_expressions": "2/3", "fact_spans": "[[[2, 59], [85, 90], [161, 166]], [[9, 59]], [[9, 59]], [[82, 84]], [[93, 96]], [[64, 67]], [[97, 100]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[60, 84]], [[68, 84]], [[82, 102]], [[104, 159]]]", "query_spans": "[[[161, 172]]]", "process": "Combining numbers and shapes, use the second definition of the ellipse to calculate |BE| and |AB|, then compute using \\cos\\angle ABE = \\frac{|BE|}{|AB|}. [Solution] As shown in the figure, extend the line intersecting the right directrix at point C, and draw AE perpendicular to BC at point E. Given \\overrightarrow{AF} = \\frac{1}{2}\\overrightarrow{FB}, let |\\overrightarrow{AF}| = m, |\\overrightarrow{FB}| = 2m, then |AB| = 3m. From |AD| = \\frac{|\\overrightarrow{AF}|}{e} = \\frac{m}{e}, |AD| = \\frac{|\\overrightarrow{FB}|}{e} = \\frac{2m}{e}, so |BE| = |BC| - |AD| = \\frac{m}{e}. Since the slope of line AB is \\sqrt{3}, we have \\angle ABE = \\angle AFx = 60^{\\circ}. Therefore, \\cos\\angle ABE = \\frac{|BE|}{|AB|} = \\frac{1}{3e} = \\frac{1}{2} \\Rightarrow e = \\frac{2}{3}." }, { "text": "The length of the moving chord $AB$ of the parabola $y^{2}=8x$ is $6$. Then, the shortest distance from the midpoint $M$ of chord $AB$ to the $y$-axis is?", "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;Expression(G) = (y^2 = 8*x);Length(LineSegmentOf(A, B)) = 6;MidPoint(LineSegmentOf(A, B)) = M;IsChordOf(LineSegmentOf(A,B),G)", "query_expressions": "Min(Distance(M, yAxis))", "answer_expressions": "9/8", "fact_spans": "[[[0, 14]], [[17, 22]], [[17, 22]], [[38, 41]], [[0, 14]], [[17, 28]], [[31, 41]], [[0, 22]]]", "query_spans": "[[[38, 53]]]", "process": "Problem Analysis: To make the distance from the midpoint M of chord AB to the y-axis shortest, it suffices to have chord AB perpendicular to the x-axis. Then the y-coordinate of point A is 3. Substituting y=3 into y^{2}=8x gives x=\\frac{9}{8}. Therefore, the shortest distance from the midpoint M of chord AB to the y-axis is \\frac{9}{8}." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $O$ the origin. A circle with diameter $O F_{2}$ intersects an asymptote of the hyperbola $C$ at points $O$ and $P$. If $\\angle O P F_{1}=45^{\\circ}$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;O: Origin;F2: Point;P: Point;F1: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(O,F2),G);Intersection(G, OneOf(Asymptote(C))) = {O, P};AngleOf(O, P, F1) = ApplyUnit(45, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[17, 77], [109, 115], [165, 168]], [[24, 77]], [[24, 77]], [[106, 107]], [[83, 86], [123, 126]], [[9, 16]], [[127, 130]], [[1, 8]], [[24, 77]], [[24, 77]], [[17, 77]], [[1, 82]], [[1, 82]], [[92, 107]], [[106, 132]], [[134, 163]]]", "query_spans": "[[[165, 174]]]", "process": "As shown in the figure, the asymptote equation is y=\\frac{b}{a}x, then the distance from focus F_{2}(c,0) to the asymptote is d=\\frac{|bc|}{\\sqrt{1+(\\frac{b}{a})^{2}}}=b, that is, PF_{2}=b, then OP=\\sqrt{F_{2}O^{2}-F_{2}P^{2}}=a. Draw PA\\bot x-axis at point A, then PA=\\frac{OP\\cdot PF_{2}}{OF_{2}}=\\frac{ab}{c}, by the Pythagorean theorem: OA=\\frac{a^{2}}{c}, so P(\\frac{a^{2}}{c},\\frac{ab}{c}), then the equation of line PF is: \\frac{y-0}{x+c}=\\frac{\\frac{ab}{c}-0}{\\frac{a^{2}}{c}+c}, that is, y=\\frac{ab}{a^{2}+c^{2}}(x+c). Draw OB\\bot PF_{1} from point O to point B, then OB=\\frac{abc}{\\sqrt{1+(\\frac{b}{a^{2}+c})^{2}}}, and \\angle OPF_{1}=45^{\\circ}, so OP=\\sqrt{2}OB, that is, \\frac{\\frac{\\sqrt{2}abc}{a^{2}+c^{2}}}{\\sqrt{1+(\\frac{ab}{2})^{2}}}=a, solving gives: \\frac{c}{a}=\\sqrt{5}, thus the eccentricity of C is \\sqrt{5}." }, { "text": "A line with slope $1$ passing through the focus of the parabola $C$: $y^{2}=2 p x$ $(p>0)$ intersects $C$ at points $A$ and $B$. If $|A B|=8$ and $O$ is the origin, then the area of $\\triangle A O B$ is?", "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;O: Origin;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(Focus(C), G);Slope(G) = 1;Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, B)) = 8", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[1, 27], [40, 43]], [[9, 27]], [[37, 39]], [[45, 48]], [[68, 71]], [[49, 52]], [[9, 27]], [[1, 27]], [[0, 39]], [[30, 39]], [[37, 54]], [[56, 65]]]", "query_spans": "[[[78, 100]]]", "process": "From the problem, we know that the equation of line AB is: $ y = x - \\frac{p}{2} $. Substituting into $ y^2 = 2px $ and simplifying, using Vieta's formulas and the definition of the parabola with $ |AB| = 8 $, we can find the value of $ p $, thus obtaining the equation of the parabola. By solving the system of equations of the line and the parabola, and using the area formula, the solution can be obtained. [Detailed Solution] From the problem, the focus is $ F\\left(\\frac{p}{2}, 0\\right) $, so the equation of line AB is: $ y = x - \\frac{p}{2} $. Substituting into $ y^2 = 2px $, 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 $. Since $ |AB| = 8 $, we have $ x_1 + x_2 + p = 8 $, solving gives $ p = 2 $. Therefore, the equation of the parabola is: $ y^2 = 4x $. Thus, the equation of line AB is: $ y = x - 1 $. Solving the system \\begin{cases} y = x - 1 \\\\ y^2 = 4x \\end{cases} yields $ y^2 - 4y - 4 = 0 $. Hence, $ y_1 + y_2 = 4 $, $ y_1 y_2 = -4 $. Therefore, the area $ S $ of $ \\Delta AOB $ is $ \\frac{1}{2} \\times 1 \\times \\sqrt{(y_1 + y_2)^2 - 4y_1 y_2} = 2\\sqrt{2} $." }, { "text": "Given that the line $y=\\frac{b}{3}$ intersects the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $B$ and $C$, and the circle with diameter $BC$ passes exactly through the right focus $F$ of the ellipse, what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;I: Line;B: Point;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(I) = (y = b/3);Intersection(I, G) = {B, C};IsDiameter(LineSegmentOf(B,C),H);RightFocus(G)=F;PointOnCurve(F,H);F:Point", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(14)/4", "fact_spans": "[[[20, 72], [99, 101], [111, 113]], [[22, 72]], [[22, 72]], [[94, 95]], [[2, 19]], [[74, 77]], [[78, 81]], [[22, 72]], [[22, 72]], [[20, 72]], [[2, 19]], [[2, 83]], [[85, 95]], [[99, 108]], [[94, 108]], [[105, 108]]]", "query_spans": "[[[111, 119]]]", "process": "According to the problem, draw a sketch. Given that the line $ y = \\frac{b}{3} $ intersects the ellipse $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $ at points $ B $ and $ C $, and the circle with diameter $ BC $ passes exactly through the right focus $ F $ of the ellipse, it follows that $ \\overrightarrow{FB} \\perp \\overrightarrow{CF} $, from which the answer can be obtained. [Detailed solution] Draw a sketch according to the problem. Since the line $ y = \\frac{b}{3} $ intersects the ellipse $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $ at points $ B $ and $ C $, and the circle with diameter $ BC $ passes exactly through the right focus $ F $, and $ \\overrightarrow{FB} \\perp \\overrightarrow{CF} $. Solve the system of equations formed by $ y = \\frac{b}{3} $ and $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $: \n$$\n\\begin{cases}\ny = \\frac{b}{3} \\\\\n\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1\n\\end{cases}\n$$\nSolving gives: $ x = \\pm \\frac{2\\sqrt{2}}{3}a $ \n$ \\therefore B\\left(-\\frac{2\\sqrt{2}}{3}a, \\frac{b}{3}\\right), C\\left(\\frac{2\\sqrt{2}}{3}a, \\frac{b}{3}\\right) $ \nThen we get $ \\overrightarrow{FB} = \\left(-\\frac{2\\sqrt{2}}{3}a - c, \\frac{b}{3}\\right), \\overrightarrow{FC} = \\left(\\frac{2\\sqrt{2}}{3}a - c, \\frac{b}{3}\\right) $ \nFrom $ \\overrightarrow{FB} \\perp \\overrightarrow{CF} $, we have $ \\overrightarrow{FB} \\cdot \\overrightarrow{CF} = 0 $, \ni.e., $ \\left(-\\frac{2\\sqrt{2}}{3}a - c, \\frac{b}{3}\\right) \\cdot \\left(\\frac{2\\sqrt{2}}{3}a - c, \\frac{b}{3}\\right) = 0 $ \n$ \\Rightarrow -\\frac{8}{9}a^{2} + c^{2} + \\frac{b^{2}}{9} = 0 $, i.e., $ \\frac{8}{9}c^{2} - \\frac{7}{9}a^{2} = 0 $ \n$ \\therefore c^{2} = \\frac{7}{8}a^{2} $, hence $ e = \\frac{c}{a} = \\frac{\\sqrt{14}}{4} $" }, { "text": "Given the line $l$: $\\sqrt{3} x - y - \\sqrt{3} = 0$ intersects the parabola $\\Gamma$: $y^{2} = 4 x$ at points $A$, $B$, and intersects the $x$-axis at $F$. If $O F = \\lambda O A + \\mu O B$ ($\\lambda \\leq \\mu$), then $\\frac{\\lambda}{\\mu} = $?", "fact_expressions": "l: Line;Gamma: Parabola;O: Origin;F: Point;A: Point;B: Point;Expression(Gamma) = (y^2 = 4*x);Expression(l)=(sqrt(3)*x - y - sqrt(3) = 0);Intersection(l,Gamma) = {A, B};Intersection(l,xAxis)=F;LineSegmentOf(O,F)=lambda*LineSegmentOf(O,A)+mu*LineSegmentOf(O,B);lambda<=mu;lambda:Number;mu:Number", "query_expressions": "lambda/mu", "answer_expressions": "1/3", "fact_spans": "[[[2, 34]], [[35, 59]], [[83, 126]], [[78, 81]], [[61, 64]], [[65, 68]], [[35, 59]], [[2, 34]], [[2, 70]], [[2, 81]], [[83, 126]], [[83, 126]], [[127, 150]], [[127, 150]]]", "query_spans": "[[[127, 152]]]", "process": "" }, { "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. Point $P$ lies on the right branch of the hyperbola such that $|P F_{2}|=|F_{1} F_{2}|$, and the equations of the asymptotes of the hyperbola are $3x \\pm 4y=0$. Then $\\cos \\angle P F_{1} F_{2} = ?$", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Expression(Asymptote(G)) = (3*x + pm*(4*y) = 0)", "query_expressions": "Cos(AngleOf(P, F1, F2))", "answer_expressions": "9/10", "fact_spans": "[[[19, 75], [87, 90], [122, 125]], [[19, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[1, 8]], [[9, 16]], [[1, 81]], [[1, 81]], [[82, 86]], [[82, 93]], [[96, 121]], [[122, 147]]]", "query_spans": "[[[149, 178]]]", "process": "Let the semi-focal length of the hyperbola be c. Since the asymptotes of the hyperbola are given by 3x±4y=0, it follows that \\frac{b}{a}=\\frac{3}{4}. Then c=\\sqrt{a^{2}+b^{2}}=\\sqrt{a^{2}+\\frac{9}{16}a^{2}}=\\frac{5}{4}a. In \\triangle PF_{1}F_{2}, |PF_{2}|=|F_{1}F_{2}|=2c, |PF_{1}|=2c+2a. By the law of cosines, we have \\cos\\angle PF_{1}F_{2}=\\frac{(2c)^{2}+(2c+2a)^{2}-(2c)^{2}}{2\\times2c(2c+2a)}=\\frac{a+c}{2c}=\\frac{a+\\frac{5}{4}a}{\\frac{5}{2}a}=\\frac{9}{10}" }, { "text": "Given line $l_{1}$: $4x - 3y + 16 = 0$ and line $l_{2}$: $x = -1$, a moving point $P$ on the parabola $y^{2} = 4x$ has distance $d_{1}$ to line $l_{1}$ and distance $d_{2}$ to line $l_{2}$. Then the minimum value of $d_{1} + d_{2}$ is?", "fact_expressions": "l1: Line;Expression(l1) = (4*x - 3*y + 16 = 0);l2: Line;Expression(l2) = (x = -1);G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);d1: Number;Distance(P, l1) = d1;d2: Number;Distance(P, l2) = d2", "query_expressions": "Min(d1 + d2)", "answer_expressions": "4", "fact_spans": "[[[2, 27], [68, 77]], [[2, 27]], [[28, 45], [95, 104]], [[28, 45]], [[46, 60]], [[46, 60]], [[64, 67], [91, 94]], [[46, 67]], [[81, 88]], [[64, 88]], [[108, 115]], [[91, 115]]]", "query_spans": "[[[117, 136]]]", "process": "Problem Analysis: The focus of the parabola y^{2}=4x is F(1,0). By the definition of the parabola, |PF|=d_{2}. Therefore, the minimum value of d_{1}+d_{2} is the distance from point F to the line 4x-3y+16=0. Hence, the minimum value of d_{1}+d_{2}=\\frac{|4\\cdot1-3\\cdot0+16|}{5}=4" }, { "text": "The number of intersection points of the curves $y^{2}=2 x$ and $(x-2)^{2}+y^{2}=4$ is?", "fact_expressions": "G: Curve;H: Curve;Expression(G) = (y^2 = 2*x);Expression(H) = (y^2 + (x - 2)^2 = 4)", "query_expressions": "NumIntersection(G, H)", "answer_expressions": "3", "fact_spans": "[[[0, 13]], [[14, 35]], [[0, 13]], [[14, 35]]]", "query_spans": "[[[0, 42]]]", "process": "" }, { "text": "The coordinates of the right focus $F$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are? If the parabola $C$ with vertex at the origin also has focus $F$, then its standard equation is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);C: Parabola;F: Point;RightFocus(G) = F;O: Origin;Vertex(C) = O;Focus(C) = F", "query_expressions": "Coordinate(F);Expression(C)", "answer_expressions": "(3, 0)\ny^2 = 12*x", "fact_spans": "[[[0, 39]], [[0, 39]], [[58, 64], [74, 75]], [[43, 46]], [[0, 46]], [[55, 57]], [[52, 64]], [[58, 72]]]", "query_spans": "[[[43, 51]], [[74, 81]]]", "process": "" }, { "text": "If $m \\neq 0$, then the range of eccentricity of the hyperbola $\\frac{x^{2}}{m^{2}+1}+\\frac{y^{2}}{m}=1$ is?", "fact_expressions": "m: Number;Negation(m=0);G: Hyperbola;Expression(G) = (x^2/(m^2 + 1) + y^2/m = 1)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,sqrt(6)/2]", "fact_spans": "[[[1, 11]], [[1, 11]], [[13, 57]], [[13, 57]]]", "query_spans": "[[[13, 68]]]", "process": "Since the equation $\\frac{x^2}{m^{2}+1}+\\frac{y^{2}}{m}=1$ represents a hyperbola, then $m<0$, and the equation can be written as: $\\frac{x^{2}}{m^{2}+1}-\\frac{y^{2}}{m}=1$. Let the semi-focal distance of the hyperbola be $c$, then $c^{2}=m^{2}+1+(-m)$. The eccentricity $e$ of the hyperbola satisfies: $e^{2}=\\frac{m^{2}+1+(-m)}{m^{2}+1}=1+\\frac{-m}{m^{2}+1}=1+\\frac{1}{-m+\\frac{1}{m}}\\leqslant1+\\frac{1}{2\\sqrt{-m\\cdot\\frac{1}{m}}}=\\frac{3}{2}$, with equality if and only if $m=-1$, so the range of the eccentricity is $(1,\\frac{\\sqrt{6}}{2}]$." }, { "text": "If the coordinates of a focus of the hyperbola $k x^{2}-y^{2}=1$ are $(2,0)$, then $k=?$", "fact_expressions": "G: Hyperbola;Expression(G) = (k*x^2 - y^2 = 1);k: Number;Coordinate(OneOf(Focus(G))) = (2, 0)", "query_expressions": "k", "answer_expressions": "1/3", "fact_spans": "[[[1, 21]], [[1, 21]], [[39, 42]], [[1, 37]]]", "query_spans": "[[[39, 44]]]", "process": "Test Analysis: According to the given conditions, the foci of the hyperbola lie on the x-axis. Convert the equation of the hyperbola into standard form, and find a, b, c. Solve the equation involving k to obtain the required value. Since the foci of the hyperbola lie on the x-axis, the standard form of the hyperbola is \\frac{x^{2}}{k}-y^{2}=1, (k>0), so that a^{2}=\\frac{1}{k}, b^{2}=1, c^{2}=1+\\frac{1}{k}. Given that the coordinates of one focus are (2,0), solving gives k=\\frac{1}{3}." }, { "text": "The eccentricity of the hyperbola $x y=2014$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x*y = 2014)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 13]], [[0, 13]]]", "query_spans": "[[[0, 19]]]", "process": "" }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ is $\\sqrt{5}$, then $m=?$", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);Eccentricity(G) = sqrt(5)", "query_expressions": "m", "answer_expressions": "16", "fact_spans": "[[[0, 38]], [[56, 59]], [[0, 38]], [[0, 54]]]", "query_spans": "[[[56, 61]]]", "process": "" }, { "text": "Let $F_{1}$ be the left focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, and let $P$ be a given point on the ellipse $C$. A circle $\\Gamma$ is constructed with $P F_{1}$ as diameter, and let $Q$ be a moving point on the circle $\\Gamma$. Then, what is the maximum distance from the origin $O$ to $Q$?", "fact_expressions": "C: Ellipse;Gamma: Circle;P: Point;F1: Point;Q: Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C)=F1;PointOnCurve(P, C);IsDiameter(LineSegmentOf(P,F1),Gamma);PointOnCurve(Q, Gamma);O:Origin", "query_expressions": "Max(Distance(O, Q))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[9, 41], [50, 55]], [[75, 84], [90, 99]], [[46, 49]], [[1, 8]], [[85, 89], [113, 116]], [[9, 41]], [[1, 45]], [[46, 60]], [[61, 84]], [[85, 103]], [[105, 112]]]", "query_spans": "[[[105, 125]]]", "process": "Let P(x,y). Since F₁ is the left focus of the ellipse C: \\frac{x^{2}}{2}+y^{2}=1, denote its right focus as F₂. Then F₁(-1,0), F₂(1,0). Let M be the midpoint of PF₁, so M(\\frac{x-1}{2},\\frac{y}{2}) is the center of circle Γ, and the radius of circle Γ is r=\\frac{|PF_{1}|}{2}. Then, by properties of circles and the definition of an ellipse, the result follows. Let P(x,y). Since F₁ is the left focus of the ellipse C: \\frac{x^{2}}{2}+y^{2}=1, denote its right focus as F₂. Then F₁(-1,0), F₂(1,0), F₂(1,0). Since point Q is a moving point on circle Γ, by the property of circles, the maximum distance from the origin O to Q is |OQ|_{\\max}=|OM|+r=\\sqrt{(\\frac{x-1}{2})^{2}+\\frac{y^{2}}{4}}+\\frac{|PF_{1}|}{2}=\\frac{\\sqrt{(x-1)^{2}+y^{2}+|PF_{1}|}}{2}=\\frac{|PF_{2}|+|PF_{1}|}{2}" }, { "text": "Given that the line $l$ intersects the ellipse $4 x^{2}+9 y^{2}=36$ at points $A$ and $B$, and the midpoint of chord $AB$ has coordinates $(1,1)$, find the equation of line $l$.", "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;Expression(G) = (4*x^2 + 9*y^2 = 36);Intersection(l, G) = {A, B};IsChordOf(LineSegmentOf(A,B),G);Coordinate(MidPoint(LineSegmentOf(A,B)))=(1,1)", "query_expressions": "Expression(l)", "answer_expressions": "4*x+9*y-13=0", "fact_spans": "[[[2, 7], [65, 70]], [[8, 30]], [[33, 36]], [[37, 40]], [[8, 30]], [[2, 42]], [[8, 49]], [[44, 62]]]", "query_spans": "[[[65, 75]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), from the given conditions: 4x_{1}^{2}+9y_{1}^{2}=36, 4x_{2}^{2}+9y_{2}^{2}=36. Subtracting these two equations yields: 4(x_{1}+x_{2})(x_{1}-x_{2})+9(y_{1}+y_{2})(y_{1}-y_{2})=0. Then: 4(x_{1}+x_{2})+9(y_{1}+y_{2})k_{AB}=0. Combining with the midpoint coordinates gives: 4+9k_{AB}=0 \\Rightarrow k_{AB}=-\\frac{4}{9}. Therefore, the required line equation is: y-1=-\\frac{4}{9}(x-1), which in general form is: 4x+9y-13=0" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, and the circle $E$: $(x-2)^{2}+y^{2}=1$ whose center is the right focus of the hyperbola $C$. If the circle $E$ is tangent to the asymptotes of the hyperbola $C$, then what is the equation of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;E: Circle;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(E) = (y^2 + (x - 2)^2 = 1);Center(E)=RightFocus(C);IsTangent(E, Asymptote(C))", "query_expressions": "Expression(C)", "answer_expressions": "x^2/3-y^2=1", "fact_spans": "[[[2, 53], [84, 90], [102, 108], [116, 122]], [[10, 53]], [[10, 53]], [[56, 80], [97, 101]], [[2, 53]], [[56, 80]], [[56, 94]], [[97, 114]]]", "query_spans": "[[[116, 127]]]", "process": "\\because c=2 \\Rightarrow a^{2}+b^{2}=4.\\textcircled{1} Take the asymptote bx-ay=0, and \\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}=1 \\Rightarrow a^{2}=3b^{2}.\\textcircled{2} From \\textcircled{1}\\textcircled{2} we get a^{2}=3, b^{2}=1, \\therefore the equation of hyperbola C is \\frac{x^{2}}{3}-y^{2}=1." }, { "text": "Given that the perimeter of $\\triangle A B C$ is $20$, and vertices $B(0,-4)$, $C(0,4)$, then the equation of the trajectory of vertex $A$ is?", "fact_expressions": "B: Point;C: Point;A: Point;Coordinate(B) = (0,-4);Coordinate(C) = (0, 4);Perimeter(TriangleOf(A,B,C))=20", "query_expressions": "LocusEquation(A)", "answer_expressions": "(x^2/20+y^2/36=1)&Negation(x=0)", "fact_spans": "[[[31, 40]], [[42, 50]], [[54, 57]], [[31, 40]], [[42, 50]], [[2, 27]]]", "query_spans": "[[[54, 64]]]", "process": "|BC|=8, so |AB|+|AC|=12>8, 2a=12, a=6, c=4, b^{2}=20. Then the trajectory equation of vertex A is \\frac{x2}{20}+\\frac{y^{2}}{36}=1 (x\\neq0)" }, { "text": "Let one of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ be given by the equation $y=\\frac{\\sqrt{2}}{2} x$. Then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Expression(OneOf(Asymptote(G))) = (y = x*(sqrt(2)/2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[1, 48], [84, 87]], [[4, 48]], [[4, 48]], [[1, 48]], [[1, 81]]]", "query_spans": "[[[84, 93]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the eccentricity $e$ lies in the interval $[\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{2}}]$. The line $y=-x+1$ intersects the ellipse at points $M$ and $N$, $O$ is the origin and $OM \\perp ON$. Then the range of the length of the major axis of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;O: Origin;M: Point;N: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 1 - x);Eccentricity(G)=e;Intersection(H,G)={M,N};Range(e)=[1/sqrt(3),1/sqrt(2)];IsPerpendicular(LineSegmentOf(O,M),LineSegmentOf(O,N));e:Number", "query_expressions": "Range(Length(MajorAxis(G)))", "answer_expressions": "[sqrt(5),sqrt(6)]", "fact_spans": "[[[2, 54], [121, 123], [159, 161]], [[4, 54]], [[4, 54]], [[110, 120]], [[133, 136]], [[124, 128]], [[129, 132]], [[4, 54]], [[4, 54]], [[2, 54]], [[110, 120]], [[2, 61]], [[110, 132]], [[2, 109]], [[142, 157]], [[58, 61]]]", "query_spans": "[[[159, 171]]]", "process": "Solve the system \\begin{cases} y = -x + 1 \\\\ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\end{cases}, simplify to obtain (a^{2} + b^{2})x^{2} - 2a^{2}x + a^{2} - a^{2}b^{2} = 0. Let M(x_{1}, y_{1}), N(x_{2}, y_{2}), then x_{1} + x_{2} = \\frac{2a^{2}}{a^{2} + b^{2}}, x_{1}x_{2} = \\frac{a^{2} - a^{2}b^{2}}{a^{2} + b^{2}}. Given OM \\perp ON, then \\overrightarrow{OM} \\cdot \\overrightarrow{ON} = x_{1}x_{2} + y_{1}y_{2} = x_{1}x_{2} + (1 - x_{1})(1 - x_{2}) = 2x_{1}x_{2} - (x_{1} + x_{2}) + 1 = 0, so \\frac{a^{2} - a^{2}b^{2}}{a^{2} + b^{2}} - \\frac{2a^{2}}{a^{2} + b^{2}} + 1 = 0, simplifying gives b^{2} = \\frac{a^{2}}{2a^{2} - 1}. Since e = \\frac{c}{a} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}}, then e^{2} = 1 - \\frac{b^{2}}{a^{2}} = 1 - \\frac{1}{2a^{2} - 1}. Given e \\in \\left[\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{2}}\\right], e^{2} \\in \\left[\\frac{1}{3}, \\frac{1}{2}\\right], so 1 - \\frac{1}{2a^{2} - 1} \\in \\left[\\frac{1}{3}, \\frac{1}{2}\\right] \\Rightarrow \\frac{1}{2a^{2} - 1} \\in \\left[\\frac{1}{2}, \\frac{2}{3}\\right] \\Rightarrow 2a^{2} - 1 \\in \\left[\\frac{3}{2}, 2\\right] \\Rightarrow a^{2} \\in \\left[\\frac{5}{4}, \\frac{3}{2}\\right]. Solve: a \\in \\left[\\frac{\\sqrt{5}}{2}, \\frac{\\sqrt{6}}{2}\\right], thus 2a \\in \\left[\\sqrt{5}, \\sqrt{6}\\right]. Therefore, the range of the major axis length of the ellipse is \\left[\\sqrt{5}, \\sqrt{6}\\right]." }, { "text": "The asymptotes of a hyperbola are $x \\pm 2 y=0$, then its eccentricity is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G))=(x+pm*2*y=0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{sqrt(5),sqrt(5)/2}", "fact_spans": "[[[0, 3], [25, 26]], [[0, 23]]]", "query_spans": "[[[25, 32]]]", "process": "The two asymptotes of the hyperbola are x±2y=0, y=±\\frac{1}{2}x. Depending on the position of the foci of the hyperbola, there are two possible cases: \\frac{b}{a}=\\frac{1}{2} or \\frac{a}{b}=\\frac{1}{2}. Then, from c^{2}=a^{2}+b^{2}, the eccentricity is \\frac{c}{a}=\\frac{\\sqrt{5}}{2} or \\sqrt{5}." }, { "text": "If the asymptotes of a hyperbola are given by $y = \\pm 4x$, and one of its foci is $(\\sqrt{17}, 0)$, then what is the equation of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(4*x));Coordinate(OneOf(Focus(G))) = (sqrt(17), 0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/16=1", "fact_spans": "[[[1, 4], [23, 24], [48, 51]], [[1, 22]], [[23, 46]]]", "query_spans": "[[[48, 56]]]", "process": "From the problem, we know: the hyperbola's foci are at $(\\sqrt{17},0)$, so the foci lie on the x-axis. Since the asymptotes of the hyperbola are $y=\\pm4x$, assume the hyperbola equation is $x^{2}-\\frac{y^{2}}{16}=\\lambda$ ($\\lambda>0$). Therefore, $a^{2}+b^{2}=c^{2}$, that is, $17\\lambda=17 \\Rightarrow \\lambda=1$. Thus, the hyperbola equation is: $x^{2}-\\frac{y^{2}}{16}=1$." }, { "text": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1(b>0)$ has foci $F_{1}$ and $F_{2}$. Let $P$ be a point on this hyperbola. If $|P F_{1}|=6$, then $|P F_{2}|=$?", "fact_expressions": "G: Hyperbola;b: Number;P: Point;F1: Point;F2: Point;b>0;Expression(G) = (x^2/9 - y^2/b^2 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);Abs(LineSegmentOf(P, F1)) = 6", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "12", "fact_spans": "[[[0, 47], [72, 75]], [[3, 47]], [[67, 70]], [[51, 58]], [[59, 66]], [[3, 47]], [[0, 47]], [[0, 66]], [[67, 79]], [[81, 94]]]", "query_spans": "[[[96, 109]]]", "process": "According to the problem, the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ gives $a^{2}=9$, so $a=3$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{9+b^{2}}$. Given that point $P$ is a point on this hyperbola and $|PF|=6$, since $a+c=3+\\sqrt{9+b^{2}}>6$, point $P$ lies on the left branch of the hyperbola. By the definition of a hyperbola, $|PF_{1}|-|PF|=2a=6$, so $|PF_{1}|=|PF|+6=12$. The answer is: 1?" }, { "text": "Given that the distance from the point $(2 , 3)$ to the focus of the parabola $y^{2}=2 p x(p>0)$ is $5$, then $p$=?", "fact_expressions": "G: Parabola;p: Number;H: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (2, 3);Distance(H, Focus(G)) = 5", "query_expressions": "p", "answer_expressions": "12", "fact_spans": "[[[13, 34]], [[47, 50]], [[2, 12]], [[16, 34]], [[13, 34]], [[2, 12]], [[2, 44]]]", "query_spans": "[[[47, 52]]]", "process": "" }, { "text": "Given that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$, and that an asymptote of the hyperbola has the equation $3x - y = 0$. Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola, respectively. If $|PF_{2}|=3$, then $|PF_{1}|=$?", "fact_expressions": "G:Hyperbola;a: Number;P: Point;F2: Point;F1: Point;Expression(G) = (-y^2/9 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));Expression(OneOf(Asymptote(G))) = (3*x - y = 0);LeftFocus(G) =F1;RightFocus(G)=F2;Abs(LineSegmentOf(P, F2)) = 3", "query_expressions": "Abs(LineSegmentOf(P, F1))", "answer_expressions": "5", "fact_spans": "[[[6, 48], [55, 58], [98, 101]], [[9, 48]], [[2, 5]], [[87, 95]], [[77, 86]], [[6, 48]], [[2, 54]], [[55, 76]], [[77, 107]], [[77, 107]], [[109, 121]]]", "query_spans": "[[[123, 135]]]", "process": "" }, { "text": "The left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $F_{1}$, $F_{2}$, respectively. Chord $AB$ passes through point $F_{1}$. If the circumference of the incircle of $\\triangle A B F_{2}$ is $\\pi$, and the coordinates of points $A$, $B$ are $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, respectively, then $|y_{1}-y_{2}|$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;PointOnCurve(F1, LineSegmentOf(A, B)) = True;Perimeter(InscribedCircle(TriangleOf(A, B, F2))) = pi;Coordinate(A) = (x1, y1);x1: Number;y1: Number;Coordinate(B) = (x2, y2);x2: Number;y2: Number", "query_expressions": "Abs(y1 - y2)", "answer_expressions": "5/3", "fact_spans": "[[[0, 39]], [[0, 39]], [[53, 60]], [[45, 52], [67, 75]], [[0, 60]], [[0, 60]], [[111, 115]], [[116, 119]], [[0, 66]], [[62, 75]], [[77, 110]], [[111, 157]], [[124, 140]], [[124, 140]], [[111, 157]], [[141, 157]], [[141, 157]]]", "query_spans": "[[[159, 176]]]", "process": "" }, { "text": "If the coordinates of the two foci of an ellipse are $F_{1}(-1 , 0)$, $F_{2}(5 , 0)$, and the length of the major axis is $10$, then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (5, 0);Focus(G) = {F1,F2};Length(MajorAxis(G)) = 10", "query_expressions": "Expression(G)", "answer_expressions": "(x-2)^2/25+y^2/16=1", "fact_spans": "[[[1, 3], [55, 57]], [[11, 26]], [[29, 43]], [[11, 26]], [[29, 43]], [[1, 43]], [[1, 53]]]", "query_spans": "[[[55, 62]]]", "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 $F_{1}$ and $F_{2}$ respectively. A tangent line is drawn from $F_{2}$ to the circle $x^{2}+y^{2}=a^{2}$, intersecting the left branch of the hyperbola at point $M$. If $\\angle F_{1} M F_{2}=45^{\\circ}$, then what is the equation of the asymptotes 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;H: Circle;Expression(H) = (x^2 + y^2 = a^2);L: Line;TangentOfPoint(F2,H) = L;Intersection(L,LeftPart(G)) = M;M: Point;AngleOf(F1, M, F2) = ApplyUnit(45, degree)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*sqrt(2)*x", "fact_spans": "[[[2, 58], [117, 120], [164, 167]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[66, 73]], [[74, 82], [84, 91]], [[2, 82]], [[2, 82]], [[92, 112]], [[92, 112]], [], [[83, 115]], [[83, 127]], [[123, 127]], [[129, 162]]]", "query_spans": "[[[164, 175]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{2}$ intersects the ellipse at points $M$ and $N$, and the minimum length of segment $MN$ is $3$. If $|\\overrightarrow{N F_{2}}|=\\lambda|\\overrightarrow{F_{2} M}|$ ($\\lambda \\in[1,2]$), then what is the range of possible values for the chord length $MN$?", "fact_expressions": "G: Ellipse;b: Number;H: Line;M: Point;N: Point;F2: Point;F1: Point;lambda: Number;Expression(G) = (x^2/4 + y^2/b^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);Intersection(H, G) = {M, N};Min(Length(LineSegmentOf(M, N))) = 3;Abs(VectorOf(N, F2)) = lambda*Abs(VectorOf(F2, M));In(lambda, [1, 2]);IsChordOf(LineSegmentOf(M, N), G)", "query_expressions": "Range(LineSegmentOf(M, N))", "answer_expressions": "[3, 27/8]", "fact_spans": "[[[2, 43], [85, 88]], [[4, 43]], [[82, 84]], [[88, 91]], [[92, 95]], [[62, 69], [73, 81]], [[52, 59]], [[118, 198]], [[2, 43]], [[2, 69]], [[2, 69]], [[72, 84]], [[82, 97]], [[98, 115]], [[118, 198]], [[118, 198]], [[85, 207]]]", "query_spans": "[[[202, 214]]]", "process": "It is easy to see that point $ F_{2}(c,0) $, where $ b^{2}+c^{2}=a^{2}=4 $. If line $ MN $ coincides with the $ x $-axis, then $ |MN|=2a=4 $. Let the equation of line $ MN $ be $ x=my+c $. Let points $ M(x_{1},y_{1}) $, $ N(x_{2},y_{2}) $. Solving the system \n\\[\n\\begin{cases}\nx=my+c \\\\\nb^{2}x^{2}+4y^{2}=4b^{2}\n\\end{cases}\n\\]\nyields \n\\[\n(m^{2}b^{2}+4)y^{2}+2mb^{2}cy-b^{4}=0,\n\\]\n\\[\n\\Delta=4m^{2}b^{4}c^{2}+4b^{4}(m^{2}b^{2}+4)=16b^{4}(m^{2}+1).\n\\]\nBy Vieta's formulas, we get\n\\[\ny_{1}+y_{2}=-\\frac{2mb^{2}c}{m^{2}b^{2}+4}, \\quad y_{1}y_{2}=-\\frac{b^{4}}{m^{2}b^{2}+4},\n\\]\n\\[\n|MN|=\\sqrt{1+m^{2}}\\cdot|y_{1}-y_{2}|=\\sqrt{1+m^{2}}\\cdot\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=\\sqrt{1+m^{2}}\\sqrt{\\left(-\\frac{2mb^{2}c}{m^{2}b^{2}+4}\\right)^{2}+\\frac{4b^{4}}{m^{2}b^{2}+4}}=\\frac{4b^{2}(m^{2}+1)}{m^{2}b^{2}+4}=4-\\frac{4c^{2}}{m^{2}b^{2}+4}.\n\\]\nTherefore, when $ m=0 $, $ |MN|_{\\min}=\\frac{4b^{2}}{4}=b^{2}=3 $. Hence, the equation of the ellipse is $ \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 $. From the condition, $ \\overrightarrow{NF}_{2}=\\lambda\\overrightarrow{F_{2}M} $, i.e., $ (1-x_{2},-y_{2})=\\lambda(x_{1}-1,y_{1}) $, then $ y_{2}=-\\lambda y_{1} $. By Vieta's formulas, we have $ y_{1}+y_{2}=-\\frac{2mb^{2}c}{m^{2}b^{2}+4} $, which gives $ y_{1}=\\frac{6m}{(\\lambda-1)(3m^{2}+4)} $, and $ y_{1}y_{2}=-\\frac{9}{3m^{2}+4}=-\\lambda y_{1}^{2} $, so\n\\[\n\\frac{36\\lambda m^{2}}{(\\lambda-1)^{2}(3m^{2}+4)^{2}}=\\frac{9}{3m^{2}+4}.\n\\]\nWhen $ \\lambda=1 $, point $ F_{2} $ is the midpoint of segment $ MN $, then $ m=0 $. When $ \\lambda\\in(1,2] $, we get\n\\[\n\\frac{1}{m^{2}}=\\frac{\\lambda}{(\\lambda-1)^{2}}-\\frac{3}{4}.\n\\]\nSince function $ f(\\lambda)=\\lambda+\\frac{1}{\\lambda}-2 $ is monotonically increasing on $ (1,2] $, when $ \\lambda\\in(1,2] $, $ f(\\lambda)\\in(0,\\frac{1}{2}] $. Thus,\n\\[\n\\frac{1}{m^{2}}=\\frac{1}{\\lambda+\\frac{1}{\\lambda}-2}-\\frac{3}{4}\\geqslant\\frac{5}{4},\n\\]\nthen $ 0 0 $), $ B(x_{2},y_{2}) $. Substituting $ x = 2 $ into the parabola equation gives $ y_{1} = 2\\sqrt{2} $. The equation of line AF is $ y = 2\\sqrt{2}(x-1) $. Solving the system \n$$\n\\begin{cases}\ny = 2\\sqrt{2}(x-1) \\\\\ny^{2} = 4x\n\\end{cases}\n$$\nEliminating $ x $ yields $ y^{2} - \\sqrt{2}y - 4 = 0 $, solving gives $ y = 2\\sqrt{2} $, $ y_{2} = -\\sqrt{2} $. Therefore, the area of $ \\Delta AOB $ is $ S = \\frac{1}{2} \\times |OF| \\times |y_{1} - y_{2}| = \\frac{1}{2} \\times 1 \\times 3\\sqrt{2} = \\frac{3\\sqrt{2}}{2} $." }, { "text": "A point $M(1, m)$ $(m>0)$ on the parabola $y^{2}=2 px(p>0)$ is at a distance of $5$ from its focus. The left vertex of the hyperbola $\\frac{x^{2}}{a}-y^{2}=1$ is $A$. If one asymptote of the hyperbola is parallel to the line $AM$, then the real number $a$ equals?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*p*x);M: Point;a: Real;m:Number;m>0;p: Number;p>0;Distance(M,Focus(H))=5;G: Hyperbola;Expression(G) = (-y^2 + x^2/a=1);A:Point;LeftVertex(G)=A;Coordinate(M) = (1, m);PointOnCurve(M, H);IsParallel(OneOf(Asymptote(G)),LineOf(A,M))", "query_expressions": "a", "answer_expressions": "1/9", "fact_spans": "[[[0, 20], [40, 41]], [[0, 20]], [[23, 39]], [[112, 117]], [[23, 39]], [[23, 39]], [[3, 20]], [[3, 20]], [[23, 50]], [[51, 81], [92, 95]], [[51, 81]], [[86, 89]], [[51, 89]], [[23, 39]], [[0, 39]], [[92, 110]]]", "query_spans": "[[[112, 120]]]", "process": "" }, { "text": "A point $P$ on the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$ has distance $\\frac{\\sqrt{3}}{2}$ to the left focus. What is the distance from $P$ to the left directrix?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/3 + y^2/2 = 1);PointOnCurve(P, G);Distance(P, LeftFocus(G)) = sqrt(3)/2", "query_expressions": "Distance(P, LeftDirectrix(G))", "answer_expressions": "3/2", "fact_spans": "[[[0, 37]], [[40, 43], [73, 76]], [[0, 37]], [[0, 43]], [[0, 71]]]", "query_spans": "[[[0, 85]]]", "process": "From $\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$, we get $a=\\sqrt{3}, c=1$. Let the left and right foci of the ellipse be $F_{1}$ and $F_{2}$ respectively, and let $d$ be the distance from a point $P$ on the ellipse to the right directrix. Then $\\frac{|PF_{1}|}{d}=\\frac{c}{a}=\\frac{1}{\\sqrt{3}}$, so $d=\\sqrt{3}|PF_{1}|=\\sqrt{3}\\times\\frac{\\sqrt{3}}{2}=\\frac{3}{2}$." }, { "text": "The shortest distance from points on the parabola $x^{2}=-2 p y(p>0)$ to the line $3 x+4 y-12=0$ is $1$. Find $p$.", "fact_expressions": "G: Parabola;p: Number;H: Line;P: Point;p>0;Expression(G) = (x^2 = -2*p*y);Expression(H) = (3*x + 4*y - 12 = 0);PointOnCurve(P, G);Min(Distance(P, H)) = 1", "query_expressions": "p", "answer_expressions": "56/9", "fact_spans": "[[[0, 22]], [[53, 56]], [[26, 42]], [], [[3, 22]], [[0, 22]], [[26, 42]], [[0, 25]], [[0, 51]]]", "query_spans": "[[[53, 58]]]", "process": "According to the problem, the line $3x + 4y - 12 = 0$ does not intersect the parabola $x^{2} = -2py$ ($p > 0$), which gives $0 < p < \\frac{32}{3}$. Let point $P(x, y)$ be a point on the parabola $x^{2} = -2py$ ($p > 0$); then $y = -\\frac{1}{2p}x^{2}$. Using the point-to-line distance formula, find the distance $d$ from point $P$ to the line $3x + 4y - 12 = 0$. According to the problem, $d_{\\min} = 1$, and thus $p$ can be determined. According to the problem, the line $3x + 4y - 12 = 0$ does not intersect the parabola $x^{2} = -2py$ ($p > 0$). From\n$$\n\\begin{cases}\n3x + 4y - 12 = 0 \\\\\nx^{2} = -2py\n\\end{cases}\n$$\nwe obtain $x^{2} - \\frac{3p}{2}x + 6p = 0$, then $\\Delta = \\left(-\\frac{3p}{2}\\right)^{2} - 4 \\times 6p < 0$, $\\therefore 0 < p < \\frac{32}{3}$. Let point $P(x, y)$ be a point on the parabola $x^{2} = -2py$ ($p > 0$), $\\therefore y = -\\frac{1}{2p}x^{2}$. The distance from point $P$ to the line $3x + 4y - 12 = 0$ is\n$$\nd = \\frac{\\frac{2}{p}\\left(x - \\frac{3p}{4}\\right)^{2} - \\left(\\frac{9p}{8} - 1\\right)}{\\sqrt{5}}\n$$" }, { "text": "Given that the asymptotes of the hyperbola $c$ are $x \\pm \\sqrt{3}y=0$, and the right focus of the hyperbola $c$ lies on the circle $x^{2}+y^{2}-8 x-2 y+16=0$, then the standard equation of the hyperbola $c$ is?", "fact_expressions": "c: Hyperbola;Expression(Asymptote(c)) = (x + pm*sqrt(3)*y = 0);G: Circle;Expression(G) = (-2*y - 8*x + x^2 + y^2 + 16 = 0);PointOnCurve(RightFocus(c), G)", "query_expressions": "Expression(c)", "answer_expressions": "x^2/12-y^2/4=1", "fact_spans": "[[[2, 8], [36, 42], [77, 83]], [[2, 34]], [[47, 74]], [[47, 74]], [[36, 75]]]", "query_spans": "[[[77, 90]]]", "process": "" }, { "text": "The center of the ellipse is at the origin, the eccentricity is $\\frac{1}{2}$, and the directrix $l$ of the parabola $y^{2}=-4 x$ passes through one of its foci. Then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;Eccentricity(G) = 1/2;H: Parabola;Expression(H) = (y^2 = -4*x);l: Line;Directrix(H) = l;PointOnCurve(OneOf(Focus(G)), l)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4+y^2/3=1", "fact_spans": "[[[0, 2], [52, 53], [60, 62]], [[6, 10]], [[0, 10]], [[0, 29]], [[30, 45]], [[30, 45]], [[48, 51]], [[30, 51]], [[48, 58]]]", "query_spans": "[[[60, 66]]]", "process": "" }, { "text": "The line $l$ passing through the right focus $F$ of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ intersects the minor axis of the ellipse at point $Q$, and intersects the ellipse at points $M$ and $N$. If $\\overrightarrow{F M}=\\lambda \\overrightarrow{M Q}$, $\\overrightarrow{F N}=\\mu \\overrightarrow{N Q}$, then the minimum value of $\\lambda \\mu$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);F: Point;RightFocus(G) = F;l: Line;PointOnCurve(F, l);Q: Point;Intersection(l, MinorAxis(G)) = Q;M: Point;N: Point;Intersection(l, G) = {M, N};lambda: Number;mu: Number;VectorOf(F, M) = lambda*VectorOf(M, Q);VectorOf(F, N) = mu*VectorOf(N, Q)", "query_expressions": "Min(lambda*mu)", "answer_expressions": "1/4", "fact_spans": "[[[1, 28], [41, 43], [52, 54]], [[1, 28]], [[31, 34]], [[1, 34]], [[35, 40]], [[0, 40]], [[46, 50]], [[35, 50]], [[55, 58]], [[59, 62]], [[35, 64]], [[168, 181]], [[168, 181]], [[66, 117]], [[119, 166]]]", "query_spans": "[[[168, 187]]]", "process": "Given that $ F(\\sqrt{3},0) $, the slope of line $ l $ exists, let its equation be $ y = k(x - \\sqrt{3}) $, then $ Q(0, -\\sqrt{3}k) $. Since point $ Q $ lies on the minor axis, it follows that $ -1 < -\\sqrt{3}k < 1 $, i.e., $ -\\frac{\\sqrt{3}}{3} < k < \\frac{\\sqrt{3}}{3} $. Solving the system of equations\n$$\n\\begin{cases}\n\\frac{x^{2}}{4} + y^{2} = 1 \\\\\ny = k(x - \\sqrt{3})\n\\end{cases}\n$$\nyields $ (4k^{2} + 1)x^{2} - 8\\sqrt{3}k^{2}x + 12k^{2} - 4 = 0 $. Let $ M(x_{1}, y_{1}) $, $ N(x_{2}, y_{2}) $, so\n$$\nx_{1} + x_{2} = \\frac{8\\sqrt{3}k^{2}}{|4k^{2} + 1|}, \\quad x_{1}x_{2} = \\frac{12k^{2} - 4}{4k^{2} + 1},\n$$\n$$\n\\overrightarrow{FM} = (x_{1} - \\sqrt{3}, y_{1}), \\quad \\overrightarrow{MQ} = (-x_{1}, -\\sqrt{3}k - y_{1}),\n$$\n$$\n\\overrightarrow{FN} = (x_{2} - \\sqrt{3}, y_{2}), \\quad \\overrightarrow{NQ} = (-x_{2}, -\\sqrt{3}k - y_{2}).\n$$\nSince $ \\overrightarrow{FM} = \\lambda \\overrightarrow{MQ} $, $ \\overrightarrow{FN} = \\mu \\overrightarrow{NQ} $, it follows that\n$$\n\\lambda = \\frac{\\sqrt{3}}{x_{1}} - 1, \\quad \\mu = \\frac{\\sqrt{3}}{x_{2}} - 1,\n$$\nso\n$$\n\\lambda\\mu = \\left( \\frac{\\sqrt{3}}{x_{1}} - 1 \\right)\\left( \\frac{\\sqrt{3}}{x_{2}} - 1 \\right) = -\\left( \\frac{\\sqrt{3}}{x_{1}} + \\frac{\\sqrt{3}}{x_{2}} \\right) + 1 = \\frac{3 - \\sqrt{3}(x_{1} + x_{2})}{x_{1}x_{2}} + 1 = \\frac{3 - 12k^{2}}{12k^{2} - 4} + 1 = -\\frac{1}{12k^{2} - 4}.\n$$\nThus $ -\\frac{3}{12k^{2} - 4} \\geqslant \\frac{1}{4} $, hence $ 12k^{2} - 4 \\in [-4, 0) $." }, { "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 $y=\\pm \\sqrt{2} x$, then what is its eccentricity?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(Asymptote(G)) = (y = pm*(sqrt(2)*x))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[1, 57], [84, 85]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 82]]]", "query_spans": "[[[84, 91]]]", "process": "" }, { "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have its right focus at $F$, and let point $P$ lie on the right branch of $C$, with $O$ being the origin. If there exists a point $P$ such that $|P F|=|O F|$ and $\\cos \\angle O F P=\\frac{1}{4}$, 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;RightFocus(C) = F;P: Point;PointOnCurve(P, RightPart(C));O: Origin;Abs(LineSegmentOf(P, F)) = Abs(LineSegmentOf(O, F));Cos(AngleOf(O, F, P)) = 1/4", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[1, 62], [76, 79], [150, 153]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[67, 70]], [[1, 70]], [[71, 75], [96, 100]], [[71, 83]], [[84, 87]], [[102, 115]], [[117, 148]]]", "query_spans": "[[[150, 159]]]", "process": "In the focal triangle, use the law of cosines to find the relationship between a and c, then find the eccentricity. Let the left focus of the hyperbola be E. In \\triangleEFP, |EF|=2c, |PF|=c, |PE|=2a+c, \\cos\\angleEFP=\\frac{1}{4}. By the law of cosines, \\cos\\angleEFP=\\frac{4c^{2}+c^{2}-(c+2a)^{2}}{2\\cdot2c\\cdotc}=\\frac{1}{4}, we obtain e=\\frac{c}{a}=2." }, { "text": "Point $P$ moves on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and points $Q$, $R$ move on the circles $(x+1)^{2}+y^{2}=1$ and $(x-1)^{2}+y^{2}=1$, respectively. Then the maximum value of $|PQ|+| PR |$ is?", "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G) = True;H1: Circle;Expression(H1) = (y^2 + (x + 1)^2 = 1);H2: Circle;Expression(H2) = (y^2 + (x - 1)^2 = 1);Q: Point;R: Point;PointOnCurve(Q, H1) = True;PointOnCurve(R, H2) = True", "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(P, R)))", "answer_expressions": "6", "fact_spans": "[[[0, 4]], [[5, 42]], [[5, 42]], [[0, 45]], [[58, 77]], [[58, 77]], [[78, 97]], [[78, 97]], [[46, 49]], [[50, 53]], [[46, 100]], [[46, 100]]]", "query_spans": "[[[102, 121]]]", "process": "" }, { "text": "The standard equation of the parabola with vertex at the origin, the $x$-axis as its axis of symmetry, and passing through $P(-2,-4)$ is?", "fact_expressions": "O: Origin;Vertex(G) = O;SymmetryAxis(G) = xAxis;G: Parabola;P: Point;Coordinate(P) = (-2, -4);PointOnCurve(P, G)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=-8*x", "fact_spans": "[[[1, 3]], [[0, 34]], [[7, 34]], [[31, 34]], [[20, 30]], [[20, 30]], [[18, 34]]]", "query_spans": "[[[31, 41]]]", "process": "Let the standard equation of the parabola be \\( y^2 = 2px \\). Substituting point \\( P(-2, -4) \\) gives \\( (-4)^2 = 2p(-2) \\Rightarrow p = -4 \\). Hence, the parabola is \\( y^2 = -8x \\)." }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\frac{2 \\sqrt{6}}{3}$, and its vertices coincide with the foci of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{5}=1$, what are the coordinates of the foci of this hyperbola? What is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Ellipse;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/8 + y^2/5 = 1);Eccentricity(G) = 2*sqrt(6)/3;Vertex(G) = Focus(H)", "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))", "answer_expressions": "(pm*2*sqrt(2), 0)\ny = pm*(sqrt(15)/3)*x", "fact_spans": "[[[2, 58], [135, 138]], [[5, 58]], [[5, 58]], [[89, 126]], [[5, 58]], [[5, 58]], [[2, 58]], [[89, 126]], [[2, 85]], [[2, 131]]]", "query_spans": "[[[135, 145]], [[135, 152]]]", "process": "" }, { "text": "Ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has left and right vertices $A_{1}$, $A_{2}$ respectively. If there exists a point $P$ on the ellipse such that $\\angle A_{1} P A_{2}=120^{\\circ}$, then the range 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;LeftVertex(C) = A1;RightVertex(C) = A2;A1: Point;A2: Point;PointOnCurve(P, C) = True;P: Point;AngleOf(A1, P, A2) = ApplyUnit(120, degree)", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[sqrt(6/3),1)", "fact_spans": "[[[0, 57], [83, 85], [130, 135]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[0, 81]], [[0, 81]], [[66, 73]], [[74, 81]], [[83, 92]], [[88, 92]], [[94, 128]]]", "query_spans": "[[[130, 146]]]", "process": "As shown in the figure, when P is at the upper vertex, ∠A₁PA₂ is maximized; at this time, it suffices that ∠A₁PA₂⩾120°, so ∠A₂PO⩾60°, tan∠A₂PO= a/b ⩾tan60°=√3, i.e., a⩾√3b. a²⩾3b², a²⩾3(a²−c²), 2a²⩽3c², c/a ⩾√6/3, e∈[√6/3,1)" }, { "text": "In the plane, the difference of distances from a moving point $P$ to two fixed points $F_{1}(-2,0)$, $F_{2}(2,0)$ is $m$. If the locus of the moving point $P$ is a hyperbola, then the range of values for $m$ is?", "fact_expressions": "G:Hyperbola;F1: Point;F2: Point;m:Number;P:Point;Coordinate(F1)=(-2,0);Coordinate(F2)=(2, 0);Distance(P,F1)-Distance(P,F2)=m;Locus(P)=G", "query_expressions": "Range(m)", "answer_expressions": "(-4,0)+(0,4)", "fact_spans": "[[[58, 61]], [[12, 25]], [[26, 38]], [[44, 47], [63, 66]], [[5, 8], [51, 54]], [[12, 25]], [[26, 38]], [[5, 47]], [[51, 61]]]", "query_spans": "[[[63, 73]]]", "process": "In a hyperbola, satisfying a < c, from the given condition we have |P₁F₁ - P₁F₂| = 2a < 4, that is, |m| < 4 and m ≠ 0; then the range of values for m is (-4, 0) ∪ (0, 4)." }, { "text": "If the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ have eccentricities $e_{1}$, $e_{2}$ respectively, then what is the range of values of $e_{1} e_{2}$?", "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2", "query_expressions": "Range(e1*e2)", "answer_expressions": "(0, 1)", "fact_spans": "[[[1, 53]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[54, 100]], [[54, 100]], [[107, 114]], [[117, 124]], [[1, 124]], [[1, 124]]]", "query_spans": "[[[126, 146]]]", "process": "" }, { "text": "A circle has its center at the right focus of an ellipse and passes through the center of the ellipse, intersecting the ellipse at point $P$. The line $PF_{1}$ ($F_{1}$ being the left focus of the ellipse) is tangent to this circle. Find the eccentricity of the ellipse.", "fact_expressions": "G:Ellipse;H:Circle;P:Point;F1:Point;Center(H)=RightFocus(G);PointOnCurve(Center(G),H);Intersection(H,G)=P;LeftFocus(G)=F1;IsTangent(LineOf(P,F1),H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[6, 8], [16, 18], [22, 24], [63, 65], [49, 51]], [[2, 3], [14, 15], [58, 59]], [[25, 28]], [[40, 47]], [[2, 11]], [[14, 20]], [[14, 28]], [[40, 54]], [[29, 61]]]", "query_spans": "[[[63, 70]]]", "process": "" }, { "text": "Given that the line $l$: $x - y + 1 = 0$ intersects the ellipse $\\frac{x^{2}}{16} + \\frac{y^{2}}{9} = 1$ at two points $A$ and $B$, if there exists a point $P$ on the ellipse such that the area of $\\Delta PAB$ is maximized, then the coordinates of point $P$ are?", "fact_expressions": "l: Line;G: Ellipse;P: Point;A: Point;B: Point;Expression(G) = (x^2/16 + y^2/9 = 1);Expression(l)=(x-y+1=0);Intersection(l,G)={A,B};PointOnCurve(P,G);WhenMax(Area(TriangleOf(P,A,B)))", "query_expressions": "Coordinate(P)", "answer_expressions": "(16/5,-9/5)", "fact_spans": "[[[2, 18]], [[19, 57], [70, 72]], [[77, 80], [101, 105]], [[59, 62]], [[63, 66]], [[19, 57]], [[2, 18]], [[2, 68]], [[70, 80]], [[82, 100]]]", "query_spans": "[[[101, 110]]]", "process": "" }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{75}=1$, $F_{1}$ and $F_{2}$ are the foci of the ellipse, and $\\angle F_{1} P F_{2}=60^{\\circ}$, find the area of $\\triangle F_{1} P F_{2}$.", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/75 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "25*sqrt(3)/3", "fact_spans": "[[[6, 45], [65, 67]], [[6, 45]], [[2, 5]], [[2, 48]], [[49, 56]], [[57, 64]], [[49, 70]], [[71, 104]]]", "query_spans": "[[[106, 136]]]", "process": "Using the cosine law and the definition of an ellipse, we obtain $ PF_{1} \\cdot PF_{2} = \\frac{100}{3} $. Then, using the triangle area formula, we get the result. [Detailed solution] In $ \\triangle F_{1}PF_{2} $, \n$ F_{1}F_{2}^{2} = PF_{1}^{2} + PF_{2}^{2} - 2PF_{1} \\cdot PF_{2} \\cos 60^{\\circ} $, \nthat is, \n$ 200 = PF_{1}^{2} + PF_{2}^{2} - PF_{1} \\cdot PF_{2} $, ① \nby the definition of the ellipse, $ 2a = PF_{1} + PF_{2} $, \n$ 300 = PF_{1}^{2} + PF_{2}^{2} + 2PF_{1} \\cdot PF_{2} $, ② \nfrom ① and ②, we get $ PF_{1} \\cdot PF_{2} = \\frac{100}{3} $, \ntherefore, \n$ S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2} PF_{1} \\cdot PF_{2} \\sin 60^{\\circ} = \\frac{25\\sqrt{3}}{3} $. \nAnswer: $ \\frac{25\\sqrt{3}}{3} $." }, { "text": "The line passing through the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ and perpendicular to the $x$-axis intersects the two asymptotes of this hyperbola at points $A$ and $B$. Then $|AB|$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);H: Line;PointOnCurve(RightFocus(G), H);IsPerpendicular(H, xAxis);A: Point;B: Point;l1: Line;l2: Line;Asymptote(G) = {l1, l2};Intersection(H, l1) = A;Intersection(H, l2) = B", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[1, 29], [47, 50]], [[1, 29]], [[42, 44]], [[0, 44]], [[34, 44]], [[57, 60]], [[61, 64]], [], [], [[47, 56]], [[42, 66]], [[42, 66]]]", "query_spans": "[[[68, 77]]]", "process": "From the hyperbola equation, the right focus is found to be F(2,0). Then substitute x=2 into the asymptote equation of the hyperbola to find the value of y, which gives the y-coordinates of points A and B, thus allowing the calculation of |AB|. [Detailed solution] The right focus of the hyperbola is F(2,0). The line passing through F and perpendicular to the x-axis is x=2. The asymptote equations are x^{2}-\\frac{y^{2}}{3}=0. Substituting x=2 into x^{2}-\\frac{y^{2}}{3}=0 yields y^{2}=12, y=\\pm2\\sqrt{3}, hence |AB|=4\\sqrt{3}. Therefore, the answer is: 4\\sqrt{3}. This question examines the properties of hyperbolas, involving basic problem-solving skills." }, { "text": "A line passing through the focus of the parabola $y=\\frac{1}{4} x^{2}$ intersects the parabola at two points $A(x_{1} , y_{1})$ and $B(x_{2}, y_{2})$. If $y_{1}+y_{2}=5$, then the length of segment $AB$ is equal to?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y = x^2/4);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);y1 + y2 = 5", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "7", "fact_spans": "[[[2, 26], [33, 36]], [[30, 32]], [[37, 55]], [[56, 75]], [[37, 55]], [[37, 55]], [[56, 75]], [[56, 75]], [[2, 26]], [[0, 32]], [[30, 77]], [[37, 55]], [[56, 75]], [[79, 94]]]", "query_spans": "[[[96, 108]]]", "process": "" }, { "text": "An ellipse $5 x^{2}+k y^{2}=5$ has a focus at $(0,2)$. Then $k=$?", "fact_expressions": "G: Ellipse;Expression(G) = (k*y^2 + 5*x^2 = 5);k: Number;F1: Point;OneOf(Focus(G)) = F1;Coordinate(F1) = (0, 2)", "query_expressions": "k", "answer_expressions": "1", "fact_spans": "[[[0, 21]], [[0, 21]], [[37, 40]], [[27, 34]], [[0, 34]], [[27, 34]]]", "query_spans": "[[[37, 42]]]", "process": "Test analysis: The standard equation of the ellipse is $x^{2}+\\frac{y^{2}}{\\frac{5}{k}}=1$, and one focus is at $(0,2)$, so $\\frac{5}{k}-1=4 \\Rightarrow k=1$." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has left and right foci $F_{1}$, $F_{2}$ respectively. Let $A$ be a moving point on $C$, and point $P$ lies on the extension of line segment $F_{1} A$, such that $(\\overrightarrow{A P}+\\overrightarrow{A F_{2}}) \\cdot \\overrightarrow{F_{2} P}=0$. Then the maximum distance from $P$ to the $y$-axis is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;PointOnCurve(A, C) = True;P: Point;PointOnCurve(P, OverlappingLine(LineSegmentOf(F1, A))) = True;DotProduct(VectorOf(A, F2) + VectorOf(A, P), VectorOf(F2, P)) = 0", "query_expressions": "Max(Distance(P, yAxis))", "answer_expressions": "5", "fact_spans": "[[[0, 42], [71, 74]], [[0, 42]], [[51, 58]], [[59, 66]], [[0, 66]], [[0, 66]], [[67, 70]], [[67, 78]], [[79, 83], [186, 189]], [[79, 100]], [[102, 184]]]", "query_spans": "[[[186, 202]]]", "process": "Take the midpoint Q of F_{2}P, connect AQ, \\overrightarrow{AP}+\\overrightarrow{AF}_{2}=2\\overrightarrow{AQ}, then \\overrightarrow{AQ}\\cdot\\overrightarrow{F_{2}P}=0, it follows that |AP|=|AF_{2}|, by the definition of ellipse we know |AF_{1}|+|AF_{2}|=|AF_{1}|+|AP|=|F_{1}P|=4, thus the trajectory of point P is a circle with center F_{1} and radius 4, from the figure we can see that when point P coincides with point M, the maximum distance to the y-axis is 5." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular from its right focus $F$ to an asymptote, with foot of perpendicular at point $B$. This perpendicular intersects the $y$-axis at point $P$ and intersects the other asymptote at point $A$, with point $P$ lying between points $A$ and $B$. Given that $O$ is the origin and $|OA|=\\frac{5}{3} a$, find the eccentricity of the hyperbola?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;A: Point;B:Point;P: Point;F:Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);l:Line;l1:Line;l2:Line;RightFocus(C)=F;PointOnCurve(F, l);IsPerpendicular(l,l1);FootPoint(l,l1)=B;Intersection(l,yAxis)=P;Intersection(l,l2)=A;Between(P,A,B)=True;Abs(LineSegmentOf(O, A)) = (5/3)*a;OneOf(Asymptote(C)) = l1;OneOf(Asymptote(C)) = l2;Negation(l1=l2)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 63], [165, 168], [65, 66]], [[10, 63]], [[10, 63]], [[134, 137]], [[120, 124], [106, 111]], [[125, 128], [83, 86]], [[93, 97], [114, 118]], [[69, 72]], [[10, 63]], [[10, 63]], [[2, 63]], [], [], [], [[65, 72]], [[64, 79]], [[64, 79]], [[64, 86]], [[64, 97]], [[64, 111]], [[114, 130]], [[142, 163]], [63, 73], [63, 102], [63, 102]]", "query_spans": "[[[165, 174]]]", "process": "" }, { "text": "In the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line with slope $k$ $(k>0)$ intersects the ellipse at the left vertex $A$ and another point $B$. The projection of point $B$ onto the $x$-axis coincides exactly with the right focus $F$. If the eccentricity of the ellipse is $e=\\frac{1}{3}$, then what is the value of $k$?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A:Point;B: Point;F: Point;e:Number;k:Number;a > b;b > 0;k>0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C)=F;Slope(G)=k;LeftVertex(C)=A;Intersection(G,C)={A,B};Projection(B, xAxis) = F;Eccentricity(C) = e ;e= 1/3", "query_expressions": "k", "answer_expressions": "2/3", "fact_spans": "[[[1, 58], [76, 78], [117, 119]], [[8, 58]], [[8, 58]], [[73, 75]], [[82, 85]], [[89, 92], [93, 97]], [[112, 115]], [[122, 137]], [[140, 143]], [[8, 58]], [[8, 58]], [[64, 72]], [[1, 58]], [[76, 115]], [[61, 75]], [[76, 85]], [[73, 92]], [[93, 115]], [[117, 137]], [[122, 137]]]", "query_spans": "[[[140, 147]]]", "process": "Test analysis: Since the eccentricity of the ellipse is $ e = \\frac{1}{3} $, it follows that $ \\frac{c}{a} = \\frac{1}{3} $, so $ c = \\frac{1}{3}a $, $ b = \\sqrt{a^{2} - c^{2}} = \\frac{2\\sqrt{2}}{3}a $. Since the projection of point $ B $ on the x-axis is exactly the right focus $ F $, the coordinates of point $ B $ are $ \\left(c, \\frac{b^{2}}{a}\\right) $. Given $ A(-a, 0) $, then $ k = \\frac{b^{2}}{c + a} = \\frac{\\frac{8}{9}a}{\\frac{1}{3}a + a} = \\frac{2}{3} $." }, { "text": "Given that $A$ is the left vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $O$ is the origin. If there exists a point $P$ on the line $l$: $x=2 c$ such that $\\angle A P O=45^{\\circ}$, then the maximum value of the ellipse's eccentricity is?", "fact_expressions": "l: Line;G: Ellipse;b: Number;a: Number;A: Point;P: Point;O: Origin;c:Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G)=A;Expression(l)=(x=2*c);PointOnCurve(P,l);AngleOf(A,P,O)=ApplyUnit(45,degree)", "query_expressions": "Max(Eccentricity(G))", "answer_expressions": "(sqrt(2)-1)/4", "fact_spans": "[[[73, 87]], [[6, 58], [123, 125]], [[8, 58]], [[8, 58]], [[2, 5]], [[90, 94]], [[63, 66]], [[80, 87]], [[8, 58]], [[8, 58]], [[6, 58]], [[2, 62]], [[73, 87]], [[73, 94]], [[96, 121]]]", "query_spans": "[[[123, 134]]]", "process": "Let P(2c,t), then using the inclinations of lines AP and OP and the tangent of the difference of two angles, we obtain that \\frac{at}{t^{2}+2c(2c+a)}=1 has a solution in R. This fractional equation can be transformed into a quadratic equation; using the discriminant, an inequality involving a and c is obtained, from which the maximum eccentricity can be found. Let P(2c,t), and let the right focus be F. Since there exists a point P on the line l: x=2c such that \\angle APO = 45^{\\circ}, it follows that \\frac{at}{t^{2}+2c(2c+a)}=1 has a solution in R, i.e., t^{2}-at+2c(2c+a)=0 has a solution in R. Therefore, a^{2}-8c(2c+a)\\geqslant0, i.e., 16e^{2}+8e-1\\leqslant0, hence 00$, $b>0$), the right vertex of the hyperbola is $A$. A circle centered at $A$ with radius $b$ intersects an asymptote of the hyperbola $C$ at points $M$ and $N$. If $\\overrightarrow{O M}=\\frac{3}{2} \\overrightarrow{O N}$ ($O$ is the 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;A: Point;RightVertex(C) = A;A1: Circle;Center(A1) = A;Radius(A1) = b;M: Point;N: Point;Intersection(A1, OneOf(Asymptote(C))) = {M, N};O: Origin;VectorOf(O, M) = (3/2)*VectorOf(O, N)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(30)/5", "fact_spans": "[[[2, 63], [94, 100], [187, 193]], [[2, 63]], [[80, 83]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [73, 76]], [[2, 71]], [[87, 88], [89, 93]], [[72, 88]], [[80, 88]], [[109, 112]], [[113, 116]], [[89, 118]], [[176, 179]], [[120, 175]]]", "query_spans": "[[[187, 199]]]", "process": "Hyperbola C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) has its right vertex at A(a,0). Taking A as the center and b as the radius, draw circle A. Circle A intersects an asymptote of hyperbola C at points M and N. Then the distance from point A to the asymptote bx-ay=0 is AB=\\frac{ab}{c}. \\because\\overrightarrow{OM}=\\frac{3}{2}\\overrightarrow{ON}, \\therefore OB=5BN=\\frac{5b^{2}}{c}, \\because OA=a \\therefore a^{2}=\\frac{25b^{4}}{c^{2}}+\\frac{a^{2}b^{2}}{c^{2}}, a^{2}c^{2}=25b^{4}+a^{2}b^{2}, \\therefore a^{2}(c^{2}-b^{2})=25b^{4}, \\therefore a^{2}=5b^{2}=5c^{2}-5a^{2}, i.e., 6a^{2}=5c^{2}, i.e., \\sqrt{6}a=\\sqrt{5}c, \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{6}}{\\sqrt{5}}=\\frac{\\sqrt{30}}{5}" }, { "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 $F_{1}$ and $F_{2}$ respectively. A line is drawn from the right focus $F_{2}$ perpendicular to one of the asymptotes of the hyperbola, with foot of perpendicular at $P$. Connect $P F_{1}$. If $|P F_{1}|=2|P F_{2}|$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;L: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, L);IsPerpendicular(L, OneOf(Asymptote(G)));FootPoint(L, OneOf(Asymptote(G))) = P;Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(21)/3", "fact_spans": "[[[2, 58], [95, 98], [154, 157]], [[5, 58]], [[5, 58]], [[112, 115]], [[67, 74]], [[75, 82], [87, 94]], [], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 108]], [[83, 108]], [[83, 115]], [[129, 151]]]", "query_spans": "[[[154, 163]]]", "process": "" }, { "text": "Through the focus $F$ of the parabola $y^{2}=4x$, draw a line perpendicular to the $x$-axis, intersecting 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;Focus(G)=F;Expression(G) = (y^2 = 4*x);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, 21]], [[1, 15]], [[0, 32]], [[22, 32]], [[30, 47]], [[49, 67]], [[57, 67]]]", "query_spans": "[[[66, 72]]]", "process": "Since the focus F of the parabola y^{2}=4x is (1,0) and the length of the latus rectum is |AB|=2p=4, the equation of the circle with F as center and AB as diameter is (x-1)^{2}+y^{2}=4." }, { "text": "A line $ l $ passing through the focus $ F $ of the parabola $ C: y^2 = 2px $ ($ p > 0 $) intersects $ C $ at points $ A $ and $ B $, where point $ A $ lies in the first quadrant. If $ |AF| = 3|BF| $, then what is the inclination angle of line $ l $?", "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;F: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Quadrant(A)=1;Abs(LineSegmentOf(A, F)) = 3*Abs(LineSegmentOf(B, F))", "query_expressions": "Inclination(l)", "answer_expressions": "ApplyUnit(60,degree)", "fact_spans": "[[[33, 38], [80, 85]], [[1, 27], [39, 42]], [[8, 27]], [[43, 46], [53, 57]], [[29, 32]], [[47, 50]], [[8, 27]], [[1, 27]], [[1, 32]], [[0, 38]], [[33, 52]], [[53, 62]], [[64, 78]]]", "query_spans": "[[[80, 91]]]", "process": "Draw AA, BB perpendicular from A, B to the directrix x = -\\frac{p}{2}, with feet at A, B respectively. Draw a perpendicular from B to AA, with foot at C. Using the definition of the parabola, the answer can be obtained. Draw AA, BB perpendicular from A, B to the directrix x = -\\frac{p}{2}, with feet at A, B respectively. Draw a perpendicular from B to AA, with foot at C. By the definition of the parabola, |BF| = |BB|, |AF| = |AA|, |BF| = \\frac{1}{4}|AB|. Therefore, |AC| = |AA| - |BB| = |AF| - |BF| = 2|BF| = \\frac{1}{2}|AB|. Since BC \\bot AA, it follows that \\angle BAC = 60^{\\circ}, so the inclination angle of line l is 60^{\\circ}." }, { "text": "If the parabola $C$: $y^{2}=2x$ has focus $F$, and two points $P_{1}$, $P_{2}$ on $C$ have $y$-coordinates $y_{1}$, $y_{2}$ respectively, with $y_{1}^{2}+y_{2}^{2}=8$, then $|P_{1}F|+|P_{2}F|$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);F: Point;Focus(C) = F;P1: Point;P2: Point;PointOnCurve(P1, C) ;PointOnCurve(P2, C) ;YCoordinate(P1) = y1;YCoordinate(P2) = y2;y1^2 + y2^2 = 8;y1:Number;y2:Number", "query_expressions": "Abs(LineSegmentOf(P1, F)) + Abs(LineSegmentOf(P2, F))", "answer_expressions": "5", "fact_spans": "[[[1, 20], [29, 32]], [[1, 20]], [[24, 27]], [[1, 27]], [[35, 42]], [[43, 50]], [[29, 50]], [[29, 50]], [[35, 73]], [[35, 73]], [[75, 98]], [[57, 64]], [[66, 73]]]", "query_spans": "[[[100, 123]]]", "process": "Let the x-coordinates of $P_{1}$ and $P_{2}$ be $x_{1}$, $x_{2}$. Since \n\\[\n\\begin{cases}\ny_{1}=2x_{1}\\\\\ny_{2}=2x_{2}\n\\end{cases}\n\\]\nit follows that $2(x_{1}+x_{2})=y_{1}^{2}+y_{2}^{2}=8$, so $x_{1}+x_{2}=4$. Also, since $|P_{1}F|+|P_{2}F|=(x_{1}+\\frac{p}{2})+(x_{2}+\\frac{p}{2})=x_{1}+x_{2}+p=4+1=5$" }, { "text": "Given that $A$ and $B$ are two points on the parabola $y^{2}=4x$, $O$ is the origin, $|OA|=|OB|$, and the orthocenter of $\\Delta AOB$ is exactly the focus of the parabola, then the area of $\\Delta AOB$ is?", "fact_expressions": "G: Parabola;A: Point;O: Origin;B: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(A,G);PointOnCurve(B,G);Abs(LineSegmentOf(O, A)) = Abs(LineSegmentOf(O, B));Orthocenter(TriangleOf(A,O,B))=Focus(G)", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "10*sqrt(5)", "fact_spans": "[[[10, 24], [74, 77]], [[2, 5]], [[30, 33]], [[6, 9]], [[10, 24]], [[2, 29]], [[2, 29]], [[40, 53]], [[55, 80]]]", "query_spans": "[[[82, 101]]]", "process": "" }, { "text": "The equation of the directrix of the parabola $x^{2}=\\frac{1}{3} y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = y/3)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-1/12", "fact_spans": "[[[0, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "According to the equation of the parabola, find the value of p, then obtain its directrix equation. Since the parabolic equation is: x^{2}=\\frac{1}{3}y, it follows that 2p=\\frac{1}{3}, so p=\\frac{1}{6}. Thus, the directrix equation is y=-\\frac{p}{2}=-\\frac{1}{12}" }, { "text": "Let $F$ be the focus of the parabola $C_{1}$: $y^{2}=2 p x(p>0)$, and let point $A$ be a common point of the parabola $C_{1}$ and an asymptote of the hyperbola $C_{2}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, such that $A F \\perp x$-axis. Then the eccentricity of the hyperbola is?", "fact_expressions": "C1: Parabola;Expression(C1) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(C1) = F;C2: Hyperbola;Expression(C2) = (x^2/a^2 - y^2/b^2 = 1);a: Number;b: Number;a>0;b>0;OneOf(Intersection(C1, OneOf(Asymptote(C2)))) = A;A: Point;IsPerpendicular(LineSegmentOf(A, F), xAxis) = True", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[5, 35], [44, 54]], [[5, 35]], [[17, 35]], [[17, 35]], [[1, 4]], [[1, 38]], [[55, 118], [148, 151]], [[55, 118]], [[67, 118]], [[67, 118]], [[67, 118]], [[67, 118]], [[39, 130]], [[39, 43]], [[132, 146]]]", "query_spans": "[[[148, 157]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, and let point $P$ lie on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/16 - y^2/9 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "9*sqrt(3)", "fact_spans": "[[[17, 56], [67, 70]], [[1, 8]], [[62, 66]], [[9, 16]], [[17, 56]], [[1, 61]], [[62, 71]], [[75, 108]]]", "query_spans": "[[[110, 137]]]", "process": "Let |PF₁| = m, |PF₂| = n (m > n), ∴ m - n = 2a = 8. ∵ (2c)² = m² + n² - 2mn cos60°, ∴ mn = 36, then the area of △F₁PF₂ is S = ½mn sin60° = 9√3." }, { "text": "The equation of the directrix of the parabola $y=4 x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-1/16", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "Test analysis: The standard equation of the parabola is x^{2}=\\frac{1}{4}y, so the directrix equation is y=-\\frac{1}{16}." }, { "text": "The standard equation of a circle with center at the focus of the parabola $y^{2}=-4 x$ and radius $\\sqrt{7}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -4*x);H: Circle;Focus(G) = Center(H);Radius(H) = sqrt(7)", "query_expressions": "Expression(H)", "answer_expressions": "(x+1)^2+y^2=7", "fact_spans": "[[[1, 16]], [[1, 16]], [[37, 38]], [[0, 38]], [[23, 38]]]", "query_spans": "[[[37, 45]]]", "process": "Since the focus of the parabola y^{2}=-4x is (-1,0), the standard equation of the required circle is (x+1)^{2}+y^{2}=7" }, { "text": "Given the parabola equation $y^{2}=4x$, and the line $l$: $x+y+\\sqrt{2}=0$, find the minimum distance from a moving point $P$ on the parabola to the line $l$.", "fact_expressions": "l: Line;G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);Expression(l) = (x + y + sqrt(2) = 0);PointOnCurve(P, G)", "query_expressions": "Min(Distance(P, l))", "answer_expressions": "(2 - sqrt(2))/2", "fact_spans": "[[[20, 43], [55, 60]], [[2, 5], [44, 47]], [[51, 54]], [[2, 19]], [[20, 43]], [[44, 54]]]", "query_spans": "[[[51, 69]]]", "process": "Let the equation of the line parallel to line $ l $ and tangent to the parabola be $ x + y + m = 0 $. From \n\\[\n\\begin{cases}\nx + y + m = 0 \\\\\ny^2 = 4x\n\\end{cases}\n\\]\nwe obtain $ y^{2} + 4y + 4m = 0 $. Then $ \\triangle = 16 - 16m = 0 $, so $ m = 1 $. Therefore, the tangent line equation is $ x + y + 1 = 0 $. Hence, the minimum distance from a moving point $ P $ on the parabola to line $ l $ is $ d = \\frac{|\\sqrt{2} - 1|}{\\sqrt{2}} = \\frac{2 - \\sqrt{2}}{2} $." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$, what is its eccentricity?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/4 - y^2/2 = 1)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[2, 45]], [[2, 45]]]", "query_spans": "[[[2, 51]]]", "process": "Test analysis: From the hyperbola equation, we know that a^{2}=4, b^{2}=2, \\therefore c^{2}=a^{2}+b^{2}=6, \\therefore a=2, c=\\sqrt{6}, \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{6}}{2}" }, { "text": "Given that the point $P(1,1)$ is the midpoint of a chord of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, then the general equation of the line containing this chord is?", "fact_expressions": "G: Ellipse;H: LineSegment;P:Point;Expression(G) = (x^2/4 + y^2/3=1);MidPoint(H)=P;Coordinate(P) = (1,1);IsChordOf(H,G)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "3*x+4*y-7=0", "fact_spans": "[[[12, 49]], [], [[2, 11]], [[12, 49]], [[2, 55]], [[2, 11]], [[12, 52]]]", "query_spans": "[[[12, 71]]]", "process": "Let the line passing through point P(1,1) intersect the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at two points A$(x_{1},y_{1})$, B$(x_{2},y_{2})$. Then $\\frac{x_{1}^{2}}{4}+\\frac{y_{1}^{2}}{3}=1$, $\\frac{x_{2}^{2}}{4}+\\frac{y_{2}^{2}}{3}=1$. Subtracting these two equations gives $\\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{4}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{3}=0$. Simplifying yields $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{3}{4}$, that is, $k_{AB}=-\\frac{3}{4}$. The equation of line AB is $y-1=-\\frac{3}{4}(x-1)$. Thus, the general equation of line AB is $3x+4y-7=0$." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has left focus at $F$. A perpendicular is drawn from point $F$ to one of the asymptotes of hyperbola $C$, with foot of the perpendicular at $H$. Point $P$ lies on the hyperbola such that $\\overrightarrow{F P}=3 \\overrightarrow{F H}$. Then the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;F: Point;P: Point;H: 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, OneOf(Asymptote(C)));FootPoint(l, OneOf(Asymptote(C)))=H;PointOnCurve(P, C);VectorOf(F, P) = 3*VectorOf(F, H)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(13)/2", "fact_spans": "[[[2, 63], [78, 84], [106, 109], [159, 162]], [[10, 63]], [[10, 63]], [[68, 71], [73, 77]], [[101, 105]], [[97, 100]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [], [[72, 93]], [[72, 93]], [[72, 100]], [[101, 110]], [[112, 157]]]", "query_spans": "[[[159, 168]]]", "process": "According to the vector condition, find the coordinates of P and substitute into the hyperbola equation to reach the conclusion. From the given conditions, let P(x, y), the equation of line FH is y = \\frac{a}{b}(x + c). Together with the asymptote y = -\\frac{b}{a}x, solving simultaneously gives the coordinates of H as \\left(-\\frac{c^{2}}{a}, \\frac{ab}{c}\\right). \n\\because \\overrightarrow{FP} = 3\\overrightarrow{FH}, that is, \n\\begin{cases} x + c = -\\frac{a^{2}}{c} + c \\\\ y = \\frac{3ab}{c} \\end{cases} \n\\therefore \\begin{cases} x = -\\frac{3a^{2}}{c} + 2c \\\\ y = \\frac{3ab}{c} \\end{cases} \nSubstituting into the hyperbola equation yields \n\\frac{\\left(-\\frac{3a^{2}}{c} + 2c\\right)^{2}}{a^{2}} - \\frac{9a^{2}}{c^{2}} = 1, \nsimplifying gives \\frac{4c^{2}}{a^{2}} = 13, \n\\therefore e = \\frac{c}{a} = \\frac{\\sqrt{13}}{2}" }, { "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ $(a>0)$ has eccentricity $2$, then what is the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/3 + x^2/a^2 = 1);a: Number;a>0;Eccentricity(G) = 2", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(3)*x)", "fact_spans": "[[[1, 48], [58, 61]], [[1, 48]], [[4, 48]], [[4, 48]], [[1, 56]]]", "query_spans": "[[[58, 69]]]", "process": "" }, { "text": "Given that point $F$ is the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and the line $y=2x-4$ passing through point $F$ intersects the hyperbola at exactly one point, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 2*x - 4);RightFocus(G)=F;PointOnCurve(F, H);NumIntersection(H,G)=1", "query_expressions": "Expression(G)", "answer_expressions": "5*x^2/4-5*y^2/16=1", "fact_spans": "[[[7, 63], [88, 91], [99, 102]], [[10, 63]], [[10, 63]], [[75, 86]], [[2, 6], [70, 74]], [[10, 63]], [[10, 63]], [[7, 63]], [[75, 86]], [[2, 67]], [[69, 86]], [[75, 97]]]", "query_spans": "[[[99, 107]]]", "process": "2x-4=0, we get x=2, so F(2,0). Since the line y=2x-4 passing through point F has only one intersection with this hyperbola, the slope of the asymptote is \\frac{b}{a}=2, c=2, a^{2}+b^{2}=c^{2}, so a^{2}=\\frac{4}{5}, b^{2}=\\frac{16}{5}, the equation of the hyperbola is \\frac{5x^{2}}{4}-\\frac{5y^{2}}{16}=1" }, { "text": "Given $ m>0 $, $ A(0, \\sqrt{m+1}) $, $ B(0, -\\sqrt{m+1}) $. If there exists a point $ P $ on the line $ y=3x $ such that $ |PB| - |PA| = 2 $, then what is the range of values for $ m $?", "fact_expressions": "G: Line;A: Point;B: Point;P: Point;Expression(G) = (y = 3*x);Coordinate(A) = (0, sqrt(m + 1));Coordinate(B) = (0, -sqrt(m + 1));m > 0;PointOnCurve(P, G);-Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, B)) = 2;m: Number", "query_expressions": "Range(m)", "answer_expressions": "(1/9, +oo)", "fact_spans": "[[[52, 61]], [[9, 28]], [[31, 49]], [[66, 69]], [[52, 61]], [[9, 28]], [[31, 49]], [[2, 7]], [[52, 69]], [[72, 87]], [[89, 92]]]", "query_spans": "[[[89, 99]]]", "process": "Since |PB| - |PA| = 2, the locus of P is the upper branch of a hyperbola with foci A and B and a transverse axis length of 2. The asymptotes of this hyperbola have equations y = \\pm\\frac{1}{\\sqrt{m}}x. Since there exists a point P on the line y = 3x such that |PB| - |PA| = 2, it follows that 3 > \\frac{1}{\\sqrt{m}}, solving which gives m > \\frac{1}{9}." }, { "text": "Given that the distance from a point on the parabola $y^{2}=4x$ to the focus is $5$, then the coordinates of this point are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);D: Point;PointOnCurve(D, G) = True;Distance(D, Focus(G)) = 5", "query_expressions": "Coordinate(D)", "answer_expressions": "(4,pm*4)", "fact_spans": "[[[2, 16]], [[2, 16]], [[35, 36]], [[2, 19]], [[2, 32]]]", "query_spans": "[[[35, 41]]]", "process": "By the definition of a parabola, we have x + 1 = 5, ∴ x = 4, ∴ y² = 4 × 4 ⇒ y = ±4, so the coordinates of this point are (4, ±4)." }, { "text": "The line $ l $ passing through the focus of the parabola $ E $: $ y^{2} = 4x $ intersects $ E $ at points $ A $ and $ B $. The tangents to $ E $ at points $ A $ and $ B $ intersect the $ y $-axis at points $ C $ and $ D $, respectively. What is the maximum value of $ 4 \\sqrt{2}|CD| - |AB| $?", "fact_expressions": "l: Line;E: Parabola;C: Point;D: Point;A: Point;B: Point;Expression(E) = (y^2 = 4*x);PointOnCurve(Focus(E), l);Intersection(l, E) = {A, B};Intersection(TangentOnPoint(A,E), yAxis) = C;Intersection(TangentOnPoint(B,E),yAxis)=D", "query_expressions": "Max(4*(sqrt(2)*Abs(LineSegmentOf(C, D)))-Abs(LineSegmentOf(A, B)))", "answer_expressions": "8", "fact_spans": "[[[23, 28]], [[1, 20], [29, 32], [44, 47]], [[69, 72]], [[73, 76]], [[34, 37], [48, 52]], [[38, 41], [53, 56]], [[1, 20]], [[0, 28]], [[23, 43]], [[44, 78]], [[44, 78]]]", "query_spans": "[[[80, 109]]]", "process": "From $ y^{2} = 4x $, $ y = 2\\sqrt{x} $, differentiate to get $ y' = \\frac{1}{\\sqrt{x}} $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. The slope of the tangent line at point $ A $ is $ k = \\frac{1}{\\sqrt{x_{1}}} $, so the tangent equation is $ y - y_{1} = \\frac{1}{\\sqrt{x_{1}}}(x - x_{1}) $. Let $ x = 0 $, solving gives $ y = \\sqrt{x_{1}} $, then $ C(0, \\sqrt{x_{1}}) $. Similarly, $ D(0, -\\sqrt{x_{2}}) $, so $ |CD| = \\sqrt{x_{1}} + \\sqrt{x_{2}} $. Let the equation of line $ AB $ be $ y = k(x - 1) $. Solving the system \n\\[\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\ngives $ k^{2}x^{2} - 2(k^{2} + 2)x + k^{2} = 0 $, so $ x_{1}x_{2} = 1 $. Therefore, $ |AB| = x_{1} + x_{2} + 2 = (\\sqrt{x_{1}} + \\sqrt{x_{2}})^{2} $. Then $ 4\\sqrt{2}|CD| - |AB| = 4\\sqrt{2}(\\sqrt{x_{1}} + \\sqrt{x_{2}}) - (\\sqrt{x_{1}} + \\sqrt{x_{2}})^{2} $. Let $ \\sqrt{x_{1}} + \\sqrt{x_{2}} = t $, $ t \\geqslant 2 $. Let $ f(t) = 4\\sqrt{2}t - t^{2} = -(t - 2\\sqrt{2})^{2} + 8 $, $ t \\geqslant 2 $. Therefore, when $ t = 2\\sqrt{2} $, $ f(t) $ reaches its maximum value, which is 8. Therefore, the maximum value of $ 4\\sqrt{2}|CD| - |AB| $ is 8. The answer is: 8." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. Let $M$ be any point on the ellipse $C$, and $N$ be any point on the circle $E$: $(x-4)^{2}+(y-3)^{2}=1$. Then the minimum value of $|M N|-|M F_{1}|$ is?", "fact_expressions": "C: Ellipse;E: Circle;M: Point;N: Point;F1:Point;F2:Point;LeftFocus(C)=F1;RightFocus(C)=F2;Expression(C) = (x^2/4+y^2/3=1);Expression(E)=((x-4)^2+(y-3)^2=1);PointOnCurve(M,C);PointOnCurve(N,E)", "query_expressions": "Min(-Abs(LineSegmentOf(M, F1))+Abs(LineSegmentOf(M, N)))", "answer_expressions": "3*sqrt(2)-5", "fact_spans": "[[[2, 44], [73, 78]], [[88, 116]], [[69, 72]], [[84, 87]], [[53, 60]], [[61, 68]], [[2, 68]], [[2, 68]], [[2, 44]], [[88, 116]], [[69, 83]], [[84, 121]]]", "query_spans": "[[[123, 146]]]", "process": "As shown in the figure, M is any point on the ellipse C, N is any point on the circle E: (x−4)^{2}+(y−3)^{2}=1, then |MF_{1}|+|MF_{2}|=4, |MN|⩾|ME|−1 (equality holds if and only if M, N, E are collinear). \n∴ \n|MN|−|MF_{1}|=|MN|−(4−|MF_{2}|)=|MN|+|MF_{2}|−4⩾|ME|+|MF_{2}|−5⩾|EF_{2}|−5, equality holds if and only if M, N, E, F_{2} are collinear. \n∵ F_{2}(1,0), E(4,3), then |EF_{2}|=\\sqrt{(4−1)^{2}+(3−0)^{2}}=3\\sqrt{2}, \n∴ the minimum value of |MN|−|MF_{1}| is 3\\sqrt{2}−5." }, { "text": "The minimum distance from a point on the parabola $y^{2}=4 x$ to the line $x-y+4=0$ is?", "fact_expressions": "G: Parabola;H: Line;P0: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x - y + 4 = 0);PointOnCurve(P0, G)", "query_expressions": "Min(Distance(P0, H))", "answer_expressions": "3*sqrt(2)/2", "fact_spans": "[[[0, 14]], [[18, 29]], [[16, 17]], [[0, 14]], [[18, 29]], [[0, 17]]]", "query_spans": "[[[16, 36]]]", "process": "Let the coordinates of point P on the parabola \\( y^{2} = 4x \\) be \\( \\left( \\frac{t^{2}}{4}, t \\right) \\). Then the distance from point P to the line \\( x - y + 4 = 0 \\) is \n\\[ d = \\frac{\\left| \\frac{1}{4}(t - 2)^{2} + 3 \\right|}{\\sqrt{2}} = \\frac{\\frac{1}{4}(t - 2)^{2} + 3}{\\sqrt{2}} \\geqslant \\frac{3\\sqrt{2}}{2} \\]" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$. A line $l$ passing through the origin $O$ intersects the ellipse $C$ at points $A$ and $B$, and $2|F O|=|A B|$. If $\\angle B A F=\\frac{\\pi}{6}$, then the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;F: Point;LeftFocus(C) = F;O: Origin;l: Line;PointOnCurve(O, l);Intersection(l, C) = {A, B};A: Point;B: Point;2*Abs(LineSegmentOf(F, O)) = Abs(LineSegmentOf(A, B));AngleOf(B, A, F) = pi/6", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3) - 1", "fact_spans": "[[[2, 59], [81, 86], [143, 148]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[64, 67]], [[2, 67]], [[69, 74]], [[75, 80]], [[68, 80]], [[75, 95]], [[87, 91]], [[92, 95]], [[97, 111]], [[113, 141]]]", "query_spans": "[[[143, 154]]]", "process": "Let the right focus be F, connect AF, BF. Since 2|OF| = |AB| = 2c, that is, |FF| = |AB|, it follows that quadrilateral AFBF is a rectangle. In right triangle \\triangle ABF, \n\\begin{matrix}\\\\AF\\end{matrix} = 2c \\cdot \\cos\\angle BAF = 2c \\cdot \\frac{\\sqrt{3}}{2} = \\sqrt{3}c, |BF| = 2c \\cdot \\sin\\angle BAF = 2c \\cdot \\frac{1}{2} = c. By the definition of the ellipse, |AF| + |AF| = 2a, so 2a = (\\sqrt{3} + 1)c, thus the eccentricity e = \\frac{c}{a} = \\frac{2}{\\sqrt{3} + 1} = \\sqrt{3} - 1" }, { "text": "The line $ l $ passes through the focus of the parabola $ C: y^{2} = 2 p x $ ($ p > 0 $), and intersects $ C $ at points $ A $ and $ B $, with $ |A B| = 4 $. If the distance from the midpoint of $ A B $ to the $ y $-axis is $ 1 $, then the value of $ p $ is?", "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(Focus(C), l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, B)) = 4;Distance(MidPoint(LineSegmentOf(A, B)), yAxis) = 1", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[0, 5]], [[6, 31], [37, 40]], [[85, 88]], [[42, 45]], [[46, 49]], [[13, 31]], [[6, 31]], [[0, 34]], [[0, 51]], [[52, 61]], [[63, 83]]]", "query_spans": "[[[85, 92]]]", "process": "According to the problem, the parabola $ C: y^{2} = 2px $ has focus $ F\\left(\\frac{p}{2}, 0\\right) $. Let the equation of line $ l $ be $ x = my + \\frac{p}{2} $, and let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system of equations\n$$\n\\begin{cases}\nx = my + \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n$$\nsubstituting $ x = my + \\frac{p}{2} $ into $ y^{2} = 2px $, we obtain $ y^{2} - 2pmy - p^{2} = 0 $. Thus, $ y_{1} + y_{2} = 2pm $, and then $ x_{1} + x_{2} = m(y_{1} + y_{2}) + p = 2pm^{2} + p $. Since $ |AB| = 4 $, we have $ |AB| = x_{1} + x_{2} + p = 2pm^{2} + 2p = 4 $, that is, $ p(m^{2} + 1) = 2 $. Also, since the distance from the midpoint of $ AB $ to the $ y $-axis is 1, we get $ \\frac{x_{1} + x_{2}}{2} = \\frac{2pm^{2} + p}{2} = 1 $, i.e., $ p(2m^{2} + 1) = 2 $. Solving the system of equations, we obtain $ m = 0 $, and thus $ p = 2 $." }, { "text": "Given that the vertex of the parabola is at the origin, the axis of symmetry is a coordinate axis, and it passes through the point $M(-5, -10)$, then the standard equation of the parabola is?", "fact_expressions": "G: Parabola;M: Point;O: Origin;Coordinate(M) = (-5, -10);Vertex(G) = O;SymmetryAxis(G)=axis;PointOnCurve(M,G)", "query_expressions": "Expression(G)", "answer_expressions": "{x^2=(-5/2)*y,y^2=-20*x}", "fact_spans": "[[[2, 5], [39, 42]], [[24, 37]], [[9, 11]], [[24, 37]], [[2, 11]], [[2, 19]], [[2, 37]]]", "query_spans": "[[[39, 49]]]", "process": "The solution process is omitted" }, { "text": "The ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Point $P$ lies on the ellipse. If $|P F_{1}|=4$, then $\\angle F_{1} P F_{2}$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) = True;Abs(LineSegmentOf(P, F1)) = 4", "query_expressions": "AngleOf(F1, P, F2)", "answer_expressions": "ApplyUnit(90, degree)", "fact_spans": "[[[0, 37], [66, 68]], [[0, 37]], [[45, 52]], [[53, 60]], [[0, 60]], [[0, 60]], [[61, 65]], [[61, 69]], [[71, 84]]]", "query_spans": "[[[86, 110]]]", "process": "According to the problem, the ellipse is given by $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, where $a=3$, $b=2$, then $c=\\sqrt{5}$. Point $P$ lies on the ellipse. If $|PF_{1}|=4$, then $|PF_{2}|=2a-|PF_{1}|=6-4=2$. In $\\triangle F_{1}PF_{2}$, $|PF_{1}|=4$, $|PF_{2}|=2$, $|F_{1}F_{2}|=2c=2\\sqrt{5}$, then $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}$, hence $\\angle F_{1}PF_{2}=90^{\\circ}$." }, { "text": "The equation of the ellipse with the foci of $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ as its vertices and the vertices of $C$ as its foci is?", "fact_expressions": "G: Ellipse;C: Curve;Expression(C) = (x^2/4 - y^2/5 = 1) ;Focus(C) = Vertex(G);Vertex(C) = Focus(G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 + y^2/5 = 1", "fact_spans": "[[[53, 55]], [[1, 40]], [[1, 40]], [[0, 55]], [[0, 55]]]", "query_spans": "[[[53, 60]]]", "process": "" }, { "text": "The hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ $(m>0)$ has eccentricity $2$. Then, the distance from one of its foci to an asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/m = 1);m: Number;m>0;Eccentricity(G) = 2", "query_expressions": "Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 33], [43, 44]], [[0, 33]], [[3, 33]], [[3, 33]], [[0, 41]]]", "query_spans": "[[[43, 60]]]", "process": "Given the eccentricity of the hyperbola is 2, we have \\sqrt{1+m}=2, so m=3. Thus, the equation of the hyperbola is x^{2}-\\frac{y^{2}}{3}=1, the coordinates of the foci are (2,0), and the equations of the asymptotes are y=\\pm\\sqrt{3}x. Therefore, the distance from a focus to an asymptote is d=\\frac{|2\\sqrt{3}+0|}{1+(\\sqrt{3})^{2}}=\\sqrt{3}." }, { "text": "The distance from the focus of the parabola $y^{2}=4 m x(m>0)$ to an asymptote of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ is $3$. What is the equation of this parabola?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 4*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], [80, 83]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 64]], [[25, 64]], [[0, 77]]]", "query_spans": "[[[80, 87]]]", "process": "The focus of the parabola y^{2}=4mx (m>0) is F(m,0). The asymptotes of the hyperbola \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1 are given by 3x\\pm4y=0. Then the distance from F(m,0) to the asymptote is \\frac{|3m|}{\\sqrt{3^{2}+4^{2}}}=\\frac{3m}{5}=3\\Rightarrow m=5. Therefore, the equation of the parabola is y^{2}=20x." }, { "text": "The equation of the directrix of the parabola $x^{2}=16 y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 16*y)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-4", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "The focus of the parabola \\( x^{2} = 16y \\) lies on the y-axis, and \\( \\frac{p}{2} = 4 \\), so its directrix equation is: \\( y = -4 \\)." }, { "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 $8$, the right focus is $F$, $M$ is a point on one of the asymptotes of hyperbola $C$, and $OM \\perp MF$, where $O$ is the origin. If $S_{\\triangle OMF}=6$, then the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;M: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Length(RealAxis(C)) = 8;RightFocus(C)=F;PointOnCurve(M,OneOf(Asymptote(C)));IsPerpendicular(LineSegmentOf(O, M), LineSegmentOf(M, F));Area(TriangleOf(O,M,F)) = 6", "query_expressions": "Eccentricity(C)", "answer_expressions": "5/4", "fact_spans": "[[[2, 63], [84, 90], [152, 158]], [[10, 63]], [[10, 63]], [[117, 120]], [[80, 83]], [[76, 79]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[2, 79]], [[80, 99]], [[101, 116]], [[127, 150]]]", "query_spans": "[[[152, 164]]]", "process": "Given the length of the real axis is 8, then a=4, OM\\botMF, so MF=b. In the right triangle \\triangleOFM, OF=c, then OM=a, so S_{\\triangleOFM}=\\frac{1}{2}ab=6, from which b can be found, and thus the eccentricity of the hyperbola can be obtained. The length of the real axis of the hyperbola is 8, so a=4, OM\\botMF, meaning MF is the distance from the focus to the asymptote, hence MF=b, and OF=c, so in the right triangle \\triangleOFM, OM=a, then S_{\\triangleOFM}=\\frac{1}{2}ab=6, yielding b=3, c=: therefore e=\\frac{c}{a}=\\frac{5}{4}" }, { "text": "Suppose the foci of a hyperbola lie on the coordinate axes, and the equations of its two asymptotes are $y = \\pm \\frac{1}{3}x$. Then the eccentricity $e$ of this hyperbola is?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G),axis) = True;Expression(Asymptote(G)) = (y = pm*(x/3));e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "(sqrt(10),sqrt(10)/3)", "fact_spans": "[[[1, 4], [45, 48]], [[1, 12]], [[1, 42]], [[52, 55]], [[45, 55]]]", "query_spans": "[[[52, 57]]]", "process": "Analysis: Based on the asymptote equations of the hyperbola, assume the standard equation of the hyperbola is \\frac{x^{2}}{9}-y^{2}=m (m\\neq0). By classifying and discussing the value of m, it can be concluded that the eccentricities of the hyperbola are \\sqrt{10} or \\frac{\\sqrt{10}}{3}. Specifically, since the asymptote equations of the hyperbola are y=\\pm\\frac{1}{3}x, the standard equation of the hyperbola can be assumed as \\frac{x^{2}}{9}-y^{2}=m (m\\neq0). When m>0, the foci of the hyperbola lie on the x-axis, in which case e^{2}=\\frac{c^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=\\frac{10}{9}, so e=\\frac{\\sqrt{10}}{3}. When m<0, the foci of the hyperbola lie on the y-axis, in which case e^{2}=\\frac{c^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}=10, so e=\\sqrt{10}. In conclusion, the eccentricity of the hyperbola is \\sqrt{10} or \\frac{\\sqrt{10}}{3}." }, { "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively, a line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Then the maximum value of $|\\overrightarrow{B F_{2}}|+|\\overrightarrow{A F_{2}}|$ is?", "fact_expressions": "l: Line;G: Ellipse;B: Point;F2: Point;A: Point;F1: Point;Expression(G) = (x^2/9 + y^2/4 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, G) = {A, B}", "query_expressions": "Max(Abs(VectorOf(A, F2)) + Abs(VectorOf(B, F2)))", "answer_expressions": "28/3", "fact_spans": "[[[72, 77]], [[2, 39], [78, 80]], [[85, 88]], [[55, 62]], [[81, 84]], [[47, 54], [64, 71]], [[2, 39]], [[2, 62]], [[2, 62]], [[63, 77]], [[72, 90]]]", "query_spans": "[[[92, 153]]]", "process": "Test analysis: From the ellipse equation, c = \\sqrt{9 - 4} = \\sqrt{5}. Also, according to the definition of the ellipse, |BF_{2}| + |AF_{2}| + |BF_{1}| + |AF_{1}| = 6, that is, |BF_{2}| + |AF_{2}| + |AB| = 12. Therefore, when |AB| is minimum, i.e., when AB is perpendicular to the x-axis, |\\overrightarrow{BF_{2}}| + |\\overrightarrow{AF_{2}}| is maximum. At this time, it is easy to see that |AB| = \\frac{8}{3}, so the maximum value of |\\overrightarrow{BF_{2}}| + |\\overrightarrow{AF_{2}}| is 12 - \\frac{8}{3} = \\frac{28}{3}. The answer should be filled in as: \\frac{28}{3}" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with left focus $F$, a line passing through the origin $O$ intersects $C$ at points $A$ and $B$. Connect $BF$. If $|OA|=|OF|=5$, $|BF|=8$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;B: Point;F: Point;O: Origin;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(O, A)) = Abs(LineSegmentOf(O, F));Abs(LineSegmentOf(O, F)) = 5;Abs(LineSegmentOf(B, F)) = 8", "query_expressions": "Eccentricity(C)", "answer_expressions": "5", "fact_spans": "[[[2, 63], [82, 85], [135, 138]], [[10, 63]], [[10, 63]], [[79, 81]], [[92, 95]], [[68, 71]], [[73, 78]], [[88, 91]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[72, 81]], [[79, 97]], [[107, 122]], [[107, 122]], [[124, 133]]]", "query_spans": "[[[135, 144]]]", "process": "Denote the right focus of the hyperbola as F_{1}, connect AF_{1} and BF_{1}. Since points A and B are symmetric about the origin O and both lie on the hyperbola C, we have |OA| = |OB| = 5, |AF| = |BF_{1}|; also |OF| = |OF_{1}| = 5. Therefore, |F_{1}F_{2}| = |AB|, so quadrilateral AF_{1}BF_{2} is a rectangle; hence AF \\bot BF, so |AF| = \\sqrt{10^{2} - 8^{2}} = 6, thus |AF_{1}| - |AF| = 2a = 2, i.e., a = 1. Also c = |OF| = 5, so the eccentricity is \\frac{c}{a} = 5." }, { "text": "If the line $y = kx + 1$ and the hyperbola $x^2 - y^2 = 1$ have exactly one common point, then the value of $k$ is?", "fact_expressions": "G: Hyperbola;H: Line;k: Number;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = k*x + 1);NumIntersection(H, G) = 1", "query_expressions": "k", "answer_expressions": "{sqrt(2),1}", "fact_spans": "[[[13, 31]], [[1, 12]], [[42, 45]], [[13, 31]], [[1, 12]], [[1, 40]]]", "query_spans": "[[[42, 49]]]", "process": "" }, { "text": "Given point $A(0,-2)$, the eccentricity of ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$, $F$ is the focus of the ellipse, the slope of line $AF$ is $\\frac{2 \\sqrt{3}}{3}$, what is the standard equation of $E$?", "fact_expressions": "E: Ellipse;b: Number;a: Number;A: Point;F: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (0,-2);Eccentricity(E) = sqrt(3)/2;Focus(E) = F;Slope(LineOf(A,F))=2*sqrt(3)/3", "query_expressions": "Expression(E)", "answer_expressions": "x^2/4+y^2=1", "fact_spans": "[[[13, 69], [101, 103], [143, 146]], [[19, 69]], [[19, 69]], [[2, 12]], [[97, 100]], [[19, 69]], [[19, 69]], [[13, 69]], [[2, 12]], [[13, 94]], [[97, 106]], [[107, 140]]]", "query_spans": "[[[143, 153]]]", "process": "According to the problem, F(c,0), then \n\\begin{cases}\\frac{c}{a}=\\frac{\\sqrt{3}}{2}\\\\\\frac{-2-0}{0-c}=\\frac{2\\sqrt{3}}{3}\\\\a^{2}b^{2}+c^{3}\\end{cases} \nSolving gives a=2, b=1, c=\\sqrt{3}. Therefore, the equation of the ellipse is \\frac{x^{2}}{4}+y^{2}=1." }, { "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{12-m}=1$ $(0\\sqrt{2}$. What is the range of values for $m$?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/(12 - m) + x^2/m = 1);m: Number;0 < m;m < 12;k: Number;Slope(OneOf(Asymptote(G))) = k;k > sqrt(2)", "query_expressions": "Range(m)", "answer_expressions": "(0, 4)", "fact_spans": "[[[0, 49]], [[0, 49]], [[72, 75]], [[3, 49]], [[3, 49]], [[58, 70]], [[0, 70]], [[58, 70]]]", "query_spans": "[[[72, 82]]]", "process": "Since the slope $ k > \\sqrt{2} $ of an asymptote of $ \\frac{x^{2}}{m} - \\frac{y^{2}}{12-m} = 1 $ $ (0 < m < 12) $, it follows that $ \\frac{b^{2}}{a^{2}} = \\frac{12-m}{m} > 2 $, so $ m < 4 $. Since $ 0 < m < 12 $, the range of values for $ m $ is $ (0, 4) $." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left focus $F$ and right vertex $A$. A point $P$ moves on $C$. If the maximum value of $|P F|$ is $3$ and the minimum value is $1$, then what is the maximum value of $\\overrightarrow{P A} \\cdot \\overrightarrow{P F}$?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F: Point;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;RightVertex(C) = A;PointOnCurve(P, C);Max(Abs(LineSegmentOf(P, F))) = 3;Min(Abs(LineSegmentOf(P, F))) = 1", "query_expressions": "Max(DotProduct(VectorOf(P, A), VectorOf(P, F)))", "answer_expressions": "4", "fact_spans": "[[[0, 57], [79, 82]], [[7, 57]], [[7, 57]], [[74, 78]], [[62, 65]], [[70, 73]], [[7, 57]], [[7, 57]], [[0, 57]], [[0, 65]], [[0, 73]], [[74, 86]], [[88, 103]], [[88, 111]]]", "query_spans": "[[[113, 168]]]", "process": "Find the equation of the ellipse as \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1, let P(x,y), find \\overrightarrow{PA}\\cdot\\overrightarrow{PF}=\\frac{1}{4}(x-2)^{2}, then use the graph and properties of quadratic functions to solve. And solve] From the problem we get \\begin{cases}a+c=3\\\\a-c=1\\end{cases}, \\therefore a=2,c=1, \\therefore b=\\sqrt{3}, \\therefore \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1, \\therefore y^{2}=3-\\frac{3}{4}x^{2}. Let P(x,y), since F(-1,0), A(2,0), so \\overrightarrow{PA}=(2-x,-y), \\overrightarrow{PF}=(-1-x,-y). Therefore \\overrightarrow{PA}\\cdot\\overrightarrow{PF}=(2-x)(-1-x)+y^{2}=x^{2}+y^{2}-x-2=x^{2}+3-\\frac{3}{4}x^{2}-x-2=\\frac{1}{4}x^{2}-x+1=(\\frac{1}{2}x-1)^{2}=\\frac{1}{4}(x-2)^{2}. Since -2\\leqslant x\\leqslant 2, \\therefore when x=-2, \\overrightarrow{PA}\\cdot\\overrightarrow{PF} has maximum value 4." }, { "text": "Given that the distance from the point $(-2 , 3)$ to the focus of the parabola $y^{2}=2 p x(p>0)$ is $5$, find $p$.", "fact_expressions": "G: Parabola;p: Number;H: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (-2, 3);Distance(H, Focus(G)) = 5", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[14, 35]], [[47, 50]], [[2, 13]], [[17, 35]], [[14, 35]], [[2, 13]], [[2, 45]]]", "query_spans": "[[[47, 52]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the distance from point $P$ to the $x$-axis equals?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1);F1: Point;F2: Point;LeftFocus(G)=F1;RightFocus(G)=F2;P: Point;PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)", "query_expressions": "Distance(P, xAxis)", "answer_expressions": "3", "fact_spans": "[[[2, 41], [70, 73]], [[2, 41]], [[50, 57]], [[58, 65]], [[2, 65]], [[2, 65]], [[66, 69], [111, 115]], [[66, 74]], [[76, 109]]]", "query_spans": "[[[111, 126]]]", "process": "Assume point P lies on the right branch of the hyperbola, |PF_{1}| = m, |PF_{2}| = n, then \n\\begin{cases}m - n = 4 \\\\ m^{2} + n^{2} = 64\\end{cases} \nmn = 24. Let h be the distance from point P to the x-axis. Using equal areas, we get \n\\frac{1}{2} \\times 8 \\times h = \\frac{1}{2} mn \n\\therefore h = 3" }, { "text": "If the length of a chord passing through the focus of a hyperbola and perpendicular to the real axis of the hyperbola is equal to $2$ times the distance from the focus to the asymptote, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;L: LineSegment;IsChordOf(L, G);PointOnCurve(Focus(G), L);IsPerpendicular(L, RealAxis(G));Length(L) = 2*Distance(Focus(G), Asymptote(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 5], [9, 12], [39, 42]], [], [[9, 19]], [[1, 19]], [[8, 19]], [[9, 36]]]", "query_spans": "[[[39, 48]]]", "process": "Let the equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0). Without loss of generality, assume F(c,0), and one asymptote is ay-bx=0. Then the distance from the focus to the asymptote is d=\\frac{|0-bc|}{\\sqrt{a^{2}+b^{2}}}=b. The length of the chord passing through the focus of the hyperbola and perpendicular to the real axis is \\frac{2b^{2}}{a}. \\because the length of the chord passing through the focus of the hyperbola and perpendicular to the real axis equals twice the distance from the focus to the asymptote, \\therefore \\frac{2b^{2}}{a}=2b, solving gives a=b. \\therefore c=\\sqrt{a^{2}+b^{2}}=\\sqrt{2}a, \\therefore e=\\frac{c}{a}=\\frac{\\sqrt{2}a}{a}=\\sqrt{2}. The problem provides conditions satisfied by a hyperbola and asks for the eccentricity of the hyperbola. It mainly examines standard equations of hyperbolas and simple geometric properties, and is a basic problem." }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2$, and its foci coincide with the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, then what is the standard equation of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/25 + y^2/9 = 1);Eccentricity(G) = 2;Focus(G) = Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4 - y^2/12=1", "fact_spans": "[[[2, 58], [116, 119]], [[5, 58]], [[5, 58]], [[70, 108]], [[5, 58]], [[5, 58]], [[2, 58]], [[70, 108]], [[2, 66]], [[2, 113]]]", "query_spans": "[[[116, 126]]]", "process": "From the ellipse equation $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, the foci are $(4,0)$, $(-4,0)$. Let the semi-focal length of the hyperbola be $c$, then $c=4$. Since the eccentricity of the hyperbola is $2$, we have $e=\\frac{c}{a}=\\frac{4}{a}=2$, hence $a=2$, and so $b=\\sqrt{c^{2}-a^{2}}=2\\sqrt{3}$. Therefore, the standard equation of the hyperbola is: $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$." }, { "text": "The parabola $C$: $y^{2}=2 px(p>0)$ has focus $F$ and directrix $l$. $M$ is a moving point on the parabola $C$, $A(0 , 3)$. Draw $MN$ perpendicular to the directrix $l$, with foot of perpendicular at $N$. If the minimum value of $|MN|+|MA|$ is $2$, then the equation of the parabola $C$ is?", "fact_expressions": "C: Parabola;p: Number;M: Point;N: Point;A: Point;F: Point;l: Line;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(A) = (0, 3);Focus(C) = F;Directrix(C) = l;PointOnCurve(M, C);PointOnCurve(M,l);IsPerpendicular(LineSegmentOf(M,N),l);Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, N))) = 2;FootPoint(LineSegmentOf(M,N),l)=N", "query_expressions": "Expression(C)", "answer_expressions": "", "fact_spans": "[[[0, 25], [46, 52], [114, 120]], [[8, 25]], [[41, 45], [69, 72]], [[88, 91]], [[57, 67]], [[29, 32]], [[36, 39], [81, 84]], [[8, 25]], [[0, 25]], [[57, 67]], [[0, 32]], [[0, 39]], [[42, 56]], [[68, 77]], [[73, 84]], [[93, 112]], [[73, 91]]]", "query_spans": "[[[114, 125]]]", "process": "" }, { "text": "The standard equation of an ellipse with eccentricity $e=\\frac{2}{3}$ and focal distance $2c=16$ is?", "fact_expressions": "G: Ellipse;FocalLength(G) = 2*c ;2*c = 16;c: Number;e: Number;Eccentricity(G) = e;e = 2/3", "query_expressions": "Expression(G)", "answer_expressions": "x^2/144 + y^2/80 = 1, x^2/80 + y^2/144 = 1", "fact_spans": "[[[30, 32]], [[19, 32]], [[21, 29]], [[21, 29]], [[3, 18]], [[0, 32]], [[3, 18]]]", "query_spans": "[[[30, 39]]]", "process": "Test analysis: ∴ e = \\frac{c}{a} = \\frac{2}{3}, 2c = 16 ∴ c = 8, a = 12 ∴ b^{2} = 80, the equation of the ellipse is \\frac{x^{2}}{144} + \\frac{y^{2}}{80} = 1 or \\frac{x^{2}}{80} + \\frac{y^{2}}{144} = 1" }, { "text": "The line $ l $ with slope $ 1 $ passes through the focus of the parabola $ y^2 = 4x $ and intersects the parabola at points $ A $ and $ B $. Then $ |AB| = $?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;PointOnCurve(Focus(G), l);Slope(l) = 1;Intersection(l, G) = {A, B};A: Point;B: Point", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[1, 15], [32, 35]], [[1, 15]], [[25, 30]], [[0, 30]], [[18, 30]], [[25, 47]], [[38, 41]], [[42, 45]]]", "query_spans": "[[[49, 58]]]", "process": "\\because the line passes through the focus (1,0) of the parabola y^{2}=4x, and has slope 1, \\therefore the equation of the line is y=x-1. Let A(x_{1},y_{1}), B(x_{2},y_{2}), and let the focus of the parabola be F. \\therefore according to the definition of the parabola: |AB|=|AF|+|BF|=x_{1}+1+x_{2}+1. Solving the system of equations \\begin{cases} y^{2}=4x \\\\ y=x-1 \\end{cases}, simplifying yields x^{2}-6x+1=0. \\therefore x_{1}+x_{2}=6. \\therefore |AB|=x_{1}+1+x_{2}+1=8" }, { "text": "Given that $B$ and $C$ are two fixed points, $|BC| = 6$, and the perimeter of $\\triangle ABC$ is equal to $16$, then the equation of the locus of vertex $A$ is?", "fact_expressions": "B: Point;C: Point;Abs(LineSegmentOf(B, C)) = 6;Perimeter(TriangleOf(A, B, C)) = 16;A: Point", "query_expressions": "LocusEquation(A)", "answer_expressions": "(x^2/25+y^2/16=1)&Negation(y=0)", "fact_spans": "[[[2, 5]], [[6, 9]], [[15, 24]], [[26, 52]], [[56, 59]]]", "query_spans": "[[[56, 66]]]", "process": "" }, { "text": "The line passing through the focus $F$ of the parabola $y^{2}=4x$ with an inclination angle of $\\frac{\\pi}{4}$ intersects the parabola at points $A$ and $B$. $||FB|-|FA||$=?", "fact_expressions": "G: Parabola;H: Line;F: Point;B: Point;A: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, H);Inclination(H) = pi/4;Intersection(H, G) = {A, B}", "query_expressions": "Abs(-Abs(LineSegmentOf(F, A)) + Abs(LineSegmentOf(F, B)))", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[1, 15], [45, 48]], [[42, 44]], [[18, 21]], [[53, 56]], [[49, 52]], [[1, 15]], [[1, 21]], [[0, 44]], [[22, 44]], [[42, 58]]]", "query_spans": "[[[59, 77]]]", "process": "It is easy to obtain that the focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, and the directrix is $ x = -1 $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From $ \\begin{cases} y = x - 1 \\\\ y^{2} = 4x \\end{cases} $, we get $ x^{2} - 6x + 1 = 0 $. Solving gives $ x_{1} = 3 + 2\\sqrt{2} $, $ x_{2} = 3 - 2\\sqrt{2} $. By the definition of the parabola, we have $ |FB| - |FA| = |x_{1} + 1 - (x_{2} + 1)| = 4\\sqrt{2} $." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ has its right focus at $F$. If a line passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the right branch of the hyperbola at exactly one point, then what is the range of values for 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;F: Point;RightFocus(G) = F;H: Line;PointOnCurve(F, H);Inclination(H) = ApplyUnit(60, degree);NumIntersection(H, RightPart(G)) = 1", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[2, +oo)", "fact_spans": "[[[2, 61], [97, 100], [114, 117]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[66, 69], [72, 76]], [[2, 69]], [[94, 96]], [[71, 96]], [[77, 96]], [[94, 111]]]", "query_spans": "[[[114, 128]]]", "process": "" }, { "text": "Given that 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;m: Number;H: Line;Expression(G) = (y^2 = m*x);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], [35, 38]], [[4, 15]], [[19, 26]], [[1, 15]], [[19, 26]], [[1, 33]]]", "query_spans": "[[[35, 43]]]", "process": "" }, { "text": "The equation $\\frac{x^{2}}{2 m}-\\frac{y^{2}}{m-1}=1$ represents an ellipse with foci on the $y$-axis. What is the range of values for $m$?", "fact_expressions": "G: Ellipse;Expression(G) = (-y^2/(m - 1) + x^2/(2*m) = 1);PointOnCurve(Focus(G), yAxis);m:Number", "query_expressions": "Range(m)", "answer_expressions": "(0,1/3)", "fact_spans": "[[[52, 54]], [[0, 54]], [[43, 54]], [[56, 59]]]", "query_spans": "[[[56, 66]]]", "process": "" }, { "text": "Given that the focus of the parabola $C$: $y^{2}=m x$ is the left focus of the ellipse $E$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, what is the equation of the directrix of the parabola $C$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = m*x);m: Number;E: Ellipse;Expression(E) = (x^2/9 + y^2/5 = 1);Focus(C) = LeftFocus(E)", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "x=2", "fact_spans": "[[[2, 21], [73, 79]], [[2, 21]], [[10, 21]], [[25, 67]], [[25, 67]], [[2, 71]]]", "query_spans": "[[[73, 86]]]", "process": "In the ellipse E: \\frac{x^2}{9} + \\frac{y^{2}}{5} = 1, a^{2} = 9, b^{2} = 5, c^{2} = 9 - 5 = 4, c = 2. Thus the focus of the parabola C is (-2, 0), so its directrix equation is x = 2." }, { "text": "If the equation $\\frac{x^{2}}{m+3}+\\frac{y^{2}}{4-m}=1$ represents an ellipse with foci on the $x$-axis, then what is the range of values for $m$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m + 3) + y^2/(4 - m) = 1);PointOnCurve(Focus(G), xAxis);m:Number", "query_expressions": "Range(m)", "answer_expressions": "(1/2,4)", "fact_spans": "[[[53, 55]], [[1, 55]], [[44, 55]], [[57, 60]]]", "query_spans": "[[[57, 67]]]", "process": "From the given conditions, we have \n\\begin{cases}m+3>0\\\\4-m>0\\\\m+3>4-m\\end{cases}, \nsolving yields $\\frac{1}{2}0)$, a line passing through its focus with slope $1$ intersects the parabola at points $A$ and $B$. If the ordinate of the midpoint of segment $AB$ is $2$, then what is the equation of the directrix of this parabola?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Line;PointOnCurve(Focus(G), H);Slope(H) = 1;A: Point;B: Point;Intersection(H, G) = {A, B};YCoordinate(MidPoint(LineSegmentOf(A, B))) = 2", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-1", "fact_spans": "[[[2, 23], [39, 42], [75, 78], [25, 26]], [[2, 23]], [[5, 23]], [[5, 23]], [[36, 38]], [[24, 38]], [[29, 38]], [[43, 46]], [[47, 50]], [[36, 52]], [[54, 72]]]", "query_spans": "[[[75, 85]]]", "process": "" }, { "text": "Given the parabola $ C $: $ y^{2} = 4x $, a line passing through the focus $ F $ with an inclination angle of $ 60^{\\circ} $ intersects the parabola $ C $ at points $ A $ and $ B $, and $ |A F| > |B F| $. Then $ \\frac{|A F|}{|B F|} $ = ?", "fact_expressions": "C: Parabola;l: Line;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Focus(C)=F;PointOnCurve(F, l);Inclination(l)=ApplyUnit(60,degree);Intersection(l,C) = {A, B};Abs(LineSegmentOf(A, F)) > Abs(LineSegmentOf(B, F))", "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))", "answer_expressions": "3", "fact_spans": "[[[2, 20], [48, 54]], [[45, 47]], [[55, 58]], [[24, 27]], [[59, 62]], [[2, 20]], [[2, 27]], [[21, 47]], [[28, 47]], [[45, 64]], [[66, 79]]]", "query_spans": "[[[81, 104]]]", "process": "The focus of the parabola y^{2}=2px (p>0) has coordinates (\\frac{p}{2},0). Since the inclination angle of line l is 60^{\\circ}, the equation of line l is: y-0=\\sqrt{3}(x-\\frac{p}{2}). Let the intersection points of the line and the parabola be A(x_{1},y_{1}) and B(x_{2},y_{2}). Therefore, |AF|=x_{1}+\\frac{p}{2}, |BF|=x_{2}+\\frac{p}{2}. Solving the system of equations, eliminating y and simplifying, we obtain 12x^{2}-20px+3p^{2}=0. Solving this equation gives x_{1}=\\frac{3p}{2}, x_{2}=\\frac{p}{6}. Therefore, |AF|=x_{1}+\\frac{p}{2}=2p, |BF|=x_{2}+\\frac{p}{2}=\\frac{2p}{3}. Hence, |AF|:|BF|=3:1, and the value of \\frac{|AF|}{|BF|} is 3." }, { "text": "Given the hyperbola $\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$ has foci $F_{1}$, $F_{2}$, and point $M$ lies on the hyperbola such that $M F_{1} \\perp M F_{2}$, find the distance from point $M$ to the $x$-axis.", "fact_expressions": "G: Hyperbola;M: Point;F1: Point;F2: Point;Expression(G) = (x^2/5 - y^2/4 = 1);Focus(G) = {F1,F2};PointOnCurve(M, G);IsPerpendicular(LineSegmentOf(M, F1), LineSegmentOf(M, F2))", "query_expressions": "Distance(M, xAxis)", "answer_expressions": "4/3", "fact_spans": "[[[2, 40], [65, 68]], [[60, 64], [95, 99]], [[44, 51]], [[52, 59]], [[2, 40]], [[2, 59]], [[60, 69]], [[70, 93]]]", "query_spans": "[[[95, 109]]]", "process": "" }, { "text": "If the minimum distance from a point on an ellipse to one of its foci is $2$ and the maximum distance is $8$, then what is the length of the minor axis of the ellipse?", "fact_expressions": "G: Ellipse;P:Point;PointOnCurve(P,G);Min(Distance(P,OneOf(Focus(G))))=2;Max(Distance(P,OneOf(Focus(G))))=8", "query_expressions": "Length(MinorAxis(G))", "answer_expressions": "8", "fact_spans": "[[[1, 3], [7, 8], [34, 36]], [[5, 6]], [[1, 6]], [[5, 23]], [[5, 31]]]", "query_spans": "[[[34, 42]]]", "process": "Without loss of generality, assume the equation of the ellipse is: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). From the given conditions, we have \\begin{cases}a+c=8\\\\a-c=2\\end{cases}, solving yields \\begin{cases}a=5\\\\c=3\\end{cases}, then the length of the minor axis of the ellipse is: 2b=2\\sqrt{5^{2}-3^{2}}=8." }, { "text": "Given that the point $(-4,0)$ is a focus of the ellipse $k x^{2}+3 k y^{2}=1$, then $k$=?", "fact_expressions": "G: Ellipse;Expression(G) = (k*x^2 + 3*(k*y^2) = 1);Coordinate(OneOf(Focus(G))) = (-4,0);k: Number", "query_expressions": "k", "answer_expressions": "1/24", "fact_spans": "[[[12, 35]], [[12, 35]], [[2, 40]], [[42, 45]]]", "query_spans": "[[[42, 47]]]", "process": "The point (-4,0) is a focus of the ellipse kx^{2}+3ky^{2}=1. The standard form of the ellipse kx^{2}+3ky^{2}=1 is \\frac{x^{2}}{k}+\\frac{y^{2}}{3k}=1, which gives: \\frac{1}{k}-\\frac{1}{3k}=16. Solving for k yields k=\\frac{1}{24}." }, { "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=8x$ intersects the parabola at points $A$ and $B$, if $|A B|=16$ and $|A F|<|B F|$, then $|A F|=$?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 16;Abs(LineSegmentOf(A, F)) < Abs(LineSegmentOf(B, F))", "query_expressions": "Abs(LineSegmentOf(A, F))", "answer_expressions": "8 - 4*sqrt(2)", "fact_spans": "[[[3, 17], [27, 30]], [[24, 26]], [[31, 34]], [[35, 38]], [[20, 23]], [[3, 17]], [[3, 23]], [[2, 26]], [[24, 40]], [[42, 52]], [[54, 67]]]", "query_spans": "[[[68, 77]]]", "process": "According to the problem, the equation of the line passing through the focus $ F $ of the parabola $ y^{2} = 8x $ can be written as $ y = k(x - 2) $. \nSolving the system \n\\[\n\\begin{cases}\ny = k(x - 2) \\\\\ny^{2} = 8x\n\\end{cases}\n\\] \ngives \n\\[\nk^{2}x^{2} - (4k^{2} + 8)x + 4k^{2} = 0.\n\\] \nLet $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then \n\\[\nx_{1} + x_{2} = \\frac{4k^{2} + 8}{k^{2}}.\n\\] \nSince $ |AB| = 16 $, \n\\[\nx_{1} + 2 + x_{2} + 2 = 16,\n\\] \nthat is, \n\\[\n\\frac{4k^{2} + 8}{k^{2}} = 12.\n\\] \nThus, \n\\[\nk^{2} = 1,\n\\] \nthen \n\\[\nx^{2} - 12x + 4 = 0.\n\\] \nHence, \n\\[\nx = \\frac{12 \\pm 8\\sqrt{2}}{2} = 6 \\pm 4\\sqrt{2}.\n\\] \nTherefore, \n\\[\nx_{1} = 6 - 4\\sqrt{2}, \\quad x_{2} = 6 + 4\\sqrt{2}.\n\\] \nThus, \n\\[\n|AF| = 6 - 4\\sqrt{2} + 2 = 8 - 4\\sqrt{2}.\n\\]" }, { "text": "The equations of the two asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(3/4)*x", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 49]]]", "process": "" }, { "text": "If $AB$ is a chord passing through the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $F_{1}$ is a focus of the ellipse, then the maximum area of $\\triangle F_{1} A B$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Center(G), LineSegmentOf(A, B));F1: Point;OneOf(Focus(G) )= F1", "query_expressions": "Max(Area(TriangleOf(F1, A, B)))", "answer_expressions": "12", "fact_spans": "[[[8, 47], [60, 62]], [[8, 47]], [[1, 6]], [[1, 6]], [[1, 51]], [[1, 51]], [[52, 59]], [[52, 65]]]", "query_spans": "[[[67, 96]]]", "process": "Let the coordinates of point A be (x, y). By symmetry, B is (-x, -y). Express the area of quadrilateral $ F_1AB $; from the figure, it is known that when point A is at a vertex of the ellipse, the area of $ F_1AB $ is maximized. Combining with the standard equation of the ellipse, the maximum area of $ F_1AB $ can be found. Let the coordinates of A be (x, y), then by symmetry, B is (-x, -y). The area $ S $ of $ F_1AB $ is \n$$\nS = \\frac{1}{2} OF \\times |y_A - y_B| = \\frac{1}{2} OF \\times |2y| = c|y|.\n$$ \nTherefore, the area of $ F_1AB $ is maximized when $ |y| $ is maximized. From the figure, it is known that when point A is at a vertex of the ellipse, the area of $ F_1AB $ is maximized. Thus, the maximum area of $ F_1AB $ is \n$$\ncb = \\sqrt{25 - 16} \\times 4 = 12.\n$$" }, { "text": "Given a parabola $C$: $x^{2}=a y (a>0)$, the distance from a point $P(2 a , 4 a)$ on the parabola to the focus $F$ is $17$. Then the equation of the line $P F$ is?", "fact_expressions": "C: Parabola;a: Number;P: Point;F: Point;a>0;Expression(C) = (x^2 = a*y);Coordinate(P) = (2*a, 4*a);PointOnCurve(P, C);Focus(C) = F;Distance(P, F) = 17", "query_expressions": "Expression(LineOf(P,F))", "answer_expressions": "y=(15/8)*x+1", "fact_spans": "[[[2, 28]], [[10, 28]], [[31, 45]], [[48, 51]], [[10, 28]], [[2, 28]], [[31, 45]], [[2, 45]], [[2, 51]], [[31, 59]]]", "query_spans": "[[[61, 73]]]", "process": "The distance from point P(2a, 4a) on the parabola C: x^{2} = ay (a > 0) to the focus F is 17, so 17 = 4a + \\frac{a}{4} = \\frac{17a}{4}, solving gives a = 4. The equation of the parabola is x^{2} = 4y, then F(0, 1), P(8, 16), k_{PF} = \\frac{16 - 1}{8} = \\frac{15}{8}, thus the equation of line PF is y = \\frac{15}{9}x + 1." }, { "text": "Given that the equations of the two asymptotes of a hyperbola are $2x - y = 0$ and $2x + y = 0$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;L1: Line;L2: Line;Asymptote(G) = {L1, L2};Expression(L1) = (2*x - y = 0);Expression(L2) = (2*x + y = 0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{(sqrt(5), sqrt(5)/2)}", "fact_spans": "[[[2, 5], [37, 40]], [], [], [[2, 11]], [[2, 35]], [[2, 35]]]", "query_spans": "[[[37, 46]]]", "process": "" }, { "text": "The equations of the asymptotes of the hyperbola $2 x^{2}-3 y^{2}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - 3*y^2 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(6)/3)*x", "fact_spans": "[[[0, 22]], [[0, 22]]]", "query_spans": "[[[0, 30]]]", "process": "The hyperbola $2x^{2}-3y^{2}=1$ rewritten in standard form is $\\frac{x^{2}}{2}-\\frac{y^{2}}{3}=1$, then the asymptotes of the hyperbola are $y=\\pm\\frac{\\sqrt{\\frac{1}{3}}}{\\sqrt{\\frac{1}{2}}}x=\\pm\\frac{\\sqrt{6}}{3}x$." }, { "text": "Given the parabola $C$: $y^{2}=16 x$ with focus $F$ and directrix $l$, point $P$ is a moving point on the curve $C$, the distance from point $P$ to the directrix $l$ is $d$, and point $A(1,6)$. Then the minimum value of $|P A|+d$ is?", "fact_expressions": "C: Parabola;A: Point;P: Point;F: Point;l: Line;d: Number;Expression(C) = (y^2 = 16*x);Coordinate(A) = (1, 6);Focus(C) = F;PointOnCurve(P, C);Directrix(C) = l;Distance(P, l) = d", "query_expressions": "Min(d + Abs(LineSegmentOf(P, A)))", "answer_expressions": "3*sqrt(5)", "fact_spans": "[[[2, 22], [42, 47]], [[70, 79]], [[37, 41], [52, 56]], [[26, 29]], [[33, 36], [59, 62]], [[66, 69]], [[2, 22]], [[70, 79]], [[2, 29]], [[37, 51]], [[2, 36]], [[52, 69]]]", "query_spans": "[[[81, 96]]]", "process": "The parabola has focus F(4,0) and directrix x = -4. By the definition of a parabola, |PF| = d, so |PA| + d = |PA| + |PF| \\geqslant |AF| = 3\\sqrt{5}. The equality holds if and only if points A, P, and F are collinear, at which point P lies on the parabola. Therefore, the minimum value of |PA| + d is 3\\sqrt{5}." }, { "text": "If the two foci of an ellipse are $F_{1}(-4 , 0)$, $F_{2}(4 , 0)$, point $P$ lies on the ellipse, and the maximum area of $\\Delta PF_{1} F_{2}$ is $12$, then what is the equation of this ellipse?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Max(Area(TriangleOf(P, F1, F2))) = 12", "query_expressions": "Expression(G)", "answer_expressions": "x^2/25+y^2/9=1", "fact_spans": "[[[1, 3], [44, 46], [85, 87]], [[7, 22]], [[24, 38]], [[7, 22]], [[24, 38]], [[1, 38]], [[39, 43]], [[39, 47]], [[49, 82]]]", "query_spans": "[[[85, 92]]]", "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 $F_{2}$, respectively. Point $P$ lies on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and $\\tan \\angle P F_{1} F_{2}=\\frac{\\sqrt{3}}{3}$. Find the eccentricity of this ellipse?", "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);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Tan(AngleOf(P, F1, F2)) = sqrt(3)/3", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[1, 53], [82, 84], [196, 198]], [[3, 53]], [[3, 53]], [[77, 81]], [[61, 68]], [[69, 76]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 76]], [[77, 85]], [[87, 146]], [[147, 193]]]", "query_spans": "[[[196, 204]]]", "process": "From $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$, we know $PF_{1}\\bot PF_{2}$. From $\\tan\\angle PF_{1}F_{2}=\\frac{\\sqrt{3}}{3}$, we know $\\angle PF_{1}F_{2}=30^{\\circ}$. In right triangle $\\triangle PF_{1}F_{2}$, $|PF_{1}|=|F_{1}F_{2}|\\cos30^{\\circ}$, $|PF_{2}|=|F_{1}F_{2}|\\sin30^{\\circ}$. Therefore, $|PF_{1}|+|PF_{2}|=|F_{1}F_{2}|(\\cos30^{\\circ}+\\sin30^{\\circ})=(\\sqrt{3}+1)c=2a$, so $e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}+1}=\\sqrt{3}-1$." }, { "text": "Given that $A$ is the left vertex of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, $F$ is the right focus of the hyperbola $C$, a circle with diameter equal to the length of the transverse axis intersects one of its asymptotes at point $P$ (point $P$ lies in the second quadrant), $PA$ is parallel to the other asymptote, and $S_{\\Delta OPA}=\\sqrt{3}$, then $PF=$?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;A: Point;LeftVertex(C) = A;F: Point;RightFocus(C) = F;G: Circle;Diameter(G) = Length(RealAxis(C));l1: Line;l2: Line;Intersection(G,l1) = P;P: Point;Quadrant(P) = 2;IsParallel(LineSegmentOf(P,A) , l2) = True;O: Origin;Area(TriangleOf(O,P,A)) = sqrt(3);OneOf(Asymptote(C)) = l1;OneOf(Asymptote(C)) = l2;Negation(l1=l2)", "query_expressions": "LineSegmentOf(P, F)", "answer_expressions": "2*sqrt(7)", "fact_spans": "[[[6, 67], [76, 82]], [[6, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[14, 67]], [[2, 5]], [[2, 71]], [[72, 75]], [[72, 86]], [[95, 96]], [[76, 96]], [], [], [[76, 109]], [[105, 109], [110, 114]], [[110, 119]], [[76, 135]], [[137, 164]], [[137, 164]], [74, 101], [74, 132], [74, 132]]", "query_spans": "[[[166, 173]]]", "process": "As shown in the figure, connect PF intersecting the other asymptote at point Q. Since PA // OQ, ∠PAO = ∠QOF = ∠POA, so △PAO is an equilateral triangle, hence ∠POA = 60°, then \\frac{b}{a} = \\tan60^{\\circ} = \\sqrt{3}, that is, b = \\sqrt{3}a; also since S_{\\triangleOPA} = \\sqrt{3}, we have \\frac{\\sqrt{3}}{4}a^{2} = \\sqrt{3}, solving gives a = 2, b = 2\\sqrt{3}, c = \\sqrt{a^{2} + b^{2}} = 4. In △PAF, PA = 2, AF = 6, ∠PAF = 60^{\\circ}, by the cosine law, PF = \\sqrt{4 + 36 - 2 \\times 2 \\times 6 \\cos60^{\\circ}} = 2\\sqrt{7}. The answer is: 2\\sqrt{7}" }, { "text": "If the coordinates of the foci of an ellipse are $(\\pm 3,0)$ and it passes through the point $(4,0)$, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;Coordinate(Focus(G)) = (pm*3,0);I: Point;Coordinate(I) = (4, 0);PointOnCurve(I,G) = True", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16+y^2/7=1", "fact_spans": "[[[1, 3], [34, 36]], [[1, 20]], [[24, 32]], [[24, 32]], [[1, 32]]]", "query_spans": "[[[34, 43]]]", "process": "From the problem, we know that the foci of the ellipse lie on the x-axis, so we assume the equation of the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). Then we have \\begin{cases}a^{2}-b^{2}=9\\\\\\frac{16}{a^{2}}=1\\end{cases}, solving gives \\begin{cases}a^{2}=16\\\\b^{2}=7\\end{cases}, thus the equation is: \\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1" }, { "text": "$F$ is the focus of the parabola $y^{2}=2 p x (p>0)$. A line passing through point $F$ intersects the parabola at point $A$ and intersects the directrix of the parabola at point $B$. If $\\overrightarrow{B A}=2 \\overrightarrow{A F}$ and $|\\overrightarrow{B A}|=4$, then $p=$?", "fact_expressions": "F: Point;G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;Focus(G) = F;PointOnCurve(F, H) = True;H: Line;OneOf(Intersection(H, G)) = A;A: Point;Intersection(H, Directrix(G)) = B;B: Point;VectorOf(B, A) = 2*VectorOf(A, F);Abs(VectorOf(B, A)) = 4", "query_expressions": "p", "answer_expressions": "3", "fact_spans": "[[[0, 3], [32, 36]], [[4, 27], [40, 43], [54, 57]], [[4, 27]], [[141, 144]], [[7, 27]], [[0, 30]], [[31, 39]], [[37, 39]], [[37, 52]], [[49, 52]], [[37, 64]], [[61, 64]], [[66, 111]], [[113, 139]]]", "query_spans": "[[[141, 146]]]", "process": "Since F is the focus of the parabola y^{2}=2px, we have F(\\frac{p}{2},0), and the directrix is x=-\\frac{p}{2}. Let the line passing through F be y=k(x-\\frac{p}{2}), intersecting the parabola at points A(x_{1},y_{1}) and C(x_{2},y_{2}). Draw AM and CN perpendicular from A and C to the directrix, meeting it at M and N respectively. Thus |CN|=|CF|, |AM|=|AF|. Since \\overrightarrow{BA}=2\\overrightarrow{AF}, it follows that |\\overrightarrow{BA}|=2|\\overrightarrow{AF}|, so |BA|=2|AM|, hence \\angle ABM=30^{\\circ}. Given |\\overrightarrow{BA}|=4, we get |AM|=|AF|=2. Also, 2|CN|=|CB|=|BA|-|AF|-|FC|=|BA|-|AM|-|CN|, so 2|CN|=6+|CN|, thus |CN|=6=|CF|. Solving the system of equations for the line and the parabola: \\begin{cases}y=k(x-\\frac{p}{2})\\\\y^{2}=2px\\end{cases}, eliminating y gives k^{2}(x^{2}-px+\\frac{p^{2}}{4})=2px, so k^{2}x^{2}-(k^{2}p+2p)x+\\frac{k^{2}p^{2}}{4}=0. Therefore, x_{1}+x_{2}=\\frac{k^{2}p+2p}{k^{2}}, x_{1}x_{2}=\\frac{p^{2}}{4}. Since x_{1}>0, x_{2}>0, and x_{1}+\\frac{p}{2}=|AM|=2, x_{2}+\\frac{p}{2}=|CN|=6, we have x_{1}x_{2}=(2-\\frac{p}{2})(6-\\frac{p}{2})=12-4p+\\frac{p^{2}}{4}=\\frac{p^{2}}{4}, so p=3" }, { "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a>0, b>0)$ is $y = \\sqrt{3}x$, what is the eccentricity of the 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 = sqrt(3)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[2, 60], [84, 87]], [[5, 60]], [[5, 60]], [[5, 60]], [[5, 60]], [[2, 60]], [[2, 81]]]", "query_spans": "[[[84, 93]]]", "process": "Since one asymptote of the hyperbola is $ y = \\sqrt{3}x $, it follows that $ \\frac{b}{a} = \\sqrt{3} $. Therefore, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{1 + \\left( \\frac{b}{a} \\right)^2} = 2 $." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, a line $l$ passing through the point $P(1 , 1)$ intersects the ellipse $C$ at points $A$ and $B$. If point $P$ is exactly the midpoint of chord $AB$, then what is the slope of line $l$?", "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;P: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Coordinate(P) = (1, 1);PointOnCurve(P, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B))=P;IsChordOf(LineSegmentOf(A,B),C)", "query_expressions": "Slope(l)", "answer_expressions": "-3/4", "fact_spans": "[[[58, 63], [98, 103]], [[2, 44], [64, 69]], [[71, 74]], [[75, 78]], [[82, 86], [46, 57]], [[2, 44]], [[46, 57]], [[45, 63]], [[58, 80]], [[82, 96]], [[64, 94]]]", "query_spans": "[[[98, 107]]]", "process": "First, assume the coordinates of points and substitute into the ellipse equation, subtract to obtain \\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{(x_{1}+x_{2})(x_{1}-x_{2})}=-\\frac{3}{4}, then use the midpoint and slope formula to find the slope. (Detailed solution) According to the problem, let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\begin{cases}\\frac{x_{1}}{4}+\\frac{y_{1}}{3}=1\\\\\\frac{x_{2}}{4}+\\frac{y_{2}}{3}=1\\end{cases}. Subtracting the two equations gives \\frac{y_{1}-y_{2}}{x_{1}^{2}-x_{2}^{2}}=-\\frac{3}{4}, that is, \\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{(x_{1}+x_{2})(x_{1}-x_{2})}=-\\frac{3}{4}. Since the midpoint of chord AB is P(1,1), it follows that x_{1}+x_{2}=2x_{P}=2, y_{1}+y_{2}=2y_{P}=2. Also, 1=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, hence 1\\times k_{AB}=-\\frac{3}{4}, that is, k_{AB}=-\\frac{3}{4}. Therefore, the slope of line l is -\\frac{3}{4}." }, { "text": "Given that the directrix of the parabola $x^{2}=2 p y(p>0)$ passes through a vertex of the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1$, find the coordinates of the focus of the parabola?", "fact_expressions": "H: Parabola;Expression(H) = (x^2 = 2*(p*y));p: Number;p>0;G: Hyperbola;Expression(G) = (-x^2/16 + y^2/9 = 1);PointOnCurve(OneOf(Vertex(G)), Directrix(H)) = True", "query_expressions": "Coordinate(Focus(H))", "answer_expressions": "(0, 4)", "fact_spans": "[[[2, 23], [73, 76]], [[2, 23]], [[5, 23]], [[5, 23]], [[27, 66]], [[27, 66]], [[2, 71]]]", "query_spans": "[[[73, 83]]]", "process": "" }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, the directrix is $l$. From a point $M$ on the parabola, draw a perpendicular to $l$, with foot $N$. Let $A(\\frac{7}{2} p, 0)$, and let $AN$ and $MF$ intersect at point $B$. If $|A F|=2|M F|$, and the area of $\\Delta A B M$ is $\\frac{9 \\sqrt{2}}{2}$, then what is the value of $p$?", "fact_expressions": "G: Parabola;p: Number;A: Point;N: Point;M: Point;F: Point;B: Point;l: Line;l1: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = ((7/2)*p, 0);Focus(G) = F;Directrix(G) = l;PointOnCurve(M, l1);IsPerpendicular(l, l1);FootPoint(l, l1) = N;Intersection(LineSegmentOf(A, N), LineSegmentOf(M, F)) = B;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(M, F));Area(TriangleOf(A, B, M)) = (9*sqrt(2))/2;PointOnCurve(M, G)", "query_expressions": "p", "answer_expressions": "3", "fact_spans": "[[[0, 21], [37, 40]], [[166, 169]], [[63, 84]], [[57, 60]], [[43, 46]], [[25, 28]], [[101, 105]], [[32, 35]], [], [[3, 21]], [[0, 21]], [[63, 84]], [[0, 28]], [[0, 35]], [[36, 53]], [[36, 53]], [[36, 60]], [[87, 105]], [[107, 122]], [[124, 164]], [[37, 46]]]", "query_spans": "[[[166, 173]]]", "process": "From the given conditions, |MF| = \\frac{1}{2}(\\frac{7p}{2} - \\frac{p}{2}) = \\frac{3}{2}p, so x_{M} + \\frac{p}{2} = \\frac{3}{2}p. Solving gives x_{M} = p, hence |y_{M}| = \\sqrt{2}p, therefore \\frac{1}{2} \\times 3p \\times \\frac{2\\sqrt{2}}{3}p = \\frac{9\\sqrt{2}}{2}, which yields p = 3, fill in 3." }, { "text": "The equation of the directrix of the parabola $y=x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y = -1/4", "fact_spans": "[[[0, 12]], [[0, 12]]]", "query_spans": "[[[0, 19]]]", "process": "Given the standard equation of the parabola \\( x^{2} = y \\), the parabola has its focus on the positive \\( y \\)-axis, and \\( 2p = 1 \\). Therefore, its directrix equation is \\( y = -\\frac{p}{2} \\), so \\( y = -\\frac{1}{4} \\)." }, { "text": "The standard equation of a parabola passing through the point $(1 , 2)$ is?", "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (1, 2);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=4*x, x^2=(1/2)*y}", "fact_spans": "[[[13, 16]], [[2, 12]], [[2, 12]], [[0, 16]]]", "query_spans": "[[[13, 23]]]", "process": "Let the standard equation of the parabola be y^{2}=mx or x^{2}=my. Substituting (1,2) gives m=4 or \\frac{1}{2}, so the required standard equations are y^{2}=4x and x^{2}=\\frac{1}{2}y." }, { "text": "Given the vertices of $\\triangle A B C$ are $A(-5,0)$, $B(5,0)$, and its incenter lies on the line $x=3$, what is the trajectory equation of vertex $C$?", "fact_expressions": "H: Line;A: Point;B: Point;C: Point;PointOnCurve(Center(InscribedCircle(TriangleOf(A,B,C))),H);Expression(H) = (x = 3);Coordinate(A) = (-5, 0);Coordinate(B)=(5,0)", "query_expressions": "LocusEquation(C)", "answer_expressions": "(x^2/9-y^2/16=1)&(x>3)", "fact_spans": "[[[49, 56]], [[22, 31]], [[33, 41]], [[61, 64]], [[2, 57]], [[49, 56]], [[22, 31]], [[33, 41]]]", "query_spans": "[[[61, 70]]]", "process": "As shown in the figure, the points of tangency between triangle ABC and the circle are E, F, G respectively, then |AE| = |AG| = 8, |BF| = |BG| = 2, |CE| = |CF|. Therefore, |CA| - |CB| = 8 - 2 = 6. According to the definition of a hyperbola, the required locus is the right branch of a hyperbola with foci at A and B and a major axis length of 6. ∴ c = 5, a = 3. This problem examines trajectory equations using the definition method: if the conditions of a moving point's trajectory match the definition of a basic locus (such as an ellipse, hyperbola, parabola, circle, etc.), the equation can be directly determined by definition. It is a basic problem." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{10}-\\frac{y^{2}}{6}=1$, a parabola has its vertex at the origin, its axis of symmetry along the $x$-axis, and its focus at the left focus of the hyperbola. Then, what is the standard equation of the parabola?", "fact_expressions": "C: Hyperbola;G: Parabola;O: Origin;Expression(C) = (x^2/10 - y^2/6 = 1);Vertex(G) = O;SymmetryAxis(G)=xAxis;Focus(G)=LeftFocus(C)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=-16*x", "fact_spans": "[[[2, 46], [69, 72]], [[47, 50], [78, 81]], [[54, 56]], [[2, 46]], [[47, 56]], [[47, 65]], [[47, 76]]]", "query_spans": "[[[78, 88]]]", "process": "The left focus of the hyperbola is (-4, 0), so the standard equation of the parabola is y^{2} = -4 \\times 4x, y^{2} = -16x." }, { "text": "A 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 $M$ and $N$, and $F_{2}$ is its right focus. Then the value of $|M F_{2}|+|N F_{2}|-|M N|$ is?", "fact_expressions": "G: Hyperbola;H: Line;M: Point;F2: Point;N: 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)) = {M, N}", "query_expressions": "Abs(LineSegmentOf(M, F2)) + Abs(LineSegmentOf(N, F2)) - Abs(LineSegmentOf(M, N))", "answer_expressions": "8", "fact_spans": "[[[1, 39], [53, 56], [78, 79]], [[50, 52]], [[60, 63]], [[70, 77]], [[64, 67]], [[42, 49]], [[1, 39]], [[1, 49]], [[70, 82]], [[0, 52]], [[50, 69]]]", "query_spans": "[[[84, 115]]]", "process": "By the definition of a hyperbola: |MF₂| - |MF₁| = 2a ①, |NF₂| - |NF₁| = 2a ②, ① + ② gives |MF₂| + |NF₂| - |MN| = 4a, and since a = 2, therefore |MF₂| + |NF₂| - |MN| = 8." }, { "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and let point $P$ lie on the hyperbola such that $\\angle F_{1} P F_{2}=90^{\\circ}$. Then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(90, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "1", "fact_spans": "[[[17, 45], [56, 59]], [[1, 8]], [[51, 55]], [[9, 16]], [[17, 45]], [[1, 50]], [[51, 60]], [[64, 97]]]", "query_spans": "[[[100, 127]]]", "process": "\\because point P lies on the right branch of the hyperbola and satisfies \\angle F_{1}PF_{2}=90^{\\circ}, \\begin{cases}|PF_{1}|-|PF_{2}|=4\\textcircled{1}\\\\|PF_{1}|^{2}+|PF_{2}|^{2}=20\\textcircled{2}\\end{cases} \\textcircled{2}-\\textcircled{1}^{2} yields |PF_{1}|+|PF_{2}|=2.\\therefore the area of \\triangle F_{1}PF_{2} is S=\\frac{1}{2}|PF_{1}||PF_{2}|=1. Hence, the result is 1." }, { "text": "Given an ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right vertex of the ellipse is $A$. A line $l$ passing through the origin intersects the ellipse $C$ at points $P$ and $Q$. If $|P Q|=a$, $A P \\perp P Q$, then what is the eccentricity of the ellipse $C$?", "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;A: Point;P: Point;Q: Point;O:Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;PointOnCurve(O, l);Intersection(l, C) = {P, Q};Abs(LineSegmentOf(P, Q)) = a;IsPerpendicular(LineSegmentOf(A, P), LineSegmentOf(P, Q))", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(5)/5", "fact_spans": "[[[74, 79]], [[2, 59], [80, 85], [127, 132]], [[9, 59]], [[9, 59]], [[64, 68]], [[86, 89]], [[90, 93]], [[71, 73]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[69, 79]], [[74, 95]], [[97, 107]], [[109, 124]]]", "query_spans": "[[[127, 138]]]", "process": "Let point P be in the first quadrant. By symmetry, |OP| = \\frac{a}{2}. Using the definition of acute angle trigonometric functions, we can find that \\angle AOP = 60^{\\circ}, thus the coordinates of point P can be determined. Substituting the coordinates of point P into the equation of ellipse C yields a relationship between a and b, allowing us to find the eccentricity of ellipse C. Without loss of generality, assume point P is in the first quadrant and O is the origin. By symmetry, |OP| = \\frac{1}{2}|PQ| = \\frac{a}{2}. \\sin \\angle POA = \\frac{|OP|}{|OA|} = \\frac{1}{2}, hence \\angle POA = 60^{\\circ}. Let point P(x_{0}, y_{0}), then x_{0} = \\frac{a}{2} \\cos 60^{\\circ} = \\frac{a}{4}, y_{0} = \\frac{a}{2} \\sin 60^{\\circ} = \\frac{\\sqrt{3}}{4}a, that is, point P\\left(\\frac{a}{4}, \\frac{\\sqrt{3}}{4}a\\right). Substituting the coordinates of point P into the equation of ellipse C gives \\frac{a}{a} = 1^{\\frac{1}{1}}, yielding \\frac{b^{2}}{a^{2}} = \\frac{1}{5}. Let the focal distance of the ellipse be 2c (c > 0), then the eccentricity of ellipse C is e = \\frac{c}{a} = \\sqrt{\\frac{a^{2}-b^{2}}{a^{2}}} = \\sqrt{1 - \\frac{b^{2}}{a^{2}}} = \\frac{2\\sqrt{5}}{5}." }, { "text": "If the line $m x+n y=4$ and the circle $x^{2}+y^{2}=4$ have no intersection points, then the number of intersection points between the line passing through the point $(m, n)$ and the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is?", "fact_expressions": "G: Ellipse;H: Circle;l: Line;m: Number;n: Number;J: Point;Expression(G) = (x^2/9 + y^2/4 = 1);Expression(H) = (x^2 + y^2 = 4);Expression(l) = (m*x + n*y = 4);Coordinate(J) = (m, n);NumIntersection(l, H)=0;PointOnCurve(J, L);L:Line", "query_expressions": "NumIntersection(L,G)", "answer_expressions": "2", "fact_spans": "[[[51, 88]], [[15, 31]], [[1, 14]], [[39, 47]], [[39, 47]], [[38, 47]], [[51, 88]], [[15, 31]], [[1, 14]], [[38, 47]], [[1, 35]], [[37, 50]], [[48, 50]]]", "query_spans": "[[[48, 95]]]", "process": "The line mx + ny = 4 and the circle x^{2} + y^{2} = 4 have no intersection points, hence \\frac{4}{\\sqrt{m^{2}+n^{2}}} > 2 \\therefore 0 < m^{2} + n^{2} < 4. Point P(m,n) lies inside the circle centered at the origin with radius 2. Therefore, the circle m^{2} + n^{2} = 4 is internally tangent to the ellipse, so point P(m,n) lies inside the ellipse. Then the number of intersection points between the line passing through point (m,n) and the ellipse \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1 is 2." }, { "text": "Given that $P$ is a point on the parabola $y^{2}=4x$, let $d_{1}$ denote the distance from $P$ to the directrix of this parabola, and $d_{2}$ denote the distance from $P$ to the line $x+2y-12=0$. Then the minimum value of $d_{1}+d_{2}$ is?", "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x + 2*y - 12 = 0);PointOnCurve(P, G);d1: Number;d2: Number;Distance(P, Directrix(G)) = d1;Distance(P, H) = d2", "query_expressions": "Min(d1 + d2)", "answer_expressions": "11*sqrt(5)/5", "fact_spans": "[[[6, 20], [30, 33]], [[54, 68]], [[2, 5], [25, 28], [50, 53]], [[6, 20]], [[54, 68]], [[2, 23]], [[40, 47]], [[72, 79]], [[25, 47]], [[50, 79]]]", "query_spans": "[[[81, 100]]]", "process": "\\because the distance from point P to the directrix equals the distance from point P to the focus F, \\therefore draw a perpendicular line from focus F to the line x+2y-12=0, at which point the value of d_{1}+d_{2} is minimized. \\because F(1,0)=\\frac{|1-12|}{\\sqrt{1^{2}+2^{2}}}=\\frac{11\\sqrt{5}}{5}" }, { "text": "A point $A$ on the parabola $x^{2}=3y$ has a $y$-coordinate of $\\frac{5}{4}$. What is the distance from point $A$ to the focus of this parabola?", "fact_expressions": "G: Parabola;A: Point;Expression(G) = (x^2 = 3*y);PointOnCurve(A,G);YCoordinate(A)=5/4", "query_expressions": "Distance(A, Focus(G))", "answer_expressions": "2", "fact_spans": "[[[0, 14], [46, 49]], [[17, 20], [40, 44]], [[0, 14]], [[0, 20]], [[17, 38]]]", "query_spans": "[[[40, 56]]]", "process": "Problem Analysis: From the given conditions, the directrix of the parabola is y = -\\frac{3}{4}. Therefore, the distance from point A to the directrix is \\frac{5}{4} - (-\\frac{3}{4}) = 2. According to the definition of a parabola, the distance from point A to the focus of 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 this parabola is 2." }, { "text": "If the left and right vertices of the hyperbola $x^{2}-y^{2}=a^{2}(a>0)$ are $A$ and $B$ respectively, and point $P$ is a point on the hyperbola in the first quadrant, and if the angles of inclination of lines $PA$ and $PB$ are $\\alpha$ and $\\beta$ respectively, and $\\beta=m \\alpha(m>1)$, then what is the value of $\\alpha$?", "fact_expressions": "G: Hyperbola;a: Number;A: Point;P: Point;B: Point;a>0;m:Number;m>1;alpha:Number;beta:Number;Expression(G) = (x^2 - y^2 = a^2);LeftVertex(G) = A;RightVertex(G) = B;Quadrant(P) = 1;PointOnCurve(P, G);Inclination(LineOf(P,A))=alpha;Inclination(LineOf(P,B))=beta;beta=m*alpha", "query_expressions": "alpha", "answer_expressions": "pi/(2*m+2)", "fact_spans": "[[[1, 28], [55, 58]], [[4, 28]], [[37, 40]], [[45, 49]], [[41, 44]], [[4, 28]], [[105, 126]], [[105, 126]], [[85, 93], [129, 137]], [[96, 103]], [[1, 28]], [[1, 44]], [[1, 44]], [[45, 61]], [[45, 61]], [[63, 103]], [[63, 103]], [[105, 126]]]", "query_spans": "[[[129, 141]]]", "process": "" }, { "text": "The equation of the directrix of a parabola is $y=\\frac{1}{2}$. Then its standard equation is?", "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (y = 1/2)", "query_expressions": "Expression(G)", "answer_expressions": "x^2=-2*y", "fact_spans": "[[[0, 3], [26, 27]], [[0, 24]]]", "query_spans": "[[[26, 33]]]", "process": "The equation of the directrix of the parabola is y=\\frac{1}{2}, so \\frac{p}{2}=\\frac{1}{2} \\therefore p=1, and the focus lies on the negative y-axis; therefore, the standard equation is x^{2}=-2y." }, { "text": "Given that the hyperbola $C$ passes through the point $(4,2 \\sqrt{3})$ and has asymptotes with equations $y=\\pm x$, then the standard equation of $C$ is?", "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (4, 2*sqrt(3));PointOnCurve(G, C);Expression(Asymptote(C)) = (y = pm*x)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4-y^2/4=1", "fact_spans": "[[[2, 8], [45, 48]], [[9, 26]], [[9, 26]], [[2, 26]], [[2, 43]]]", "query_spans": "[[[45, 55]]]", "process": "Given that the asymptotes of the hyperbola are y = \\pm x, assume the equation of the hyperbola is x^{2} - y^{2} = \\lambda (\\lambda \\neq 0). Then substitute the point (4, 2\\sqrt{3}) to solve. Since the asymptotes of the hyperbola are y = \\pm x, assume the equation of the hyperbola is x^{2} - y^{2} = \\lambda (\\lambda \\neq 0). Since the hyperbola C passes through the point (4, 2\\sqrt{3}), it follows that \\lambda = 4^{2} - (2\\sqrt{3})^{2} = 4. Therefore, the standard equation of the hyperbola is \\frac{x^{2}}{4} - \\frac{y^{2}}{4} = 1." }, { "text": "Given the parabola $C$: $x^{2}=2 p y$ ($p>0$) with focus $F$, its directrix intersects the $y$-axis at point $D$. A line passing through point $F$ intersects the parabola $C$ at points $A$ and $B$. If $A B \\perp A D$ and $|B F|=|A F|+4$, then the value of $p$ is?", "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;B: Point;D: Point;F: Point;p>0;Expression(C) = (x^2 = 2*(p*y));Focus(C) = F;Intersection(Directrix(C), yAxis) = D;PointOnCurve(F, G);Intersection(G, C) = {A, B};IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(A, D));Abs(LineSegmentOf(B, F)) = Abs(LineSegmentOf(A, F)) + 4", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 28], [36, 37], [60, 66]], [[112, 115]], [[57, 59]], [[67, 70]], [[71, 74]], [[46, 50]], [[32, 35], [52, 56]], [[9, 28]], [[2, 28]], [[2, 35]], [[36, 50]], [[51, 59]], [[57, 76]], [[78, 93]], [[95, 110]]]", "query_spans": "[[[112, 119]]]", "process": "Find the coordinates of the focus of the parabola. Let A(x_{1},y_{1}), B(x_{2},y_{2}), and assume the equation of line AB is: y = kx + \\frac{p}{2}. By solving this equation together with the equation of the parabola, we obtain x_{1}x_{2} = -p^{2}, y_{1}y_{2} = \\frac{p^{2}}{4}. Using AB \\perp AD, we know \\overrightarrow{AD} \\perp \\overrightarrow{AF}, which translates to their dot product being 0. This allows us to solve for y_{1} = \\frac{\\sqrt{5}-2}{2}p, y_{2} = \\frac{\\sqrt{5}+2}{2}p. Then, using the definition of the parabola, express |BF| = |AF| + 4 in terms of p to solve for the value of p. (Solution) The parabola C: x^{2} = 2py (p > 0) has focus F(0, \\frac{p}{2}), D(0, -\\frac{p}{2}). Let A(x_{1},y_{1}), B(x_{2},y_{2}). Assume the slope of line AB exists, and let the equation of line AB be: y = kx + \\frac{p}{2}. From \\begin{cases} y = kx + \\frac{p}{2} \\\\ x^{2} = 2py \\end{cases}, we get: x^{2} - 2pkx - p^{2} = 0, so x_{1}x_{2} = -p^{2}, y_{1}y_{2} = \\frac{p^{2}}{4}, \\overrightarrow{AD} = (-x_{1}, -\\frac{p}{2} - y_{1}), \\overrightarrow{AF} = (-x_{1}, \\frac{p}{2} - y_{1}). Since AB \\perp AD, then AD \\perp AF, so \\overrightarrow{AD} \\cdot \\overrightarrow{AF} = x_{1}^{2} - (\\frac{p}{2} + y_{1})(\\frac{p}{2} - y_{1}) = 0, that is, x_{1}^{2} + y_{1}^{2} = \\frac{p^{2}}{4}, so y_{1}^{2} + 2py_{1} = \\frac{p^{2}}{4}, yielding y_{1} = \\frac{\\sqrt{5}-2}{2}p, y_{2} = \\frac{\\sqrt{5}+2}{2}p. Therefore, |BF| - |AF| = (y_{2} + \\frac{p}{2}) - (y_{1} + \\frac{p}{2}) = 2p = 4, so p = 2." }, { "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 $F_{1}$ and $F_{2}$ respectively. Point $A$ lies on the right branch of the hyperbola, $|A F_{1}|=\\frac{11}{2} a-c$ ($c$ is the semi-focal length), and the line $A F_{2}$ is parallel to the line $y=-\\frac{b}{a} x$. What is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F2: Point;A: Point;F1: Point;c:Number;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));Abs(LineSegmentOf(A,F1))=(11/2)*a-c;HalfFocalLength(G)=c;IsParallel(LineOf(A,F2),H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[2, 58], [88, 91], [88, 91]], [[5, 58]], [[5, 58]], [[150, 170]], [[75, 82]], [[83, 87]], [[67, 74]], [[128, 131]], [[5, 58]], [[5, 58]], [[2, 58]], [[150, 170]], [[2, 82]], [[2, 82]], [[83, 98]], [[99, 127]], [[88, 136]], [[138, 172]]]", "query_spans": "[[[175, 184]]]", "process": "According to the problem, since |AF₁| - |AF₂| = 2a and |AF₁| = (11/2)a - c, it follows that |AF₂| = (7/2)a - c. Since line AF₂ is parallel to the line y = -(b/a)x, we have tan∠F₁F₂A = b/a. Therefore, cos∠F₁F₂A = a/c = [4c² + ((7/2)a - c)² - ((11/2)a - c)²] / [2 × 2c × ((7/2)a - c)]. Thus, 8a² - 2ac - c² = 0, which factors as (2a - c)(4a + c) = 0, yielding e = c/a = 2." }, { "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, a line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$. Perpendiculars are drawn from points $A$ and $B$ to the line $x=-2$, with feet of perpendiculars $E$ and $H$, respectively. If $|AE|=2|BH|$, then the slope $k$ of line $l$ is?", "fact_expressions": "l: Line;C: Parabola;G: Line;A: Point;E: Point;B: Point;H: Point;F: Point;k: Number;Expression(C) = (y^2 = 8*x);Expression(G) = (x = -2);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};L1:Line;L2:Line;PointOnCurve(A,L1);PointOnCurve(B,L2);IsPerpendicular(L1,G);IsPerpendicular(L2,G);FootPoint(L1,G)=E;FootPoint(L2,G)=H;Abs(LineSegmentOf(A, E)) = 2*Abs(LineSegmentOf(B, H));Slope(l) = k", "query_expressions": "k", "answer_expressions": "pm*2*sqrt(2)", "fact_spans": "[[[35, 40], [115, 120]], [[2, 21], [41, 47]], [[72, 80]], [[49, 52], [62, 65]], [[89, 92]], [[53, 56], [66, 69]], [[93, 96]], [[25, 28], [30, 34]], [[124, 127]], [[2, 21]], [[72, 80]], [[2, 28]], [[29, 40]], [[35, 58]], [], [], [[59, 83]], [[59, 83]], [[59, 83]], [[59, 83]], [[59, 96]], [[59, 96]], [[99, 113]], [[115, 127]]]", "query_spans": "[[[123, 129]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From |AE| = 2|BF|, using the definition of the parabola, we obtain x_{1}+2 = 2(x_{2}+2). Then by drawing AA_{1} perpendicular to the x-axis, and BB_{1} perpendicular to the x-axis, from \\triangle AA_{1}F \\sim \\triangle BB_{1}F, we get 2(2-x_{2}) = x_{1}-2, then solve for the coordinates of point A or B, and subsequently find the solution using the focus coordinates. Solution: As shown in the figure, let the equation of line l be y = k(x-2), A(x_{1},y_{1}), B(x_{2},y_{2}). Since |AE| = 2|BF|, it follows that x_{1}+2 = 2(x_{2}+2). Draw AA_{1} perpendicular to the x-axis with foot A_{1}, and draw BB_{1} perpendicular to the x-axis with foot B_{1}. Then \\triangle AA_{1}F \\sim \\triangle BB_{1}F. Thus \\frac{|B_{1}F|}{|A_{1}F|} = \\frac{|BF|}{|AF|} = \\frac{1}{2}, that is, 2(2-x_{2}) = x_{1}-2. Solving gives x_{1} = 4. Since A(x_{1},y_{1}) lies on the parabola C, we have y_{1} = \\pm4\\sqrt{2}, then k = \\frac{\\pm4\\sqrt{2}}{4-2} = \\pm2\\sqrt{2}" }, { "text": "Given that the eccentricity of hyperbola $C$ is $e=\\frac{2 \\sqrt{3}}{3}$. A line is drawn from the focus $F$ perpendicular to one of the asymptotes of hyperbola $C$, with the foot of the perpendicular at $M$. The line $MF$ intersects the other asymptote at $N$. Then $\\frac{M F}{N F}$=?", "fact_expressions": "C: Hyperbola;e: Number;Eccentricity(C) = e;e = 2*sqrt(3)/3;F: Point;OneOf(Focus(C)) = F;Z: Line;PointOnCurve(F, Z);L1: Line;L2: Line;OneOf(Asymptote(C)) = L1;OneOf(Asymptote(C)) = L2;Negation(L1 = L2);IsPerpendicular(L1, Z);M: Point;FootPoint(L1, Z) = M;N: Point;Intersection(LineOf(M, F), L2) = N", "query_expressions": "LineSegmentOf(M, F)/LineSegmentOf(N, F)", "answer_expressions": "1/2", "fact_spans": "[[[2, 8], [44, 50]], [[12, 36]], [[2, 36]], [[12, 36]], [[40, 43]], [[2, 43]], [], [[37, 59]], [], [], [[44, 56]], [[44, 81]], [[44, 81]], [[37, 59]], [[63, 66]], [[37, 66]], [[82, 85]], [[44, 85]]]", "query_spans": "[[[87, 106]]]", "process": "From the given condition, the eccentricity of the hyperbola is: $ e = \\frac{2\\sqrt{3}}{3} $, we get $ \\frac{c}{a} = \\frac{2\\sqrt{3}}{3} $, from which $ \\frac{a^2 + b^{2}}{a^{2}} = \\frac{4}{3} $, so $ \\frac{b}{a} = \\frac{\\sqrt{3}}{3} $, the asymptotes are: $ y = \\pm\\frac{\\sqrt{3}}{3}x $. As shown in the figure, $ \\angle MOF = 30^{\\circ} $, $ F(c,0) $, then $ MF = \\frac{\\sqrt{3}c}{\\sqrt{3^{2}+(\\sqrt{3})^{2}}} = b $, $ OM = a $, so $ MN = \\sqrt{3}a $, therefore, $ \\frac{|MF|}{|NF|} = \\frac{b}{\\sqrt{3}a - b} = \\frac{\\frac{\\sqrt{3}}{3}a}{\\sqrt{3}a - \\frac{\\sqrt{3}}{3}a} = \\frac{1}{2} $." }, { "text": "Given that $ c $ is the semi-focal distance of the ellipse $ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $, then the range of $ \\frac{b+c}{a} $ is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;c: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);HalfFocalLength(G) = c", "query_expressions": "Range((b + c)/a)", "answer_expressions": "(1,sqrt(2)]", "fact_spans": "[[[6, 58]], [[8, 58]], [[8, 58]], [[2, 5]], [[8, 58]], [[8, 58]], [[6, 58]], [[2, 62]]]", "query_spans": "[[[64, 86]]]", "process": "Analysis: According to the given conditions, we have a^{2}=b^{2}+c^{2}, so (\\frac{b+c}{a})^{2}=\\frac{b^{2}+2bc+c^{2}}{a^{2}}=\\frac{a^{2}+2bc}{a^{2}}=1+\\frac{2bc}{a^{2}}. Since 0<2bc\\leqslant b^{2}+c^{2}=a^{2} (equality holds if and only if b=c), it follows that 1<1+\\frac{2bc}{a^{2}}\\leqslant1+\\frac{a^{2}}{a^{2}}=2, hence 1<\\frac{b+c}{a}\\leqslant\\sqrt{2}." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ with left and right foci $F_{1}$, $F_{2}$. A line passing through a focus and parallel to an asymptote intersects the hyperbola at point $M$. Find the area of $\\Delta M F_{1} F_{2}$.", "fact_expressions": "C: Hyperbola;G: Line;M: Point;F1: Point;F2: Point;Expression(C) = (x^2/9 - y^2/16 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(Focus(C),G);IsParallel(G,Asymptote(C));Intersection(G,C)=M", "query_expressions": "Area(TriangleOf(M, F1, F2))", "answer_expressions": "32/3", "fact_spans": "[[[2, 46], [82, 85]], [[79, 81]], [[88, 92]], [[52, 59]], [[60, 67]], [[2, 46]], [[2, 67]], [[2, 67]], [[2, 81]], [[2, 81]], [[79, 92]]]", "query_spans": "[[[94, 121]]]", "process": "Analysis: First find the asymptote equations, then find the equation of the line passing through one focus and parallel to an asymptote, substitute into the hyperbola equation to obtain the coordinates of intersection point M, and thus find the area of the triangle. The foci of the hyperbola are F_{1}(-5,0), F_{2}(5,0), and the asymptote equations are y=\\pm\\frac{4}{3}x. The equation of the line passing through F_{2} and parallel to one asymptote is y=\\frac{4}{3}(x-5). Solving \\begin{cases}y=\\frac{4}{3}(x-5)\\\\\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1\\end{cases} gives \\begin{cases}x=\\frac{17}{5}\\\\y=-\\frac{32}{15}\\end{cases}, so M(\\frac{17}{5},-\\frac{32}{15}). Therefore, S_{AF_{1}MF_{2}}=\\frac{1}{2}\\times10\\times\\frac{32}{15}=\\frac{32}{3}." }, { "text": "Given that one focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$ is $(1,0)$, what is the eccentricity of $C$?", "fact_expressions": "C: Ellipse;a: Number;G: Point;Expression(C) = (y^2/3 + x^2/a^2 = 1);Coordinate(G) = (1, 0);OneOf(Focus(C)) = G", "query_expressions": "Eccentricity(C)", "answer_expressions": "1/2", "fact_spans": "[[[2, 48], [63, 66]], [[9, 48]], [[54, 61]], [[2, 48]], [[54, 61]], [[2, 61]]]", "query_spans": "[[[63, 72]]]", "process": "The ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{3} = 1 $ has a focus at $ (1,0) $. Thus, $ a^{2} - 3 = 1 $, solving gives $ a = 2 $. Therefore, the eccentricity of the ellipse is: $ e = \\frac{c}{a} = \\frac{1}{2} $." }, { "text": "The equation of a parabola with the coordinate axes as axes of symmetry, the origin as vertex, and passing through the center of the circle $x^{2}+y^{2}-2 x+6 y+9=0$ is?", "fact_expressions": "G: Parabola;SymmetryAxis(G) = axis;O: Origin;Vertex(G) = O;H: Circle;Expression(H) = (6*y - 2*x + x^2 + y^2 + 9 = 0);PointOnCurve(Center(H), G)", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=9*x, x^2=(-1/3)*y}", "fact_spans": "[[[46, 49]], [[0, 49]], [[9, 11]], [[9, 49]], [[17, 43]], [[17, 43]], [[16, 49]]]", "query_spans": "[[[46, 53]]]", "process": "Test analysis: The center of the circle has coordinates (1, -3), located in the fourth quadrant. Since the parabola passes through the center, it can be expressed as y^{2}=2px (p>0) or x^{2}=2py (p>0). Substituting (1, -3) into these equations respectively yields the parabola equations y^{2}=9x or x^{2}=-\\frac{1}{3}y." }, { "text": "Given that a line passing through the left focus $F$ of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ intersects $C$ at points $A$ and $B$, if $|A F|+2|B F| \\geq k$ always holds, then the maximum value of $k$ is?", "fact_expressions": "C: Ellipse;G: Line;A: Point;F: Point;B: Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};k: Number;Abs(LineSegmentOf(A,F))+2*Abs(LineSegmentOf(B,F)) >= k", "query_expressions": "Max(k)", "answer_expressions": "1+(3*sqrt(2))/4", "fact_spans": "[[[3, 35], [46, 49]], [[43, 45]], [[50, 53]], [[39, 42]], [[54, 57]], [[3, 35]], [[3, 42]], [[2, 45]], [[43, 59]], [[87, 90]], [[61, 82]]]", "query_spans": "[[[87, 96]]]", "process": "As shown in the figure, let the equation of line AB be x = my - 1, and let points A(x_{1}, y_{1}), B(x_{2}, y_{2}). Solving the system of equations consisting of the line AB and the ellipse C: \n\\begin{cases}x = my - 1 \\\\ \\frac{x^{2}}{2} + y^{2} = 1\\end{cases}, \neliminating x gives (m^{2} + 2)y^{2} - 2my - 1 = 0, \\Delta = 4m^{2} + 4(m^{2} + 2) = 8(m^{2} + 1) > 0. By Vieta's formulas, y_{1} + y_{2} = \\frac{2m}{m^{2} + 2}, y_{1}y_{2} = -\\frac{1}{m^{2} + 2}. Then \\frac{|y_{1}| + |y_{2}|}{\\sqrt{1 + m^{2}}} = \\frac{\\sqrt{8(1 + m^{2})}}{\\sqrt{1 + m^{2}}} = 2\\sqrt{2}. \\therefore 2\\sqrt{2}(|AF| + 2|BF|) = \\left(\\frac{1}{|AF|} + \\frac{2}{|BF|}\\right)(|AF| + 2|BF|) = 3 + \\frac{2|BF|}{|AF|} + \\frac{|AF|}{|BF|}. The equality holds if and only if |AF| = \\sqrt{2}|BF|. Therefore, |AF| + 2|BF| \\geqslant \\frac{3 + 2\\sqrt{2}}{2\\sqrt{2}} = 1 + \\frac{3\\sqrt{2}}{4}. \\therefore k \\leqslant 1 + \\frac{3\\sqrt{2}}{4}. Hence, the maximum value of k is 1 + \\frac{3\\sqrt{2}}{4}." }, { "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": "K: Point;G: Point;Coordinate(G) = (2, 0);Distance(K, yAxis) = Distance(K, G) - 2", "query_expressions": "LocusEquation(K)", "answer_expressions": "{y^2=8*x, (y=0)&(x<=0)}", "fact_spans": "[[[1, 3], [33, 35]], [[13, 23]], [[13, 23]], [[1, 30]]]", "query_spans": "[[[33, 42]]]", "process": "" }, { "text": "Through the focus of the parabola $y^{2}=2 px(p>0)$, draw a line $l$ intersecting the parabola at points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. What is the value of $\\frac{y_{1} y_{2}}{x_{1} x_{2}}$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p > 0;PointOnCurve(Focus(G), l) = True;l: Line;Intersection(l, G) = {A, B};A: Point;Coordinate(A) = (x1, y1);x1: Number;y1: Number;B: Point;Coordinate(B) = (x2, y2);x2: Number;y2: Number", "query_expressions": "y1*y2/(x1*x2)", "answer_expressions": "-4", "fact_spans": "[[[2, 22], [33, 36]], [[2, 22]], [[5, 22]], [[5, 22]], [[0, 32]], [[27, 32]], [[27, 73]], [[37, 54]], [[37, 54]], [[37, 54]], [[37, 54]], [[56, 73]], [[56, 73]], [[56, 73]], [[56, 73]]]", "query_spans": "[[[76, 113]]]", "process": "" }, { "text": "If the equation $\\frac{x^{2}}{k+2}-\\frac{y^{2}}{5-k}=1$ represents a hyperbola, then what is the range of values for $k$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(k + 2) - y^2/(5 - k) = 1);k: Number", "query_expressions": "Range(k)", "answer_expressions": "(-2, 5)", "fact_spans": "[[[44, 47]], [[1, 47]], [[49, 52]]]", "query_spans": "[[[49, 59]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ has foci $F_{1}$ and $F_{2}$, with focal distance $2 c$. There exists a point $P$ on the hyperbola such that the line $P F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at the midpoint $M$ of $P F_{1}$. Then the eccentricity 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;F1: Point;F2: Point;Focus(G) = {F1,F2};c: Number;FocalLength(G) = 2*c;P: Point;PointOnCurve(P, G) = True;M: Point;H: Circle;Expression(H) = (x^2 + y^2 = a^2);TangentPoint(LineOf(P,F1),H) = M;MidPoint(LineSegmentOf(P,F1)) = M", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 61], [92, 95], [157, 160]], [[2, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[5, 61]], [[67, 74]], [[75, 82]], [[2, 82]], [[86, 91]], [[2, 91]], [[100, 103]], [[92, 103]], [[152, 155]], [[117, 137]], [[117, 137]], [[105, 155]], [[140, 155]]]", "query_spans": "[[[157, 166]]]", "process": "Let point P be a point on the right branch of a hyperbola. The line PF_{1} is tangent to the circle x^{2}+y^{2}=a^{2} at the midpoint M of PF_{1}. Using the property of the midline and the definition of the hyperbola, we obtain |PF_{2}|=2a, |PF_{1}|=4a. Then, using the Pythagorean theorem and the eccentricity formula, we can solve the problem. Let point P be a point on the right branch of the hyperbola. Since the line PF_{1} is tangent to the circle x^{2}+y^{2}=a^{2} at the midpoint M of PF_{1}, it follows that OM\\bot PF_{1}, |OM|=\\frac{1}{2}|PF_{2}|=a, hence |PF_{2}|=2a. By the definition of the hyperbola, |PF_{1}|-|PF_{2}|=2a, so |PF_{1}|=4a. Also, since PF_{1}\\bot PF_{2}, we have |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}, which gives 16a^{2}+4a^{2}=4c^{2}, thus c^{2}=5a^{2}, so e=\\frac{c}{a}=\\sqrt{5}." }, { "text": "If the equation $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{3-m}=1$ represents an ellipse, then what is the range of values for $m$?", "fact_expressions": "G: Ellipse;m:Number;Expression(G)=(x^2/(m-1)+y^2/(3-m)=1)", "query_expressions": "Range(m)", "answer_expressions": "(1,2)+(2,3)", "fact_spans": "[[[44, 46]], [[48, 51]], [[1, 46]]]", "query_spans": "[[[48, 58]]]", "process": "From $ m-1>0 $, $ 3-m>0 $, and $ m-1 \\neq 3-m $, we obtain. The equation $ \\frac{x^{2}}{m-1} + \\frac{y^{2}}{3-m} = 1 $ represents an ellipse $ \\Rightarrow \\begin{cases} m-1>0 \\\\ 3-m>0 \\\\ m-1 \\neq 3-m \\end{cases} $, solving gives $ 10)$ has coordinates $(1,0)$, then the range of distances from a point on this parabola to the focus is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;Coordinate(Focus(G)) = (1, 0);H: Point;PointOnCurve(H, G)", "query_expressions": "Range(Distance(H, Focus(G)))", "answer_expressions": "[1, +oo)", "fact_spans": "[[[2, 23], [39, 42]], [[2, 23]], [[5, 23]], [[5, 23]], [[2, 36]], [], [[39, 45]]]", "query_spans": "[[[39, 58]]]", "process": "Since the coordinates of the focus of the parabola \\( y^{2} = 2px \\) (\\( p > 0 \\)) are \\( (1, 0) \\), we have \\( \\frac{p}{2} = 1 \\), solving gives \\( p = 2 \\). Let an arbitrary point on the parabola be \\( M(x_{1}, y_{1}) \\) (\\( x_{1} \\geqslant 0 \\)), and the focus be \\( F(1, 0) \\). By the definition of a parabola, we obtain \\( |MF| = x_{1} + 1 \\). Since \\( x_{1} \\geqslant 0 \\), it follows that \\( |MF| \\geqslant 1 \\). Therefore, the range of distances from a point on the parabola to the focus is \\( [1, +\\infty) \\)." }, { "text": "Given that the line $y = kx$ ($k \\neq 0$) intersects the hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a > 0$, $b > 0$) at points $A$ and $B$, and the circle with diameter $AB$ passes exactly through the right focus $F$ of the hyperbola. If the area of triangle $ABF$ is $4a^{2}$, then what is the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;I: Line;k: Number;A: Point;B: Point;F:Point;a>0;b>0;Negation(k=0);Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(I) = (y = k*x);Intersection(I, G) = {A, B};IsDiameter(LineSegmentOf(A,B),H);PointOnCurve(F,H);RightFocus(G)=F;Area(TriangleOf(A,B,F))=4*a^2", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[22, 78], [105, 108], [142, 145]], [[25, 78]], [[25, 78]], [[100, 101]], [[2, 21]], [[4, 21]], [[80, 83]], [[84, 87]], [[112, 115]], [[25, 78]], [[25, 78]], [[4, 21]], [[22, 78]], [[2, 21]], [[2, 89]], [[90, 101]], [[100, 115]], [[105, 115]], [[117, 140]]]", "query_spans": "[[[142, 153]]]", "process": "Since the circle with diameter AB passes exactly through the right focus F of the hyperbola, the equation of the circle with diameter AB is x^{2}+y^{2}=c^{2}. The circle also passes through the left focus F_{1} because AB is equal and bisected by F_{1}F_{2}. Therefore, the quadrilateral AF_{1}BF is a rectangle, so |AF|=|BF_{1}|. Let |AF|=m, |BF|=n, then |AF|-|BF|=|BF_{1}|-|BF|=m-n=2a, so m^{2}+n^{2}-2mn=4a^{2}. Since AF\\bot BF, m^{2}+n^{2}=|AB|^{2}=4c^{2}. Since the area of triangle ABF is 4a^{2}, \\frac{1}{2}mn=4a^{2}, thus mn=8a^{2}. Therefore, 4c^{2}-16a^{2}=4a^{2}, so c^{2}=5a^{2}. Hence a^{2}+b^{2}=5a^{2}, so b^{2}=4a^{2}, giving b=2a. Thus, the asymptotes of the hyperbola are y=\\pm\\frac{b}{a}x=\\pm2x." }, { "text": "A line passing through a focus $F_1$ of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ intersects the ellipse at points $A$ and $B$. What is the perimeter of triangle $\\triangle A B F_{2}$ formed by $A$, $B$, and the other focus $F_{2}$ of the ellipse?", "fact_expressions": "E: Ellipse;Expression(E) = (x^2/3 + y^2 = 1);F1: Point;Focus(E) = {F1, F2};l: Line;PointOnCurve(F1, l);A: Point;B: Point;Intersection(l, E) = {A, B};F2: Point", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[1, 28], [44, 46], [67, 69]], [[1, 28]], [[33, 40]], [[1, 81]], [[41, 43]], [[0, 43]], [[48, 51], [59, 62]], [[52, 55], [63, 66]], [[41, 57]], [[74, 81]]]", "query_spans": "[[[59, 110]]]", "process": "According to the problem, by the definition of an ellipse, the sum of the distances from points A and B to the two foci is equal to the length of the major axis of the ellipse, so the perimeter of triangle ABF_{2} can be found. Solution: From the ellipse \\frac{x^{2}}{3}+y^{2}=1, we have a=\\sqrt{3}, b=1, c=\\sqrt{2}. Since AB passes through focus F_{1}, by the definition of an ellipse, |AF_{1}|+|AF_{2}|=2a=2\\sqrt{3}, |BF_{1}|+|BF_{2}|=2\\sqrt{3}. Therefore, the perimeter of triangle ABF_{2} = |AF_{1}|+|AF_{2}|+|BF_{1}|+|BF_{2}|=4\\sqrt{3}." }, { "text": "If the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$, $F_{2}$, the line $y=k x(k>0)$ intersects the hyperbola at points $M$, $N$, and $|O M|=|O F_{2}|$, where $O$ is the origin, and $S_{\\Delta M F_{2} N}=\\frac{(|M F_{2}|+|N F_{2}|)^{2}}{16}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;k: Number;O: Origin;M: Point;F1:Point;F2: Point;N: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);k>0;Expression(H) = (y = k*x);LeftFocus(G) =F1;RightFocus(G)=F2;Intersection(H, G) = {M, N};Abs(LineSegmentOf(O, M)) = Abs(LineSegmentOf(O, F2));Area(TriangleOf(M,F2,N))=(Abs(LineSegmentOf(M,F2))+Abs(LineSegmentOf(N,F2)))^2/16", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[1, 57], [95, 98], [201, 204]], [[4, 57]], [[4, 57]], [[80, 94]], [[82, 94]], [[129, 132]], [[100, 103]], [[64, 71]], [[72, 79]], [[104, 107]], [[4, 57]], [[4, 57]], [[1, 57]], [[82, 94]], [[80, 94]], [[1, 79]], [[1, 79]], [[80, 109]], [[111, 128]], [[139, 198]]]", "query_spans": "[[[201, 210]]]", "process": "By the symmetry of the line y = kx (k > 0) and the hyperbola, points M and N are symmetric with respect to the origin. In triangle \\triangle MF_{2}N, |OM| = |OF_{2}|, so MF_{2} \\bot NF_{2}. M and N are the intersection points of the circle with diameter F_{1}F_{2} and the hyperbola. Assume M(x_{1}, y_{1}) lies in the first quadrant. S_{\\triangle NF_{2}}M = S_{\\triangle MF_{1}F_{2}}. Since the circle has diameter F_{1}F_{2}, its radius is c. Since point M(x_{1}, y_{1}) lies on both the circle and the hyperbola, we have\n\\begin{cases}\n\\frac{x^{2}}{a^{2}} - \\frac{y_{1}^{2}}{b^{2}} = 1 \\\\\nx^{2} + y^{2} = c^{2}\n\\end{cases}\nSolving and simplifying yields b^{2}(c^{2} - y_{1}^{2}) - a^{2}y_{1} = a^{2}b^{2}, rearranging gives b^{2}c^{2} - a^{2}b^{2} = b^{2}y_{1}^{2} + a^{2}y_{1}, b^{4} = c^{2}y^{2}, y_{1} = \\frac{b^{2}}{c}. Thus S_{\\triangle MF_{1}F_{2}} = \\frac{1}{2} \\cdot 2c \\cdot y_{1} = b^{2}. From S_{\\triangle MF_{2}N} = \\frac{(|MF_{1}| + |NF_{2}|)^{2}}{}, solving the system\n\\begin{cases}\nMF_{1} + MF_{2} = 4b \\\\\nMF_{1} - MF_{2} = 2a\n\\end{cases}\ngives MF_{1} = 2b + a, MF_{2} = 2b - a. Since F_{1}F_{2} is the diameter of the circle, MF_{1}^{2} + MF_{2}^{2} = F_{1}F_{2}^{2}, i.e., (2b + a)^{2} + (2b - a)^{2} = 4c^{2}, 8b^{2} + 2a^{2} = 4c^{2}, 4b^{2} + a^{2} = 2c^{2}. Then 4c^{2} - 4a^{2} + a^{2} = 2c^{2}, 2c^{2} = 3a^{2}, \\frac{c^{2}}{a^{2}} = \\frac{3}{2}, so the eccentricity e = \\frac{c}{a} = \\frac{\\sqrt{6}}{2}" }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$ has focus $F$, point $P$ lies on the parabola, and point $Q\\left(\\frac{9}{2} p, 0\\right)$. If $|Q F|=2|P F|$ and the area of $\\Delta P Q F$ is $8 \\sqrt{3}$, then $p=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);Q: Point;Coordinate(Q) = (9*p/2, 0);Abs(LineSegmentOf(Q, F)) = 2*Abs(LineSegmentOf(P, F));Area(TriangleOf(P, Q, F)) = 8*sqrt(3)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 23], [36, 39]], [[2, 23]], [[114, 117]], [[5, 23]], [[27, 30]], [[2, 30]], [[31, 35]], [[31, 40]], [[41, 63]], [[41, 63]], [[66, 80]], [[82, 112]]]", "query_spans": "[[[114, 119]]]", "process": "From the given condition, F(\\frac{p}{2},0), so |QF|=4p, hence |PF|=\\frac{1}{2}|QF|=2p. From the directrix of the parabola being x=-\\frac{p}{2}, and by the definition of the parabola, the x-coordinate of point P is 2p-\\frac{p}{2}=\\frac{3}{2}p. Assume without loss of generality that point P lies above the x-axis, then the y-coordinate of P is \\sqrt{3}p; thus S_{\\trianglePQF}=\\frac{1}{2}\\times4p\\times\\sqrt{3}p=8\\sqrt{3}, solving gives p=2." }, { "text": "Given the parabola $C$: $y^{2}=16 x$ with focus $F$ and directrix $l$, $P$ is a point on $l$, and $Q$ is an intersection point of line $P F$ and $C$. If $\\overrightarrow{F P}=4 \\overrightarrow{F Q}$, then $|F Q|$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 16*x);F: Point;Focus(C) = F;l: Line;Directrix(C) = l;P: Point;PointOnCurve(P, l);Q: Point;OneOf(Intersection(LineOf(P, F), C)) = Q;VectorOf(F, P) = 4*VectorOf(F, Q)", "query_expressions": "Abs(LineSegmentOf(F, Q))", "answer_expressions": "6", "fact_spans": "[[[2, 22], [62, 65]], [[2, 22]], [[26, 29]], [[2, 29]], [[33, 36], [43, 46]], [[2, 36]], [[39, 42]], [[39, 49]], [[50, 53]], [[50, 70]], [[72, 117]]]", "query_spans": "[[[119, 128]]]", "process": "Let the distance from Q to 1 be d, then |FQ| = d. Since \\overrightarrow{FP} = 4\\overrightarrow{FQ}, \\therefore |PQ| = 3d, \\therefore the slope of line PF is -\\frac{\\sqrt{(3d)^{2}-d^{2}}}{d}. Since F(4,0), \\therefore the equation of line PF is y = -2\\sqrt{2}(x-4). Solving together with y^{2} = 16x gives x = 2, \\therefore |FO| = d = 2 + 4 = 6." }, { "text": "The distance from a moving point $P$ to the point $F(2,0)$ is equal to its distance to the line $x+2=0$. Then, what is the equation of the trajectory of $P$?", "fact_expressions": "G: Line;Expression(G) = (x + 2 = 0);P: Point;F: Point;Coordinate(F) = (2, 0);Distance(P, F) = Distance(P, G)", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[21, 30]], [[21, 30]], [[2, 5], [19, 20], [37, 40]], [[6, 15]], [[6, 15]], [[2, 35]]]", "query_spans": "[[[37, 47]]]", "process": "By the definition of a parabola, the locus of point P is a parabola with focus F and directrix the line x = -2, so its standard equation is y^{2} = 8x." }, { "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": "This question examines the standard equation of a parabola and the application of its simple properties, classified as an easy problem." }, { "text": "If the line $x - y = 1$ intersects the ellipse $\\frac{x^{2}}{3} + \\frac{y^{2}}{2} = 1$ at points $A$ and $B$, then the coordinates of the midpoint of segment $AB$ are?", "fact_expressions": "H: Line;Expression(H) = (x - y = 1);G: Ellipse;Expression(G) = (x^2/3 + y^2/2 = 1);Intersection(H,G) = {A,B};A: Point;B: Point", "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "(3/5,-2/5)", "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 48]], [[11, 48]], [[1, 59]], [[50, 53]], [[54, 57]]]", "query_spans": "[[[61, 75]]]", "process": "" }, { "text": "Given that the foci of ellipse $C$ are $F_{1}(-1,0)$ and $F_{2}(1,0)$, a line passing through $F_{2}$ intersects $C$ at points $A$ and $B$. If $|A F_{2}|=\\frac{3}{2}|B F_{2}|$, $|B F_{1}|=2|B F_{2}|$, then the equation of ellipse $C$ is?", "fact_expressions": "C: Ellipse;G: Line;F2: Point;A: Point;B: Point;F1: Point;Coordinate(F2) = (1, 0);Coordinate(F1)=(-1, 0) ;Focus(C) = {F1,F2};PointOnCurve(F2, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F2)) = (3/2)*Abs(LineSegmentOf(B, F2));Abs(LineSegmentOf(B, F1)) = 2*Abs(LineSegmentOf(B, F2))", "query_expressions": "Expression(C)", "answer_expressions": "x^2/5+y^2/4=1", "fact_spans": "[[[2, 7], [52, 55], [126, 131]], [[49, 51]], [[27, 39], [41, 48]], [[57, 60]], [[61, 64]], [[11, 25]], [[27, 39]], [[11, 25]], [[2, 39]], [[40, 51]], [[49, 66]], [[68, 100]], [[102, 124]]]", "query_spans": "[[[126, 136]]]", "process": "First, analyze to obtain that point A is the upper (lower) vertex of the ellipse. From \\overrightarrow{AF_{2}}=\\frac{3}{2}\\overrightarrow{F_{2}B}, we get B(\\frac{5}{3},\\pm\\frac{2}{3}b). Substituting the coordinates of point B into the ellipse equation gives the value of a, thus obtaining the equation of the ellipse. Let |BF_{2}|=2m, then |AF_{2}|=3m, |BF_{1}|=4m. By the definition of the ellipse, |BF_{1}|+|BF_{2}|=|AF_{1}|+|AF_{2}|=6m, so |AF_{1}|=6m-3m=3m, hence |AF_{1}|=|AF_{2}|, therefore point A is the upper (lower) vertex of the ellipse. Let A(0,\\pmb). From \\overrightarrow{AF}_{2}=\\frac{3}{2}\\overrightarrow{F_{2}B}, we get B(\\frac{5}{3},\\pm\\frac{2}{3}b). Since point B lies on the ellipse, \\frac{25}{a^{2}}+\\frac{\\frac{4}{9}b^{2}}{b^{2}}=1, solving gives a^{2}=5. Given c=1, we obtain b, hence the ellipse equation is \\frac{x^{2}}{5}+\\frac{y^{2}}{4}=" }, { "text": "Given a point $P(3 , y1)$ on the parabola $y^{2}=4 x$, what is the distance from point $P$ to the focus of the parabola?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (3, y1);y1:Number;PointOnCurve(P,G)", "query_expressions": "Distance(P, Focus(G))", "answer_expressions": "4", "fact_spans": "[[[2, 16], [37, 40]], [[32, 36], [19, 30]], [[2, 16]], [[19, 30]], [[19, 30]], [[2, 30]]]", "query_spans": "[[[32, 47]]]", "process": "\\because the parabola y^{2}=4x, comparing with the standard equation of the parabola y^{2}=2px, we get p=2. \\therefore the directrix of the parabola is x=-1. By the definition of the 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": "If the product of the distances from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to its two foci is $m$, then the maximum value of $m$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G) = True;Focus(G) = {F1,F2};F1: Point;F2: Point;Distance(P,F1) * Distance(P,F2) = m;m:Number", "query_expressions": "Max(m)", "answer_expressions": "25", "fact_spans": "[[[1, 39]], [[1, 39]], [[42, 45]], [[1, 45]], [[1, 49]], [], [], [[1, 58]], [[55, 58], [60, 63]]]", "query_spans": "[[[60, 69]]]", "process": "According to the definition of the ellipse, |PF₁| + |PF₂| = 10. Combining this with the basic inequality, we obtain |PF₁|⋅|PF₂| ≤ ( (|PF₁| + |PF₂|)/2 )², which can be used to solve the problem. \nSolution: Let the left and right foci of the ellipse x²/25 + y²/9 = 1 be F₁ and F₂, respectively. Then m = |PF₁|⋅|PF₂|. According to the definition of the ellipse, we have |PF₁| + |PF₂| = 2a = 10. Also, by the inequality |PF₁|⋅|PF₂| ≤ ( (|PF₁| + |PF₂|)/2 )² = (10/2)² = 25, equality holds if and only if |PF₁| = |PF₂|, that is, when |PF₁| = |PF₂| = 5. Therefore, the maximum value of m is 25." }, { "text": "Let $P$ be a moving point on the parabola $y^{2}=4x$, and let $F$ be the focus of the parabola. Denote by $M$ the minimum value of the sum of the distance from point $P$ to point $A(-1,1)$ and the distance from point $P$ to the line $x=-1$. If $B(3,2)$, denote by $N$ the minimum value of $|PB|+|PF|$, then $M+N=$?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;P: Point;F: Point;M:Number;N:Number;Expression(G) = (y^2 = 4*x);Expression(H) = (x = -1);Coordinate(A) = (-1, 1);Coordinate(B) = (3, 2);PointOnCurve(P, G);Focus(G) = F;Min(Distance(P,A)+Distance(P,H))=M;Min(Abs(LineSegmentOf(P, B)) + Abs(LineSegmentOf(P, F))) = N", "query_expressions": "M + N", "answer_expressions": "sqrt(5)+4", "fact_spans": "[[[5, 19], [30, 33]], [[62, 70]], [[43, 53]], [[85, 93]], [[1, 4], [38, 42], [57, 61]], [[26, 29]], [[80, 83]], [[111, 114]], [[5, 19]], [[62, 70]], [[43, 53]], [[85, 93]], [[1, 25]], [[26, 36]], [[38, 83]], [[95, 114]]]", "query_spans": "[[[116, 123]]]", "process": "We obtain that the focus of the parabola $ y^{2} = 4x $ is $ F(1,0) $, and the equation of the directrix is $ x = -1 $. Therefore, the sum of the distance from point $ P $ to point $ A(-1,1) $ and the distance from point $ P $ to the line $ x = -1 $ equals the sum of the distance from point $ P $ to point $ A(-1,1) $ and the distance from point $ P $ to the focus $ F $. When points $ P $, $ A $, and $ F $ are collinear, this sum is minimized, and $ M = |AF| $. By the distance formula between two points, we get $ M = |AF| = \\sqrt{(} $. From the definition of the parabola, $ |PF| $ equals the distance from $ P $ to the directrix $ x = -1 $, hence $ |PB| + |PF| $ equals the sum of $ |PB| $ and the distance from $ P $ to the directrix $ x = -1 $. It follows that when points $ P $, $ B $, and $ F $ are collinear, the sum is minimized, and the minimum distance $ N $ is $ 3.(-1) = 4 $. Therefore, $ M + N = \\sqrt{5} + 4 $." }, { "text": "Let the ellipse $C$: $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ have left and right foci $F_{1}$ and $F_{2}$, respectively, and let $A$ be any point on $C$. Then the perimeter of $\\Delta A F_{1} F_{2}$ is?", "fact_expressions": "C: Ellipse;A: Point;F1: Point;F2: Point;Expression(C) = (x^2/36 + y^2/9 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(A, C)", "query_expressions": "Perimeter(TriangleOf(A, F1, F2))", "answer_expressions": "12+6*sqrt(3)", "fact_spans": "[[[1, 44], [77, 80]], [[73, 76]], [[53, 60]], [[63, 70]], [[1, 44]], [[1, 70]], [[1, 70]], [[72, 85]]]", "query_spans": "[[[87, 114]]]", "process": "According to the problem, the ellipse $ C: \\frac{x^{2}}{36} + \\frac{y^{2}}{9} = 1 $, where $ a = 6 $, $ b = 3 $, then $ c = \\sqrt{36 - 9} = 3\\sqrt{3} $. $ A $ is any point on $ C $, then the perimeter $ l $ of triangle $ AF_{1}F_{2} $ is $ |AF_{1}| + |AF_{2}| + |F_{1}F_{2}| = 2a + 2c = 12 + 6\\sqrt{3} $. The correct answer to this problem is: $ 12 + 6\\sqrt{3} $. This problem examines geometric properties of ellipses, and the key is recognizing that the perimeter of the focal triangle is a constant value: $ 2a + 2c $." }, { "text": "Given that the vertex of the parabola is at the origin, the axis of symmetry is the $x$-axis, and the common chord length of the intersection with the circle $x^{2}+y^{2}=4$ is $2\\sqrt{3}$, then the equation of the parabola is?", "fact_expressions": "G: Parabola;O: Origin;H: Circle;Vertex(G) = O;SymmetryAxis(G) = xAxis;Expression(H) = (x^2 + y^2 = 4);Length(InterceptChord(G, H)) = 2*sqrt(3)", "query_expressions": "Expression(G)", "answer_expressions": "{(y^2=3*x), (y^2=-3*x)}", "fact_spans": "[[[2, 5], [63, 66]], [[9, 13]], [[25, 41]], [[2, 13]], [[2, 22]], [[25, 41]], [[2, 61]]]", "query_spans": "[[[63, 71]]]", "process": "Let the equation of the desired parabola be y^{2}=2px (p>0) or y^{2}=-2px (p>0), and let the intersection points with the circle be A(x_{1},y_{1}) and B(x_{2},y_{2}) (y_{1}>0, y_{2}<0). By symmetry, we have x_{1}=x_{2}, y_{2}=-y_{1}, then |y_{1}|+|y_{2}|=2\\sqrt{3}, that is, y_{1}-y_{2}=2y_{1}=2\\sqrt{3}, so y_{1}=\\sqrt{3}. Substituting y_{1}=\\sqrt{3} into x^{2}+y^{2}=4, we get x_{1}=\\pm1. Therefore, the point (1,\\sqrt{3}) lies on the parabola y^{2}=2px, and the point (-1,\\sqrt{3}) lies on the parabola y^{2}=-2px. We obtain p=\\frac{3}{2}. Thus, the equations of the desired parabolas are y^{2}=3x or y^{2}=-3x." }, { "text": "If the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ coincides with the center of the circle $x^{2}+y^{2}-4 x=0$, then $a$=?", "fact_expressions": "G: Hyperbola;H: Circle;a: Number;a>0;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (-4*x + x^2 + y^2 = 0);RightFocus(G) = Center(H)", "query_expressions": "a", "answer_expressions": "sqrt(3)", "fact_spans": "[[[1, 38]], [[43, 63]], [[70, 73]], [[4, 38]], [[1, 38]], [[43, 63]], [[1, 68]]]", "query_spans": "[[[70, 75]]]", "process": "From $\\frac{x^{2}}{a^{2}}-y^{2}=1$, we obtain $c^{2}=a^{2}+1$, so $c=\\sqrt{a^{2}+1}$. Thus, the coordinates of the right focus of the hyperbola are $(\\sqrt{a^{2}+1},0)$. From $x^{2}+y^{2}-4x=0$, we obtain $(x-2)^{2}+y^{2}=4$, so the center of the circle is at $(2,0)$. According to the problem, $\\sqrt{a^{2}+1}=2$, solving gives $a=\\sqrt{3}$ or $a=-\\sqrt{3}$ (discarded)." }, { "text": "The left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $F$, and $A(-a, 0)$, $B(0, b)$ are two vertices. If the distance from $F$ to the line $AB$ equals $\\frac{b}{\\sqrt{7}}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;B: Point;A: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (-a, 0);Coordinate(B) = (0, b);LeftFocus(G) = F;In(A, Vertex(G));In(B, Vertex(G));Distance(F, LineOf(A, B)) = b/sqrt(7)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[0, 52], [129, 131]], [[2, 52]], [[2, 52]], [[74, 83]], [[62, 72]], [[57, 60], [91, 94]], [[2, 52]], [[2, 52]], [[0, 52]], [[62, 72]], [[74, 83]], [[0, 60]], [[0, 88]], [[0, 88]], [[91, 127]]]", "query_spans": "[[[129, 137]]]", "process": "" }, { "text": "If the coordinates of the two foci of an ellipse are $F_{1}(-1,0)$, $F_{2}(5,0)$, and the length of the major axis is $10$, then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (5, 0);Focus(G) = {F1, F2};Length(MajorAxis(G)) = 10", "query_expressions": "Expression(G)", "answer_expressions": "(x-2)^2/25 + y^2/16 = 1", "fact_spans": "[[[1, 3], [51, 53]], [[11, 24]], [[27, 39]], [[11, 24]], [[27, 39]], [[1, 39]], [[1, 49]]]", "query_spans": "[[[51, 58]]]", "process": "According to the problem: the center of the ellipse is (2,0), a=5, c=3, thus the equation of the ellipse can be found. Since the coordinates of the two foci of the ellipse are F_{1}(-1,0), F_{2}(5,0), and the length of the major axis is 10, the center of the ellipse is (2,0), a=5, c=3, so b^{2}=a^{2}-c^{2}=25-9=16. Therefore, the equation of the ellipse is: \\frac{(x-2)^{2}}{25}+\\frac{y^{2}}{16}=1" }, { "text": "If the eccentricity of the ellipse $x^{2}+m y^{2}=1$ is $\\frac{\\sqrt{2}}{2}$, then what is the length of the major axis of the ellipse?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (m*y^2 + x^2 = 1);Eccentricity(G) = sqrt(2)/2", "query_expressions": "Length(MajorAxis(G))", "answer_expressions": "{2*sqrt(2), 2}", "fact_spans": "[[[1, 20], [47, 49]], [[3, 20]], [[1, 20]], [[1, 45]]]", "query_spans": "[[[47, 54]]]", "process": "The ellipse $ x^{2} + m y^{2} = 1 $, i.e., $ x^{2} + \\frac{y^{2}}{1} = 1 $. When the foci of the ellipse are on the $ y $-axis, $ \\therefore a^{2} = \\frac{1}{m}, b^{2} = 1 $, by $ c^{2} = a^{2} - b^{2} $, we get $ c^{2} = \\frac{1 - m}{m} $. Since $ \\frac{c^{2}}{a^{2}} = 1 - m = \\frac{1}{2} $, solving gives $ m = \\frac{1}{2} $, $ \\therefore a = \\sqrt{2} $, i.e., the major axis length is $ 2\\sqrt{2} $. When the foci of the ellipse are on the $ x $-axis, $ b^{2} = \\frac{1}{m}, a^{2} = 1 $, $ \\therefore a = 1 $, i.e., the major axis length is $ 2 $. In summary, the major axis length of the ellipse is $ 2\\sqrt{2} $ or $ 2 $." }, { "text": "If $\\sqrt{(x-\\sqrt{3})^{2}+y^{2}}+\\sqrt{(x+\\sqrt{3})^{2}+y^{2}}=4$, then the maximum value of $|3 x-4 y+10|$ is?", "fact_expressions": "x: Number;y: Number;sqrt(y^2 + (x - sqrt(3))^2) + sqrt(y^2 + (x + sqrt(3))^2) = 4", "query_expressions": "Max(Abs(3*x - 4*y + 10))", "answer_expressions": "10 + 2*sqrt(13)", "fact_spans": "[[[1, 64]], [[1, 64]], [[1, 64]]]", "query_spans": "[[[66, 86]]]", "process": "From $\\sqrt{(x-\\sqrt{3})^{2}+y^{2}}+\\sqrt{(x+\\sqrt{3})^{2}+y^{2}}=4$, it follows that the sum of distances from point $(x,y)$ to $(\\sqrt{3},0)$ and $(-\\sqrt{3},0)$ is 4. Noting that $4>2\\sqrt{3}$, the trajectory of $(x,y)$ lies on an ellipse with foci at $(\\pm\\sqrt{3},0)$, major axis length 4, and center at the origin. The equation of the ellipse is $\\frac{x^{2}}{4}+y^{2}=1$. Using trigonometric substitution, let $x=2\\cos\\alpha$, $y=\\sin\\alpha$ $(0\\leqslant\\alpha<2\\pi)$. Then $|3x-4y+10|=|6\\cos\\alpha-4\\sin\\alpha+10|$. Using the auxiliary angle formula, $|3x-4y+10|=|2\\sqrt{13}\\cos(\\alpha+\\varphi)+10|$, where $\\tan\\varphi=\\frac{2}{3}$. Since $|\\cos(\\alpha+\\varphi)|\\leqslant1$, $|3x-4y+10|=|2\\sqrt{13}\\cos(\\alpha+\\varphi)+10|\\in[10-2\\sqrt{13},10+2\\sqrt{13}]$, and the maximum value is $10+2\\sqrt{13}$." }, { "text": "If the directrix of the parabola $x^{2}=2 p y(p>0)$ passes through a focus of the hyperbola $y^{2}-x^{2}=1$, then $p$=?", "fact_expressions": "H: Parabola;Expression(H) = (x^2 = 2*p*y);p: Number;p>0;G: Hyperbola;Expression(G) = (-x^2 + y^2 = 1);PointOnCurve(OneOf(Focus(G)), Directrix(H)) = True", "query_expressions": "p", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[1, 22]], [[1, 22]], [[52, 55]], [[4, 22]], [[27, 45]], [[27, 45]], [[1, 50]]]", "query_spans": "[[[52, 57]]]", "process": "According to the problem, the equation of the hyperbola is: $ y^{2} - x^{2} = 1 $, so its foci lie on the $ y $-axis, and $ c = \\sqrt{2} $. The focus coordinates of the parabola $ x^{2} = 2py $ ($ p > 0 $) are $ (0, \\pm\\frac{p}{2}) $, so $ \\frac{p}{2} = \\sqrt{2} $, which gives $ p = 2\\sqrt{2} $." }, { "text": "If point $P$ is on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{64}=1$, $F_{1}$, $F_{2}$ are the foci, and $\\angle F_{1} PF_{2}=60^{\\circ}$, then the area of $\\triangle F_{1} PF_{2}$ is?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/100 + y^2/64 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "64*sqrt(3)/3", "fact_spans": "[[[6, 46]], [[51, 58]], [[1, 5]], [[61, 68]], [[6, 46]], [[1, 50]], [[6, 71]], [[73, 105]]]", "query_spans": "[[[107, 136]]]", "process": "" }, { "text": "The distance from the point $P\\left(\\frac{3}{2}, y_{0}\\right)$ on the parabola $y^{2}=a x$ to the focus $F$ of the parabola is $2$. Let $O$ be the origin. Then the area of $\\triangle P O F$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = a*x);a: Number;P: Point;Coordinate(P) = (3/2, y0);y0: Number;Focus(G) = F;F: Point;Distance(P, F) = 2;O: Origin;PointOnCurve(P, G) = True", "query_expressions": "Area(TriangleOf(P, O, F))", "answer_expressions": "sqrt(3)/4", "fact_spans": "[[[0, 14], [42, 45]], [[0, 14]], [[3, 14]], [[16, 40]], [[16, 40]], [[17, 40]], [[42, 50]], [[47, 50]], [[16, 57]], [[59, 62]], [[0, 40]]]", "query_spans": "[[[69, 91]]]", "process": "Using the coordinates of point P, we obtain a > 0, thus obtaining the equation of the directrix and the coordinates of the focus. Then, using the focal radius formula, we find the value of a. Substituting the coordinates of point P into the parabola equation gives y_{0}, and using the triangle area formula yields the answer. Since point P(\\frac{3}{2}, y_{0}) lies on the parabola, \\therefore a > 0, \\therefore the directrix of the parabola is x = -\\frac{a}{4}, and the focus is F(\\frac{a}{4}, 0). \\therefore |PF| = \\frac{3}{2} + \\frac{a}{4} = 2 \\Rightarrow a = 2. Without loss of generality, assume y_{0} > 0, \\therefore y_{0} = \\sqrt{3}, \\therefore s = \\frac{1}{2} \\cdot |OF| \\cdot y_{0} = \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\sqrt{3} = \\frac{\\sqrt{3}}{4}" }, { "text": "Given that the focus of the parabola $y^{2}=4 x$ is $F$, line $m$ is the tangent to the parabola at a point $P$ in the first quadrant, line $n$ is drawn through $P$ parallel to the $x$-axis, and a line through focus $F$ parallel to $m$ intersects $n$ at point $M$. If $|P M |=4$, then what are the coordinates of point $P$?", "fact_expressions": "m: Line;G: Parabola;n: Line;P: Point;M: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Quadrant(P) = 1;TangentOnPoint(P, G) = m;PointOnCurve(P,n);IsParallel(xAxis, n);C:Line;PointOnCurve(F,C);IsParallel(C,m);Intersection(C,n)=M;Abs(LineSegmentOf(P,M))=4", "query_expressions": "Coordinate(P)", "answer_expressions": "(3,2*sqrt(3))", "fact_spans": "[[[24, 29], [77, 80]], [[2, 16], [30, 33]], [[62, 67], [84, 87]], [[41, 44], [50, 53], [107, 111]], [[88, 92]], [[71, 74], [20, 23]], [[2, 16]], [[2, 23]], [[34, 44]], [[24, 48]], [[49, 67]], [[54, 67]], [[81, 83]], [[68, 83]], [[74, 83]], [[81, 92]], [[95, 105]]]", "query_spans": "[[[107, 116]]]", "process": "" }, { "text": "A line with slope $2$ intersects a conic section at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If the chord length $|A B|=2 \\sqrt{5}$, then $|y_{1}-y_{2}|=$?", "fact_expressions": "G: ConicSection;H: Line;A: Point;B: Point;x1:Number;x2:Number;y1:Number;y2:Number;Slope(H)=2;Intersection(H,G)={A,B};Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Abs(LineSegmentOf(A,B))=2*sqrt(5);IsChordOf(LineSegmentOf(A,B),G)", "query_expressions": "Abs(y1 - y2)", "answer_expressions": "4", "fact_spans": "[[[10, 14]], [[7, 9]], [[16, 33]], [[36, 53]], [[16, 33]], [[16, 33]], [[36, 53]], [[36, 53]], [[0, 9]], [[7, 55]], [[16, 33]], [[36, 53]], [[59, 77]], [[10, 77]]]", "query_spans": "[[[79, 96]]]", "process": "Let the slope of line AB be k, k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, AB=\\sqrt{(x_{1}-x_{2})^{2}+(y_{1}} to obtain the chord length formula: |AB|=\\sqrt{1+(\\frac{1}{k})^{2}}|y_{1}-y_{1}, k=2, thus we have |AB|=\\sqrt{1+(\\frac{1}{2})^{2}}|y,-y_{1}|=2\\sqrt{5}\\Rightarrow\\frac{\\sqrt{5}}{2}|y,-y_{1}|=2\\sqrt{5}\\Rightarrow|y_{1}-y_{1}|=4" }, { "text": "Given the ellipse equation $C$ is $\\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1$, $F_{1}$, $F_{2}$ are the two foci of the ellipse, point $P$ lies on $C$ and $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. Then the area of triangle $F_{1} P F_{2}$ is?", "fact_expressions": "C: Ellipse;F1: Point;P: Point;F2: Point;Expression(C) = (x^2/16+y^2/8=1);Focus(C)={F1,F2};PointOnCurve(P, C);AngleOf(F1, P, F2) = pi/3", "query_expressions": "Area(TriangleOf(F1, P,F2))", "answer_expressions": "8*sqrt(3)/3", "fact_spans": "[[[2, 4], [6, 9], [67, 69], [80, 83]], [[48, 56]], [[75, 79]], [[59, 66]], [[2, 46]], [[48, 74]], [[75, 84]], [[85, 121]]]", "query_spans": "[[[124, 147]]]", "process": "From $\\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1$ we get $a^{2}=16$, $b^{2}=8$, $\\therefore c^{2}=a^{2}-b^{2}=8$, let $|PF_{1}|=t_{1}$, $|PF_{2}|=t_{2}$, from the definition of ellipse we obtain $t_{1}+t_{2}=8$, $\\textcircled{1}$, by the law of cosines we have $t_{1}^{2}+t_{2}^{2}-2t_{1}t_{2}\\cdot\\cos60^{\\circ}=32$, $\\textcircled{2}$, subtracting $\\textcircled{2}$ from $\\textcircled{1}$ squared yields $t_{1}t_{2}=\\frac{32}{3}$, $\\therefore S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}\\times\\frac{32}{3}\\times\\sin60^{\\circ}=\\frac{8\\sqrt{3}}{3}$," }, { "text": "Given that point $P$ is an intersection point of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, then $\\cos \\angle F_{1} P F_{2}$=?", "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2 - y^2/2 = 1);Expression(H) = (x^2/4 + y^2 = 1);LeftFocus(H) = F1;RightFocus(H) = F2;OneOf(Intersection(H, G)) = P", "query_expressions": "Cos(AngleOf(F1, P, F2))", "answer_expressions": "-1/3", "fact_spans": "[[[35, 63]], [[7, 34], [85, 87]], [[69, 76]], [[2, 6]], [[77, 84]], [[35, 63]], [[7, 34]], [[69, 92]], [[69, 92]], [[2, 68]]]", "query_spans": "[[[94, 124]]]", "process": "" }, { "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, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "From the parabola equation x^{2}=8y, we know the focus of the parabola lies on the y-axis. From 2p=8, we get \\frac{p}{2}=2, so the coordinates of the focus are (0,2)." }, { "text": "Given $F_{1}(-c, 0)$, $F_{2}(c, 0)$ $(c>0)$ are the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. A line passing through $F_{1}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $M$, and intersects the right branch of $C$ at point $N$. If $O$ is the origin and $|O N|=c$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;c:Number;G: Circle;H: Line;F1: Point;O: Origin;N: Point;M: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = a^2);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(C)=F1;RightFocus(C) = F2;PointOnCurve(F1, H);TangentPoint(H, G) = M;Intersection(H,RightPart(C))=N;Abs(LineSegmentOf(O, N)) = c;F2:Point;c>0", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[39, 100], [148, 151], [182, 185]], [[47, 100]], [[47, 100]], [[17, 36]], [[119, 139]], [[116, 118]], [[2, 16], [108, 115]], [[161, 164]], [[156, 160]], [[142, 146]], [[47, 100]], [[47, 100]], [[39, 100]], [[119, 139]], [[2, 16]], [[17, 36]], [[2, 106]], [[2, 106]], [[107, 118]], [[116, 146]], [[116, 160]], [[171, 180]], [[17, 36]], [[17, 36]]]", "query_spans": "[[[182, 191]]]", "process": "Since |ON| = c, it follows that |ON| = \\frac{1}{2}|F_{1}F_{2}|, so NF_{1}\\bot NF_{2}. Also, the line NF_{1} is tangent to the circle x^{2} + y^{2} = a^{2} at point M, so OM\\bot NF_{1}, hence OM/\\!/NF_{2}. Since O is the midpoint of F_{1}F_{2}, OM is the midline of \\triangle NF_{1}F_{2}, so |NF_{2}| = 2|OM| = 2a. According to the definition of hyperbola, |NF_{1}| - |NF_{2}| = 2a, so |NF_{1}| = 4a. In right triangle \\triangle NF_{1}F_{2}, |NF_{1}|^{2} + |NF_{2}|^{2} = |F_{1}F_{2}|^{2}, that is (4a)^{2} + (2a)^{2} = (2c)^{2}, yielding c^{2} = 5a^{2}. Therefore, e = \\frac{c}{a} = \\sqrt{5}." }, { "text": "Given the parabola $y^{2}=8x$ with focus $F$, a line passing through $F$ intersects the parabola at points $A$ and $B$. Find the minimum value of $|FA|+4|FB|$.", "fact_expressions": "G: Parabola;H: Line;F: Point;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(F,H);Intersection(H,G)={A,B}", "query_expressions": "Min(Abs(LineSegmentOf(F, A)) + 4*Abs(LineSegmentOf(F, B)))", "answer_expressions": "18", "fact_spans": "[[[2, 16], [31, 34]], [[28, 30]], [[19, 22], [24, 27]], [[36, 39]], [[42, 45]], [[2, 16]], [[2, 22]], [[23, 30]], [[28, 47]]]", "query_spans": "[[[49, 69]]]", "process": "Solving the system of equations by elimination, from the relationship between roots and coefficients we obtain the x-coordinates of A and B satisfying x_{1}x_{2}=4. Using the properties of the parabola, we derive |FA|+4|FB|=\\frac{4}{x_{2}}+4x_{2}+10. According to the basic inequality, we find the minimum value. The focus of the parabola y^{2}=8x is F(2,0). Let A(x_{1},y_{1}), B(x_{2},y_{2}), then |FA|+4|FB|=x_{1}+2+4(x_{2}+2)=x_{1}+4x_{2}+10. When the slope of line AB does not exist, |FA|+4|FB|=2+4\\times2+10=20. When the slope of line AB exists, let the equation of line AB be y=k(x-2). Substituting into y^{2}=8x gives k^{2}x^{2}-(4k^{2}+8)x+4k^{2}=0, \\therefore x_{1}x_{2}=4, \\therefore |FA|+4|FB|=\\frac{4}{x_{2}}+4x_{2}+10\\geqslant2\\sqrt{\\frac{4}{x_{2}}\\times4x_{2}}+10=18, with equality if and only if x_{1}=1. The minimum value of |FA|+4|FB| is 18." }, { "text": "Let the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ be $F_{1}$ and $F_{2}$, respectively. If point $P$ lies on the ellipse and $\\cos \\angle F_{1} P F_{2}=\\frac{1}{2}$, then $|PF_{1}|\\cdot| P F_{2} |$=?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(P, G);Cos(AngleOf(F1, P, F2)) = 1/2", "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))", "answer_expressions": "4/3", "fact_spans": "[[[1, 39], [68, 70]], [[47, 54]], [[64, 68]], [[55, 62]], [[1, 39]], [[1, 62]], [[64, 71]], [[73, 111]]]", "query_spans": "[[[113, 141]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and as point $P$ moves on the ellipse $C$, $I$ is the incenter of $\\Delta P F_{1} F_{2}$, then the range of $|I O|$ ($O$ being the origin) is?", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;I: Point;O: Origin;Expression(C) = (x^2/4 + y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Incenter(TriangleOf(P, F1, F2)) = I", "query_expressions": "Range(Abs(LineSegmentOf(I,O)))", "answer_expressions": "[sqrt(3)/3,1)", "fact_spans": "[[[18, 60], [72, 77]], [[67, 71]], [[2, 9]], [[10, 17]], [[82, 85]], [[127, 130]], [[18, 60]], [[2, 66]], [[2, 66]], [[67, 78]], [[82, 111]]]", "query_spans": "[[[113, 139]]]", "process": "In $\\frac{x}{4}+\\frac{y^{2}}{3}=1$, $a=2$, $b=\\sqrt{3}$, $c=1$. Let point $P(x_{0},y_{0})$, without loss of generality, assume point $P$ is above the $x$-axis, then $y_{0}\\in(0,\\sqrt{3}]$, and $\\frac{x_{0}^{2}}{4}+\\frac{y_{0}^{2}}{3}=1$, we get $y_{0}^{2}=3-\\frac{3}{4}x_{0}^{2}$. Let the radius of circle $I$ be $r$, then $S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}\\cdot2cy_{0}=\\frac{1}{2}(2a+2c)r$, that is $cy_{0}=3r$, we obtain $r=\\frac{y_{0}}{3}$. Let the inclination angle of line $IF_{1}$ be $\\theta$, from the figure we know $\\theta$ is acute, and $\\angle PF_{1}F_{2}=2\\theta$, $\\frac{\\tan\\theta}{\\tan^{2}\\theta}=\\frac{y_{0}}{x_{0}+1}$, we get $y_{0}\\tan^{2}\\theta+2(x_{0}+1)\\tan\\theta-y_{0}=0$. Therefore, the equation of line $IF_{1}$ is $y=-2\\frac{2y_{0}}{0}1)$, from $y=\\frac{2-}{2y}\\frac{x_{0}}{v_{0}}(x+1)=\\frac{y_{0}}{3}$, we get $x=\\frac{2y_{0}^{2}}{3(2-x_{0})}-1=\\frac{}{6(}\\frac{4y_{0}^{2}}{2-x_{0}}$. So, $|IO|=\\sqrt{\\frac{x^{2}}{4}}+\\frac{y_{0}^{2}}{9}=\\sqrt{1-}\\frac{0}{3}\\frac{3(-x_{0}^{2}y_{0}}{6(2-x_{0})}-1=\\frac{2+x_{0}}{9}-1=\\frac{x_{0}}{2}$, hence point $t(\\frac{x_{0}}{2},\\frac{y_{0}}{3})$." }, { "text": "The equation of a hyperbola that shares common asymptotes with $\\frac{x^{2}}{5}-\\frac{y^{2}}{3}=1$ and has a focal length of $8$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/5 - y^2/3 = 1);Z: Hyperbola;Asymptote(G) = Asymptote(Z);FocalLength(Z) = 8", "query_expressions": "Expression(Z)", "answer_expressions": "{x^2/10-y^2/6=1, x^2/10-y^2/6=-1}", "fact_spans": "[[[1, 39]], [[1, 39]], [[54, 57]], [[0, 57]], [[47, 57]]]", "query_spans": "[[[54, 61]]]", "process": "" }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively, and asymptotes $l_{1}$, $l_{2}$. Point $P$ lies in the first quadrant and on $l_{1}$. If $l_{2} \\perp P F_{1}$, $l_{2} \\| P F_{2}$, 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;l1: Line;l2: Line;Asymptote(G) = {l1, l2};P: Point;Quadrant(P) = 1;PointOnCurve(P, l1);IsPerpendicular(l2, LineSegmentOf(P, F1));IsParallel(l2, LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[0, 56], [168, 171]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[65, 72]], [[73, 80]], [[0, 80]], [[0, 80]], [[87, 94], [116, 123]], [[96, 103]], [[0, 103]], [[104, 108]], [[104, 113]], [[104, 124]], [[126, 147]], [[148, 166]]]", "query_spans": "[[[168, 177]]]", "process": "Test analysis: From the given conditions, it is clear that PF_{1} \\bot PF_{2}, hence |PO| = \\frac{1}{2}|F_{1}F_{2}| = c. Since point P lies on the asymptote l_{1}: y = \\frac{b}{a}x, the coordinates of point P can be found as (a, b). Then, from l_{2} // PF_{2}, we obtain \\frac{b}{a - c} = -\\frac{b}{a} \\Rightarrow c = 2a \\Rightarrow e = 2." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the right focus is $F$. A line $l$ with slope $2$ is drawn through $F$. If the line $l$ intersects the right branch of the hyperbola at exactly one point, what is the range of the eccentricity of the hyperbola?", "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;F: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;Slope(l)=2;PointOnCurve(F,l);NumIntersection(l, RightPart(G)) = 1", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(5),oo)", "fact_spans": "[[[69, 74], [76, 81]], [[2, 48], [82, 85], [99, 102]], [[5, 48]], [[5, 48]], [[53, 56], [58, 61]], [[2, 48]], [[2, 56]], [[62, 74]], [[57, 74]], [[76, 97]]]", "query_spans": "[[[99, 109]]]", "process": "Since the line $ l $ with slope 2 passing through $ F $ intersects the right branch of the hyperbola at exactly one point, we have $ \\frac{b}{a} \\geqslant 2 $, so $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} \\geqslant \\sqrt{5} $. Moreover, since $ e > 1 $, it follows that $ e \\in [\\sqrt{5}, +\\infty) $." }, { "text": "Given that $P$ is an arbitrary point on the parabola $y = x^{2}$, when the distance from point $P$ to the line $x + y + 2 = 0$ is minimized, what is the distance from point $P$ to the directrix of this parabola?", "fact_expressions": "G: Parabola;H: Line;P: Point;Expression(G) = (y = x^2);Expression(H) = (x + y + 2 = 0);PointOnCurve(P, G);WhenMin(Distance(P, H))", "query_expressions": "Distance(P, Directrix(G))", "answer_expressions": "1/2", "fact_spans": "[[[6, 18], [55, 58]], [[31, 42]], [[26, 30], [2, 5], [49, 53]], [[6, 18]], [[31, 42]], [[2, 23]], [[25, 48]]]", "query_spans": "[[[49, 66]]]", "process": "First, find the coordinates of point P according to the given conditions, then use the properties of the parabola to find the distance from point P to its directrix. When the line y = -x + b is tangent to the parabola at point P, the distance to the line x + y + 2 = 0 is minimized. Substitute y = -x + b into y = x^{2} to obtain x^{2} + x - b = 0. Since the line is tangent, \\therefore A = 1 - 4b = 0, so b = -\\frac{1}{4}. Therefore, P(-\\frac{1}{2}, \\frac{1}{4}), and the distance from this point to the directrix y = -\\frac{1}{4} is \\frac{1}{2}." }, { "text": "The center of a hyperbola is at the origin, its eccentricity is $2$, and the coordinates of one focus are $(2 , 0)$. What is the equation of the asymptotes of this hyperbola?", "fact_expressions": "G: Hyperbola;O:Origin;Center(G)=O;Eccentricity(G)=2;Coordinate(OneOf(Focus(G))) = (2, 0)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(3)*x", "fact_spans": "[[[0, 3], [43, 46]], [[7, 11]], [[0, 11]], [[0, 21]], [[0, 40]]]", "query_spans": "[[[43, 55]]]", "process": "" }, { "text": "Given that the asymptotes of a hyperbola with foci on the $x$-axis pass through the intersection points of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{16}=1$ and the ellipse $\\frac{a x^{2}}{16}+\\frac{y^{2}}{4}=1 (a \\leq 1)$, then the range of the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;H: Ellipse;D:Ellipse;a: Number;a<=1;Expression(H) = (x^2/4 + y^2/16 = 1);Expression(D) = (y^2/4 + (a*x^2)/16 = 1);PointOnCurve(Focus(G),xAxis);PointOnCurve(Intersection(D,H),Asymptote(G))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(2),sqrt(21)/3)", "fact_spans": "[[[11, 14], [116, 119]], [[19, 57]], [[58, 110]], [[60, 110]], [[60, 110]], [[19, 57]], [[58, 110]], [[2, 14]], [[11, 114]]]", "query_spans": "[[[116, 130]]]", "process": "" }, { "text": "The standard equation of a parabola with the center of the ellipse $x^{2}+\\frac{y^{2}}{5}=1$ as its vertex and the right vertex of the ellipse as its focus is?", "fact_expressions": "G: Parabola;H: Ellipse;Expression(H) = (x^2 + y^2/5 = 1);Vertex(G) = Center(H);Focus(G) = RightVertex(H)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[41, 44]], [[1, 28]], [[1, 28]], [[0, 44]], [[0, 44]]]", "query_spans": "[[[41, 51]]]", "process": "The center of the ellipse is at the origin, and the right vertex is (1,0). It is the focus of the parabola, so \\frac{p}{2}=1, p=2. \\therefore the standard equation of the parabola is y^2=4x." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse. A line passing through $F_{1}$ intersects the ellipse at points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$. If the inradius of $\\triangle A B F_{2}$ is $1$, $|F_{1} F_{2}|=2$, and $|y_{1}-y_{2}|=3$, 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;F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;L: Line;x1: Number;x2: Number;y1: Number;y2: Number;PointOnCurve(F1,L) = True;Intersection(L,G) = {A,B};Coordinate(A) = (x1,y1);Coordinate(B) = (x2,y2);Radius(InscribedCircle(TriangleOf(A,B,F2))) = 1;Abs(LineSegmentOf(F1,F2)) = 2;Abs(y1-y2) = 3;A: Point;B: Point", "query_expressions": "Eccentricity(G)", "answer_expressions": "2/3", "fact_spans": "[[[2, 54], [73, 75], [94, 96], [216, 218]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[65, 72]], [[57, 64], [83, 90]], [[57, 81]], [[57, 81]], [[91, 93]], [[97, 114]], [[117, 134]], [[97, 114]], [[117, 134]], [[82, 93]], [[82, 136]], [[97, 115]], [[117, 134]], [[138, 170]], [[171, 188]], [[190, 207]], [[97, 114]], [[117, 134]]]", "query_spans": "[[[209, 216]]]", "process": "" }, { "text": "The midpoint of chord $AB$ of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ is $M(1,1)$. What is the equation of line $AB$?", "fact_expressions": "G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2/16 + y^2/4 = 1);Coordinate(M) = (1, 1);IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M", "query_expressions": "Expression(LineOf(A,B))", "answer_expressions": "x+4*y-5=0", "fact_spans": "[[[0, 38]], [[40, 45]], [[40, 45]], [[48, 56]], [[0, 38]], [[48, 56]], [[0, 45]], [[40, 56]]]", "query_spans": "[[[58, 69]]]", "process": "Let the coordinates of A and B be given, use the point difference method to find the slope of line AB, then use the point-slope form of the line equation to find the equation of line AB, and finally convert it into the general form. Let A(x_{1},y_{1}), B(x_{2},y_{2}), so \n\\begin{cases}x_{1}^{2}+4y_{1}^{2}=16\\\\x_{2}^{2}+4y_{2}^{2}=16\\end{cases}, \nthus \n-\\frac{1}{4}\\cdot\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}. \nAlso, since \n\\begin{cases}x_{1}+x_{2}=1\\times2=2\\\\y_{1}+y_{2}=1\\times2=2\\end{cases}, \nwe have \n-\\frac{1}{4}\\cdot\\frac{2}{2}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=k_{AB}, \nso \nk_{AB}=-\\frac{1}{4}, \nthus \nl_{AB}:y-1=-\\frac{1}{4}(x-1), \nthat is, \nx+4y-5=0." }, { "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$, and let a line $l$ passing through $F$ intersect the parabola at two distinct points $A$ and $B$. Let $M$ be the point of intersection of the directrix of parabola $C$ and the $x$-axis. If $\\tan \\angle AMB = 2\\sqrt{6}$, then $|AB|=$?", "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};Negation(A=B);M: Point;Intersection(Directrix(C), xAxis) = M;Tan(AngleOf(A, M, B)) = 2*sqrt(6)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "6", "fact_spans": "[[[1, 20], [39, 42], [61, 67]], [[1, 20]], [[24, 27], [29, 32]], [[1, 27]], [[33, 38]], [[28, 38]], [[49, 52]], [[53, 56]], [[33, 56]], [[44, 56]], [[57, 60]], [[57, 78]], [[80, 110]]]", "query_spans": "[[[112, 121]]]", "process": "" }, { "text": "A point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is at a distance of $6$ from one focus of the ellipse. What is the distance from point $P$ to the other focus?", "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;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 6", "query_expressions": "Distance(P,F2)", "answer_expressions": "4", "fact_spans": "[[[0, 39], [47, 49]], [[42, 45], [63, 67]], [[0, 39]], [[0, 45]], [], [], [[47, 54]], [[47, 73]], [[47, 73]], [[42, 61]]]", "query_spans": "[[[47, 78]]]", "process": "The foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ are $F_{1}, F_{2}$, with semi-major axis length $a=5$, so the major axis length is $2a=10$. Without loss of generality, let $|PF_{1}|=6$. By the definition of an ellipse, $|PF_{1}|+|PF_{2}|=2a=10$, then $|PF_{2}|=4$. Therefore, the distance from point $P$ to the other focus is 4. Hence, the answer is: 4" }, { "text": "If a point $P$ on the parabola $y^{2}=4 x$ is at a distance of $4$ from the $y$-axis, then what is the distance from point $P$ to the focus of this parabola?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Distance(P, yAxis) = 4", "query_expressions": "Distance(P, Focus(G))", "answer_expressions": "5", "fact_spans": "[[[1, 15], [41, 44]], [[18, 21], [35, 39]], [[1, 15]], [[1, 21]], [[18, 33]]]", "query_spans": "[[[35, 51]]]", "process": "Since the equation of the parabola is y^{2}=4x, the equation of the directrix is x=-1, so the distance from point P to the directrix is 4+1=5, hence the distance from point P to the focus of the parabola is 5." }, { "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1$ is $(3 , 0)$, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/5 + x^2/a^2 = 1);a: Number;H: Point;Coordinate(H) = (3, 0);RightFocus(G) = H", "query_expressions": "Eccentricity(G)", "answer_expressions": "3/2", "fact_spans": "[[[2, 44], [61, 64]], [[2, 44]], [[5, 44]], [[49, 58]], [[49, 58]], [[2, 58]]]", "query_spans": "[[[61, 70]]]", "process": "" }, { "text": "The parabola $y^{2}=8x$ intersects the curve $xy=k$ ($k>0$) at point $M$. If the distance from $M$ to the focus $F$ of the parabola is $4$, then $k=$?", "fact_expressions": "G: Parabola;H: Curve;k: Number;M: Point;F: Point;Expression(G) = (y^2 = 8*x);k>0;Expression(H) = (x*y = k);Intersection(G, H) = M;Focus(G) = F;Distance(M, F) = 4", "query_expressions": "k", "answer_expressions": "8", "fact_spans": "[[[0, 14], [41, 44]], [[15, 29]], [[58, 61]], [[31, 35], [37, 40]], [[46, 49]], [[0, 14]], [[17, 29]], [[15, 29]], [[0, 35]], [[41, 49]], [[37, 56]]]", "query_spans": "[[[58, 63]]]", "process": "Problem Analysis: \nSolving \\begin{cases}y^{2}=8x\\\\xy=k\\end{cases} gives the x-coordinate of point M as \\sqrt[3]{\\frac{k^{2}}{8}}. If the distance from M to the focus F of the parabola is 4, then the distance from M to the directrix x=-2 is also 4, that is, \\sqrt[3]{\\frac{k^{2}}{8}}+2=4. It follows easily that: k=8" }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{36}=1$ are given by?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/36 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*3*x", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 47]]]", "process": "Let $\\frac{4}{4}-\\frac{36}{36}=0$, we obtain the asymptote equations $y=\\pm3x$." }, { "text": "It is known that the axis of symmetry of a hyperbola is the coordinate axis, and one asymptote is $2x - y = 0$. Then, the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (2*x - y = 0);SymmetryAxis(G) = axis", "query_expressions": "Eccentricity(G)", "answer_expressions": "{sqrt(5), sqrt(5)/2}", "fact_spans": "[[[2, 5], [31, 34]], [[2, 29]], [[2, 13]]]", "query_spans": "[[[31, 40]]]", "process": "" }, { "text": "Given that the asymptotes of hyperbola $ C $, whose foci lie on the $ y $-axis, are $ y = \\pm 2x $, what is the eccentricity of this hyperbola?", "fact_expressions": "C: Hyperbola;PointOnCurve(Focus(C), yAxis);Expression(Asymptote(C)) = (y = pm*(2*x))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[11, 17], [38, 41]], [[2, 17]], [[11, 35]]]", "query_spans": "[[[38, 47]]]", "process": "Since the asymptotes of the hyperbola C centered at the origin with foci on the y-axis are given by $ y = \\pm\\frac{a}{b}x $, it follows that $ \\frac{a}{b} = 2 $. Therefore, $ e = \\frac{c}{a} = \\frac{\\sqrt{a^{2}+b^{2}}}{a} = \\frac{\\sqrt{5}b}{2b} = \\frac{\\sqrt{5}}{2} $." }, { "text": "A moving point is such that its distance to the point $F(0, -4)$ is greater by $1$ than its distance to the line $y - 3 = 0$. What is the equation of the trajectory of the moving point?", "fact_expressions": "G: Line;F: Point;Expression(G) = (y - 3 = 0);Coordinate(F) = (0, -4);P:Point;Distance(P,F)=Distance(P,G)+1", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2=-16*y", "fact_spans": "[[[20, 29]], [[5, 16]], [[20, 29]], [[5, 16]], [[2, 4], [38, 40]], [[2, 36]]]", "query_spans": "[[[38, 47]]]", "process": "" }, { "text": "The left and right foci of the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is any point on the ellipse $M$. The range of the maximum value of $| P F_{1}|\\cdot| P F_{2} |$ is $[2 c^{2} , 3 c^{2}]$, where $c=\\sqrt{a^{2}-b^{2}}$. What is the range of the eccentricity $e$ of the ellipse $M$?", "fact_expressions": "M: Ellipse;b: Number;a: Number;c:Number;P: Point;F1: Point;F2: Point;e: Number;a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(M)=F1;RightFocus(M)=F2;PointOnCurve(P,M);Range(Abs(LineSegmentOf(P,F1))*Abs(LineSegmentOf(P,F2)))=[2*c^2,3*c^2];c=sqrt(a^2-b^2);Eccentricity(M)=e", "query_expressions": "Range(e)", "answer_expressions": "[sqrt(3)/3,sqrt(2)/2]", "fact_spans": "[[[0, 57], [86, 91], [183, 188]], [[6, 57]], [[6, 57]], [[135, 156]], [[82, 85]], [[66, 73]], [[74, 81]], [[192, 195]], [[6, 57]], [[6, 57]], [[0, 57]], [[0, 81]], [[0, 81]], [[82, 95]], [[97, 156]], [[159, 181]], [[183, 195]]]", "query_spans": "[[[192, 202]]]", "process": "" }, { "text": "Given that the hyperbola $C$ shares common asymptotes with the curve $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ and passes through the point $A(-3,4 \\sqrt{2})$, then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;G: Curve;A: Point;Expression(G) = (x^2/9 - y^2/16 = 1);Coordinate(A) = (-3, 4*sqrt(2));Asymptote(C) = Asymptote(G);PointOnCurve(A,C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/16-x^2/9=1", "fact_spans": "[[[2, 8], [79, 82]], [[9, 47]], [[58, 77]], [[9, 47]], [[58, 77]], [[2, 54]], [[2, 77]]]", "query_spans": "[[[79, 87]]]", "process": "" }, { "text": "Let the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have eccentricity $\\sqrt{3}$, and suppose one of its directrices coincides with the directrix of the parabola $y^{2}=4x$. Then the equations of the asymptotes of this hyperbola are?", "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 = 4*x);Eccentricity(G) = sqrt(3);OneOf(Directrix(G))=Directrix(H)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(2)*x", "fact_spans": "[[[1, 57], [74, 75], [103, 106]], [[4, 57]], [[4, 57]], [[81, 95]], [[4, 57]], [[4, 57]], [[1, 57]], [[81, 95]], [[1, 72]], [[74, 100]]]", "query_spans": "[[[103, 114]]]", "process": "" }, { "text": "If the left and right foci of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are $F_{1}$, $F_{2}$, and if there exists a point $P$ on its asymptote such that $||P F_{1}|-| P F_{2}||=2 b$, then the range of the eccentricity of hyperbola $E$ is?", "fact_expressions": "E: Hyperbola;b: Number;a: Number;P: Point;F2: Point;F1: Point;a>0;b>0;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(P,Asymptote(E));Abs(Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2))) = 2*b", "query_expressions": "Range(Eccentricity(E))", "answer_expressions": "(1,sqrt(2))", "fact_spans": "[[[1, 62], [87, 88], [133, 139]], [[9, 62]], [[9, 62]], [[96, 99]], [[77, 84]], [[69, 76]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 84]], [[1, 84]], [[86, 99]], [[103, 131]]]", "query_spans": "[[[133, 150]]]", "process": "Without loss of generality, assume point P is in the first quadrant, then |PF_{1}| > |PF_{2}|. Let segment PF_{2} intersect the right branch of hyperbola E at point M. Then 2b = |PF_{1}| - |PF_{2}| = |PF_{1}| - |PM| - |MF_{2}| < |MF_{1}| - |MF_{2}| = 2a, hence 0 < \\frac{b}{a} < 1. Thus e = \\frac{c}{a} = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{a^{2}+b^{2}}{a^{2}}} = \\sqrt{1+(\\frac{b}{a})^{2}} \\in (1,\\sqrt{2})." }, { "text": "If the directrix of the parabola $y^{2}=8 x$ is tangent to the circle $x^{2}+y^{2}+6 x+m=0$, then the value of the real number $m$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);H: Circle;Expression(H) = (m + 6*x + x^2 + y^2 = 0);m: Real;IsTangent(Directrix(G), H)", "query_expressions": "m", "answer_expressions": "8", "fact_spans": "[[[1, 15]], [[1, 15]], [[19, 41]], [[19, 41]], [[45, 50]], [[1, 43]]]", "query_spans": "[[[45, 54]]]", "process": "The circle $ x^{2}+y^{2}+6x+m=0 $ has center $(-3,0)$ and radius $\\sqrt{9-m}$. The directrix of the parabola $ y^{2}=8x $ is the line $ x=-2 $, so $-2+3=\\sqrt{9-m}$, giving $ m=8 $." }, { "text": "Given point $M$ $(\\sqrt{3}, 0)$, the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and the line $y=k(x+\\sqrt{3})$ intersect at points $A$ and $B$, then the perimeter of $\\triangle A B M$ is?", "fact_expressions": "M: Point;Coordinate(M) = (sqrt(3), 0);G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);H: Line;Expression(H) = (y = k*(x + sqrt(3)));A: Point;B: Point;Intersection(G,H) = {A,B};k:Number", "query_expressions": "Perimeter(TriangleOf(A, B, M))", "answer_expressions": "8", "fact_spans": "[[[2, 22]], [[2, 22]], [[23, 50]], [[23, 50]], [[51, 70]], [[51, 70]], [[72, 76]], [[77, 80]], [[23, 80]], [[51, 70]]]", "query_spans": "[[[82, 104]]]", "process": "" }, { "text": "Given that a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ is at a distance of $3$ from one focus of the ellipse, what is the distance from $P$ to the other focus?", "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;OneOf(Focus(G)) = F2;Negation(F1 = F2);Distance(P, F1) = 3", "query_expressions": "Distance(P, F2)", "answer_expressions": "7", "fact_spans": "[[[2, 41], [49, 51]], [[45, 48], [64, 67]], [[2, 41]], [[2, 48]], [], [], [[49, 55]], [[49, 72]], [[49, 72]], [[45, 62]]]", "query_spans": "[[[49, 76]]]", "process": "Test analysis: By the definition of an ellipse, |PF| + |PF₂| = 2a, so the distance from P to the other focus is 7." }, { "text": "The distance from point $P(4,4)$ on the parabola $y^{2}=2 p x (p>0)$ to the focus $F$ is?", "fact_expressions": "G: Parabola;p: Number;P: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(P) = (4, 4);PointOnCurve(P, G);Focus(G) = F", "query_expressions": "Distance(P, F)", "answer_expressions": "5", "fact_spans": "[[[0, 23]], [[3, 23]], [[24, 33]], [[36, 39]], [[3, 23]], [[0, 23]], [[24, 33]], [[0, 33]], [[0, 39]]]", "query_spans": "[[[0, 44]]]", "process": "From the given condition, 8p = 16, we get p = 2, ∴ the equation of the parabola is y² = 4x, hence the focus coordinates are (1, 0). ∴ |PF| = √((4 - 1)² + 4²) = 5." }, { "text": "If the line $y=x+k$ and the curve $x=\\sqrt{1-y^{2}}$ have exactly one common point, then what is the range of values for $k$?", "fact_expressions": "G: Line;Expression(G) = (y = k + x);k: Number;H: Curve;Expression(H) = (x = sqrt(1 - y^2));NumIntersection(G, H)=1", "query_expressions": "Range(k)", "answer_expressions": "{(-1, 1], -sqrt(2)}", "fact_spans": "[[[1, 10]], [[1, 10]], [[40, 43]], [[11, 31]], [[11, 31]], [[1, 38]]]", "query_spans": "[[[40, 50]]]", "process": "" }, { "text": "If two points $A$ and $B$ on the parabola $y^{2}=4x$ with focus $F$ satisfy $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then what is the distance from the midpoint of chord $AB$ to the directrix?", "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);VectorOf(A, F) = 2*(VectorOf(F, B));IsChordOf(LineSegmentOf(A, B), G)", "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), Directrix(G))", "answer_expressions": "9/4", "fact_spans": "[[[9, 23]], [[27, 30]], [[33, 36]], [[2, 5]], [[9, 23]], [[1, 23]], [[9, 36]], [[9, 36]], [[38, 83]], [[9, 90]]]", "query_spans": "[[[9, 101]]]", "process": "" }, { "text": "Given that the two foci of a hyperbola are the endpoints of the major axis of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, and its directrix passes through the focus of the ellipse, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/16 + y^2/7 = 1);Focus(G) = Endpoint(MajorAxis(H));PointOnCurve(Focus(H),Directrix(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 5], [56, 57], [68, 71]], [[11, 49], [60, 62]], [[11, 49]], [[2, 55]], [[56, 65]]]", "query_spans": "[[[68, 77]]]", "process": "" }, { "text": "The standard equation of the hyperbola with asymptotes $3 x \\pm 4 y=0$ and passing through the point $M(-2,3)$ is?", "fact_expressions": "Expression(Asymptote(G)) = (3*x + pm*4*y = 0);G: Hyperbola;M: Point;Coordinate(M) = (-2, 3);PointOnCurve(M, G) = True", "query_expressions": "Expression(G)", "answer_expressions": "y^2/(27/4)-x^2/12=1", "fact_spans": "[[[0, 37]], [[34, 37]], [[23, 33]], [[23, 33]], [[22, 37]]]", "query_spans": "[[[34, 44]]]", "process": "Since the asymptotes are 3x±4y=0, assume the hyperbola equation is 9x^{2}-16y^{2}=\\lambda(\\lambda\\neq0). It passes through the point M(-2,3); substituting gives 9×4-16×9=\\lambda, so \\lambda=-108. Thus, 9x^{2}-16y^{2}=-108, which simplifies to \\frac{y^{2}}{4}-\\frac{x^{2}}{12}=1." }, { "text": "The equations of 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 point $A(x_{1}, y_{1})$ lies on the hyperbola $r_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, point $B(x_{2}, y_{2})$ lies on the hyperbola $r_{2}$: $\\frac{y^{2}}{b^{2}}-\\frac{x^{2}}{a^{2}}=1$, and satisfies $x_{1}^{2}-x_{2}^{2}=a^{2}$, the product of the slopes of lines $O A$ and $O B$ is $\\frac{1}{3}$, then the product of the eccentricities of $r_{1}$ and $r_{2}$ equals?", "fact_expressions": "r1: Hyperbola;r2:Hyperbola;O: Origin;A: Point;B: Point;x1:Number;x2:Number;y1:Number;y2:Number;a:Number;b:Number;a>0 ;b>0;Expression(r1) = (x^2/a^2-y^2/b^2=1);Expression(r2) = (y^2/b^2-x^2/a^2=1);PointOnCurve(A,r1);PointOnCurve(B,r2);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);x1^2 - x2^2 = a^2;Slope(LineOf(O, A))*Slope(LineOf(O,B)) = 1/3", "query_expressions": "Eccentricity(r2)*Eccentricity(r1)", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[21, 86], [230, 237]], [[106, 161], [238, 245]], [[196, 202], [204, 209]], [[2, 20]], [[88, 105]], [[3, 20]], [[88, 105]], [[3, 20]], [[88, 105]], [[33, 86]], [[33, 86]], [[33, 86]], [[33, 86]], [[21, 86]], [[106, 161]], [[2, 87]], [[88, 162]], [[3, 20]], [[88, 105]], [[166, 193]], [[194, 228]]]", "query_spans": "[[[230, 254]]]", "process": "Since points A, B lie on the hyperbolas r_{1}, r_{2}, substituting their coordinates into the equations and multiplying the two equations yields a^{4}y_{1}^{2}y_{2}^{2}=b^{4}x_{1}^{2}x_{2}^{2}+a^{2}(x_{1}^{2}-x_{2}^{2})-a^{4}, combined with the condition that the product of the slopes of lines OA, OB is \\frac{1}{3} and x_{1}^{2}-x_{2}^{2}=a^{2}, we can find \\frac{b}{a}, and then determine the product of the eccentricities of r_{1}, r_{2}. [Detailed solution] From the given conditions, we have \\begin{cases} b^{2}x_{1}^{2}-a^{2}y_{1}^{2}-a^{2}b^{2}=0 \\\\ b^{2}x_{2}^{2}-a^{2}y_{2}^{2}+a^{2}b^{2}=0 \\end{cases}, i.e., \\begin{cases} a^{2}y_{1}^{2}=b^{2}x_{1}^{2}-a^{2}b^{2} \\\\ a^{2}y_{2}^{2}=b^{2}x_{2}^{2}+a^{2}b^{2} \\end{cases}. Multiplying these two equations gives a^{4}y_{1}^{2}y_{2}^{2}=b^{4}[x_{1}^{2}x_{2}^{2}+a^{2}(x_{1}^{2}-x_{2}^{2})-a^{4}]. Since the product of the slopes of lines OA, OB is \\frac{1}{3}, we have \\frac{y_{1}y_{2}}{x_{1}x_{2}}=\\frac{1}{3}, i.e., y_{1}^{2}y_{2}^{2}=\\frac{1}{9}x_{1}^{2}x_{2}^{2}, and with x_{1}^{2}-x_{2}^{2}=a^{2}, substituting into the above equation yields \\frac{b^{4}}{a^{4}}=\\frac{1}{9}, so \\frac{b}{a}=\\frac{\\sqrt{3}}{3}. Thus, the product of the eccentricities of r_{1}, r_{2} is e_{1}\\cdot e_{2}=\\frac{c}{a}\\cdot\\frac{c}{b}=\\frac{a^{2}+b^{2}}{ab}=\\frac{b}{a}+\\frac{a}{b}=\\frac{4\\sqrt{3}}{3}." }, { "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ has eccentricity $\\sqrt{5}$, then what is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(5)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "2*x+pm*y=0", "fact_spans": "[[[1, 60], [77, 78]], [[4, 60]], [[4, 60]], [[4, 60]], [[4, 60]], [[1, 60]], [[1, 75]]]", "query_spans": "[[[77, 85]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, let $A$ be its upper vertex, $F$ its left focus, and $C$ the midpoint of $OA$ ($O$ being the origin). Point $P$ lies in the first quadrant such that $|C P|=\\frac{\\sqrt{3}}{2}|O C|$. Then the equation of the trajectory of point $P$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/8 + y^2/4 = 1);A: Point;UpperVertex(G) = A;F: Point;LeftFocus(G) = F;C: Point;MidPoint(LineSegmentOf(O, A)) = C;O: Origin;P: Point;Quadrant(P) = 1;Abs(LineSegmentOf(C, P)) = (sqrt(3)/2)*Abs(LineSegmentOf(O, C))", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x^2+(y-1)^2=3/4)&(x>0,y>0)", "fact_spans": "[[[2, 39]], [[2, 39]], [[44, 47]], [[2, 47]], [[52, 55]], [[2, 55]], [[56, 59]], [[56, 68]], [[70, 73]], [[80, 84], [128, 132]], [[80, 93]], [[95, 126]]]", "query_spans": "[[[128, 139]]]", "process": "From the given conditions, F(-2,0), A(0,2). Since C is the midpoint of OA, C(0,1). Because |CP| = \\frac{\\sqrt{3}}{2}|OC|, it follows that |CP| = \\frac{\\sqrt{3}}{2}, so point P lies on a circle with center C and radius \\frac{\\sqrt{3}}{2}. Also, point P is in the first quadrant, so the trajectory equation of point P is x^{2} + (y-1)^{2} = \\frac{3}{4} (x>0, y>0)." }, { "text": "Let the left and right vertices of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be $A_{1}$ and $A_{2}$, respectively. If point $P$ is on the left branch of the hyperbola, and the slopes of lines $P A_{1}$ and $P A_{2}$ are $-1$ and $-\\frac{1}{3}$, respectively, then what is the equation of the asymptotes 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;A1: Point;A2: Point;LeftVertex(G) = A1;RightVertex(G) = A2;P: Point;PointOnCurve(P, LeftPart(G));Slope(LineOf(P, A1)) = -1;Slope(LineOf(P, A2)) = -1/3", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(3)/3)*x", "fact_spans": "[[[1, 57], [88, 91], [148, 151]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[66, 73]], [[74, 81]], [[1, 81]], [[1, 81]], [[83, 87]], [[83, 97]], [[99, 146]], [[99, 146]]]", "query_spans": "[[[148, 159]]]", "process": "The equation of $ PA_{1} $ is $ y = -(x + a) $, the equation of $ PA_{2} $ is $ y = -\\frac{1}{3}(x - a) $, then $ P(-2a, a) $. Substituting the coordinates of point $ P $ into the hyperbola yields $ \\frac{4a^{2}}{a^{2}} - \\frac{a^{2}}{b^{2}} = 1 $, then $ \\frac{b^{2}}{a^{2}} = \\frac{1}{3} $, then $ \\frac{b}{a} = \\frac{\\sqrt{3}}{3} $; thus, the equations of the asymptotes of the hyperbola are $ y = \\pm \\frac{\\sqrt{3}}{3}x $." }, { "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has a real axis length of $2$ and an imaginary axis length of $4$, what is the focal distance 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);Length(RealAxis(G))=2;Length(ImageinaryAxis(G)) = 4", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 58], [77, 80]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[2, 74]]]", "query_spans": "[[[77, 85]]]", "process": "By the given condition, 2a=2, 2b=4, we get a=1, b=2. \\therefore c=\\sqrt{a^{2}+b^{2}}=\\sqrt{5}. \\therefore the focal distance of the hyperbola is 2\\sqrt{5}" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right focus is point $F$, point $B$ is an endpoint of the imaginary axis, and point $P$ is a moving point on the left branch of the hyperbola $C$. If the minimum perimeter of $\\triangle B P F$ equals 4 times the length of the real axis, then the equations of the asymptotes of hyperbola $C$ are?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;B: Point;P: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;OneOf(Endpoint(ImageinaryAxis(C))) = B;PointOnCurve(P, LeftPart(C));Min(Perimeter(TriangleOf(B, P, F))) = 4*Length(RealAxis(C))", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y = pm*2*x", "fact_spans": "[[[2, 63], [91, 97], [141, 147]], [[10, 63]], [[10, 63]], [[73, 77]], [[86, 90]], [[68, 72]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 72]], [[2, 85]], [[86, 104]], [[91, 139]]]", "query_spans": "[[[141, 155]]]", "process": "Transform the distance from P to the right focus into the distance from P to the left focus. Based on the minimum value, establish an equation in a, b, c, then find \\frac{b}{a}, and obtain the asymptotic equations. Let F_{1} be the left focus of the hyperbola, as shown in the figure, then |PF|=|PF_{1}|+2a, |BF|+|BP|+|PF|=|BF|+|BP|+|PF_{1}|+2a. Clearly, |BP|+|PF_{1}|\\geqslant|BF_{1}|=\\sqrt{b^{2}+c^{2}}, with equality if and only if points B, P, F_{1} are collinear. \\therefore the minimum value of \\Delta BPF is 2\\sqrt{b^{2}+c^{2}}+2a, \\therefore 2\\sqrt{b^{2}+c^{2}}+2a=4\\times2a, 9a^{2}=b^{2}+c^{2}=b^{2}+b^{2}+a^{2}b^{2}=4a^{2}, \\frac{b}{a}=2, \\therefore the asymptotic equations are y=\\pm2x" }, { "text": "If a line with slope $2$ passes through the focus of the parabola $x^{2}=4 y$ and intersects the parabola at points $A$ and $B$, then $|A B|$=?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (x^2 = 4*y);Slope(H)=2;PointOnCurve(Focus(G),H);Intersection(H,G)={A,B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "20", "fact_spans": "[[[12, 26], [32, 35]], [[8, 10]], [[38, 41]], [[42, 45]], [[12, 26]], [[1, 10]], [[8, 29]], [[8, 47]]]", "query_spans": "[[[49, 58]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). For the parabola x^{2}=4y, the length of the focal chord |AB| = p + (y_{1}+y_{2}). Since the focus of the parabola is at (0,1) and k_{AB}=2, the equation of line AB is y=2x+1, or equivalently x=\\frac{y-1}{2}. Substituting x=\\frac{y-1}{2} into the parabola equation yields y^{2}-18y+1=0, so y_{1}+y_{2}=18. Therefore, |AB|=2+18=20" }, { "text": "Given that $M$ is a point on the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and the distance from the midpoint $P$ of segment $MF$ to the $y$-axis is $2$, then $|PF|=$?", "fact_expressions": "M: Point;PointOnCurve(M, G) = True;G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;MidPoint(LineSegmentOf(M,F)) = P;Distance(P, yAxis) = 2", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "2", "fact_spans": "[[[2, 5]], [[2, 24]], [[6, 20], [29, 32]], [[6, 20]], [[25, 28]], [[25, 35]], [[45, 48]], [[36, 48]], [[45, 60]]]", "query_spans": "[[[62, 72]]]", "process": "" }, { "text": "The focus of the parabola $y^{2}=4 x$ is $F$, and point $P(x, y)$ is a moving point on this parabola. Given a fixed point $A(2,2)$, what is the minimum value of $|P A|+|P F|$?", "fact_expressions": "G: Parabola;P: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (x1, y1);Coordinate(A) = (2, 2);Focus(G) = F;x1:Number;y1:Number;PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "3", "fact_spans": "[[[0, 14], [34, 37]], [[22, 32]], [[47, 55]], [[18, 21]], [[0, 14]], [[22, 32]], [[47, 55]], [[0, 21]], [[23, 32]], [[23, 32]], [[22, 41]]]", "query_spans": "[[[57, 76]]]", "process": "The focus of the parabola y^{2}=4x is F(1,0), and the directrix is l:x=-1, as shown in the figure below: the perpendicular from point P to l has foot B. By the definition of a parabola, |PB|=|PF|, so |PA|+|PF|=|PA|+|PB|. The minimum value of |PA|+|PB| is attained if and only if points A, P, B are collinear, that is, when AB\\botl, and this minimum value is 2+1=3." }, { "text": "Given circle $O_{1}$: $x^{2}+y^{2}=1$ and circle $O_{2}$: $(x-4)^{2}+y^{2}=4$, from point $P(x , y)$ draw tangents $P A$ and $P B$ to $O_{1}$ and $O_{2}$ respectively, where $A$ and $B$ are the points of tangency, and $|P A|=|P B|$. Then the equation of the locus of the moving point $P$ is?", "fact_expressions": "O1: Circle;Expression(O1) = (x^2 + y^2 = 1);O2: Circle;Expression(O2) = (y^2 + (x - 4)^2 = 4);P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;TangentOfPoint(P, O1) = LineOf(P, A);TangentOfPoint(P, O2) = LineOf(P, B);TangentPoint(LineOf(P, A), O1) = A;TangentPoint(LineOf(P, B), O2) = B;A: Point;B: Point;Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, B))", "query_expressions": "LocusEquation(P)", "answer_expressions": "x=13/8", "fact_spans": "[[[2, 26], [71, 78]], [[2, 26]], [[27, 55], [79, 86]], [[27, 55]], [[57, 68], [132, 135]], [[57, 68]], [[58, 68]], [[58, 68]], [[56, 100]], [[56, 100]], [[71, 113]], [[71, 113]], [[103, 106]], [[107, 110]], [[115, 128]]]", "query_spans": "[[[132, 142]]]", "process": "Let P(x, y), then from |PA| = |PB|, we get |PA|^2 = |PB|^2, so x^2 + y^2 - 1 = (x - 4)^2 + y^2 - 4, simplifying yields: x = \\frac{13}{8}, which is the equation of the locus of P." }, { "text": "If a point $P(m, n)$ on the parabola $y^{2}=8x$ is at a distance of $8m$ from its focus, then $m=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;m: Number;n: Number;Coordinate(P) = (m, n);PointOnCurve(P, G);Distance(P, Focus(G)) = 8*m", "query_expressions": "m", "answer_expressions": "2/7", "fact_spans": "[[[1, 15], [28, 29]], [[1, 15]], [[18, 27]], [[42, 45]], [[18, 27]], [[18, 27]], [[1, 27]], [[18, 40]]]", "query_spans": "[[[42, 47]]]", "process": "According to the problem, the directrix of the parabola is given by the equation x = -2. Since the distance from point P(m, n) to the focus is 8m, it follows that |PF| = m + 2 = 8m. Solving this equation yields m = \\frac{2}{7}." }, { "text": "The line $y = x - 1$ intersects the ellipse $\\frac{x^{2}}{4} + \\frac{y^{2}}{2} = 1$ at points $A$ and $B$. Then $|AB| =$?", "fact_expressions": "H: Line;Expression(H) = (y = x - 1);G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "4*sqrt(5)/3", "fact_spans": "[[[0, 9]], [[0, 9]], [[10, 47]], [[10, 47]], [[50, 53]], [[54, 57]], [[0, 59]]]", "query_spans": "[[[61, 70]]]", "process": "Substitute y = x - 1 into the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ and simplify to obtain $3x^{2}-4x-2=0$, therefore $x_{1}+x_{2}=\\frac{4}{3}$, $x_{1}x_{2}=-\\frac{2}{3}$. By the chord length formula, $|AB|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\sqrt{1+k^{2}}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\frac{4\\sqrt{5}}{3}$." }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $e \\in (1 , 2)$, then the range of the slope of one of its asymptotes is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;e: Number;Eccentricity(G) = e;In(e, (1, 2))", "query_expressions": "Range(Slope(OneOf(Asymptote(G))))", "answer_expressions": "(0, sqrt(3))", "fact_spans": "[[[0, 46]], [[0, 46]], [[3, 46]], [[3, 46]], [[50, 65]], [[0, 65]], [[50, 65]]]", "query_spans": "[[[0, 83]]]", "process": "" }, { "text": "Let the right vertex $ A $ of the hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0 $, $ b>0 $) be the center of a circle with radius $ \\sqrt{2} a $. The circle intersects the right branch of the hyperbola at two points $ P $ and $ Q $. If $ \\angle P A Q = \\frac{\\pi}{2} $, then the eccentricity of the hyperbola $ C $ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;A: Point;Q: Point;H:Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;Center(H)=A;Radius(H)=sqrt(2)*a;Intersection(H,RightPart(C))={P,Q};AngleOf(P, A, Q) = pi/2", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[1, 62], [92, 95], [140, 146]], [[9, 62]], [[73, 85]], [[99, 102]], [[66, 69]], [[103, 106]], [[89, 90]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 69]], [[66, 90]], [[73, 90]], [[89, 108]], [[110, 138]]]", "query_spans": "[[[140, 152]]]", "process": "From the given conditions, it is clear that: |PA| = \\sqrt{2}a, \\angle PAO = 45^{\\circ}, thus P(2a, a). Substituting into the hyperbola equation, we get: 4 - \\frac{a^{2}}{b^{2}} = 1, so a^{2} = 3b^{2}, 4a^{2} = 3c^{2}, hence e = \\frac{2\\sqrt{3}}{3}." }, { "text": "Given the parabola $C$: $y^{2}=6x$ with focus $F$, a line $l$ passing through point $F$ intersects $C$ at points $A$ and $B$, and the directrix of $C$ intersects the $x$-axis at point $M$. If $|AF|=2|MF|$, then $|AB|=$?", "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;M: Point;B: Point;Expression(C) = (y^2 = 6*x);Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Intersection(Directrix(C), xAxis) = M;Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(M, F))", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[35, 40]], [[2, 21], [41, 44], [57, 60]], [[46, 49]], [[25, 28], [30, 34]], [[69, 73]], [[50, 53]], [[2, 21]], [[2, 28]], [[29, 40]], [[35, 55]], [[57, 73]], [[76, 90]]]", "query_spans": "[[[92, 101]]]", "process": "From the parabola equation, find the coordinates of the focus, the equation of the directrix, |AF| = 2|MF| = 6. Using the focal radius formula and the parabola equation, find point A, then determine the equation of line AB. Solve simultaneously with the parabola equation to find the coordinates of point B, compute |BF|, and thus solve the problem. The parabola C: y^{2} = 6x has focus F, and its directrix intersects the x-axis at point M; |MF| = 6. Let point A(x_{1}, y_{1}), B(x_{2}, y_{2}). Without loss of generality, assume A is in the first quadrant due to symmetry, i.e., y_{1} > 0. According to the definition of the parabola, |AF| = x_{1} + \\frac{3}{2} = 6, \\therefore x_{1} = \\frac{9}{2}, y_{1} = \\sqrt{\\frac{9}{2} \\times 6} = 3\\sqrt{3}. A(\\frac{9}{2}, 3\\sqrt{3}). The equation of line AB is y = \\sqrt{3}(x - \\frac{3}{2}). Solving simultaneously with the parabola equation: \n\\begin{cases} y = \\sqrt{3}(x - \\frac{3}{2}) \\\\ y^{2} = 6x \\end{cases}, \neliminating x and simplifying gives \\sqrt{3}y^{2} - 6y - 9\\sqrt{3} = 0. Solving yields y_{1} = 3\\sqrt{3}, y_{2} = -\\sqrt{3}, x_{2} = \\frac{1}{2}, \\therefore |BF| = \\frac{1}{2} + \\frac{3}{2} = 2. \\therefore |AB| = |AF| + |BF| = 8. The answer is 8. This question examines the relationship between a parabola and a line, and the application of the parabola definition to find focal chords, classified as a medium-difficulty problem." }, { "text": "The coordinates of the foci of the ellipse $2 x^{2}+3 y^{2}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (2*x^2 + 3*y^2 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*(sqrt(6)/6),0)", "fact_spans": "[[[0, 21]], [[0, 21]]]", "query_spans": "[[[0, 28]]]", "process": "Test analysis: $2x^{2}+3y^{2}=1 \\therefore \\frac{x^{2}}{1}+\\frac{y^{2}}{1}=1 \\therefore a^{2}=\\frac{1}{2}, b^{2}=\\frac{1}{3} \\therefore c^{2}=\\frac{1}{6} \\therefore c=\\frac{\\sqrt{6}}{6}$, the coordinates of the foci are $(\\pm\\frac{\\sqrt{6}}{6}, 0)$" }, { "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, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "Test analysis: Dividing the coefficient of the linear term by 4 gives the x-coordinate or y-coordinate of the focus, so the focus is (2,0)." }, { "text": "If the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{2}=1$ and the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{a^{2}}=1$ have the same foci, then the value of $a$ is?", "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;Expression(G) = (-y^2/2 + x^2/a = 1);Expression(H) = (x^2/4 + y^2/a^2 = 1);Focus(G)=Focus(H)", "query_expressions": "a", "answer_expressions": "1", "fact_spans": "[[[1, 39]], [[89, 92]], [[40, 81]], [[1, 39]], [[40, 81]], [[1, 87]]]", "query_spans": "[[[89, 96]]]", "process": "The foci of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{a>0}=1$ lie on the $x$-axis. According to the given conditions, we have\n$$\n\\begin{cases}\na\\\\\n00 , b>0)$, let $P$ be a moving point on the $x$-axis. The line $y=2 x+m$ $(m\\neq 0)$ passing through point $P$ intersects the hyperbola $C$ at exactly one point. Find the eccentricity of the hyperbola $C$.", "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, xAxis) = True;G: Line;Expression(G) = (y = m + 2*x);m: Number;Negation(m = 0);NumIntersection(G, C) = 1;PointOnCurve(P, G) = True", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[2, 66], [110, 116], [126, 132]], [[2, 66]], [[10, 67]], [[10, 67]], [[10, 67]], [[10, 67]], [[69, 72], [84, 88]], [[69, 81]], [[89, 109]], [[89, 109]], [[91, 109]], [[91, 109]], [[89, 124]], [[82, 109]]]", "query_spans": "[[[126, 138]]]", "process": "" }, { "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4 x$. A line passing through the point $P(1 , 0)$ intersects the parabola at points $A$ and $B$, and $2 \\overrightarrow{B P}=\\overrightarrow{P A}$. Then $|A F|+2|B F|$=?", "fact_expressions": "C: Parabola;G: Line;P: Point;B: Point;A: Point;F: Point;Expression(C) = (y^2 = 4*x);Coordinate(P) = (1, 0);Focus(C) = F;PointOnCurve(P, G);Intersection(G, C) = {A, B};2*VectorOf(B, P) = VectorOf(P, A)", "query_expressions": "Abs(LineSegmentOf(A, F)) + 2*Abs(LineSegmentOf(B, F))", "answer_expressions": "6", "fact_spans": "[[[5, 24], [45, 48]], [[42, 44]], [[30, 41]], [[54, 57]], [[50, 53]], [[1, 4]], [[5, 24]], [[30, 41]], [[1, 27]], [[28, 44]], [[42, 59]], [[61, 106]]]", "query_spans": "[[[108, 124]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), since P(1,0), we have \\overrightarrow{BP}=(1-x_{2},-y_{2}), \\overrightarrow{PA}=(x_{1}-1,y_{1}). Since 2\\overrightarrow{BP}=\\overrightarrow{PA}, it follows that 2(1-x_{2},-y_{2})=(x_{1}-1,y_{1}), which gives x_{1}+2x_{2}=3, -2y_{2}=y_{1}. Substituting A(x_{1},y_{1}), B(x_{2},y_{2}) into the parabola y^{2}=4x, we obtain y_{1}^{2}=4x_{1}, y_{2}^{2}=4x_{2}. Since -2y_{2}=y_{1}, we have 4y_{2}^{2}=y_{1}^{2}, which yields 16x_{2}=4x_{1}, or 4x_{2}=x_{1}. Together with x_{1}+2x_{2}=3, solving simultaneously gives x_{1}=2, x_{2}=\\frac{1}{2}. By the definition of the parabola, we have |AF|+2|BF|=x_{1}+1+2(x_{2}+1)=2+1+2\\times(\\frac{1}{2}+1)=6." }, { "text": "The length of the real axis of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)", "query_expressions": "Length(RealAxis(G))", "answer_expressions": "4", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 34]]]", "process": "According to the characteristics of the standard equation of a hyperbola, the solution can be obtained. [Detailed Solution] From the problem, we have a^{2}=4\\Rightarrow a=2\\Rightarrow 2a=4." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$, with foci $F_{1}$ and $F_{2}$. A line perpendicular to the $x$-axis is drawn through $F_{2}$, intersecting the hyperbola at points $A$ and $B$. The inradius of triangle $\\triangle A B F_{1}$ is $a$. Find the eccentricity of this hyperbola.", "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;B: Point;F1: Point;F2: Point;a>0;b>0;l:Line;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1,F2};PointOnCurve(F2,l);IsPerpendicular(l,xAxis);Intersection(l,G)={A,B};Radius(InscribedCircle(TriangleOf(A,B,F1)))=a", "query_expressions": "Eccentricity(G)", "answer_expressions": "(1+sqrt(5))/2", "fact_spans": "[[[2, 59], [100, 103], [149, 152]], [[5, 59]], [[143, 146]], [[104, 107]], [[108, 111]], [[64, 71]], [[73, 81], [84, 91]], [[5, 59]], [[5, 59]], [], [[2, 59]], [[2, 81]], [[83, 99]], [[83, 99]], [[83, 113]], [[115, 146]]]", "query_spans": "[[[149, 158]]]", "process": "" }, { "text": "A moving circle passes through the fixed point $ A(2,0) $ and is internally tangent to the fixed circle $ B $: $ x^{2}+4x+y^{2}-32=0 $. What is the trajectory equation of the center $ M $ of the moving circle?", "fact_expressions": "C: Circle;A: Point;Coordinate(A) = (2, 0);PointOnCurve(A, C);B: Circle;Expression(B) = (y^2 + x^2 + 4*x - 32 = 0);IsInTangent(C, B);M: Point;Center(C) = M", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2/9+y^2/5=1", "fact_spans": "[[[1, 3], [50, 52]], [[6, 14]], [[6, 14]], [[1, 14]], [[19, 46]], [[19, 46]], [[1, 48]], [[54, 57]], [[50, 57]]]", "query_spans": "[[[54, 64]]]", "process": "The equation of circle B in standard form is (x+2)^{2}+y^{2}=36, with center B(-2,0) and radius R=6. Let the coordinates of the moving circle's center M be (x,y), and its radius be r. According to the problem, |BM|=R-r, and r=|MA|, so |BM|=R-|MA|, that is, |BM|+|AM|=6. Therefore, by the definition of an ellipse, the locus of M is an ellipse with foci at B(-2,0) and A(2,0), and the midpoint O(0,0) of segment AB as the center. Let the equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). Then a=3, c=2, b=\\sqrt{a^{2}-c^{2}}=\\sqrt{5}. Thus, the required trajectory equation of the center M is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1." }, { "text": "Given that $F$ is the left focus of the ellipse $E$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, $P$ is a moving point on the ellipse $E$, and $A(1, \\sqrt{3})$ is a fixed point, then the maximum value of $|P A|+|P F|$ is?", "fact_expressions": "F: Point;LeftFocus(E) = F;E: Ellipse;Expression(E) = (x^2/16 + y^2/12 = 1);P: Point;PointOnCurve(P, E) = True;A: Point;Coordinate(A) = (1, sqrt(3))", "query_expressions": "Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "10", "fact_spans": "[[[2, 5]], [[2, 54]], [[6, 50], [59, 64]], [[6, 50]], [[55, 58]], [[55, 68]], [[69, 85]], [[69, 85]]]", "query_spans": "[[[92, 111]]]", "process": "First, according to the definition of the ellipse, we obtain |PF| = 8 - |PF₂|. Then, transform |PA| + |PF| into |PA| - |PF₂| + 8, and find the maximum value of |PA| - |PF₂| based on the graph. Let F₂ be the right focus of the ellipse, as shown in the figure: since |PF| + |PF₂| = 2a = 8, it follows that |PF| = 8 - |PF₂|, so |PA| + |PF| = |PA| - |PF₂| + 8. When point P moves to position P₁, points A, F₂, and P₁ are collinear. At this moment, |PA| - |PF₂| reaches its maximum value, which is |AF₂| = √((1 - 2)² + √3) = 2. Therefore, |PA| + |PF| = |PA| - |PF₂| + 8 ≤ 2 + 8 = 10. The maximum value of |PA| + |PF| is 10." }, { "text": "Given that the standard equation of an ellipse is $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, what is the focal distance of this ellipse?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1)", "query_expressions": "FocalLength(G)", "answer_expressions": "2", "fact_spans": "[[[2, 4], [48, 50]], [[2, 45]]]", "query_spans": "[[[48, 55]]]", "process": "" }, { "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ always pass through the fixed point $A(1,\\ 2)$. What is the minimum value of the distance from the center of the ellipse to its directrix?", "fact_expressions": "C: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);A: Point;Coordinate(A)=(1, 2);PointOnCurve(A, C)", "query_expressions": "Min(Distance(Center(C),Directrix(C)))", "answer_expressions": "", "fact_spans": "[[[1, 58], [74, 76]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[1, 58]], [[62, 72]], [[62, 72]], [[1, 72]]]", "query_spans": "[[[74, 90]]]", "process": "" }, { "text": "Given that the ellipse $E$ is centered at the origin, has coordinate axes as its axes of symmetry, one focus at $F(0,1)$, and eccentricity $e = \\frac{1}{2}$, then the equation of this ellipse is?", "fact_expressions": "E: Ellipse;F: Point;O:Origin;Center(E)=O;SymmetryAxis(E)=axis;Coordinate(F) = (0, 1);OneOf(Focus(E))=F;e:Number;e=1/2;Eccentricity(E)=e", "query_expressions": "Expression(E)", "answer_expressions": "y^2/4+x^2/3=1", "fact_spans": "[[[2, 7], [61, 63]], [[29, 37]], [[11, 13]], [[2, 13]], [[2, 22]], [[29, 37]], [[2, 37]], [[43, 58]], [[43, 58]], [[2, 58]]]", "query_spans": "[[[61, 68]]]", "process": "\\because one focus of the ellipse E is F(0,1) \\therefore c=1. Let the equation of ellipse E be: \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1. \\because e=\\frac{1}{2} \\therefore \\frac{c}{a}=\\frac{1}{2}, we get a=2. According to b=\\sqrt{a^{2}-c^{2}}=\\sqrt{2^{2}-1^{2}}=\\sqrt{3}. \\therefore the equation of ellipse E is: \\frac{y^{2}}{4}+\\frac{x^{2}}{3}=1" }, { "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{m}=1$ has foci at $(-4,0)$ and $(4,0)$. What is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;F1: Point;F2: Point;Expression(G) = (x^2/25 + y^2/m = 1);Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(G) = {F1,F2}", "query_expressions": "m", "answer_expressions": "9", "fact_spans": "[[[0, 38]], [[62, 65]], [[44, 52]], [[53, 60]], [[0, 38]], [[44, 52]], [[53, 60]], [[0, 60]]]", "query_spans": "[[[62, 69]]]", "process": "Since the coordinates of the foci are (-4,0) and (4,0), the foci lie on the x-axis, m<25, c=4, so 25-m=4^{2}, solving gives m=9. This problem examines properties of ellipses, mainly the relationship among a, b, c of an ellipse and the relationship between the foci of an ellipse and its standard equation, testing computational ability; it is a simple problem." }, { "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has left and right foci $F_{1}$, $F_{2}$. The vertices $A$, $B$ of $\\Delta A B F_{1}$ lie on the ellipse, and side $AB$ passes through the right focus $F_{2}$. Then the perimeter of $\\triangle A B F_{1}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(F2, LineSegmentOf(A, B))", "query_expressions": "Perimeter(TriangleOf(A, B, F1))", "answer_expressions": "20", "fact_spans": "[[[0, 38], [91, 93]], [[0, 38]], [[45, 52]], [[53, 60], [106, 113]], [[0, 60]], [[0, 60]], [[82, 85]], [[87, 90]], [[82, 94]], [[87, 94]], [[97, 113]]]", "query_spans": "[[[115, 141]]]", "process": "" }, { "text": "A line passing through the focus $F$ of the parabola $y=\\frac{1}{4} x^{2}$ with an inclination angle of $30^{\\circ}$ intersects the parabola at points $A$ and $B$. Then $|A B|=$?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;F: Point;Expression(G) = (y = x^2/4);Focus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(30, degree);Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16/3", "fact_spans": "[[[1, 25], [54, 57]], [[51, 53]], [[58, 61]], [[62, 65]], [[28, 31]], [[1, 25]], [[1, 31]], [[0, 53]], [[34, 53]], [[51, 67]]]", "query_spans": "[[[69, 78]]]", "process": "Let the coordinates of point A be (x_{1}, y_{1}) and point B be (x_{2}, y_{2}). The focus of the parabola y = \\frac{1}{4}x^{2} is (0, 1). According to the problem, the equation of the line is x = \\sqrt{3}y - \\sqrt{3}. Solving the parabola equation and the line equation simultaneously gives: 3(y - 1)^{2} = 4y, that is, 3y^{2} - 10y + 3 = 0 \\textcircled{1}. The y-coordinates of intersection points A and B are the two solutions of equation \\textcircled{1}. By Vieta's formulas, y_{1} + y_{2} = \\frac{10}{3}. From the properties of the parabola, the distance from point A to the focus is y_{1} + 1, and the distance from point B to the focus is y_{2} + 1. Therefore, |+1 + y_{1} + 1 = y + y_{1} + 2 = \\frac{16}{3}. Hence, the correct answer is \\frac{16}{3}." }, { "text": "If the length of the major axis of an ellipse is 3 times the length of the minor axis, and the focal distance is $8$, then what is the standard equation of this ellipse?", "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 3 * Length(MinorAxis(G));FocalLength(G) = 8", "query_expressions": "Expression(G)", "answer_expressions": "{(x^2/18+y^2/2=1),(y^2/18+x^2=1)}", "fact_spans": "[[[3, 5], [29, 31]], [[3, 18]], [[3, 25]]]", "query_spans": "[[[29, 38]]]", "process": "If the foci of the ellipse lie on the x-axis, the equation of the ellipse can be written as \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), and 2c=8, that is, c=4. Also, 2a=6b, \\therefore a=3b. Using a^{2}=b^{2}+c^{2}, we get 9b^{2}=b^{2}+16, \\therefore b^{2}=2^{n}, then a^{2}=9b^{2}=18. \\therefore The standard equation of the ellipse is \\frac{x^{2}}{18}+\\frac{y^{2}}{2}=1. If the foci of the ellipse lie on the y-axis, similarly we obtain \\frac{y^{2}}{18}+\\frac{x^{2}}{2}=1" }, { "text": "The foci are at $(-5,0)$ and $(5,0)$, and the point $B(0,12)$ lies on the ellipse. What is the standard equation of this ellipse?", "fact_expressions": "G: Ellipse;H: Point;A: Point;B: Point;Coordinate(H) = (-5, 0);Coordinate(A) = (5, 0);Coordinate(B) = (0, 12);Focus(G)={H,A};PointOnCurve(B, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/169+y^2/144=1", "fact_spans": "[[[34, 36], [34, 36]], [[5, 13]], [[14, 21]], [[23, 33]], [[5, 13]], [[14, 21]], [[23, 33]], [[0, 36]], [[23, 37]]]", "query_spans": "[[[42, 50]]]", "process": "According to the problem, c=5, b=12, so a^{2}=b^{2}+c^{2}=144+25=169, solving gives a=13. Therefore, the standard equation of the ellipse is \\frac{x^{2}}{169}+\\frac{y^{2}}{144}=1" }, { "text": "Given $A(0,7)$, $B(0,-7)$, $C(12,2)$, using $C$ as one focus, an ellipse passing through $A$ and $B$ is drawn. Then the equation of the trajectory of the other focus $F$ of the ellipse is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;Coordinate(A) = (0, 7);Coordinate(B) = (0, -7);Coordinate(C) = (12, 2);Focus(G)={C,F};PointOnCurve(A,G);PointOnCurve(B,G);F:Point", "query_expressions": "LocusEquation(F)", "answer_expressions": "(y^2-x^2/48=1)&(y<0)", "fact_spans": "[[[54, 56], [58, 60]], [[2, 10], [46, 49]], [[13, 22], [50, 53]], [[25, 34], [36, 39]], [[2, 10]], [[13, 22]], [[25, 34]], [[35, 69]], [[45, 56]], [[45, 56]], [[66, 69]]]", "query_spans": "[[[66, 76]]]", "process": "According to the definition of an ellipse, |AF| - |BF| = 2, so the trajectory equation of focus F is the lower branch of a hyperbola. From the definition of the ellipse: |AF| + |AC| = |BF| + |BC|, that is, |AF| + 13 = |BF| + 15, hence |AF| - |BF| = 2. Therefore, the trajectory equation of focus F is the lower branch of a hyperbola, with a = 1, c = 7, so b = 4\\sqrt{3}. Thus, the trajectory equation is: y^2 - \\frac{x^2}{48} = 1 (y < 0)" }, { "text": "A line passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) intersects the parabola $C$ at points $A$ and $B$. If $|A F|=6$, $|B F|=3$, then the value of $p$ is?", "fact_expressions": "C: Parabola;p: Number;G: Line;A: Point;F: Point;B: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, F)) = 6;Abs(LineSegmentOf(B, F)) = 3", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[1, 27], [37, 43]], [[78, 81]], [[34, 36]], [[44, 47]], [[30, 33]], [[48, 51]], [[9, 27]], [[1, 27]], [[1, 33]], [[0, 36]], [[34, 53]], [[56, 66]], [[67, 76]]]", "query_spans": "[[[78, 85]]]", "process": "Let the directrix $ l $ of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) intersect the $ x $-axis at point $ G $, and intersect line $ AB $ at $ C $. Draw a perpendicular from $ A $ to $ l $, with foot $ E $, and draw $ BD \\perp l $ at $ D $. By the property of similar triangles, we have $ \\frac{BD}{AE} = \\frac{BF}{AF} = \\frac{1}{2} \\Rightarrow B $ is the midpoint of $ AC $, so $ BC = 9 $, $ \\frac{FG}{AE} = \\frac{CF}{AC} \\Rightarrow \\frac{FG}{6} = \\frac{12}{18} \\Rightarrow FG = 4 $, $ \\therefore P = 4 $." }, { "text": "Given that the eccentricity of the ellipse $y^{2}+2 m x^{2}=1$ ($m>0$) is $\\frac{1}{2}$, then $m$=?", "fact_expressions": "G: Ellipse;m: Number;m>0;Expression(G) = (2*(m*x^2) + y^2 = 1);Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "{2/3,3/8}", "fact_spans": "[[[2, 28]], [[48, 51]], [[4, 28]], [[2, 28]], [[2, 46]]]", "query_spans": "[[[48, 53]]]", "process": "Transform the ellipse equation $ y^{2} + 2mx^{2} = 1 $ ($ m > 0 $) into standard form: $ y^{2} + \\frac{x^{2}}{2m} = 1 $. If $ \\frac{1}{2m} > 1 $, that is, $ 0 < m < \\frac{1}{2} $, then the eccentricity of the ellipse is $ e = \\frac{c}{a} = \\frac{\\sqrt{1 - \\frac{1}{2m}}}{1} = \\frac{1}{2} $, solving gives: $ m = \\frac{3}{8} $; if $ \\frac{1}{2m} < 1 $, that is, $ m > \\frac{1}{2} $, then the eccentricity of the ellipse is $ e = \\frac{c}{a} = \\frac{\\sqrt{1 - \\frac{1}{2m}}}{1} $, solving gives: $ m = \\frac{2}{3} $." }, { "text": "Given the ellipse $\\frac{y^{2}}{a^{2}}+x^{2}=1$ $(a>1)$ has eccentricity $e=\\frac{2 \\sqrt{5}}{5}$, and $P$ is a moving point on the ellipse, then the maximum distance from $P$ to the fixed point $B(-1,0)$ is?", "fact_expressions": "G: Ellipse;a: Number;B: Point;P: Point;e:Number;a>1;Expression(G) = (x^2 + y^2/a^2 = 1);Coordinate(B) = (-1, 0);Eccentricity(G)=e;e = (2*sqrt(5))/5;PointOnCurve(P, G)", "query_expressions": "Min(Distance(P, B))", "answer_expressions": "5/2", "fact_spans": "[[[2, 38], [72, 74]], [[4, 38]], [[88, 97]], [[68, 71], [82, 85]], [[42, 66]], [[4, 38]], [[2, 38]], [[88, 97]], [[2, 66]], [[42, 66]], [[68, 80]]]", "query_spans": "[[[82, 107]]]", "process": "The eccentricity of the ellipse $\\frac{y^{2}}{a^{2}}+x^{2}=1$ $(a>1)$ is $e=\\frac{2\\sqrt{5}}{5}$, thus $\\frac{\\sqrt{a^{2}-1}}{a}=\\frac{2\\sqrt{5}}{5}$, solving gives $a=\\sqrt{5}$. The equation of the ellipse is $\\frac{y^{2}}{5}+x^{2}=1$. Let point $P(x,y)$, then $-1\\leqslant x\\leqslant 1$, $y^{2}=5-5x^{2}$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, a line $l$ passing through the point $P(0,6)$ intersects the ellipse $C$ at points $A$ and $B$. If $A$ is the midpoint of segment $PB$, then what are the coordinates of point $A$?", "fact_expressions": "C: Ellipse;l: Line;P: Point;B: Point;A: Point;Expression(C) = (x^2/16 + y^2/12 = 1);Coordinate(P) = (0, 6);PointOnCurve(P, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(P,B)) = A", "query_expressions": "Coordinate(A)", "answer_expressions": "{(2,3),(-2,3)}", "fact_spans": "[[[2, 45], [64, 69]], [[58, 63]], [[48, 57]], [[75, 78]], [[71, 74], [82, 85], [98, 102]], [[2, 45]], [[48, 57]], [[47, 63]], [[58, 80]], [[82, 96]]]", "query_spans": "[[[98, 107]]]", "process": "① When the slope of line l does not exist, the equation of l is: y=0 ⇒ A(0,2\\sqrt{3}), B(0,-2\\sqrt{3}), which obviously does not satisfy the condition, so discard. ② When the slope of line l exists, let l: y=kx+6, substitute into the ellipse equation to get: (3+4k^{2})x^{2}+48kx+96=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=-\\frac{48k}{3+4k^{2}}, x_{1}x_{2}=\\frac{96}{3+4k^{2}} ⇒ A(2,3) or A(-2,3)" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$. A line passing through the point $(-1,0)$ intersects $C$ at points $A$ and $B$. If the minimum value of $4|F A|+|F B|$ is $19$, then what is the standard equation of the parabola $C$?", "fact_expressions": "C: Parabola;p: Number;G: Line;F: Point;A: Point;B: Point;H:Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(H) = (-1, 0);Focus(C) = F;PointOnCurve(H, G);Intersection(G, C) = {A, B};Min(4*Abs(LineSegmentOf(F, A)) + Abs(LineSegmentOf(F, B))) = 19", "query_expressions": "Expression(C)", "answer_expressions": "y^2=12*x", "fact_spans": "[[[2, 28], [50, 53], [91, 97]], [[10, 28]], [[47, 49]], [[32, 35]], [[55, 58]], [[59, 62]], [[37, 46]], [[10, 28]], [[2, 28]], [[37, 46]], [[2, 35]], [[36, 49]], [[47, 64]], [[66, 89]]]", "query_spans": "[[[91, 104]]]", "process": "Let AB: y = k(x + 1), A(x₁, y₁), B(x₂, y₂). Solving the system of equations: \n\\begin{cases} y = k(x + 1) \\\\ y^{2} = 2px \\end{cases} \n⇒ x₂x₁ = 14|FA| + |FB| = (4x₁ + x₂) + \\frac{5}{2}p \\geqslant 4 + \\frac{5}{2}p = 19 ⇒ p = 6, y^{2} = 12x." }, { "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have left and right foci $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through point $F_{1}$ intersects the left and right branches of the hyperbola at points $A$ and $B$, respectively. If the circle with diameter $AB$ passes through point $F_{2}$ and $|A F_{2}|=|B F_{2}|$, then what is the eccentricity of the hyperbola?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;G: Circle;A: Point;B: 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(F1, l);Intersection(l, LeftPart(C)) =A;Intersection(l,RightPart(C))=B;IsDiameter(LineSegmentOf(A,B),G);PointOnCurve(F2,G);Abs(LineSegmentOf(A,F2))=Abs(LineSegmentOf(B,F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[97, 102]], [[1, 62], [105, 108], [170, 173]], [[9, 62]], [[9, 62]], [[135, 136]], [[115, 119]], [[120, 123]], [[79, 86], [137, 145]], [[71, 78], [88, 96]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 86]], [[1, 86]], [[87, 102]], [[97, 123]], [[97, 123]], [[125, 136]], [[135, 145]], [[146, 167]]]", "query_spans": "[[[170, 179]]]", "process": "First, according to the given conditions, determine that $\\triangle ABF_{2}$ is an isosceles right triangle, find $|BF_{2}|=2\\sqrt{2}a$, in $\\triangle F_{1}BF_{2}$, use the cosine law to establish a homogeneous equation in terms of $a$, $c$, and find the eccentricity $e$. Since the circle with diameter $AB$ passes through point $F_{2}$, then $AF_{2}\\perp BF_{2}$, and $|AF_{2}|=|BF_{2}|$, so $\\triangle ABF_{2}$ is an isosceles right triangle, hence $\\angle ABF_{2}=45^{\\circ}$. Let $|AF_{2}|=|BF_{2}|=t$, then $|AB|=\\sqrt{2}t$. From the definition of the hyperbola, we have $|AF_{2}|-|AF_{1}|=2a$, $|BF_{1}|-|BF_{2}|=2a$. Adding these two equations gives: $|BF_{1}|-|AF_{1}|=4a$, that is, $|AB|=4a$. So $\\sqrt{2}t=4a$, solving yields: $t=2\\sqrt{2}a$. In $\\triangle F_{1}BF_{2}$, $|BF_{1}|=2a+|BF_{2}|=(2+2\\sqrt{2})a$, $|F_{1}F_{2}|=2c$, $|BF_{2}|=2\\sqrt{2}a$, $\\angle F_{1}BF_{2}=45^{\\circ}$. By the cosine law: $|F_{1}F_{2}|^{2}=|BF_{1}|^{2}+|BF_{2}|^{2}-2|BF_{1}|\\times|BF_{2}|\\cos45^{\\circ}$, that is, $4c^{2}=((2+2\\sqrt{2})a)^{2}+(2\\sqrt{2}a)^{2}-2((2+2\\sqrt{2})a)\\times(2\\sqrt{2}a)\\times\\frac{\\sqrt{2}}{2}$, simplifying yields $e=\\frac{c}{a}=\\sqrt{3}$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with left and right vertices $A$ and $B$, respectively. Point $P$ lies on the ellipse $C$ (not coinciding with $A$ or $B$). The line $AP$ intersects the $y$-axis at point $M$. Through point $A$, draw $AN \\parallel BP$, and let line $AN$ intersect the $y$-axis at point $N$. If point $S(t, 0)$ satisfies $\\overrightarrow{S M} \\cdot \\overrightarrow{S N}=5$, then $t=?$", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);A: Point;B: Point;LeftVertex(C) = A;RightVertex(C) = B;P: Point;PointOnCurve(P, C);Negation(P=A);Negation(P=B);M: Point;Intersection(LineOf(A, P), yAxis) = M;N: Point;PointOnCurve(A, LineOf(A, N));IsParallel(LineOf(A, N), LineOf(B, P));Intersection(LineOf(A, N), yAxis) = N;S: Point;t: Number;Coordinate(S) = (t, 0);DotProduct(VectorOf(S, M), VectorOf(S, N)) = 5", "query_expressions": "t", "answer_expressions": "pm*sqrt(6)", "fact_spans": "[[[2, 34], [57, 62]], [[2, 34]], [[44, 47], [66, 69], [98, 102]], [[48, 51], [70, 73]], [[2, 51]], [[2, 51]], [[52, 56]], [[52, 63]], [[52, 77]], [[52, 77]], [[92, 96]], [[78, 96]], [[130, 134]], [[97, 115]], [[103, 115]], [[116, 134]], [[136, 146]], [[201, 204]], [[136, 146]], [[148, 199]]]", "query_spans": "[[[201, 206]]]", "process": "Let P(x₀,y₀), then \\frac{x₀2}{2}+y₀^{2}=1; from the problem, A(-\\sqrt{2},0), B(\\sqrt{2},0), the equation of line AP is \\frac{y-0}{y₀-0}=\\frac{x+\\sqrt{2}}{x₀+\\sqrt{2}}, setting x=0 gives y=\\frac{\\sqrt{2}y₀}{x₀+\\sqrt{2}}, so M(0,\\frac{\\sqrt{2}y₀}{x₀+\\sqrt{2}}). Let N(0,n), since AN//BP, we have \\frac{n}{\\sqrt{2}}=\\frac{y₀}{x₀-\\sqrt{2}}\\frac{\\sqrt{2}y₀}{x₀-\\sqrt{2}}, thus N(0,\\frac{\\sqrt{2}y₀}{x₀-\\sqrt{2}}). \\overrightarrow{SM}=(-t,\\frac{\\sqrt{2}y₀}{x₀+\\sqrt{2}}), \\overrightarrow{SN}=(-t,\\frac{\\sqrt{2}y₀}{x₀-\\sqrt{2}}). Since \\overrightarrow{SM}\\cdot\\overrightarrow{SN}=5, it follows that t^{2}+\\frac{2y₀^{2}}{x₀2-2}=5. From \\frac{x₀2}{2}+y₀2=1 we get 2y₀^{2}=2-x₀^{2}, substituting yields t^{2}=6, so t=\\pm\\sqrt{6}" }, { "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and a point $P$ on the ellipse satisfies $\\angle F_{1} PF_{2}=90^{\\circ}$, then the area of $\\triangle 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);AngleOf(F1,P,F2)=ApplyUnit(90,degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "1", "fact_spans": "[[[20, 47], [55, 57]], [[2, 9]], [[58, 62]], [[12, 19]], [[20, 47]], [[2, 52]], [[55, 62]], [[64, 96]]]", "query_spans": "[[[98, 127]]]", "process": "" }, { "text": "Let the ellipse $ C $: $ \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1 $, $ F $ be the right focus, $ l $ be a line passing through point $ F $ (not parallel to the $ y $-axis), intersecting the ellipse at points $ A $ and $ B $, $ l^{\\prime} $ be the perpendicular bisector of $ A B $, intersecting the major axis of the ellipse at a point $ D $. Then the value of $ \\frac{D F}{A B} $ is?", "fact_expressions": "C: Ellipse;l: Line;l1:Line;A: Point;B: Point;D: Point;F: Point;Expression(C) = (x^2/25 + y^2/9 = 1);RightFocus(C) = F;PointOnCurve(F,l);Negation(IsParallel(l,yAxis));Intersection(l,C)={A,B};PerpendicularBisector(LineSegmentOf(A, B)) = l1;Intersection(MajorAxis(C),l1)=D", "query_expressions": "LineSegmentOf(D, F)/LineSegmentOf(A, B)", "answer_expressions": "2/5", "fact_spans": "[[[1, 43], [79, 81], [117, 119]], [[53, 56]], [[93, 105]], [[82, 85]], [[87, 90]], [[125, 128]], [[45, 48], [58, 62]], [[1, 43]], [[1, 52]], [[53, 67]], [[53, 77]], [[53, 92]], [[93, 115]], [[93, 128]]]", "query_spans": "[[[130, 151]]]", "process": "" }, { "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has its right focus at $F$, points $M$ and $N$ lie on one asymptote of the hyperbola $E$, $|F M|=|F N|=\\sqrt{3} a$ and $\\overrightarrow{F M} \\cdot \\overrightarrow{F N}=a^{2}$, then the eccentricity of the hyperbola $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;RightFocus(E) = F;F: Point;M: Point;N: Point;J: Line;OneOf(Asymptote(E)) = J;PointOnCurve(M, J) = True;PointOnCurve(N, J) = True;Abs(LineSegmentOf(F, M)) = Abs(LineSegmentOf(F, N));Abs(LineSegmentOf(F, N)) = sqrt(3)*a;DotProduct(VectorOf(F, M), VectorOf(F, N)) = a^2", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 63], [81, 87], [177, 183]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 71]], [[68, 71]], [[72, 76]], [[77, 80]], [], [[81, 93]], [[72, 94]], [[72, 94]], [[95, 119]], [[95, 119]], [[120, 175]]]", "query_spans": "[[[177, 189]]]", "process": "According to the problem, draw a schematic diagram and use the given conditions to find the distance |PF| from the focus F to the asymptote. Then, using the point-to-line distance formula, calculate the length of |PF|. Setting these two expressions equal gives the corresponding eccentricity. Since \\overrightarrow{FM}\\cdot\\overrightarrow{FN}=a^{2}, |FM|=|FN|=\\sqrt{3}a, it follows that \\cos\\angleMFN=\\frac{a^{2}}{(\\sqrt{3}a)^{2}}=\\frac{1}{3}. Because \\angleMFP=\\angleNFP, we have 2\\cos^{2}\\angleMFP-1=\\frac{1}{3}, so \\cos\\angleMFP=\\frac{\\sqrt{6}}{3}. Thus, \\frac{|PF|}{|MF|}=\\frac{\\sqrt{6}}{3}, so |PF|=\\sqrt{2}a. Take the asymptote bx-ay=0. By the point-to-line distance formula: |PF|=\\frac{|bc|}{\\sqrt{b^{2}+a^{2}}}=b. Hence, \\sqrt{2}a=b, so 2a^{2}=b^{2}=c^{2}-a^{2}, thus c^{2}=3a^{2}, so e^{2}=3, e=\\sqrt{3}" }, { "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. Let $P$ be a point on the hyperbola $C$, and $Q$ be a point on an asymptote of the hyperbola $C$, with both $P$ and $Q$ located in the first quadrant. If $2 \\overrightarrow{Q P}=\\overrightarrow{P F_{2}}$ and $\\overrightarrow{Q F_{1}} \\cdot \\overrightarrow{Q F_{2}}=0$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;Q: Point;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);PointOnCurve(Q,Asymptote(C));Quadrant(P)=1;Quadrant(Q)=1;2*VectorOf(Q, P) = VectorOf(P, F2);DotProduct(VectorOf(Q, F1), VectorOf(Q, F2)) = 0", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(13)-2", "fact_spans": "[[[2, 63], [91, 97], [105, 111], [245, 251]], [[10, 63]], [[10, 63]], [[101, 104], [122, 125]], [[87, 90], [118, 121]], [[79, 86]], [[71, 78]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]], [[2, 86]], [[87, 100]], [[101, 117]], [[118, 132]], [[118, 132]], [[134, 183]], [[184, 243]]]", "query_spans": "[[[245, 257]]]", "process": "The left and right foci of the hyperbola given by the equation $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are $F_{1}$ and $F_{2}$ respectively. Let $P$ be a point on the hyperbola $C$, and $Q$ a point on an asymptote of the hyperbola $C$, with both $P$ and $Q$ located in the first quadrant. Given that $2\\overrightarrow{QP}=\\overrightarrow{PF}_{2}$, $\\overrightarrow{QF}_{1}\\cdot\\overrightarrow{QF}_{2}=0$, it follows that $P$ is a trisection point of $QF_{2}$, and $\\overrightarrow{QF}_{1}\\bot\\overrightarrow{QF}_{2}$. The point $Q$ lies on the line $bx-ay=0$, and $|OQ|=c$, so $Q(a,b)$, $F_{2}(c,0)$. Let $P(x_{1},y_{1})$, then $2(x_{1}-a,y_{1}-b)=(c-x_{1},-y_{1})$. Solving gives $x_{1}=\\frac{2a+c}{3}, y_{1}=\\frac{2b}{3}$, so $P\\left(\\frac{2a+c}{3},\\frac{2b}{3}\\right)$. Substituting into the hyperbola equation yields $\\frac{(2a+c)^{2}}{9a^{2}}-\\frac{4}{9}=1$, solving gives $e=\\frac{c}{a}=\\sqrt{13}-2$." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$, respectively, and point $P$ is any point on the ellipse, then the range of values of $\\frac{|PF_{1}-PF_{2}|}{PF_{1}}$ is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/16 + y^2/12 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)", "query_expressions": "Range(Abs(LineSegmentOf(P, F1) - LineSegmentOf(P, F2))/LineSegmentOf(P, F1))", "answer_expressions": "[0,2]", "fact_spans": "[[[20, 59], [71, 73]], [[66, 70]], [[2, 9]], [[10, 17]], [[20, 59]], [[2, 65]], [[2, 65]], [[66, 79]]]", "query_spans": "[[[81, 121]]]", "process": "\\frac{|PF_{1}-PF_{2}|}{PF_{1}}=\\frac{|PF_{1}-(8-PF_{1})|}{PF_{1}}=\\left|\\frac{PF_{1}-(8-PF_{1})}{PF_{1}}\\right|=\\left|2-\\frac{8}{PF_{1}}\\right| because 2\\leqslant PF_{1}\\leqslant6 and the function y=2-\\frac{8}{x} is monotonically increasing on x\\in[2,6], so -2\\leqslant2-\\frac{8}{PF_{1}}\\leqslant\\frac{2}{3}, hence \\left|2-\\frac{8}{PF_{1}}\\right|\\in[0,2]" }, { "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 is given by $y=x$. The point $P(\\sqrt{3}, y_{0})$ lies on the hyperbola. Find $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}} = ?$", "fact_expressions": "G: Hyperbola;b: Number;P: Point;F1: Point;F2: Point;y0:Number;b>0;Expression(G) = (x^2/2 - y^2/b^2 = 1);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], [113, 116]], [[5, 49]], [[89, 111]], [[58, 65]], [[66, 73]], [[90, 111]], [[5, 49]], [[2, 49]], [[89, 111]], [[2, 73]], [[2, 73]], [[74, 88]], [[89, 117]]]", "query_spans": "[[[119, 178]]]", "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}$, respectively. Point $P$ lies on the ellipse such that $\\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}}=0$ and $\\tan \\angle P F_{1} F_{2}=2$. Then the eccentricity of the ellipse is equal to?", "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};P: Point;PointOnCurve(P, G);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], [177, 179]], [[1, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[61, 68]], [[69, 76]], [[1, 76]], [[77, 81]], [[77, 85]], [[87, 144]], [[146, 175]]]", "query_spans": "[[[177, 186]]]", "process": "" }, { "text": "$P$ is a point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$, $F_{2}$ are its foci, the eccentricity of the hyperbola is $\\frac{5}{4}$, and $\\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}}=0$. If the area of $\\triangle F_{1}PF_{2}$ is $9$, then $a+b=$?", "fact_expressions": "G: Hyperbola;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};Eccentricity(G) = 5/4;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Area(TriangleOf(F1, P, F2)) = 9", "query_expressions": "a + b", "answer_expressions": "7", "fact_spans": "[[[4, 57], [81, 84], [77, 78]], [[4, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[0, 3]], [[0, 60]], [[61, 68]], [[69, 76]], [[61, 80]], [[81, 102]], [[104, 161]], [[163, 194]]]", "query_spans": "[[[196, 203]]]", "process": "" }, { "text": "If point $P$ lies on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{12}=1$, and its $x$-coordinate is the same as the $x$-coordinate of the right focus of the hyperbola, then what is the distance between point $P$ and the left focus of the hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/12 = 1);P: Point;PointOnCurve(P, G);XCoordinate(P) = XCoordinate(RightFocus(G))", "query_expressions": "Distance(P, LeftFocus(G))", "answer_expressions": "11", "fact_spans": "[[[6, 46], [54, 57], [74, 77]], [[6, 46]], [[1, 5], [48, 49], [69, 73]], [[1, 47]], [[48, 67]]]", "query_spans": "[[[69, 86]]]", "process": "Let the left and right foci of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{12}=1$ be $F_{1}$ and $F_{2}$, respectively. Since point $P$ lies on the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{12}=1$, and its $x$-coordinate is the same as that of the right focus of the hyperbola, the $x$-coordinate of point $P$ is: $x_{P}=\\sqrt{16+12}=2\\sqrt{7}$. Therefore, $\\frac{28}{16}-\\frac{y_{P}^{2}}{12}=1$, solving yields $y_{P}=\\pm3$. Thus, $|PF_{2}|=3$. By the definition of a hyperbola, we have: $\\left||PF_{1}|-|PF_{2}|\\right|=2a=8$, so $|PF_{1}|=11$ or $|PF_{1}|=-5$ (discarded)." }, { "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)$, and let $P$ be any point on the right branch of the hyperbola. When the minimum value of $\\frac{|PF_{1}|^{2}}{|PF_{2}|}$ is $8a$, the range of the eccentricity $e$ of the 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));Min(Abs(LineSegmentOf(P, F1))^2/Abs(LineSegmentOf(P, F2))) = 8*a;e: Number;Eccentricity(G) = e", "query_expressions": "Range(e)", "answer_expressions": "(1, 3]", "fact_spans": "[[[17, 73], [83, 86], [139, 142]], [[20, 73]], [[20, 73]], [[79, 82]], [[1, 8]], [[9, 16]], [[20, 73]], [[20, 73]], [[17, 73]], [[1, 78]], [[1, 78]], [[79, 92]], [[94, 136]], [[145, 148]], [[139, 148]]]", "query_spans": "[[[145, 155]]]", "process": "By definition: |PF₂| − |PF₁| = 2a, ∴ |PF₂| = 2a + |PF₁|, ∴ $\\frac{|PF_{2}|^{2}}{|PF_{1}|} = \\frac{(2a+|PF_{1}|)^{2}}{|PF_{1}|} = \\frac{4a^{2}}{|PF_{1}|} + 4a + |PF_{1}| \\geqslant 8a$. The equality holds if and only if $\\frac{4a^{2}}{|PF_{1}|} = |PF_{1}|$, i.e., when |PF₁| = 2a. Then, using the focal radius formula, the range of eccentricity e of this hyperbola can be derived. By definition: |PF₂| − |PF₁| = 2a, ∴ |PF₂| = 2a + |PF₁|, $\\frac{|PF_{2}|^{2}}{|PF_{1}|} = \\frac{(2a+|PF_{1}|)^{2}}{|PF_{1}|} = \\frac{4a^{2}}{|PF_{1}|} + 4a + |PF_{1}| > 8a$, with equality if and only if $\\frac{4a^{2}}{|PF_{1}|} = |PF_{1}|$, i.e., |PF₁| = 2a. Let P(x₀, y₀), (x₀ ≤ −a). By the focal radius formula: |PF₁| = −e x₀ − a = 2a, ∴ x₀ = −3a, ∴ e = −$\\frac{3a}{x_{0}}$ ≤ 3. Also, since e > 1, it follows that e ∈ (1, 3]. Answer: (1, 3]." }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, the intersection point of the directrix and the $x$-axis is $M$, and $N$ is a point on the parabola satisfying $|MN|=2|NF|$, then $\\angle NMF=$?", "fact_expressions": "G: Parabola;M: Point;N: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;Intersection(Directrix(G),xAxis)=M;PointOnCurve(N,G);Abs(LineSegmentOf(M,N))=2*Abs(LineSegmentOf(N,F))", "query_expressions": "AngleOf(N,M,F)", "answer_expressions": "pi/3", "fact_spans": "[[[2, 16], [42, 45]], [[34, 37]], [[38, 41]], [[20, 23]], [[2, 16]], [[2, 23]], [[2, 37]], [[38, 49]], [[53, 69]]]", "query_spans": "[[[71, 85]]]", "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}$ and $F_{2}$ respectively. A perpendicular line is drawn from $F_{2}$ to an asymptote. If this perpendicular line is tangent to the circle centered at $F_{1}$ with radius $|O F_{1}|$, then the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;O: Origin;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(Asymptote(C),L);Center(G)=F1;Radius(G)=Abs(LineSegmentOf(O,F1));IsTangent(L,G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[2, 63], [146, 149]], [[10, 63]], [[10, 63]], [[140, 141]], [[125, 136]], [[72, 79], [114, 121]], [[80, 87], [89, 96]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [], [[2, 105]], [[2, 105]], [[113, 141]], [[125, 141]], [[2, 141]]]", "query_spans": "[[[146, 155]]]", "process": "Let the line $ l: y = k(x - c) $ be the perpendicular drawn from $ F_2 $ to an asymptote. Given that this perpendicular is tangent to the circle centered at $ F_1 $ with radius $ |OF_1| $, using the point-to-line distance formula we obtain $ \\frac{a^2}{b^2} $. Since $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^2}{a^2}} $, the answer can be found. \n$\\because$ 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, we have: $ F_1(-c, 0), F_2(c, 0) $. Draw a perpendicular from $ F_2 $ to an asymptote; without loss of generality, assume it is perpendicular to $ y = \\frac{b}{a}x $. Let the perpendicular from $ F_2 $ to the asymptote be $ l: y = k(x - c) $. According to the problem, drawing the graph shows that $ k $ exists. Since two lines are perpendicular, we have: $ \\frac{b}{a} \\cdot k = -1 $, hence $ k = -\\frac{a}{b} $. \nAlso, since the circle has center $ F_1 $ and radius $ |OF_1| $, its equation is $ (x + c)^2 + y^2 = c^2 $. Using the point-to-line distance formula, we get: $ \\frac{|-\\frac{a}{\\sqrt{1+(\\frac{a}{b})^2}}|}{\\sqrt{1+k^2}} = c $. Simplifying yields: $ \\frac{a^2}{b^2} = \\frac{1}{3} $, i.e., $ \\frac{b^2}{a^2} = 3 $. The eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^2}{a^2}} = \\sqrt{1 + 3} = 2 $." }, { "text": "If the line passing through the points $A(a, 0)$ and $B(0, a)$ does not intersect the parabola $y = x^{2} - 2x - 3$, then what is the range of real values of $a$?", "fact_expressions": "a: Real;A: Point;Coordinate(A) = (a, 0);B: Point;Coordinate(B) = (0, a);H: Line;PointOnCurve(A, H);PointOnCurve(B, H);G: Parabola;Expression(G) = (y = x^2 - 2*x - 3);NumIntersection(H, G) = 0", "query_expressions": "Range(a)", "answer_expressions": "(-oo, -13/4)", "fact_spans": "[[[54, 59]], [[5, 14]], [[5, 14]], [[15, 25]], [[15, 25]], [[26, 28]], [[2, 28]], [[2, 28]], [[29, 47]], [[29, 47]], [[26, 51]]]", "query_spans": "[[[54, 66]]]", "process": "" }, { "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$ $(a>\\sqrt{3})$ have eccentricity $\\frac{1}{2}$. Then the distance from one of the intersection points of the line $y=6x$ and $C$ to the $y$-axis is?", "fact_expressions": "C: Ellipse;a: Number;G: Line;a>sqrt(2);Expression(C) = (y^2/3 + x^2/a^2 = 1);Expression(G) = (y = 6*x);Eccentricity(C) = 1/2", "query_expressions": "Distance(OneOf(Intersection(G,C)),yAxis)", "answer_expressions": "2/7", "fact_spans": "[[[1, 59], [89, 92]], [[8, 59]], [[79, 88]], [[8, 59]], [[1, 59]], [[79, 88]], [[1, 77]]]", "query_spans": "[[[79, 109]]]", "process": "\\because e=\\frac{c}{a}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\sqrt{1-\\frac{3}{a^{2}}}=\\frac{1}{2},\\therefore a=2, from \\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1\\\\y=6x\\end{cases}, we get |x|=\\frac{2}{7}.\\therefore the distance from one of the intersection points of the line y=6x and C to the y-axis is \\frac{2}{5}" }, { "text": "If the distance from point $A$ on the parabola $x^{2}=4 y$ to the focus is $10$, then what is the distance from $A$ to the $x$-axis?", "fact_expressions": "G: Parabola;A: Point;Expression(G) = (x^2 = 4*y);PointOnCurve(A, G);Distance(A, Focus(G)) = 10", "query_expressions": "Distance(A, xAxis)", "answer_expressions": "9", "fact_spans": "[[[1, 15]], [[17, 21], [34, 37]], [[1, 15]], [[1, 21]], [[1, 32]]]", "query_spans": "[[[34, 47]]]", "process": "According to the parabola equation, the focus coordinates are (0,1) and the directrix equation is y = -1. Since the distance from point A on the parabola x^{2} = 4y to the focus is 10, the distance from point A to the x-axis is 10 - 1 = 9." }, { "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{P Q}=2 \\overrightarrow{F_{1} O}$, and $\\overrightarrow{F_{1} Q}=\\lambda\\left(\\frac{\\overrightarrow{F_{1} P}}{|\\overrightarrow{F_{1} P}|}+\\frac{\\overrightarrow{F_{1} O}}{| \\overrightarrow{F_{1} O}|}\\right)$ $(\\lambda>0)$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;Q: Point;F1: Point;O: Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F1;PointOnCurve(P,G);PointOnCurve(Q, RightDirectrix(G));VectorOf(P, Q) = 2*VectorOf(F1, O);lambda:Number;VectorOf(F1, Q) = lambda*(VectorOf(F1, O)/Abs(VectorOf(F1, O)) + VectorOf(F1, P)/Abs(VectorOf(F1, P)));lambda>0", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)-1)/2", "fact_spans": "[[[2, 54], [80, 82], [89, 91], [318, 320]], [[4, 54]], [[4, 54]], [[75, 79]], [[84, 88]], [[58, 65]], [[66, 69]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 65]], [[75, 83]], [[84, 96]], [[98, 147]], [[149, 316]], [[148, 316]], [[149, 316]]]", "query_spans": "[[[318, 326]]]", "process": "" }, { "text": "Given that $P$ is a moving point on the parabola $x^{2}=4 y$, the projection of point $P$ onto the $x$-axis is $M$, and the coordinates of point $A$ are $(2,0)$, then the minimum value of $|P A|+|P M|$ is?", "fact_expressions": "G: Parabola;P: Point;A: Point;M: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (2, 0);PointOnCurve(P,G);Projection(P, xAxis) = M", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "sqrt(5)-1", "fact_spans": "[[[6, 20]], [[2, 5], [25, 29]], [[43, 47]], [[39, 42]], [[6, 20]], [[43, 58]], [[2, 24]], [[25, 42]]]", "query_spans": "[[[60, 79]]]", "process": "Problem Analysis: By the definition of a parabola, PA + |PM| = |PF| - 1 + |PA| \\geqslant |AF| - 1 = \\sqrt{5} - 1." }, { "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{5}}{3}$, and the sum of the distances from a point $P$ on the ellipse to the two foci is $12$, then what is the length of the minor axis of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(5)/3;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P, F1) + Distance(P, F2) = 12", "query_expressions": "Length(MinorAxis(G))", "answer_expressions": "8", "fact_spans": "[[[2, 54], [80, 82], [103, 105]], [[4, 54]], [[4, 54]], [[85, 88]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 79]], [[80, 88]], [], [], [[80, 92]], [[80, 101]]]", "query_spans": "[[[103, 111]]]", "process": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is: $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{3}$. The sum of the distances from a point $P$ on the ellipse to the two foci is 12, that is: $2a=12$. It follows that: $a=6$, $c=2\\sqrt{5}$, $\\therefore b=\\sqrt{a^{2}-c^{2}}=\\sqrt{36-20}=4$. Then the length of the minor axis of the ellipse is: $2b=8$." }, { "text": "A point $P$ on the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{24}=1$ is at a distance of $11$ from one focus. Then, what is its distance to the other focus?", "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (x^2/25 - y^2/24 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 11", "query_expressions": "Distance(P,F2)", "answer_expressions": "21", "fact_spans": "[[[0, 40]], [[42, 46], [61, 62]], [[0, 40]], [[0, 46]], [], [], [[0, 51]], [[0, 68]], [[0, 68]], [[0, 59]]]", "query_spans": "[[[0, 73]]]", "process": "Let the left and right foci of the hyperbola be $ F_{1} $ and $ F_{2} $, respectively. Without loss of generality, assume $ PF_{1} = 11 $. According to the definition of the hyperbola, $ |PF_{1} - PF_{2}| = 2a = 10 $, $ \\therefore PF_{2} = 1 $ or $ PF_{2} = 21 $. Since $ F_{1}F_{2} = 14 $, when $ PF_{2} = 1 $, $ 1 + 11 < 14 $ (discarded), $ \\therefore PF_{2} = 21 $." }, { "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ pass through the point $P(1, \\frac{4}{3})$, what is the eccentricity of this 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);Coordinate(P) = (1, 4/3);PointOnCurve(P,Asymptote(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/3", "fact_spans": "[[[2, 58], [86, 89]], [[5, 58]], [[5, 58]], [[63, 83]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 83]], [[2, 83]]]", "query_spans": "[[[86, 95]]]", "process": "" }, { "text": "Given the parabola $E$: $y^{2}=2 p x(p>0)$ has focus $F$, $O$ is the origin, point $A$ lies on $E$, and $|A F|=2|O F|$. If $|O A|=\\sqrt{10}$, then $p=$?", "fact_expressions": "E: Parabola;p: Number;A: Point;F: Point;O: Origin;p>0;Expression(E) = (y^2 = 2*(p*x));Focus(E) = F;PointOnCurve(A, E);Abs(LineSegmentOf(A, F)) = 2*Abs(LineSegmentOf(O, F));Abs(LineSegmentOf(O, A)) = sqrt(10)", "query_expressions": "p", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[2, 28], [50, 53]], [[91, 94]], [[45, 49]], [[32, 35]], [[36, 39]], [[9, 28]], [[2, 28]], [[2, 35]], [[45, 54]], [[56, 70]], [[72, 89]]]", "query_spans": "[[[91, 96]]]", "process": "Let A(x₀, y₀), from the problem we know F(p/2, 0), |OF| = p/2. Since |AF| = 2|OF|, then |AF| = 2|OF| = p. Because point A lies on E, |AF| = x₀ + p/2 = p, solving gives x₀ = p/2. Thus A(p/2, ±p). Therefore |OA| = √10 = √(p²/4 + p²) = (√5/2)p, solving gives p = 2√2." }, { "text": "In the coordinate plane, a moving point $P$ and points $M(-2,0)$, $N(2,0)$ satisfy $|\\overrightarrow{M N}| \\cdot|\\overrightarrow{M P}|+\\overrightarrow{M N} \\cdot \\overrightarrow{N P}=0$. Then the trajectory equation of the moving point $P(x, y)$ is?", "fact_expressions": "P: Point;M: Point;Coordinate(M) = (-2, 0);N: Point;Coordinate(N) = (2, 0);DotProduct(VectorOf(M, N), VectorOf(N, P)) + DotProduct(Abs(VectorOf(M, N)), Abs(VectorOf(M, P))) = 0;Coordinate(P) = (x1, y1);x1: Number;y1: Number", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2=-8*x", "fact_spans": "[[[10, 13], [144, 153]], [[14, 24]], [[14, 24]], [[27, 35]], [[27, 35]], [[37, 139]], [[144, 153]], [[144, 153]], [[144, 153]]]", "query_spans": "[[[144, 160]]]", "process": "|\\overrightarrow{MN}|\\cdot|\\overrightarrow{MP}|+\\overrightarrow{MN}\\cdot\\overrightarrow{NP}=0, then 4\\sqrt{(x+2)^{2}+y^{2}}+(4,0)\\cdot(x-2,y)=0, that is \\sqrt{(x+2)^{2}+y^{2}}=2-x, simplifying yields y^{2}=-8x" }, { "text": "The locus of points in the plane that are equidistant from a fixed point $F$ and a fixed line $l$ (where $F$ lies on $l$) is?", "fact_expressions": "l: Line;F:Point;G:Point;PointOnCurve(F,l);Distance(F,G)=Distance(G,l)", "query_expressions": "Locus(G)", "answer_expressions": "The line passing through point F and perpendicular to l", "fact_spans": "[[[16, 19], [16, 19]], [[20, 23], [20, 23]], [[35, 36]], [[20, 28]], [[7, 36]]]", "query_spans": "[[[35, 41]]]", "process": "Since F lies on l, the locus of points in the plane that are equidistant from a fixed point F and a fixed line l (with F on l) is the line passing through F and perpendicular to l." }, { "text": "Let points $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ $(a>0)$, respectively. The line $l$ passing through point $F_{1}$ and perpendicular to the $x$-axis intersects the hyperbola $C$ at points $A$ and $B$. If the area of triangle $\\triangle A B F_{2}$ is $2 \\sqrt{6}$, then what is the equation of the asymptotes of this hyperbola?", "fact_expressions": "l: Line;C: Hyperbola;a: Number;A: Point;B: Point;F2: Point;F1: Point;a>0;Expression(C) = (-y^2/2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, l);IsPerpendicular(xAxis,l);Intersection(l, C) = {A, B};Area(TriangleOf(A, B, F2)) = 2*sqrt(6)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(sqrt(2)/2)*x", "fact_spans": "[[[97, 102]], [[20, 72], [103, 109], [162, 165]], [[28, 72]], [[111, 114]], [[115, 118]], [[10, 17]], [[1, 9], [80, 88]], [[28, 72]], [[20, 72]], [[1, 78]], [[1, 78]], [[79, 102]], [[89, 102]], [[97, 120]], [[122, 159]]]", "query_spans": "[[[162, 173]]]", "process": "First find the coordinates of points A and B, then the chord length |AB| can be obtained. S_{\\triangle ABF_{2}} = \\frac{1}{2} \\times |AB| \\times |F_{1}F_{2}| = 2\\sqrt{6}. Together with c^{2} = a^{2} + b^{2}, b^{2} = 2, the value of a can be found. The asymptotes are given by y = \\pm \\frac{b}{a}x. Let F_{1}(-c,0), b = \\sqrt{2}. Substitute x = -c into \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{2} = 1 to get: y^{2} = \\frac{4}{a^{2}}, so y = \\pm \\frac{2}{a}. Thus A(-c, \\frac{2}{a}), B(-c, -\\frac{2}{a}), |AB| = \\frac{4}{a}. Then S_{\\Delta ABF_{2}} = \\frac{1}{2} \\times |AB| \\times |F_{1}F_{2}| = \\frac{1}{2} \\times \\frac{4}{a} \\times 2c = 2\\sqrt{6}. Hence \\frac{c}{a} = \\frac{\\sqrt{6}}{2}. Since c^{2} = a^{2} + 2, solving gives a = 2. Therefore \\frac{b}{a} = \\frac{\\sqrt{2}}{2}, so the asymptotes of the hyperbola are y = \\pm \\frac{\\sqrt{2}}{2}x." }, { "text": "It is known that one focus of a hyperbola coincides with the focus of the parabola $y^{2}=4 \\sqrt{5} x$, and the asymptotes of the hyperbola are given by $y=\\pm \\frac{1}{2} x$. Find the equation of this hyperbola?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y^2 = 4*(sqrt(5)*x));OneOf(Focus(G)) = Focus(H);Expression(Asymptote(G)) = (y = pm*(x/2))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2=1", "fact_spans": "[[[2, 5], [41, 44], [75, 78]], [[11, 34]], [[11, 34]], [[2, 39]], [[41, 72]]]", "query_spans": "[[[75, 82]]]", "process": "Given: the focus of the parabola $ y^{2} = 4\\sqrt{5}x $ is $ (\\sqrt{5}, 0) $. Let the equation of the hyperbola be $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $, we obtain:\n$$\n\\begin{cases}\nc = \\sqrt{5} \\\\\n\\frac{b}{a} = \\frac{1}{2} \\\\\nc^{2} = a^{2} + b^{2}\n\\end{cases}\n$$\nSolving gives $ a = 2 $, $ b = 1 $. Therefore, the equation of the hyperbola is $ \\frac{x^{2}}{4} - y^{2} = 1 $." }, { "text": "Let $M$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and let $F_{1}$, $F_{2}$ be the foci of the ellipse. If the distance from point $M$ to point $F_{1}$ is $4$, then what is the distance from point $M$ to point $F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);M: Point;PointOnCurve(M, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(M, F1) = 4", "query_expressions": "Distance(M, F2)", "answer_expressions": "6", "fact_spans": "[[[5, 43], [65, 67]], [[5, 43]], [[1, 4], [73, 77], [96, 100]], [[1, 48]], [[49, 56], [78, 86]], [[57, 64], [101, 109]], [[49, 70]], [[73, 93]]]", "query_spans": "[[[96, 114]]]", "process": "" }, { "text": "Given that the line $l$ is the directrix of the parabola $y^{2}=2 p x(p>0)$, and a circle with radius $3$ passes through the vertex $O$ and the focus $F$ of the parabola and is tangent to $l$, then the equation of the parabola is?", "fact_expressions": "l: Line;G: Parabola;p: Number;H: Circle;O: Origin;F:Point;p>0;Expression(G) = (y^2 = 2*p*x);Directrix(G)=l;Radius(H)=3;Vertex(G)=O;Focus(G)=F;PointOnCurve(O,H);PointOnCurve(F,H);IsTangent(H,l)", "query_expressions": "Expression(G)", "answer_expressions": "y^2 = 8*x", "fact_spans": "[[[2, 7], [57, 60]], [[8, 29], [42, 45], [64, 67]], [[11, 29]], [[40, 41]], [[47, 50]], [[53, 56]], [[11, 29]], [[8, 29]], [[2, 32]], [[33, 41]], [[42, 50]], [[42, 56]], [[40, 50]], [[40, 56]], [[40, 62]]]", "query_spans": "[[[64, 72]]]", "process": "According to the problem, let the equation of the circle be: $(x-a)^{2}+(y\\cdot b)^{2}=3^{2}$ $(r>0)$. The focus of the parabola $y^{2}=2px$ $(p>0)$ is $F(\\frac{p}{2},0)$. A circle with radius 3 passes through the vertex $O$ and the focus $F$ of the parabola $y^{2}=2px$ $(p>0)$, then \n$$\n\\begin{cases}\na^{2}+b^{2}=3^{2} \\\\\n(\\frac{p}{2}-a)^{2}+b^{2}=3^{2}\n\\end{cases}\n$$\nSolving gives $p=\\frac{4}{3}\\times3=4$, therefore the equation of the parabola is: $y^{2}=8x$" }, { "text": "Given that the distance from point $P$ to point $F(-3,0)$ is greater than its distance to the line $x=2$ by $1$, what equation does point $P$ satisfy?", "fact_expressions": "P: Point;F: Point;Coordinate(F) = (-3, 0);G: Line;Expression(G) = (x = 2);Distance(P, F) = Distance(P, G) + 1", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2=-12*x", "fact_spans": "[[[2, 6], [21, 22], [39, 43]], [[7, 17]], [[7, 17]], [[23, 30]], [[23, 30]], [[2, 37]]]", "query_spans": "[[[39, 50]]]", "process": "" }, { "text": "Draw a line $l$ with slope $-\\frac{1}{3}$ passing through the point $M(1,1)$. The line $l$ intersects the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ at points $A$ and $B$. If $M$ is the midpoint of segment $AB$, then what is the eccentricity of ellipse $C$?", "fact_expressions": "M: Point;Coordinate(M) = (1, 1);Slope(l) = -1/3;PointOnCurve(M, l) = True;l: Line;C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;B: Point;A: Point", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(6)/3", "fact_spans": "[[[1, 10], [106, 109]], [[1, 10]], [[11, 34]], [[0, 34]], [[29, 34]], [[35, 92], [122, 127]], [[35, 92]], [[42, 92]], [[42, 92]], [[42, 92]], [[42, 92]], [[29, 104]], [[106, 120]], [[99, 102]], [[95, 98]]]", "query_spans": "[[[122, 133]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), from the given conditions we have \n\\begin{cases} \nb^{2}x_{1}^{2}+a^{2}y_{1}^{2}=a^{2}b^{2} \\\\ \nb^{2}x_{2}^{2}+a^{2}y_{2}^{2}=a^{2}b^{2} \n\\end{cases} \n\\therefore b^{2}(x_{1}+x_{2})(x_{1}-x_{2})+a^{2}(y_{1}+y_{2})(y_{1}-y_{2})=0 \n\\therefore 2b^{2}(x_{1}-x_{2})+2a^{2}(y_{1}-y_{2})=0 \n\\therefore b^{2}(x_{1}-x_{2})=-a^{2}(y_{1}-y_{2}) \n\\therefore \\frac{b^{2}}{a^{2}} \\cdot \\frac{2}{1^{2}} = -\\frac{y_{1}-y_{2}}{\\sqrt{6}} = \\frac{1}{3} \n\\therefore a^{2}=3b^{2} \n\\therefore a^{2}=3(a^{2}-c^{2}) \n\\therefore 2a^{2}=3c^{2} \nHence, fill in \\frac{\\sqrt{6}}{}" }, { "text": "Given the parabola $E$: $y^{2}=2 p x(p>0)$ has focus $F$, point $A$ lies on $E$, a circle centered at $A$ is tangent to the $y$-axis and intersects $AF$ at point $B$. If $|A B|=2|B F|$, and the length of the chord cut from the perpendicular bisector of segment $AF$ by circle $A$ is $\\sqrt{7}$, then $p=?$", "fact_expressions": "E: Parabola;p: Number;A: Point;F: Point;B: Point;G: Circle;p>0;Expression(E) = (y^2 = 2*(p*x));Focus(E) = F;PointOnCurve(A, E);Center(G) = A;IsTangent(G, yAxis);Intersection(G, LineSegmentOf(A, F)) = B;Abs(LineSegmentOf(A, B)) = 2*Abs(LineSegmentOf(B, F));Length(InterceptChord(PerpendicularBisector(LineSegmentOf(A, F)), G)) = sqrt(7)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 28], [41, 44]], [[128, 131]], [[36, 40], [47, 50]], [[32, 35]], [[71, 75]], [[54, 55], [93, 97]], [[9, 28]], [[2, 28]], [[2, 35]], [[36, 45]], [[46, 56]], [[54, 62]], [[54, 75]], [[77, 91]], [[93, 126]]]", "query_spans": "[[[128, 133]]]", "process": "Let A(x_{1},y_{1}), and let the circle centered at A be tangent to the y-axis, then the radius r = |x_{1}|. By the definition of the parabola, |AF| = x_{1} + \\frac{p}{2}. Since |AB| = 2|BF|, we have |AF| = x_{1} + \\frac{p}{2} = x_{1} + \\frac{1}{2}x_{1}. Solving gives x_{1} = p, then |AF| = \\frac{3p}{2}. The chord length intercepted by the perpendicular bisector of segment AF on circle A is \\sqrt{7}, that is, p^{2} - \\frac{9p^{2}}{16} = \\frac{7}{4}. Solving gives p = 2." }, { "text": "The parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F(1,0)$. A line passing through point $F$ intersects $C$ at points $A$ and $B$. The intersection point of the directrix of $C$ and the $x$-axis is $M$. If the area of $\\triangle M A B$ is $\\frac{8 \\sqrt{3}}{3}$, then $\\frac{|A F|}{|B F|}$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;Focus(C) = F;F: Point;Coordinate(F) = (1, 0);PointOnCurve(F, G) = True;G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;Intersection(Directrix(C), xAxis) = M;M: Point;Area(TriangleOf(M, A, B)) = 8*sqrt(3)/3", "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))", "answer_expressions": "{3,1/3}", "fact_spans": "[[[0, 26], [48, 51], [63, 66]], [[0, 26]], [[8, 26]], [[8, 26]], [[0, 38]], [[30, 38], [40, 44]], [[30, 38]], [[39, 47]], [[45, 47]], [[45, 62]], [[53, 56]], [[57, 60]], [[63, 81]], [[78, 81]], [[83, 126]]]", "query_spans": "[[[128, 151]]]", "process": "Since the focus of parabola C is F(1,0), then \\frac{p}{2}=1, we get p=2, so the equation of parabola C is y^{2}=4x. Let A(x_{1},y_{1}), B(x_{2},y_{2}), solve the system \\begin{cases}x=my+1\\\\y^{2}=4x\\end{cases}, eliminating x gives y^{2}-4m-4=0, \\Delta_{A}=16m^{2}+16>0, by Vieta's formulas we have \\begin{cases}y_{1}+y_{2}=4m\\\\y_{1}y_{2}=-4\\end{cases}, thus S_{\\triangle MAB}=\\frac{1}{2}|MF|\\cdot|y_{1}-y_{2}|=|y_{1}-y_{2}|=\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=4\\sqrt{m^{2}+1}=\\frac{8\\sqrt{3}}{3}, we obtain m^{2}=\\frac{1}{3}, so (y_{1}+y_{2})^{2}=y_{1}^{2}+y_{2}^{2}+2y_{1}y_{2}=16m^{2}=\\frac{16}{3}, \\therefore y_{1}^{2}+y_{2}^{2}=\\frac{40}{3}. Hence \\frac{y_{1}}{y_{2}}+\\frac{y_{2}}{y_{1}}+\\frac{10}{3}=0. Let \\frac{y_{1}}{y_{2}}=t, we get t+\\frac{1}{t}+\\frac{10}{3}=0, i.e., 3t^{2}+10t+3=0, solving gives t=-\\frac{1}{3} or t=-3. Therefore, \\frac{|AF|}{|BF|}=\\left|\\frac{y_{1}}{y_{2}}\\right|=|t|=3 or \\frac{1}{3}." }, { "text": "If the curve $C$: $m x^{2}+(2-m) y^{2}=1$ is a hyperbola with foci on the $y$-axis, then one possible value of $m$ is?", "fact_expressions": "C: Curve;Expression(C) = (m*x^2 + y^2*(2 - m) = 1);m: Number;G: Hyperbola;C=G;PointOnCurve(Focus(G),yAxis)", "query_expressions": "m", "answer_expressions": "-1", "fact_spans": "[[[1, 31]], [[1, 31]], [[46, 49]], [[41, 44]], [[1, 44]], [[32, 44]]]", "query_spans": "[[[46, 56]]]", "process": "The curve C: mx^{2} + (2 - m)y^{2} = 1 is a hyperbola with foci on the y-axis, which can be rewritten as \\frac{x^{2}}{m} - \\frac{y^{2}}{m-2} = 1, implying m < 0 and m - 2 < 0. Solving gives m < 0. Thus, m can take the value -1. This problem has non-unique answers: for example, -1" }, { "text": "Given that the hyperbola $E$ is centered at the origin, $F(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 standard equation of the hyperbola $E$ is?", "fact_expressions": "E: Hyperbola;O: Origin;Center(E) = O;F: Point;Coordinate(F) = (3, 0);Focus(E) = F;l: Line;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], [84, 90]], [[12, 16]], [[2, 16]], [[17, 25], [34, 37]], [[17, 25]], [[17, 32]], [[38, 43]], [[33, 43]], [[38, 59]], [[50, 53]], [[54, 57]], [[61, 82]], [[70, 82]], [[70, 82]]]", "query_spans": "[[[84, 97]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), substitute into the hyperbola equation and subtract to obtain \\frac{x_{1}+x_{2}}{y_{1}+y_{2}}=\\frac{a^{2}}{b^{2}}\\cdot\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, that is, \\frac{-12}{15}=\\frac{a^{2}}{b^{2}}\\cdot\\frac{0-(-15)}{3-(-12)}, \\frac{b^{2}}{a^{2}}=\\frac{5}{4}, together with c=3, c^{2}=a^{2}+b^{2}, solve to get a^{2}=4, b^{2}=5, thus the hyperbola equation is \\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1." }, { "text": "Given that $M$ is a point on the parabola $C$: $y^{2}=2 p x$ ($p>0$), $F$ is the focus of the parabola $C$, if $|M F|=p$, and $K$ is the intersection point of the directrix of the parabola $C$ and the $x$-axis, then $\\angle M K F$=?", "fact_expressions": "C: Parabola;p: Number;M: Point;F: Point;K: Point;p>0;Expression(C) = (y^2 = 2*(p*x));PointOnCurve(M, C);Focus(C) = F;Abs(LineSegmentOf(M, F)) = p;Intersection(Directrix(C), xAxis) = K", "query_expressions": "AngleOf(M, K, F)", "answer_expressions": "{ApplyUnit(45, degree), pi/4}", "fact_spans": "[[[6, 31], [39, 45], [65, 71]], [[13, 31]], [[2, 5]], [[35, 38]], [[61, 64]], [[13, 31]], [[6, 31]], [[2, 34]], [[35, 48]], [[50, 59]], [[61, 82]]]", "query_spans": "[[[84, 100]]]", "process": "Draw a perpendicular line from point M to the directrix, with foot of perpendicular at N. By the definition of the parabola, |MN| = |MF| = |FK| = p. Since MN // FK, quadrilateral FMNK is a square. Therefore, ∠MKF = π/4. Answer: 45° or π/4" }, { "text": "Given that $P$ is a point on the right branch of the hyperbola $C$: $\\frac{x^{2}}{2}-y^{2}=1$, line $l$ is an asymptote of the hyperbola, $Q$ is the projection of $P$ onto $l$, and $F_{1}$ is the left focus of the hyperbola, then the minimum value of $|P F_{1}|+|P Q|$ is?", "fact_expressions": "l: Line;C: Hyperbola;P: Point;F1: Point;Q: Point;Expression(C) = (x^2/2 - y^2 = 1);PointOnCurve(P, RightPart(C));OneOf(Asymptote(C))=l;Projection(P,l)=Q;LeftFocus(C)=F1", "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))", "answer_expressions": "1+2*sqrt(2)", "fact_spans": "[[[45, 50], [65, 68]], [[6, 39], [51, 54], [85, 88]], [[2, 5], [61, 64]], [[77, 84]], [[73, 76]], [[6, 39]], [[2, 44]], [[45, 60]], [[61, 76]], [[77, 92]]]", "query_spans": "[[[94, 117]]]", "process": "|PF_{1}|+|PQ|=2a+|PF_{2}|+|PQ|\\geqslant2a+d_{F_{2}-l}=2a+b=2\\sqrt{2}+1." }, { "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1(m>n>0)$ and the curve $x^{2}+y^{2}=|m-n|$ have common points, then what is the range of values for the eccentricity $e$ of the ellipse?", "fact_expressions": "G: Ellipse;m: Number;n: Number;H: Curve;e: Number;m > n;n > 0;Expression(G) = (y^2/n + x^2/m = 1);Expression(H) = (x^2 + y^2 = Abs(m - n));IsIntersect(G, H);Eccentricity(G) = e", "query_expressions": "Range(e)", "answer_expressions": "[sqrt(2)/2,1)", "fact_spans": "[[[1, 45], [73, 75]], [[3, 45]], [[3, 45]], [[46, 67]], [[79, 82]], [[3, 45]], [[3, 45]], [[1, 45]], [[46, 67]], [[1, 71]], [[73, 82]]]", "query_spans": "[[[79, 89]]]", "process": "" }, { "text": "Through the focus of the parabola $y^{2}=4x$, draw a line $l$ intersecting the parabola at points $A$ and $B$. If $|AB|=8$, then the horizontal coordinate of the midpoint of segment $AB$ is?", "fact_expressions": "A: Point;B: Point;l: Line;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), l);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 8", "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "3", "fact_spans": "[[[29, 32]], [[33, 36]], [[19, 24]], [[1, 15], [25, 28]], [[1, 15]], [[0, 24]], [[19, 38]], [[40, 49]]]", "query_spans": "[[[51, 65]]]", "process": "Problem Analysis: From the parabola $ y^{2} = 4x $, we obtain the focus $ F(1,0) $. If $ AB \\perp x $-axis, then $ |AB| = 2p = 4 $, which does not satisfy the condition and is therefore discarded. Let the equation of line $ l $ be: $ my = (x - 1) $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system \n\\[\n\\begin{cases}\nmy = x - 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\nyields $ y^{2} - 4my - 4 = 0 $, thus $ y_{1} + y_{2} = 4m $, $ y_{1} \\cdot y_{2} = -4 $. Hence, $ m^{2} = 1 $, solving gives $ m = \\pm 1 $. When $ m = 1 $, solving the system \n\\[\n\\begin{cases}\ny = x - 1 \\\\\ny^{2} = 4x\n\\end{cases}\n\\]\ngives $ x^{2} - 6x + 1 = 0 $, so $ x_{1} + x_{2} = 6 $, therefore $ \\frac{x_{1} + x_{2}}{2} = 3 $. Similarly, when $ m = -1 $, $ \\frac{x_{1} + x_{2}}{2} = 3 $. Thus, the x-coordinate of the midpoint of segment $ AB $ is 3." }, { "text": "Given that the center of the ellipse is at the origin, one focus is $(\\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;H: Point;Coordinate(H) = (sqrt(3), 0);OneOf(Focus(G)) = H;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4+y^2=1", "fact_spans": "[[[2, 4], [48, 50]], [[7, 9]], [[2, 9]], [[15, 31]], [[15, 31]], [[2, 31]], [[2, 45]]]", "query_spans": "[[[48, 57]]]", "process": "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. Thus, $c=\\sqrt{3}$, $a=2b$, $\\because a^{2}=b^{2}+c^{2}$ $\\therefore b^{2}=1$, $b=1$, $a=2$. Therefore, the equation of the ellipse is: $\\frac{x^{2}}{4}+y^{2}=1$." }, { "text": "The equation of a parabola with vertex at the origin, symmetric about the $y$-axis, and with distance from the vertex to the directrix equal to $4$ is?", "fact_expressions": "G: Parabola;O: Origin;Vertex(G)=O;SymmetryAxis(G)=yAxis;Distance(Vertex(G),Directrix(G))=4", "query_expressions": "Expression(G)", "answer_expressions": "x^2=pm*16*y", "fact_spans": "[[[28, 31]], [[3, 5]], [[0, 31]], [[6, 31]], [[15, 31]]]", "query_spans": "[[[28, 35]]]", "process": "" }, { "text": "Given that the length of the major axis of an ellipse is twice the focal distance, what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 2*FocalLength(G)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[2, 4], [18, 20]], [[2, 16]]]", "query_spans": "[[[18, 26]]]", "process": "Since the length of the major axis of the ellipse is twice the focal distance, and the eccentricity of the ellipse is the ratio of the focal distance to the length of the major axis, the eccentricity of the ellipse is $\\frac{1}{3}$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $O$ is the coordinate origin, $|F_{1} F_{2}|=10$, point $M$ is a point on the left branch of the hyperbola, $|O M|=\\sqrt{a^{2}+b^{2}}$, and $4|M F_{1}|=3|M F_{2}|$. Then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;LeftFocus(G) = F1;RightFocus(G) = F2;F1: Point;F2: Point;O: Origin;Abs(LineSegmentOf(F1, F2)) = 10;M: Point;PointOnCurve(M, LeftPart(G)) = True;Abs(LineSegmentOf(O, M)) = sqrt(a^2 + b^2);4*Abs(LineSegmentOf(M, F1)) = 3*Abs(LineSegmentOf(M, F2))", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/24=1", "fact_spans": "[[[2, 48], [106, 109], [170, 173]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 72]], [[2, 72]], [[57, 64]], [[65, 72]], [[73, 76]], [[82, 100]], [[101, 105]], [[101, 114]], [[116, 142]], [[145, 168]]]", "query_spans": "[[[170, 180]]]", "process": "Since |F_{1}F_{2}|=10, we have 2c=10, that is, c=5. Because |OM|=\\sqrt{a^{2}+b^{2}}=c=5 and |F_{1}F_{2}|=2c=10, it follows that MF_{1}\\bot MF_{2}. Let |MF_{1}|=3k, |MF_{2}|=4k, then (3k)^{2}+(4k)^{2}=|F_{1}F_{2}|^{2}=100, solving gives k=2, so |MF_{1}|=6, |MF_{2}|=8. By the definition of hyperbola, 2a=|MF_{2}|-|MF_{1}|=8-6=2, that is, a=1. Thus b^{2}=c^{2}-a^{2}=24, so the standard equation of the hyperbola is x^{2}-\\frac{y^{2}}{24}=1" }, { "text": "Given a point $P$ on the parabola $y^{2}=4x$, the distance from $P$ to the directrix is $d_{1}$, and the distance from $P$ to the line $l$: $4x-3y+16=0$ is $d_{2}$. Then the minimum value of $d_{1}+d_{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G) = True;Distance(P, Directrix(G)) = d1;d1: Number;l: Line;Expression(l) = (4*x - 3*y + 16 = 0);Distance(P, l) = d2;d2: Number", "query_expressions": "Min(d1 + d2)", "answer_expressions": "4", "fact_spans": "[[[2, 16]], [[2, 16]], [[19, 22]], [[2, 22]], [[2, 36]], [[29, 36]], [[38, 59]], [[38, 59]], [[19, 67]], [[60, 67]]]", "query_spans": "[[[69, 88]]]", "process": "As shown in the figure, point P is any point on the parabola. According to the definition of the parabola, d_{1}=|PF|, so d_{1}+d_{2}=|PF|+d_{2}. From the figure, the minimum value of |PF|+d_{2} is the distance from point F to the line 4x-3y+16=0. The distance from F(1,0) to the line l: 4x-3y+16=0 is d=\\frac{|4+16|}{\\sqrt{4^{2}+3^{2}}}=4. Therefore, the minimum value of d_{1}+d_{2} is 4. The answer is: 4" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola, respectively, $P$ is a point in the first quadrant on the hyperbola, point $I$ is the incenter of $\\Delta P F_{1} F_{2}$, what is the range of possible values for the area of $\\Delta I F_{1} F_{2}$?", "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;I: Point;Expression(C) = (x^2/9 - y^2/16 = 1);LeftFocus(C)=F1;RightFocus(C) = F2;Quadrant(P) = 1;PointOnCurve(P, C);Incenter(TriangleOf(P, F1, F2)) = I", "query_expressions": "Range(Area(TriangleOf(I, F1, F2)))", "answer_expressions": "(0,20)", "fact_spans": "[[[2, 47], [69, 72], [83, 86]], [[79, 82]], [[49, 57]], [[59, 66]], [[96, 100]], [[2, 47]], [[49, 78]], [[49, 78]], [[79, 95]], [[79, 95]], [[96, 126]]]", "query_spans": "[[[127, 159]]]", "process": "First, from the definition of the hyperbola, the foot of the perpendicular from point I to $ F_{1}F_{2} $ is the right vertex. Let the angle of inclination of the asymptote be $ \\theta $, then $ \\tan\\theta = \\frac{4}{3} $. Let $ \\angle IF_{2}H = \\alpha $, then $ 2\\alpha \\in (0, \\pi - \\theta) $, find $ \\tan\\alpha \\in (0, 2) $, thus obtain $ y_{1} = |HF_{2}| \\tan\\alpha \\in (0, 4) $, and find the range of the area of $ \\triangle IF_{1}F_{2} $. [Detailed solution] From the given: $ c^{2} = 9 + 16 = 25 $, hence $ |F_{1}F_{2}| = 2c = 10 $. Let point $ I(x_{1}, y_{1}) $, and let $ H $ be the foot of the perpendicular from $ I $ to $ F_{1}F_{2} $. According to the definition of the hyperbola and the tangent segment theorem: $ |PF_{1}| - |PF_{2}| = 2a = |HF_{1}| - |HF_{2}| $. Also, $ |HF_{1}| + |HF_{2}| = 2c $. Solving gives: $ |HF_{1}| = a + c $. Therefore, the coordinates of point $ H $ are $ (a, 0) $. The asymptote $ y = \\frac{4}{3}x $ has angle of inclination $ \\theta $, so $ \\tan\\theta = \\frac{4}{3} $. Denote $ \\angle IF_{2}H = \\alpha $, then $ 2\\alpha \\in (0, \\pi - \\theta) $, so $ \\alpha < \\frac{\\pi}{2} - \\frac{\\theta}{2} $, i.e., $ \\tan\\alpha < \\tan\\left( \\frac{\\pi}{2} - \\frac{\\theta}{2} \\right) = \\frac{1}{\\tan\\frac{\\theta}{2}} $. Also, $ \\frac{2\\tan\\frac{\\theta}{2}}{1 - \\tan^{2}\\frac{\\theta}{2}} = \\tan\\theta = \\frac{4}{3} $, solving gives: $ \\tan\\frac{\\theta}{2} = \\frac{1}{2} $ (negative value discarded). Thus $ \\tan\\alpha \\in (0, 2) $, then $ y_{1} = |HF_{2}| \\tan\\alpha \\in (0, 4) $. Therefore, $ S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}| \\cdot y_{1} = 5y_{1} \\in (0, 20) $." }, { "text": "Given that the line $ l $: $ x = m y + 1 $ passes through the focus $ F $ of the parabola $ C $: $ y^2 = 2 p x $, and intersects the parabola $ C $ at points $ A $ and $ B $, if $ \\overrightarrow{A F} = 2 \\overrightarrow{F B} $, then the slope of line $ l $ is?", "fact_expressions": "l: Line;Expression(l)=(x=m*y+1);m: Number;C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;F: Point;Focus(C)=F;PointOnCurve(F,l);A: Point;B: Point;Intersection(l,C)={A,B};VectorOf(A, F) = 2*VectorOf(F, B)", "query_expressions": "Slope(l)", "answer_expressions": "pm*2*sqrt(2)", "fact_spans": "[[[2, 18], [113, 118]], [[2, 18]], [[9, 18]], [[19, 40], [48, 54]], [[19, 40]], [[27, 40]], [[43, 46]], [[19, 46]], [[2, 46]], [[55, 58]], [[59, 62]], [[2, 64]], [[66, 111]]]", "query_spans": "[[[113, 123]]]", "process": "Find the line passing through the fixed point (1,0), thus p=2. Let A(x_{1},y_{1}), B(x_{2},y_{2}). From the given condition, y_{1}=-2y_{2}. Substitute the line into the parabola and use Vieta's formulas to obtain y_{2}=\\pm\\sqrt{2}, then find the slope. Since the line x=my+1 passes through (1,0), p=2. Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\overrightarrow{AF}=2\\overrightarrow{FB}, we get y_{1}=-2y_{2}. Substituting the line x=my+1 into the parabola y^{2}=4x yields y^{2}=4my, so y_{1}y_{2}=-4, giving y_{2}=\\pm\\sqrt{2}. Therefore, k=\\frac{\\pm\\sqrt{2}}{1-1}=\\pm2\\sqrt{2}" }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, a line $AB$ passing through point $F$ intersects the parabola at points $A$ and $B$, and intersects the directrix at point $C$. If $|BC|=2|BF|$, then $|AB|=$?", "fact_expressions": "G: Parabola;A: Point;B: Point;C: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, LineOf(A, B));Intersection(LineOf(A, B), G) = {A, B};Intersection(LineOf(A, B), Directrix(G)) = C;Abs(LineSegmentOf(B, C)) = 2*Abs(LineSegmentOf(B, F))", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16/3", "fact_spans": "[[[2, 16], [38, 41]], [[42, 45]], [[46, 49]], [[56, 60]], [[20, 23], [25, 29]], [[2, 16]], [[2, 23]], [[24, 37]], [[30, 51]], [[30, 60]], [[62, 76]]]", "query_spans": "[[[78, 87]]]", "process": "Draw perpendiculars from A and B to the directrix. Using the definition of the parabola, convert the distances from A and B to the focus into distances to the directrix. Use the given conditions and the similarity ratio of similar triangles to establish equations and solve for |AF| and |BF| to compute the chord length. Let y^{2}=4x=2px, so p=2. As shown in the figure, draw AM and BN perpendicular to the directrix meeting it at M and N respectively. Then |BN|=|BF|. Also |BC|=2|BN|, \\frac{|CB|}{|CF|}=\\frac{2}{3}, \\therefore\\frac{|BN|}{p}=\\frac{2}{3}, |BN|=\\frac{4}{3}, |BC|=\\frac{8}{3}, \\therefore|CF|=4. \\frac{2}{|AM|}=\\frac{|CF|}{|CA|}, \\therefore\\frac{2}{|AM|}=\\frac{|CF|}{|CA|}=\\frac{4}{4+|AM|}, solving gives |AM|=4, \\therefore|AF|=4. |AB|=|AF|+|BF|=4+\\frac{4}{3}=\\frac{16}{3}." }, { "text": "The line $y = x + 1$ intersects the hyperbola $\\frac{x^{2}}{2} - \\frac{y^{2}}{3} = 1$ at points $A$ and $B$. Find $|AB|$.", "fact_expressions": "G: Hyperbola;H: Line;A: Point;B: Point;Expression(G) = (x^2/2 - y^2/3 = 1);Expression(H) = (y = x + 1);Intersection(H, G) = {A,B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "4*sqrt(5)", "fact_spans": "[[[10, 48]], [[0, 9]], [[51, 54]], [[55, 58]], [[10, 48]], [[0, 9]], [[0, 60]]]", "query_spans": "[[[62, 71]]]", "process": "" }, { "text": "The ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{2}=1$ have common foci $F_{1}$, $F_{2}$, and $P$ is a point of intersection of the two curves. Then $\\angle F_{1} P F_{2}$=?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/8 + y^2/2 = 1);G: Hyperbola;Expression(G) = (x^2/4 - y^2/2 = 1);F1: Point;F2: Point;Focus(H) = {F1, F2};Focus(G) = {F1, F2};P: Point;OneOf(Intersection(H, G)) = P", "query_expressions": "AngleOf(F1, P, F2)", "answer_expressions": "ApplyUnit(90, degree)", "fact_spans": "[[[0, 37]], [[0, 37]], [[38, 76]], [[38, 76]], [[82, 89]], [[90, 97]], [[0, 97]], [[0, 97]], [[98, 101]], [[98, 110]]]", "query_spans": "[[[113, 137]]]", "process": "" }, { "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{169}+\\frac{y^{2}}{144}=1$, and let $F_{1}$, $F_{2}$ be the foci of the ellipse. If $|P F_{1}|=4$, then $|P F_{2}|$ equals?", "fact_expressions": "G: Ellipse;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/169 + y^2/144 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};Abs(LineSegmentOf(P, F1)) = 4", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "22", "fact_spans": "[[[5, 46], [66, 68]], [[1, 4]], [[58, 65]], [[50, 57]], [[5, 46]], [[1, 49]], [[50, 71]], [[73, 86]]]", "query_spans": "[[[88, 102]]]", "process": "According to the definition of an ellipse, |PF_{1}| + |PF_{2}| = 2a. Combining with the given conditions, the value of |PF_{2}| can be calculated. Since |PF_{1}| + |PF_{2}| = 2a = 26 and |PF_{1}| = 4, it follows that |PF_{2}| = 26 - 4 = 22." }, { "text": "The equation of the directrix of the parabola $x^{2}=-8 y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -8*y)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=2", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "Test analysis: Since the directrix equation of the parabola $ x^{2} = 2py $ is $ y = \\frac{p}{2} $, the directrix equation of the parabola $ x^{2} = -8y $ can thus be obtained. Since the directrix equation of the parabola $ x^{2} = -2py $ is $ y = \\frac{p}{2} $, then the directrix equation of the parabola $ x^{2} = -8y $ is $ y = 2 $." }, { "text": "The focal distance of the hyperbola $\\frac{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": "2*sqrt(5)", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 33]]]", "process": "Let the semi-focal length of the hyperbola \\frac{x^{2}}{4}-y^{2}=1 be c, then c^{2}=4+1=5, solving gives c=\\sqrt{5}; thus, the focal length of the hyperbola \\frac{x^{2}}{4}-y^{2}=1 is 2\\sqrt{5}." }, { "text": "The parabola is given by $y^{2}=8 x$. What is the equation of its directrix?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-2", "fact_spans": "[[[1, 15]], [[1, 15]]]", "query_spans": "[[[1, 22]]]", "process": "According to the given conditions and the standard equation of a parabola, determine its directrix equation. [Detailed solution] From the parabola equation y^{2}=8x, we have 2p=8, then \\frac{p}{2}=2, thus the directrix equation is x=-2" }, { "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $4 x-3 y=0$, find the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (4*x - 3*y = 0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "", "fact_spans": "[[[2, 48], [70, 73]], [[5, 48]], [[5, 48]], [[2, 48]], [[2, 68]]]", "query_spans": "[[[70, 79]]]", "process": "" }, { "text": "The hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1 (a>0 , b>0)$ has an asymptote with equation $y=\\frac{3}{5} x$. What is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (y = (3/5)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(34)/3", "fact_spans": "[[[0, 59], [87, 90]], [[3, 59]], [[3, 59]], [[3, 59]], [[3, 59]], [[0, 59]], [[0, 85]]]", "query_spans": "[[[87, 96]]]", "process": "Combining the standard equation of the hyperbola, the asymptotes of the hyperbola are given by y=\\pm\\frac{a}{b}x. Thus we have: \\frac{b}{a}=\\frac{5}{3}, \\therefore \\frac{b^{2}}{a^{2}}=\\frac{c^{2}-a^{2}}{a^{2}}=\\frac{25}{9}, then: e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{25}{9}+1=\\frac{34}{9}, \\therefore e=\\frac{\\sqrt{34}}{3}" }, { "text": "Given that the equation of the directrix of the parabola $y=ax^{2}$ is $y=1$, find the value of $a$.", "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;Expression(Directrix(G)) = (y = 1)", "query_expressions": "a", "answer_expressions": "-1/4", "fact_spans": "[[[2, 15]], [[2, 15]], [[28, 31]], [[2, 26]]]", "query_spans": "[[[28, 35]]]", "process": "" }, { "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 is not the right vertex of the ellipse), point $Q$ lies on the right directrix of the ellipse. If $\\overrightarrow{P Q}=2 \\overrightarrow{F_{1} O}$, $\\overrightarrow{F_{1} Q}=\\lambda\\left(\\frac{\\overrightarrow{F_{1} P}}{|\\overrightarrow{F_{1} P}|}+\\frac{\\overrightarrow{F_{1} O}}{|\\overrightarrow{F_{1} O}|}\\right)$ $(\\lambda>0)$, then the eccentricity of the ellipse is?", "fact_expressions": "C:Ellipse;a:Number;b:Number;lambda:Number;F1:Point;P:Point ;Q:Point;O:Origin;a>b ;b>0;Expression(C)=(x^2/a^2+y^2/b^2=1);LeftFocus(C)=F1;PointOnCurve(P,C);PointOnCurve(Q,RightDirectrix(C));VectorOf(P,Q)=2*VectorOf(F1,O);VectorOf(F1,Q) = lambda * (VectorOf(F1,P)/Abs(VectorOf(F1,P))+VectorOf(F1,O)/Abs(VectorOf(F1,O)));lambda>0;Negation(RightVertex(C)=P)", "query_expressions": "Eccentricity(C)", "answer_expressions": "(sqrt(5)-1)/2", "fact_spans": "[[[2, 54], [80, 82], [88, 90], [101, 103], [328, 330]], [[4, 54]], [[4, 54]], [[160, 326]], [[58, 65]], [[75, 79]], [[96, 100]], [[66, 69]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 65]], [[75, 83]], [[96, 108]], [[110, 159]], [[160, 326]], [[160, 326]], [[75, 95]]]", "query_spans": "[[[328, 336]]]", "process": "Since $\\overrightarrow{PQ}=2\\overrightarrow{F_{1}O}$, $PQ$ is parallel to the $x$-axis, and the horizontal coordinate of point $P$ is $\\frac{a^{2}}{c}-2c$. Also, from $\\overrightarrow{F_{1}Q}=\\lambda\\left(\\frac{\\overrightarrow{F_{1}P}}{|\\overrightarrow{F_{1}P}|}+\\frac{\\overrightarrow{F_{1}O}}{|\\overrightarrow{F_{1}O}|}\\right)$ ($\\lambda>0$), it follows that point $Q$ lies on the angle bisector of $\\angle PF_{1}O$, hence $\\angle PF_{1}O=2\\angle QF_{1}O$. Since $PQ \\parallel F_{1}F_{2}$, $|PQ|=|F_{1}F_{2}|$, the quadrilateral $PF_{1}F_{2}Q$ is a parallelogram. Combining this with the diagonal being an angle bisector, the quadrilateral $PF_{1}F_{2}Q$ is a rhombus, so $|PF_{1}|=2c$. Thus, $|PF_{2}|=2a-2c$. From the second definition of the ellipse, $e=\\frac{|PF_{2}|}{|PQ|}=\\frac{2a-2c}{2c}$, solving gives $e=\\frac{\\sqrt{5}-1}{2}$." }, { "text": "The equation of the hyperbola with asymptotes $y=\\pm \\sqrt{3} x$ and passing through the point $P(\\sqrt{3}, 2 \\sqrt{3})$ is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*sqrt(3)*x);P: Point;Coordinate(P) = (sqrt(3), 2*sqrt(3));PointOnCurve(P, G) = True", "query_expressions": "Expression(G)", "answer_expressions": "y^2/3-x^2=1", "fact_spans": "[[[53, 56]], [[0, 56]], [[26, 52]], [[26, 52]], [[25, 56]]]", "query_spans": "[[[53, 60]]]", "process": "If the foci of the hyperbola lie on the x-axis, let the equation of the hyperbola be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{32}=1, then \\frac{b}{a}=\\sqrt{3} and \\frac{3}{a^{2}}-\\frac{12}{b^{2}}=1, which has no solution; if the foci of the hyperbola lie on the y-axis, let the equation of the hyperbola be \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{12}=1, then \\frac{a}{b}=\\sqrt{3} and \\frac{12}{a^{2}}-\\frac{3}{b^{2}}=1, yielding a^{2}=3, b^{2}=1, so the equation of the hyperbola is \\frac{y^{2}}{3}-x^{2}=1." }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, and point $M(x_{0}, 2 \\sqrt{2})(x_{0}>\\frac{p}{2})$ is a point on the parabola $C$. A circle centered at $M$ intersects the segment $M F$ at point $A$, and the chord cut by the line $x=\\frac{p}{2}$ has length $\\sqrt{3}|M A|$. If $|M A|=2$, then $|A F|$=?", "fact_expressions": "C: Parabola;p: Number;G: Circle;H: Line;M: Point;F: Point;A: Point;p>0;x0: Number;x0 > p/2;Expression(C) = (y^2 = 2*(p*x));Expression(H) = (x = p/2);Coordinate(M) = (x0, 2*sqrt(2));Focus(C) = F;PointOnCurve(M, C);Center(G) = M;Intersection(G, LineSegmentOf(M, F)) = A;Length(InterceptChord(H, G)) = sqrt(3)*Abs(LineSegmentOf(M, A));Abs(LineSegmentOf(M, A)) = 2", "query_expressions": "Abs(LineSegmentOf(A, F))", "answer_expressions": "1", "fact_spans": "[[[2, 28], [79, 85]], [[10, 28]], [[97, 98]], [[116, 133]], [[36, 78], [90, 93]], [[32, 35]], [[109, 113]], [[10, 28]], [[37, 78]], [[37, 78]], [[2, 28]], [[116, 133]], [[36, 78]], [[2, 35]], [[36, 88]], [[89, 98]], [[97, 113]], [[97, 154]], [[156, 165]]]", "query_spans": "[[[167, 176]]]", "process": "According to the problem, point $ M(x_{0},2\\sqrt{2}) $ lies on the parabola, so $ 8 = 2px_{0} $, then $ px_{0} = 4 $, $\\textcircled{1}$ then $ |MA| = 2|AF| = \\frac{2}{3}|MF| = \\frac{2}{3}(x_{0} + \\frac{p}{2}) $. Since the chord intercepted by the line $ x = \\frac{p}{2} $ has length $ \\sqrt{3}|MA| $, then $ |DE| = \\frac{\\sqrt{3}}{2}|MA| = \\frac{\\sqrt{3}}{3}(x_{0} + \\frac{p}{2}) $. Given $ |MA| = |ME| = r $, in right triangle $ \\triangle MDE $, $ |DE|^{2} + |DM|^{2} = |ME|^{2} $, that is, $ \\frac{1}{3}(x_{0} + \\frac{p}{2})^{2} + (x_{0} - \\frac{p}{2})^{2} = \\frac{4}{9}(x_{0} + \\frac{p}{2})^{2} $. Substituting and simplifying yields $ 4x_{0}^{2} + p^{2} = 20 $, $\\textcircled{2}$ From $\\textcircled{1}$ and $\\textcircled{2}$, solving gives $ x_{0} = 2 $, $ p = 2 $, therefore $ |AF| = \\frac{1}{3}(x_{0} + \\frac{p}{2}) = 1 $." }, { "text": "Given that the hyperbola $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ $(m, n \\in \\mathbb{R})$ with eccentricity $2$ has foci coinciding with the foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, then $\\frac{m}{n}=$?", "fact_expressions": "G: Hyperbola;m: Real;n: Real;H: Ellipse;Expression(G) = (y^2/n + x^2/m = 1);Expression(H) = (x^2/5 + y^2/4 = 1);Eccentricity(G)=2;Focus(G)=Focus(H)", "query_expressions": "m/n", "answer_expressions": "-1/3", "fact_spans": "[[[10, 60]], [[13, 60]], [[13, 60]], [[64, 101]], [[10, 60]], [[64, 101]], [[2, 60]], [[10, 106]]]", "query_spans": "[[[108, 123]]]", "process": "According to the problem, for the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, we have $c=\\sqrt{a^{2}-b^{2}}=1$, so $F(\\pm1,0)$. Since the foci of the hyperbola $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ coincide with the foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, it follows that $c=1$. Given that the eccentricity of the hyperbola $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ is 2, we have $e=-\\frac{c}{m}=2$, solving gives $m=\\frac{1}{4}$. Also, from $-n=c^{2}-m=1-\\frac{1}{4}=\\frac{3}{4}$, solving gives $n=-\\frac{3}{4}$. Therefore, $\\frac{m}{n}=-\\frac{1}{3}$." }, { "text": "Given the parabola $M$: $y^{2}=16 x$ with focus $F$, and let $P$ be a point on the parabola $M$. If $|P F|=5$, then what are the coordinates of point $P$?", "fact_expressions": "M: Parabola;P: Point;F: Point;Expression(M) = (y^2 = 16*x);Focus(M)=F;PointOnCurve(P, M);Abs(LineSegmentOf(P, F)) = 5", "query_expressions": "Coordinate(P)", "answer_expressions": "{(1,-4),(1,4)}", "fact_spans": "[[[2, 22], [34, 40]], [[30, 33], [57, 61]], [[26, 29]], [[2, 22]], [[2, 29]], [[30, 43]], [[46, 55]]]", "query_spans": "[[[57, 66]]]", "process": "The focus of the parabola M: y^{2}=16x is F(4,0), the directrix is l: x=-4. If |PF|=5, then the distance from P to the directrix is 5. Let P(x,y), so x+4=5, solving gives x=1. Substituting into the parabola equation yields y=\\pm4. Therefore, the coordinates of point P are (1,4) or (1,-4)." }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$ is? The equations of the asymptotes are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/12 = 1)", "query_expressions": "Eccentricity(G);Expression(Asymptote(G))", "answer_expressions": "2\ny=pm*sqrt(3)*x", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 45]], [[0, 52]]]", "process": "" }, { "text": "A focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $(5,0)$, and one asymptote is $3 y-4 x=0$. Then what is the equation of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (5, 0);OneOf(Focus(G))=H;Expression(OneOf(Asymptote(G)))=(3*y-4*x=0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9-y^2/16=1", "fact_spans": "[[[0, 46], [81, 84]], [[3, 46]], [[3, 46]], [[52, 59]], [[0, 46]], [[52, 59]], [[0, 59]], [[0, 78]]]", "query_spans": "[[[81, 89]]]", "process": "Since the focus of the hyperbola is (5,0), c=5. Because the asymptote is 3y-4x=0, \\frac{b}{a}=\\frac{4}{3}. Also, c^{2}=a^{2}+b^{2}. Therefore, a=3, b=4, so the equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1." }, { "text": "Given line $l_{1}$: $2x - y + 6 = 0$ and line $l_{2}$: $x = -1$, $F$ is the focus of the parabola $C$: $y^{2} = 4x$. Point $P$ moves on the parabola $C$. When the sum of the distances from point $P$ to line $l_{1}$ and line $l_{2}$ is minimized, what is the length of the line segment intercepted by the parabola $C$ along line $PF$?", "fact_expressions": "l1: Line;Expression(l1) = (2*x -y + 6 = 0);l2: Line;Expression(l2) = (x = -1);C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;P: Point;PointOnCurve(P, C);WhenMin(Distance(P, l1) + Distance(P, l2))", "query_expressions": "Length(InterceptChord(LineOf(P, F), C))", "answer_expressions": "20", "fact_spans": "[[[2, 24], [92, 101]], [[2, 24]], [[25, 42], [102, 111]], [[25, 42]], [[48, 67], [76, 82], [127, 130]], [[48, 67]], [[44, 47]], [[44, 70]], [[71, 75], [87, 91]], [[71, 83]], [[86, 119]]]", "query_spans": "[[[120, 139]]]", "process": "The line $ l_{2} $ is the directrix of the parabola $ y^{2} = 4x $. By the definition of a parabola, the distance from point $ P $ to $ l_{2} $ equals the distance from $ P $ to the focus $ F(1,0) $ of the parabola. The minimum sum of distances from point $ P $ to line $ l_{1} $ and line $ l_{2} $ is thus transformed into the minimum sum of distances from $ P $ to point $ F(1,0) $ and line $ l_{1} $. When this sum is minimized, the line $ PF \\perp l_{1} $, so the equation of line $ PF $ is $ y = -\\frac{1}{2}(x - 1) $. Substituting into the equation of $ C $ gives $ x^{2} - 18x + 1 = 0 $, so $ x_{1} + x_{2} = 18 $, and thus the required segment length is $ x_{1} + x_{2} + p = 18 + 2 = 20 $." }, { "text": "Given that the focus $F$ of the parabola $y^{2}=20x$ 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{15}{4}, 3)$, then the equations of the asymptotes of this hyperbola are?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;I: Point;F: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 20*x);Coordinate(I) = (15/4, 3);Focus(H) = F;RightFocus(G) = F;PointOnCurve(I, G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*(4/3)*x", "fact_spans": "[[[25, 71], [77, 80], [104, 107]], [[28, 71]], [[28, 71]], [[2, 17]], [[81, 101]], [[19, 22]], [[25, 71]], [[2, 17]], [[81, 101]], [[2, 22]], [[19, 75]], [[77, 101]]]", "query_spans": "[[[104, 115]]]", "process": "Test analysis: Since the focus of the parabola y^{2}=20x is F(5,0), it follows that for the hyperbola, c=5. The hyperbola passes through the point (\\frac{15}{4},3). Thus, we have \\frac{15^{2}}{16a^{2}}-\\frac{3^{2}}{b^{2}}=1. Solving gives a=3, b=4. Therefore, the asymptotes of the hyperbola are y=\\pm\\frac{4}{3}x." }, { "text": "Given that point $P$ lies on the parabola $y^{2}=16x$, and the distance from point $P$ to the $y$-axis is $6$, then what is the distance from point $P$ to the focus?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 16*x);PointOnCurve(P, G);Distance(P, yAxis) = 6", "query_expressions": "Distance(P,Focus(G))", "answer_expressions": "10", "fact_spans": "[[[7, 22]], [[2, 6], [26, 30], [44, 48]], [[7, 22]], [[2, 23]], [[26, 41]]]", "query_spans": "[[[7, 56]]]", "process": "First find the coordinates of the focus, then use the first definition of the parabola to solve. As shown in the figure, from y^{2};=16x, the focus coordinates are (4,0), so the directrix of the parabola is x=-4, |PB|=6, then |PF|=|PB|+\\frac{P}{2}=6+4=10" }, { "text": "The difference between the distance from a moving point $P(x, y)$ $(x \\geq 0)$ to the point $F(1,0)$ and the distance from point $P$ to the $y$-axis is $1$. Then, the equation of the trajectory of point $P$ is?", "fact_expressions": "P: Point;F: Point;x1:Number;y1:Number;x1>=0;Coordinate(P) = (x1, y1);Coordinate(F) = (1, 0);Distance(P,F)-Distance(P,yAxis)=1", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[2, 21], [35, 39], [54, 58]], [[22, 31]], [[2, 21]], [[2, 21]], [[2, 21]], [[2, 21]], [[22, 31]], [[2, 52]]]", "query_spans": "[[[54, 65]]]", "process": "Problem Analysis: For a moving point P(x, y) (x ≥ 0), the difference between the distance from P to the point F(1, 0) and the distance from P to the y-axis is 1. Therefore, we have √((x−1)² + y²) = x + 1. Simplifying and rearranging yields y² = 4x;" }, { "text": "The equation of the directrix of the parabola $y^{2}=4 x$ is? The coordinates of the focus are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x)", "query_expressions": "Expression(Directrix(G));Coordinate(Focus(G))", "answer_expressions": "x=-1\n(1,0)", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]], [[0, 27]]]", "process": "" }, { "text": "The distance from the right focus to the left directrix of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$ is $\\frac{5}{2}$, then $b$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Number;b>0;Distance(RightFocus(G), LeftDirectrix(G)) = 5/2", "query_expressions": "b", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 37]], [[0, 37]], [[64, 67]], [[3, 37]], [[0, 62]]]", "query_spans": "[[[64, 69]]]", "process": "Solve using the focus and directrix of the hyperbola. From $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, the right focus is $F(c,0)$, and the left directrix is $x=-\\frac{a^{2}}{c}=-\\frac{1}{c}$. Therefore, $c+\\frac{1}{c}=\\frac{5}{2}$, solving gives $c=2$ or $c=\\frac{1}{2}$ (discarded). Thus, $b=\\sqrt{c^{2}-a^{2}}=\\sqrt{3}$." }, { "text": "Given the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ has its right focus at $F$, its upper vertex at $A$, point $P$ lies on the circle $x^{2}+y^{2}=8$, and point $Q$ lies on the ellipse, then the minimum value of $2|P A|+|P Q|-|Q F|$ is?", "fact_expressions": "G: Ellipse;H: Circle;P: Point;A: Point;Q: Point;F: Point;Expression(G) = (x^2/6 + y^2/2 = 1);Expression(H) = (x^2 + y^2 = 8);RightFocus(G) = F;UpperVertex(G) = A;PointOnCurve(P, H);PointOnCurve(Q, G)", "query_expressions": "Min(2*Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, Q)) - Abs(LineSegmentOf(Q, F)))", "answer_expressions": "6-2*sqrt(6)", "fact_spans": "[[[2, 39], [84, 86]], [[61, 77]], [[56, 60]], [[52, 55]], [[79, 83]], [[44, 47]], [[2, 39]], [[61, 77]], [[2, 47]], [[2, 55]], [[56, 78]], [[79, 87]]]", "query_spans": "[[[89, 115]]]", "process": "Find the ellipse's a, b, c, then obtain the coordinates of foci and vertices. Let P(2\\sqrt{2}\\cos\\theta,2\\sqrt{2}\\sin\\theta). From the distance formula between two points, 2|PA|=|PB|, which is the distance from point P to B(0,4\\sqrt{2}). By the definition of the ellipse, 2|PA|+|PQ|-|QF|=|PB|+|PQ|+|QF_{2}|-2\\sqrt{6}. When four points are collinear, the extremum is obtained. For the required ellipse \\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1, we have a=\\sqrt{6}, b=\\sqrt{2}, c=2, the right focus is F(2,0), the left focus is F_{2}(-2,0), the upper vertex is A(0,\\sqrt{2}), P(2\\sqrt{2}\\cos\\theta,2\\sqrt{2}\\sin\\theta), =2\\sqrt{10-8\\sin\\theta}=\\sqrt{40-32\\sin\\theta}-4\\sqrt{2}^{2} the distance from point P to B(0,4\\sqrt{2}). By the definition of the ellipse, -|QF|=|QF_{2}|-2a=|QF_{2}|-2\\sqrt{6}, \\frac{2|PA|+|PQ|-|QF|}{=\\sqrt{(0+2)^{2}+(4\\sqrt{2})^{2}}-2\\sqrt{6}}+|PQ|+|QF_{2}|-2\\sqrt{6}\\geqslant|BF_{2}|-2\\sqrt{6}. The equality holds if and only if B, P, Q, F_{2} are collinear. Hence, the minimum value of 2|PA|+|PQ|-|QF| is 6-2\\sqrt{6}." }, { "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{2} + y^{2} = 1 $. Point $ P $ is a moving point on the ellipse $ C $, satisfying $ \\overrightarrow{O P} \\cdot \\overrightarrow{P F_{2}} \\geq -1 $ ($ O $ is the coordinate origin, $ F_{2} $ is the right focus of the ellipse). What is the range of the horizontal coordinate of point $ P $?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);P: Point;PointOnCurve(P,C) = True;DotProduct(VectorOf(O,P),VectorOf(P,F2)) >= -1;O: Origin;F2: Point;RightFocus(C) = F2", "query_expressions": "Range(XCoordinate(P))", "answer_expressions": "[0,sqrt(2)]", "fact_spans": "[[[2, 34], [41, 46], [133, 135]], [[2, 34]], [[36, 40], [142, 146]], [[36, 52]], [[55, 115]], [[116, 119]], [[125, 132]], [[125, 139]]]", "query_spans": "[[[142, 157]]]", "process": "Let point P(x,y). According to \\overrightarrow{OP}\\cdot\\overrightarrow{PF_{2}}\\geqslant-1, obtain the relationship between x and y, substitute into the ellipse equation, derive an inequality in terms of x, solve for the range of x, and combine with the range of x-coordinates of points on the ellipse to obtain the answer. Ellipse C: \\frac{x^{2}}{2}+y^{2}=1, F_{2} is the right focus of the ellipse, so F_{2}(1,0). Let point P(x,y), so \\overrightarrow{OP}=(x,y), \\overrightarrow{PF_{2}}=(1-x,-y). From \\overrightarrow{OP}\\cdot\\overrightarrow{PF_{2}}\\geqslant-1, we get x-x^{2}-y^{2}\\geqslant-1. Since P(x,y) lies on the ellipse C: \\frac{x^{2}}{2}+y^{2}=1, we have y^{2}=1-\\frac{x^{2}}{2}, so x-x^{2}-1+\\frac{x^{2}}{2}\\geqslant-1, solving gives 0\\leqslant x\\leqslant2. Because P(x,y) lies on the ellipse C: \\frac{x^{2}}{2}+y^{2}=1, we have -\\sqrt{2}\\leqslant x\\leqslant\\sqrt{2}. Therefore, the range of the x-coordinate of point P is [0,\\sqrt{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, and one of its asymptotes is given by $y=x$. Find $b=$? If point $P( \\sqrt{3} , y_{0})$ lies on the hyperbola, then $\\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}}$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2/b^2 = 1);b: Number;b>0;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Expression(OneOf(Asymptote(G))) = (y = x);y0: Number;P: Point;Coordinate(P) = (sqrt(3), y0);PointOnCurve(P, G)", "query_expressions": "b;DotProduct(VectorOf(P, F1), VectorOf(P, F2))", "answer_expressions": "sqrt(2)\n0", "fact_spans": "[[[2, 49], [120, 123], [74, 75]], [[2, 49]], [[90, 93]], [[5, 49]], [[58, 65]], [[66, 73]], [[2, 73]], [[2, 73]], [[74, 88]], [[97, 119]], [[96, 119]], [[96, 119]], [[96, 124]]]", "query_spans": "[[[90, 95]], [[126, 183]]]", "process": "" }, { "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 line containing this chord is $2$, then $p$=?", "fact_expressions": "G: Parabola;p: Number;H: LineSegment;P: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(P) = (3, 1);IsChordOf(H,G);MidPoint(H)=P;Slope(OverlappingLine(H))=2", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[12, 33]], [[58, 61]], [], [[1, 11]], [[15, 33]], [[12, 33]], [[1, 11]], [[12, 37]], [[1, 40]], [[12, 56]]]", "query_spans": "[[[58, 63]]]", "process": "" }, { "text": "The hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ $(m>0)$ has eccentricity $\\sqrt{5}$, then $m=$?", "fact_expressions": "G: Hyperbola;m: Number;m>0;Expression(G) = (x^2 - y^2/m = 1);Eccentricity(G) = sqrt(5)", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[0, 33]], [[50, 53]], [[3, 33]], [[0, 33]], [[0, 48]]]", "query_spans": "[[[50, 55]]]", "process": "Given that a=1, so e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{1+m}=\\sqrt{5}, solving yields m=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 $\\left(\\frac{3 a^{2}}{p}, \\frac{2 b^{2}}{p}\\right)$, then the equations of the asymptotes of this hyperbola are?", "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, 22]], [[2, 22]], [[5, 22]], [[5, 22]], [[24, 27]], [[2, 27]], [[30, 76], [82, 85], [130, 133]], [[30, 76]], [[33, 76]], [[33, 76]], [[24, 80]], [[86, 127]], [[86, 127]], [[82, 127]]]", "query_spans": "[[[130, 141]]]", "process": "" }, { "text": "Given that the ellipse $\\frac{x^{2}}{3 m^{2}}+\\frac{y^{2}}{5 n^{2}}=1$ and the hyperbola $\\frac{x^{2}}{2 m^{2}}-\\frac{y^{2}}{3 n^{2}}=1$ have common foci, what is the equation of the asymptotes of the hyperbola?", "fact_expressions": "H: Ellipse;Expression(H) = (y^2/(5*n^2) + x^2/(3*m^2) = 1);m: Number;n: Number;G: Hyperbola;Expression(G) = (-y^2/(3*n^2) + x^2/(2*m^2) = 1);Focus(H) = Focus(G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(3)/4)*x", "fact_spans": "[[[2, 51]], [[2, 51]], [[4, 51]], [[4, 51]], [[52, 102], [110, 113]], [[52, 102]], [[2, 107]]]", "query_spans": "[[[110, 121]]]", "process": "" }, { "text": "Let the focus of the parabola $y^{2}=4x$ be $F$. A line passing through the point $(2,0)$ intersects the parabola at points $A$ and $B$, and intersects the directrix of the parabola at point $C$. If $\\frac{S_{\\triangle A C F}}{S_{\\triangle B C F}}=\\frac{2}{5}$, then $AF=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;K: Point;Coordinate(K) = (2, 0);H: Line;PointOnCurve(K, H);A: Point;B: Point;Intersection(H, G) = {A, B};C: Point;Intersection(H, Directrix(G)) = C;Area(TriangleOf(A, C, F))/Area(TriangleOf(B, C, F)) = 2/5", "query_expressions": "LineSegmentOf(A, F)", "answer_expressions": "2", "fact_spans": "[[[1, 15], [36, 39], [51, 54]], [[1, 15]], [[19, 22]], [[1, 22]], [[24, 32]], [[24, 32]], [[33, 35]], [[23, 35]], [[40, 43]], [[44, 47]], [[33, 49]], [[58, 62]], [[33, 62]], [[64, 125]]]", "query_spans": "[[[127, 133]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), AB: y = k(x - 2). Then from \\frac{S_{AACF}}{S_{ABCF}} = \\frac{2}{5} we get \\frac{x_{1}+1}{x_{2}+1} = \\frac{2}{5}, \\therefore x_{2} = \\frac{5}{2}x_{1} + \\frac{3}{2}. From y = k(x - 2), y^{2} = 4x, we obtain k^{2}x^{2} - (4k^{2} + 4)x + 4k^{2} = 0, so x_{1}x_{2} = 4, x_{1}(\\frac{5}{2}x_{1} + \\frac{3}{2}) = 4, \\therefore x_{1} = 1 (discarding the negative value), therefore |AF| = 2." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right vertex and right focus are denoted as $A$ and $F$, respectively. The intersection point of its left directrix with the $x$-axis is $B$. If $A$ is the midpoint of segment $BF$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "B: Point;F: Point;C: Hyperbola;a: Number;b: Number;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;RightFocus(C)=F;Intersection(LeftDirectrix(C),xAxis)=B;MidPoint(LineSegmentOf(B, F)) = A", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)+1", "fact_spans": "[[[98, 101]], [[79, 82]], [[2, 63], [83, 84], [118, 124]], [[10, 63]], [[10, 63]], [[74, 77], [103, 106]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 82]], [[2, 82]], [[83, 101]], [[103, 116]]]", "query_spans": "[[[118, 130]]]", "process": "" }, { "text": "The right focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $F$, and the intersection point of its right directrix with the $x$-axis is $A$. If there exists a point $P$ on the ellipse such that the perpendicular bisector of segment $AP$ passes through point $F$, then what is the range of values for the eccentricity of the ellipse?", "fact_expressions": "A: Point;P: Point;G: Ellipse;b: Number;a: Number;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;Intersection(RightDirectrix(G), xAxis) = A;PointOnCurve(P,G);PointOnCurve(F,PerpendicularBisector(LineSegmentOf(A,P)))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[1/2,1)", "fact_spans": "[[[73, 76]], [[83, 87]], [[0, 52], [60, 61], [78, 80], [108, 110]], [[2, 52]], [[2, 52]], [[56, 59], [102, 106]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 59]], [[60, 76]], [[78, 87]], [[91, 106]]]", "query_spans": "[[[108, 120]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{m+3}+\\frac{y^{2}}{4}=1$ has eccentricity $e=\\frac{1}{3}$, then the value of $m$ is equal to?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(m + 3) + y^2/4 = 1);Eccentricity(G) = e ;e=1/3;e:Number", "query_expressions": "m", "answer_expressions": "{3/2,5/9}", "fact_spans": "[[[2, 41]], [[62, 65]], [[2, 41]], [[2, 60]], [[45, 60]], [[45, 60]]]", "query_spans": "[[[62, 70]]]", "process": "Classify and solve according to the position of the focus. When the focus is on the x-axis, a=\\sqrt{m+3}, b=2, c=\\sqrt{m-1}, e=\\frac{c}{a}=\\frac{\\sqrt{m-1}}{\\frac{m+3}{m}}=\\frac{1}{3}, solving gives m=\\frac{3}{2}, solving gives m=\\frac{5}{9}, \\therefore m=\\frac{3}{2} or m=\\frac{5}{9}." }, { "text": "$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)$, respectively. The eccentricity of the hyperbola $C$ is $e=\\frac{2 \\sqrt{3}}{3}$. Point $P$ lies on the right branch of the hyperbola $C$ and satisfies $\\overrightarrow{P F_{2}} \\cdot \\overrightarrow{F_{1} F_{2}}=0$ and $|\\overrightarrow{P F_{2}}|=\\frac{2 \\sqrt{3}}{3}$. What is the length of the imaginary axis of the hyperbola?", "fact_expressions": "F1:Point;F2:Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;b: Number;a: Number;a>0;b>0;e:Number;Eccentricity(C)=e;P: Point;PointOnCurve(P,RightPart(C));DotProduct(VectorOf(P,F2),VectorOf(F1,F2)) =0;Abs(VectorOf(P, F2)) = (2*sqrt(3))/3;e=2*sqrt(3)/3", "query_expressions": "Length(ImageinaryAxis(C))", "answer_expressions": "4", "fact_spans": "[[[0, 7]], [[10, 17]], [[20, 81], [88, 94], [128, 134], [258, 261]], [[20, 81]], [[0, 87]], [[0, 87]], [[27, 81]], [[27, 81]], [[27, 81]], [[27, 81]], [[98, 122]], [[88, 122]], [[123, 127]], [[123, 138]], [[141, 206]], [[207, 256]], [[98, 122]]]", "query_spans": "[[[258, 267]]]", "process": "From the dot product formula, we get PF_{2}\\bot F_{1}F_{2}; combining with the eccentricity formula, latus rectum, and c^{2}=a^{2}+b^{2}, solve simultaneously to find the length of the conjugate axis of the hyperbola. \\because\\overrightarrow{PF}\\cdot\\overrightarrow{F_{1}F_{2}}=0,\\therefore PF_{2}\\bot F_{1}F_{2} \\begin{matrix} e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3} \\\\ |\\overrightarrow{PF_{2}}|=\\frac{b^{2}}{a}=\\frac{2\\sqrt{3}}{3} \\end{matrix} \\Rightarrow a=2\\sqrt{3}, then the length of the conjugate axis is 4" }, { "text": "Given that points $M$ and $N$ are two distinct points on the parabola $y = 4x^2$, $F$ is the focus of the parabola, and $\\angle MFN = 135^\\circ$. Let $P$ be the midpoint of chord $MN$, and let $d$ denote the distance from $P$ to the line $l$: $y = -\\frac{1}{16}$. If $|MN|^2 = \\lambda \\cdot d^2$, then the minimum value of $\\lambda$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = 4*x^2);M: Point;N: Point;PointOnCurve(M, G);PointOnCurve(N, G);Negation(M=N);F: Point;Focus(G) = F;AngleOf(M, F, N) = ApplyUnit(135, degree);IsChordOf(LineSegmentOf(M, N), G);P: Point;MidPoint(LineSegmentOf(M, N)) = P;l: Line;Expression(l) = (y=-1/16);d: Number;Distance(P, l) = d;lambda: Number;Abs(LineSegmentOf(M, N))^2 = (d^2)*lambda", "query_expressions": "Min(lambda)", "answer_expressions": "2+sqrt(2)", "fact_spans": "[[[11, 25], [36, 39]], [[11, 25]], [[2, 6]], [[7, 10]], [[2, 31]], [[2, 31]], [[2, 31]], [[32, 35]], [[32, 42]], [[43, 72]], [[36, 79]], [[82, 85]], [[74, 85]], [[86, 110]], [[86, 110]], [[115, 118]], [[82, 118]], [[153, 162]], [[120, 151]]]", "query_spans": "[[[153, 168]]]", "process": "As shown in the figure: $ l: y = -\\frac{1}{16} $ is the directrix of the parabola. Draw $ MA \\perp l $ at $ A $, $ PQ \\perp l $ at $ Q $, $ NB \\perp l $ at $ B $. Then $ MA + NB = 2PQ = 2d \\geqslant 2\\sqrt{MA \\cdot NB} $, hence $ MA \\cdot NB \\leqslant d^{2} $, with equality when $ MA = NB = d $. By the law of cosines: $ MN^{2} = MF^{2} + NF^{2} - 2MF \\cdot NF \\cos 135^{\\circ} = (MA + MB)^{2} - (2 - \\sqrt{2})MA \\cdot MB $. Hence $ MN^{2} \\geqslant 4d^{2} - (2 - \\sqrt{2})d^{2} = (2 + \\sqrt{2})d^{2} $, therefore $ \\lambda \\geqslant 2 + \\sqrt{2} $." }, { "text": "Given the line $ l $: $ y=1 $ intersects the $ y $-axis at point $ M $, and $ Q $ is a moving point on line $ l $ distinct from point $ M $. Denote the horizontal coordinate of point $ Q $ as $ n $. If there exists a point $ N $ on the curve $ C $: $ \\frac{x^{2}}{2}+y^{2}=1 $ such that $ \\angle M Q N = 45^{\\circ} $, then what is the range of values of $ n $? (Express in interval notation)", "fact_expressions": "l: Line;C: Curve;M: Point;Q: Point;N: Point;n:Number;Expression(C) = (x^2/2 + y^2 = 1);Expression(l)=(y=1);Intersection(l,yAxis)=M;PointOnCurve(Q,l);Negation(Q=M);XCoordinate(MidPoint(Q)) = n;PointOnCurve(N, C);AngleOf(M, Q, N) = ApplyUnit(45, degree)", "query_expressions": "Range(n)", "answer_expressions": "{[-1-sqrt(3),0)+(0,1+sqrt(3)]}", "fact_spans": "[[[2, 14], [30, 35]], [[61, 93]], [[21, 25], [38, 42]], [[26, 29], [47, 51]], [[96, 100]], [[56, 59], [130, 133]], [[61, 93]], [[2, 14]], [[2, 25]], [[26, 36]], [[26, 45]], [[47, 59]], [[61, 100]], [[103, 128]]]", "query_spans": "[[[130, 140]]]", "process": "According to the problem, Q(n,1), n\\neq0, let k_{QN}=k, then l_{QN}:y=k(x-n)+1. When Q is in the first quadrant, k=1. When l_{QN} is tangent to the ellipse, n reaches its maximum value. Solving simultaneously:\n\\begin{cases}x^{2}+2y^{2}-2=0\\\\y=(x-n)+1\\end{cases},\nwe get 3x^{2}+4(1-n)x+2(n^{2}-2n)=0. Let A=0, then n=1\\pm\\sqrt{3}; n=1-\\sqrt{3} does not satisfy the condition and is discarded. When Q is in the second quadrant, k=-1. When l_{QN} is tangent to the ellipse, n reaches its minimum value. Solving simultaneously:\n\\begin{cases}x^{2}+2y^{2}-2=0\\\\y=-(x-n)+1\\end{cases},\nwe get 3x^{2}-4(1+n)x+2(n^{2}+2n)=0. Let d=0, then n=-1\\pm\\sqrt{3}; n=-1+\\sqrt{3} does not satisfy the condition and is discarded. In conclusion, n\\in[-1-\\sqrt{3},0)\\cup(0,1+\\sqrt{3}]." }, { "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 $M$ on the right branch of the hyperbola such that $(\\overrightarrow{O M}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{F_{2} M}=0$ (where $O$ is the origin), and $|\\overrightarrow{M F_{1}}|=\\sqrt{3}|\\overrightarrow{M F_{2}}|$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Origin;M: 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(M, RightPart(G));DotProduct((VectorOf(O, F2) + VectorOf(O, M)),VectorOf(F2, M)) = 0;Abs(VectorOf(M, F1)) = sqrt(3)*Abs(VectorOf(M, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3) + 1", "fact_spans": "[[[20, 76], [84, 87], [256, 259]], [[23, 76]], [[23, 76]], [[180, 183]], [[95, 98]], [[10, 17]], [[2, 9]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 81]], [[2, 81]], [[83, 98]], [[101, 177]], [[191, 254]]]", "query_spans": "[[[256, 264]]]", "process": "Let C be the midpoint of MF₂. Since (OM + OF₂) · F₂M = 0, it follows that 2OC · F₂M = 0, which implies OC ⊥ MF₂, hence OM = OF₂. Since OC // F₁M, we have F₁M ⊥ MF₂. Given |MF₁| = √3 |MF₂|, then |MF₁| − |MF₂| = √3 |MF₂| − |MF₂| = 2a, so |MF₂| = 2a/(√3−1) = (√3+1)a, and |MF₁| = √3 |MF₂| = √3(√3+1)a. Since |MF₁|² + |MF₂|² = 4c², it follows that 4(√3+1)²a² = 4c², thus (√3+1)²a² = c², so (√3+1)a = c. Therefore, the eccentricity is e = c/a = √3+1." }, { "text": "Given that the line $a x + b y + c - 1 = 0$ $(b c > 0)$ passes through the center of the circle $x^{2} + y^{2} - 2 y - 15 = 0$, then the minimum value of $\\frac{1}{b} + \\frac{9}{c}$ is?", "fact_expressions": "H: Line;b: Number;a: Number;c: Number;b*c>0;Expression(H) = (a*x + b*y + c - 1 = 0);G: Circle;Expression(G) = (-2*y + x^2 + y^2 - 15 = 0);PointOnCurve(Center(G),H) = True", "query_expressions": "Min(1/b + 9/c)", "answer_expressions": "16", "fact_spans": "[[[2, 28]], [[4, 28]], [[4, 28]], [[4, 28]], [[4, 28]], [[2, 28]], [[30, 53]], [[30, 53]], [[2, 56]]]", "query_spans": "[[[58, 89]]]", "process": "The circle $ x^{2} + y^{2} - 2y - 15 = 0 $ is converted into standard form, yielding $ x^{2} + (y - 1)^{2} = 16 $. Therefore, the center of the circle $ x^{2} + y^{2} - 2y - 15 = 0 $ is $ C(0,1) $, and the radius $ r = 4 $. Since the line $ ax + by + c - 1 = 0 $ passes through the center $ C $, we have $ a \\times 0 + b \\times 1 + c - 1 = 0 $, that is, $ b + c = 1 $. Therefore, $ \\frac{1}{b} + \\frac{9}{c} = (b + c)\\left( \\frac{1}{b} + \\frac{9}{c} \\right) = 10 + \\frac{9b}{c} + \\frac{c}{b} \\geqslant 10 + 2\\sqrt{\\frac{9b}{c} \\cdot \\frac{c}{b}} = 16 $. It follows that when $ 3b = c $, the minimum value of $ \\frac{1}{b} + \\frac{9}{c} $ is 16." }, { "text": "The curve $y=\\sqrt{1-\\frac{x^{2}}{4}}$ and the line $y=2 x+m$ have exactly one common point; then the range of real values for $m$ is?", "fact_expressions": "H: Curve;G: Line;m: Real;Expression(H) = (y = sqrt(1 - x^2/4));Expression(G) = (y = m + 2*x);NumIntersection(H, G) = 1", "query_expressions": "Range(m)", "answer_expressions": "{(-4,4) , sqrt(17)}", "fact_spans": "[[[0, 30]], [[31, 42]], [[53, 58]], [[0, 30]], [[31, 42]], [[0, 51]]]", "query_spans": "[[[53, 65]]]", "process": "The curve $ y = \\sqrt{1 - \\frac{x^2}{4}} $ is the same as the curve $ \\frac{x^2}{4} + y^2 = 1 $ ($ y \\geqslant 0 $), which represents the semi-ellipse shown in the figure. If the line is tangent to the ellipse, then $ A = 16(17 - m^2) = 0 $, yielding $ m = \\pm\\sqrt{17} $. If the line passes through $ (2, 0) $, then $ m = -4 $; if the line passes through $ (-2, 0) $, then $ m = 4 $. Combining with the graph, it is known that $ m = \\sqrt{17} $ or $ -4 \\leqslant m < 4 $. The answer is: $ m = \\sqrt{17} $ or $ -4 \\leqslant m < 4 $." }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is $2$, and one of its foci coincides with the focus of the parabola $y^{2}=8x$, what are the coordinates of the foci of the hyperbola? What is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Eccentricity(G) = 2;H: Parabola;Expression(H) = (y^2 = 8*x);OneOf(Focus(G)) = Focus(H)", "query_expressions": "Coordinate(Focus(G));Expression(Asymptote(G))", "answer_expressions": "(pm*2, 0)\ny=pm*sqrt(3)*x", "fact_spans": "[[[2, 48], [58, 59], [87, 90]], [[2, 48]], [[5, 48]], [[5, 48]], [[2, 57]], [[65, 79]], [[65, 79]], [[58, 84]]]", "query_spans": "[[[87, 97]], [[87, 104]]]", "process": "" }, { "text": "The equation of a hyperbola centered at the origin, with one focus at $(-5 , 0)$ and one asymptote being the line $4 x-3 y=0$, is?", "fact_expressions": "G: Hyperbola;O:Origin;Expression(OneOf(Asymptote(G))) = (4*x - 3*y = 0);Coordinate(OneOf(Focus(G))) = (-5, 0);Center(G)=O", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9-y^2/16=1", "fact_spans": "[[[42, 45]], [[3, 5]], [[22, 45]], [[6, 45]], [[0, 45]]]", "query_spans": "[[[42, 49]]]", "process": "" }, { "text": "If the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=k^{2}$ and the circle $x^{2}+y^{2}=1$ have common points, then the range of real values for $k$ is?", "fact_expressions": "G: Hyperbola;k: Real;H: Circle;Expression(G) = (x^2/9 - y^2/4 = k^2);Expression(H) = (x^2 + y^2 = 1);IsIntersect(G, H)", "query_expressions": "Range(k)", "answer_expressions": "[-1/3, 0)+(0, 1/3]", "fact_spans": "[[[1, 43]], [[66, 71]], [[44, 60]], [[1, 43]], [[44, 60]], [[1, 64]]]", "query_spans": "[[[66, 78]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{64}+\\frac{y^{2}}{28}=1$ with left focus $F_{1}$, a point $P$ on the ellipse is at a distance of $6$ from the left focus. Point $M$ is the midpoint of segment $P F_{1}$, and $O$ is the origin. Then $|O M|$=?", "fact_expressions": "C: Ellipse;P: Point;F1: Point;O: Origin;M: Point;Expression(C) = (x^2/64 + y^2/28 = 1);LeftFocus(C) = F1;PointOnCurve(P, C);Distance(P,LeftFocus(C))=6;MidPoint(LineSegmentOf(P,F1)) = M", "query_expressions": "Abs(LineSegmentOf(O, M))", "answer_expressions": "5", "fact_spans": "[[[2, 46], [59, 64]], [[68, 71]], [[51, 58]], [[103, 106]], [[83, 87]], [[2, 46]], [[2, 58]], [[59, 71]], [[59, 82]], [[83, 102]]]", "query_spans": "[[[113, 122]]]", "process": "" }, { "text": "Let the common foci of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$ be $F_{1}$, $F_{2}$, and let $P$ be an intersection point of the two curves, \nthen $\\angle F_{1} P F_{2}$=?", "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/2 - y^2/2 = 1);Expression(H) = (x^2/6 + y^2/2 = 1);OneOf(Intersection(G,H))=P;Focus(G)={F1,F2};Focus(H)={F1,F2}", "query_expressions": "AngleOf(F1, P, F2)", "answer_expressions": "ApplyUnit(90,degree)", "fact_spans": "[[[39, 77]], [[1, 38]], [[83, 90]], [[99, 102]], [[91, 98]], [[39, 77]], [[1, 38]], [[99, 111]], [[1, 98]], [[1, 98]]]", "query_spans": "[[[114, 138]]]", "process": "" }, { "text": "Given that the circle $x^{2}+y^{2}=R^{2}$ and the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ have no common points, find the range of values for $R$.", "fact_expressions": "H: Circle;Expression(H) = (x^2 + y^2 = R^2);R: Number;G: Hyperbola;Expression(G) = (x^2/4 - y^2/9 = 1);NumIntersection(H, G) = 0", "query_expressions": "Range(R)", "answer_expressions": "(-2, 0)+(0, 2)", "fact_spans": "[[[2, 22]], [[2, 22]], [[67, 70]], [[23, 61]], [[23, 61]], [[2, 65]]]", "query_spans": "[[[67, 76]]]", "process": "" }, { "text": "The equation of the directrix of the parabola $y^{2}=12 x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-3", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "The parabola y^{2}=12x, \\therefore p=6, \\therefore the equation of the directrix is x=-3" }, { "text": "From the left focus $F_{1}$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, draw a tangent to the circle $x^{2}+y^{2}=4$, intersecting the right branch of the hyperbola at point $P$, with tangent point $T$. Let $M$ be the midpoint of $P F_{1}$. Then $|M O| -| M T |$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);F1: Point;LeftFocus(G) = F1;H: Circle;Expression(H) = (x^2 + y^2 = 4);L1: Line;TangentOfPoint(F1, H) = L1;Intersection(L1, RightPart(G)) = P;P: Point;TangentPoint(L1, H) = T;T: Point;MidPoint(LineSegmentOf(P, F1)) = M;M: Point;O: Origin", "query_expressions": "Abs(LineSegmentOf(M, O)) - Abs(LineSegmentOf(M, T))", "answer_expressions": "sqrt(5)-2", "fact_spans": "[[[1, 39], [72, 75]], [[1, 39]], [[43, 50]], [[1, 50]], [[52, 68]], [[52, 68]], [], [[0, 71]], [[0, 82]], [[78, 82]], [[0, 89]], [[86, 89]], [[92, 108]], [[105, 108]], [[110, 126]]]", "query_spans": "[[[110, 128]]]", "process": "Test analysis: From the given, MT = PT - PM = PF₁ - TF₁ - ½PF₁ = ½PF₁ - TF₁, then |MO| - |MT| =" }, { "text": "The equation of the directrix of the parabola $x^{2}=2 y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*y)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-1/2", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "Test analysis: Given that p=1, the equation of the directrix is y=-\\frac{1}{2}" }, { "text": "Given that 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)$, points $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola, respectively, and point $I$ is the incenter (the center of the incircle) of triangle $\\triangle P F_{1} F_{2}$. If the inequality $S_{\\Delta I P F_{1}}-S_{\\Delta I P F_{2}} \\leq \\frac{\\sqrt{2}}{2} S_{\\Delta I F_{1} F_{2}}$ always holds, 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;I: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, RightPart(G));LeftFocus(G) = F1;RightFocus(G) = F2;Incenter(TriangleOf(P, F1, F2)) = I;Center(InscribedCircle(TriangleOf(P, F1, F2))) = I;Area(TriangleOf(I, P, F1)) - Area(TriangleOf(I, P, F2)) <= (sqrt(2)/2)*Area(TriangleOf(I, F1, F2))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(2), +oo)", "fact_spans": "[[[7, 63], [88, 91], [239, 242]], [[10, 63]], [[10, 63]], [[2, 6]], [[69, 77]], [[78, 85]], [[97, 101]], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 68]], [[69, 96]], [[69, 96]], [[97, 131]], [[97, 142]], [[146, 235]]]", "query_spans": "[[[239, 252]]]", "process": "Let the inradius of $\\triangle PF_{1}F_{2}$ be $r$. By the definition of the hyperbola, we have $|PF_{1}|-|PF_{2}|=2a$, $|F_{1}F_{2}|=2c$, then $S_{\\Delta IPF_{1}}=\\frac{1}{2}|PF_{1}|r$, $S_{\\Delta IPF_{2}}=\\frac{1}{2}|PF_{2}|r$, $S_{\\Delta IF_{1}F_{2}}=\\frac{1}{2}\\cdot2c\\cdot r$. Since $S_{\\triangle IPF_{1}}-S_{\\Delta IPF_{2}}\\leqslant\\frac{\\sqrt{2}}{2}S_{\\Delta IF_{1}F_{2}}$, it follows that $\\frac{1}{2}|PF_{1}|r-\\frac{1}{2}|PF_{2}|r\\leqslant\\frac{\\sqrt{2}}{2}\\cdot c\\cdot r$. Thus, $\\sqrt{2}c\\geqslant|PF_{1}|-|PF_{2}|=2a$, hence $e=\\frac{c}{a}\\geqslant\\sqrt{2}$." }, { "text": "If one focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ has coordinates $(0,2)$, then the real number $m$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/5 + y^2/m = 1);m: Real;Coordinate(OneOf(Focus(G))) = (0, 2)", "query_expressions": "m", "answer_expressions": "9", "fact_spans": "[[[1, 38]], [[1, 38]], [[55, 60]], [[1, 53]]]", "query_spans": "[[[57, 62]]]", "process": "From the given conditions, we have m > 5. For the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$, $a=\\sqrt{m}$, $b=\\sqrt{5}$, $c=\\sqrt{m-5}$, so $\\sqrt{m-5}=2$. Solving gives $m=9$." }, { "text": "If the asymptotes of a hyperbola are given by $y=\\pm \\frac{3}{2} x$, then an equation of a hyperbola satisfying this condition is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (y = pm*(3/2)*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2/9=1", "fact_spans": "[[[1, 4], [41, 44]], [[1, 32]]]", "query_spans": "[[[41, 49]]]", "process": "From the relationship between the standard equation of a hyperbola and its asymptotes, it follows that [detailed explanation] the hyperbola with asymptotes $ y = \\pm\\frac{3}{2}x $ has the equation $ \\left(\\frac{3}{2}x + y\\right)\\left(\\frac{3}{2}x - y\\right) = \\lambda $ ($ \\lambda \\neq 0 $), then $ \\frac{x^{2}}{4} - \\frac{y^{2}}{9} = 1 $ is one such hyperbola." }, { "text": "Given that the slope of a line $l$ with inclination angle $\\alpha$ is equal to the eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, then $\\sin (\\pi-2 \\alpha)$=?", "fact_expressions": "alpha: Number;Inclination(l) = alpha;l: Line;Slope(l) = Eccentricity(G);G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1)", "query_expressions": "Sin(-2*alpha + pi)", "answer_expressions": "4/5", "fact_spans": "[[[6, 14]], [[2, 20]], [[15, 20]], [[15, 57]], [[25, 53]], [[25, 53]]]", "query_spans": "[[[59, 82]]]", "process": "\\because the eccentricity of the hyperbola \\( x^{2} - \\frac{y^{2}}{3} = 1 \\) is \\( \\frac{c}{a} = 2 \\), \\( \\therefore \\tan\\alpha = 2 \\). Since \\( \\alpha \\) is the inclination angle of the line, \\( \\alpha \\in (0, \\pi) \\), \\( \\therefore \\sin\\alpha = \\frac{2\\sqrt{5}}{5} \\), \\( \\cos\\alpha = \\frac{\\sqrt{5}}{5} \\), \\( \\therefore \\sin(\\pi - 2\\alpha) = \\sin 2\\alpha = 2\\sin\\alpha\\cos\\alpha = \\frac{4}{5} \\)" }, { "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ such that the distance from $P$ to the left focus is $\\frac{5}{2}$, then what is the distance from $P$ to the right directrix?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G);Distance(P, LeftFocus(G)) = 5/2", "query_expressions": "Distance(P, RightDirectrix(G))", "answer_expressions": "3", "fact_spans": "[[[2, 39]], [[42, 45], [69, 70]], [[2, 39]], [[2, 45]], [[2, 66]]]", "query_spans": "[[[2, 79]]]", "process": "Problem Analysis: Let the distance from point P to the right directrix be d. By the first definition of the ellipse, the distance from point P to the right focus is 4 - \\frac{5}{2} = \\frac{3}{2}, the eccentricity is e = \\frac{1}{2}. Then, according to the second definition of the ellipse, \\frac{\\frac{3}{2}}{d} = e = \\frac{1}{2}, solving gives d = 3. Therefore, the distance from point P to the right directrix is 3." }, { "text": "If $M$ is a moving point on the ellipse $C_{1}$: $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$, and $N$ is a moving point on the ellipse $C_{2}$: $\\frac{x^{2}}{64}+\\frac{y^{2}}{36}=1$, then the maximum area of $\\Delta O M N$ is?", "fact_expressions": "C1: Ellipse;C2: Ellipse;O: Origin;M: Point;N: Point;Expression(C1) = (x^2/16+y^2/9=1);Expression(C2) = (x^2/64+y^2/36=1);PointOnCurve(M,C1);PointOnCurve(N,C2)", "query_expressions": "Max(Area(TriangleOf(O, M, N)))", "answer_expressions": "12", "fact_spans": "[[[6, 53]], [[62, 110]], [[117, 131]], [[2, 5]], [[58, 61]], [[6, 53]], [[62, 110]], [[2, 57]], [[58, 114]]]", "query_spans": "[[[117, 139]]]", "process": "Let points M(4\\cos\\alpha,3\\sin\\alpha) and N(8\\cos\\beta,6\\sin\\beta). It can be calculated that S_{\\triangleOMN}=12|\\sin(\\alpha-\\beta)|. Using the boundedness of the sine function, the result can be found. Let M(4\\cos\\alpha,3\\sin\\alpha), N(8\\cos\\beta,6\\sin\\beta)\\widehat{V}|^{2}\\cos^{2}\\angleMO=\\frac{1}{2}|24(\\sin\\alpha\\cos\\beta-\\cos\\alpha\\sin\\beta)|=12|\\sin(\\alpha-\\beta)|\\leqslant12" }, { "text": "The line passing through the right focus of the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ with slope $2$ intersects the right branch of $E$ at two distinct points. What is the range of values for the eccentricity of the hyperbola?", "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;G: Line;PointOnCurve(RightFocus(E), G);Slope(G) = 2;NumIntersection(G, RightPart(E)) = 2", "query_expressions": "Range(Eccentricity(E))", "answer_expressions": "(1,sqrt(5))", "fact_spans": "[[[1, 62], [78, 81], [95, 98]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[75, 77]], [[0, 77]], [[68, 77]], [[75, 93]]]", "query_spans": "[[[95, 108]]]", "process": "From the hyperbola and its asymptotes, it is known that the line intersects the right branch of the hyperbola at two distinct points if and only if 0<\\frac{b}{a}<2; thus 0<\\frac{b^{2}}{a^{2}}<4, 1<\\frac{c^{2}}{a^{2}}=1+\\frac{b^{2}}{a^{2}}<5, hence 10 , b>0)$ is $y=\\sqrt{3}x$, and one of its foci coincides with the focus of the parabola $y^{2}=16x$. Find the equation of the hyperbola?", "fact_expressions": "E: Hyperbola;a: Number;b: Number;a > 0;b > 0;Expression(E) = (x^2/a^2 - y^2/b^2 = 1);Expression(OneOf(Asymptote(E))) = (y = sqrt(3)*x);F: Parabola;Expression(F) = (y^2 = 16*x);OneOf(Focus(E)) = Focus(F)", "query_expressions": "Expression(E)", "answer_expressions": "x^2/4-y^2/12 = 1", "fact_spans": "[[[2, 59], [81, 82], [111, 114]], [[5, 59]], [[5, 59]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 80]], [[88, 103]], [[88, 103]], [[81, 108]]]", "query_spans": "[[[111, 119]]]", "process": "" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, two mutually perpendicular lines $l_{1}$ and $l_{2}$ are drawn through $F$. The line $l_{1}$ intersects $C$ at points $A$ and $B$, and the line $l_{2}$ intersects $C$ at points $D$ and $E$. Then the value of $\\frac{1}{|A B|}+\\frac{1}{|D E|}$ is?", "fact_expressions": "C: Parabola;l1:Line;l2:Line;A: Point;B: Point;D: Point;E: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F,l1);PointOnCurve(F,l2);IsPerpendicular(l1,l2);Intersection(l1,C)={A, B};Intersection(l2,C)={D, E}", "query_expressions": "1/Abs(LineSegmentOf(D, E)) + 1/Abs(LineSegmentOf(A, B))", "answer_expressions": "1/4", "fact_spans": "[[[6, 25], [71, 74], [96, 99]], [[41, 50], [61, 70]], [[53, 60], [86, 95]], [[76, 79]], [[80, 83]], [[101, 104]], [[105, 108]], [[2, 5], [30, 33]], [[6, 25]], [[2, 28]], [[29, 60]], [[29, 60]], [[36, 60]], [[61, 85]], [[86, 110]]]", "query_spans": "[[[112, 149]]]", "process": "From the given conditions, we know: F(1,0), and the slopes of l₁ and l₂ must exist. Let l₁: y = k₁(x−1), l₂: y = k₂(x−1). Let A(x₁,y₁), B(x₂,y₂), D(x₃,y₃), E(x₄,y₄). Combining with the parabola equation: y = k₁(x−1), we obtain k₁²x² − 2(k₁² + 2)x + k₁² = 0. Clearly, Δ = 8k₁² + 16 > 0, so x₁ + x₂ = (2(k₁² + 2))/k₁². Then by the definition of the parabola: \nAB = x₁ + x₂ + 2 = (4(k₁² + 1))/k₁². \nFor the system: \n{ y = k₂(x−1) \n{ y² = 4x \nwe obtain k₂²x² − 2(k₂² + 2)x + k₂² = 0. Clearly, Δ = 8k₂² + 16 > 0, so x₃ + x₄ = (2(k₂² + 2))/k₂². Then: \nDE = x₃ + x₄ + 2 = 4(k₂² + 1)/k₂² = 4(k₂² + 1)/k₂² = 4(k₁² + 1). \n1/|AB| + 1/|DE| = k₁²/(4(k²+1)) + k₂²/(4(k²+1)) = 1/(4) + 1/(4) = 1/4" }, { "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ $(m>n>0)$ has eccentricity $\\frac{1}{2}$, and one of its foci coincides with the focus of the parabola $y^{2}=4x$, then $mn=$?", "fact_expressions": "G: Parabola;H: Ellipse;m: Number;n: Number;Expression(G) = (y^2 = 4*x);m > n;n > 0;Expression(H) = (y^2/n + x^2/m = 1);Eccentricity(H) = 1/2;OneOf(Focus(H))=Focus(G)", "query_expressions": "m*n", "answer_expressions": "12", "fact_spans": "[[[70, 84]], [[1, 45]], [[3, 45]], [[3, 45]], [[70, 84]], [[3, 45]], [[3, 45]], [[1, 45]], [[1, 63]], [[1, 89]]]", "query_spans": "[[[91, 98]]]", "process": "Given the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{n}=1$ $(m>n>0)$ has eccentricity $\\frac{1}{2}$, we obtain $\\frac{m-n}{m}=\\frac{1}{4}$, so $4n=3m$. Since the focus of the parabola $y^{2}=4x$ is $(1,0)$, and the right focus of the ellipse is $(c,0)$, it follows that $c=1$, hence $m-n=1$. Solving gives $m=4$, $n=3$, so $mn=12$." }, { "text": "The parabola $y^{2}=2 p x(p>0)$ has focus $F$ and origin $O$. A line passing through the focus of the parabola and perpendicular to the $x$-axis intersects the parabola at point $P$. If $P O=3 \\sqrt{5}$, then the value of $p$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;O: Origin;H: Line;PointOnCurve(Focus(G), H);IsPerpendicular(H, xAxis);P: Point;Intersection(H, G) = P;LineSegmentOf(P, O) = 3*sqrt(5)", "query_expressions": "p", "answer_expressions": "6", "fact_spans": "[[[0, 21], [36, 39], [52, 55]], [[0, 21]], [[81, 84]], [[3, 21]], [[24, 27]], [[0, 27]], [[31, 34]], [[49, 51]], [[35, 51]], [[41, 51]], [[57, 61]], [[49, 61]], [[63, 79]]]", "query_spans": "[[[81, 88]]]", "process": "Analysis: First, according to the problem, find the focus coordinates of the parabola. Then substitute the x-coordinate into the parabola equation to find the y-coordinate of point P, thereby obtaining the coordinates of point P. Use the distance formula between two points to obtain the equation that p satisfies, and solve for the result. According to the problem, $ F\\left(\\frac{p}{2}, 0\\right) $. Substitute $ x = \\frac{p}{2} $ into the parabola equation to get $ y = \\pm p $. Thus, $ P\\left(\\frac{p}{2}, \\pm p\\right) $. Since $ PO = 3\\sqrt{5} $, we have $ \\frac{p^{2}}{4} + p^{2} = 45 $, solving which gives $ p = 6 $." }, { "text": "The asymptotes of the hyperbola $\\frac{y^{2}}{16}-\\frac{x^{2}}{4}=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": "" }, { "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $y=2x$, and the right directrix is $x=\\frac{\\sqrt{5}}{5}$, then the focal distance 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);Expression(OneOf(Asymptote(G))) = (y = 2*x);Expression(RightDirectrix(G)) = (x = sqrt(5)/5)", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 58], [106, 109]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 74]], [[2, 103]]]", "query_spans": "[[[106, 114]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{m}-\\frac{y^{2}}{4}=1$ has eccentricity $\\sqrt{5}$, find the length of the real axis of the hyperbola $C$.", "fact_expressions": "C: Hyperbola;m: Number;Expression(C) = (-y^2/4 + x^2/m = 1);Eccentricity(C) = sqrt(5)", "query_expressions": "Length(RealAxis(C))", "answer_expressions": "2", "fact_spans": "[[[2, 45], [63, 69]], [[10, 45]], [[2, 45]], [[2, 60]]]", "query_spans": "[[[63, 75]]]", "process": "According to the eccentricity formula, since the eccentricity is \\sqrt{5}, we have \\frac{\\sqrt{m+4}}{\\sqrt{m}} = \\sqrt{5}, which gives m=1, then the length of the real axis is 2." }, { "text": "Given a hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw two tangents from one of its foci $F$ to the circle $x^{2}+y^{2}=a^{2}$, with points of tangency $A$ and $B$. If $\\angle A F B = 120^{\\circ}$, 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;OneOf(Focus(C)) = F;G: Circle;Expression(G) = (x^2 + y^2 = a^2);TangentOfPoint(F, G) = {l1, l2};l1: Line;l2: Line;TangentPoint(l1,G) = A;TangentPoint(l2,G) = B;A: Point;B: Point;AngleOf(A, F, B) = ApplyUnit(120, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[1, 62], [139, 145]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[67, 70]], [[1, 70]], [[71, 91]], [[71, 91]], [[0, 96]], [], [], [[70, 109]], [[70, 109]], [[102, 105]], [[106, 109]], [[111, 137]]]", "query_spans": "[[[139, 151]]]", "process": "Without loss of generality, let F(c,0). Since A and B are the points of tangency from F to the circle x^{2}+y^{2}=a^{2}, and \\angleAFB=120^{\\circ}, it follows that \\angleAFO=60^{\\circ}. Given |OF|=c, |OA|=a, we have \\sin\\angleAFO=\\frac{|OA|}{|OF|}=\\frac{a}{c}=\\sin60^{\\circ}=\\frac{\\sqrt{3}}{2}. Therefore, \\frac{c}{a}=\\frac{2\\sqrt{3}}{3}, so the eccentricity is \\frac{2\\sqrt{3}}{3}." }, { "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": "F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;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,LeftPart(G)) = True;Abs(LineSegmentOf(P,F2))^2/Abs(LineSegmentOf(P,F1)) = 8*a", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,3]", "fact_spans": "[[[10, 17]], [[2, 9]], [[2, 82]], [[2, 82]], [[20, 76], [84, 87], [138, 141]], [[20, 76]], [[23, 76]], [[99, 136]], [[23, 76]], [[23, 76]], [[94, 97]], [[84, 97]], [[99, 136]]]", "query_spans": "[[[138, 152]]]", "process": "" }, { "text": "The standard equation of an ellipse with foci $(\\pm 2,0)$ and passing through the point $P\\left(\\frac{5}{2},-\\frac{3}{2}\\right)$ is?", "fact_expressions": "G: Ellipse;Coordinate(Focus(G)) = (pm*2, 0);P: Point;Coordinate(P) = (5/2, -3/2);PointOnCurve(P, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/10+y^2/6=1", "fact_spans": "[[[49, 51]], [[0, 51]], [[18, 48]], [[18, 48]], [[17, 51]]]", "query_spans": "[[[49, 58]]]", "process": "From the given conditions, we have c=2, the foci are on the x-axis, $\\left(\\frac{3}{2}\\right)^{2}+\\sqrt{\\left(\\frac{5}{2}+2\\right)^{2}+\\left(-\\frac{3}{2}\\right)^{2}}=2\\sqrt{10}$, $a=\\sqrt{10}$, $b=\\sqrt{6}$, therefore the equation of the ellipse is $\\frac{x^{2}}{10}+\\frac{y^{2}}{6}=1$" }, { "text": "Given that the ellipse $\\frac{x^{2}}{m}+y^{2}=1$ $(m>0)$ with foci on the $x$-axis has a focal distance of $2$, what is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;m>0;Expression(G) = (y^2 + x^2/m = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 2", "query_expressions": "m", "answer_expressions": "2", "fact_spans": "[[[11, 43]], [[52, 55]], [[13, 43]], [[11, 43]], [[2, 43]], [[11, 50]]]", "query_spans": "[[[52, 59]]]", "process": "From the given condition, c=1, so m-1=1, m=2." }, { "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ are $y=\\pm \\frac{3}{4} x$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*(3/4)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/4", "fact_spans": "[[[2, 48], [76, 79]], [[5, 48]], [[5, 48]], [[2, 48]], [[2, 74]]]", "query_spans": "[[[76, 85]]]", "process": "According to the problem: \\frac{b}{a}=\\frac{3}{4}, thus \\frac{c}{a}=\\frac{5}{4}," }, { "text": "Given that the minor axis of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has length $2$, the upper vertex is $A$, the left vertex is $B$, the left and right foci are $F_{1}$, $F_{2}$ respectively, and the area of $\\Delta F_{1} A B$ is $\\frac{2-\\sqrt{3}}{2}$. If point $P$ is an arbitrary point on the ellipse, then the range of values of $\\frac{1}{|P F_{1}|}+\\frac{1}{|P F_{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;Length(MinorAxis(G)) = 2;A: Point;UpperVertex(G) = A;B: Point;LeftVertex(G) = B;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;Area(TriangleOf(F1, A, B)) = (2 - sqrt(3))/2;P: Point;PointOnCurve(P, G)", "query_expressions": "Range(1/Abs(LineSegmentOf(P, F2)) + 1/Abs(LineSegmentOf(P, F1)))", "answer_expressions": "[1, 4]", "fact_spans": "[[[2, 54], [154, 156]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[2, 62]], [[67, 70]], [[2, 70]], [[75, 78]], [[2, 78]], [[87, 94]], [[95, 102]], [[2, 102]], [[2, 102]], [[104, 148]], [[149, 153]], [[149, 162]]]", "query_spans": "[[[164, 212]]]", "process": "" }, { "text": "Given that a moving point $P(x, y)$ lies on the ellipse $\\frac{x^{2}}{25} + \\frac{y^{2}}{16}=1$. If $F(3,0)$, $|P F|=2$, and $M$ is the midpoint of $PF$, then $|O M|$=?", "fact_expressions": "P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;PointOnCurve(P, G);G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F: Point;Coordinate(F) = (3, 0);Abs(LineSegmentOf(P, F)) = 2;O: Origin;M: Point;MidPoint(LineSegmentOf(P, F)) = M", "query_expressions": "Abs(LineSegmentOf(O, M))", "answer_expressions": "4", "fact_spans": "[[[4, 13]], [[4, 13]], [[4, 13]], [[4, 13]], [[4, 56]], [[14, 55]], [[14, 55]], [[58, 66]], [[58, 66]], [[68, 77]], [[91, 98]], [[79, 82]], [[79, 89]]]", "query_spans": "[[[91, 100]]]", "process": "" }, { "text": "If the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ and $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$ have the same foci, then the real number $m$=?", "fact_expressions": "G: Hyperbola;C:Hyperbola;m: Real;Expression(G) = (-y^2 + x^2/m = 1);Expression(C)=(x^2/3-y^2/2=1);Focus(G)=Focus(C)", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[1, 29]], [[30, 65]], [[73, 78]], [[1, 29]], [[30, 65]], [[1, 71]]]", "query_spans": "[[[73, 80]]]", "process": "Combining the geometric properties of hyperbolas, we obtain m+1=3+2, which can be solved to get the answer. According to the problem, the hyperbola \\frac{x^{2}}{m}-y^{2}=1 and \\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1 have the same foci, so m+1=3+2, solving gives m=4" }, { "text": "Given the parabola $ C $ has equation $ x^{2} = 2 p y $ ($ p > 0 $), with focus $ F $. Let $ A B $ be a chord of the parabola $ C $ passing through the focus $ F $. Let $ l_{1} $, $ l_{2} $ be the tangent lines to the parabola at points $ A $ and $ B $, respectively. Suppose $ l_{1} $ and $ l_{2} $ intersect at point $ P $. Then $ \\overrightarrow{P A} \\cdot \\overrightarrow{P B} $ = ?", "fact_expressions": "C: Parabola;p:Number;p>0;A: Point;B: Point;P: Point;F: Point;Expression(C) = (x^2 = 2*(p*y));Focus(C) = F;IsChordOf(LineSegmentOf(A,B),C);PointOnCurve(F,LineSegmentOf(A,B));l1:Line;l2:Line;TangentOfPoint(A,C)=l1;TangentOfPoint(B,C)=l2;Intersection(l1,l2)=P", "query_expressions": "DotProduct(VectorOf(P, A), VectorOf(P, B))", "answer_expressions": "0", "fact_spans": "[[[2, 8], [31, 32], [52, 58], [72, 75]], [[12, 30]], [[12, 30]], [[62, 65]], [[66, 69]], [[113, 117]], [[35, 38], [48, 51]], [[2, 30]], [[31, 38]], [[39, 60]], [[39, 60]], [[78, 85], [95, 102]], [[86, 93], [103, 110]], [[61, 93]], [[61, 93]], [[95, 117]]]", "query_spans": "[[[120, 171]]]", "process": "Let $ A(x_{1},\\frac{x_{1}^{2}}{2p}) $, $ B(x_{2},\\frac{x_{2}^{2}}{2p}) $. Since $ F(0,\\frac{p}{2}) $, let the equation of line $ AB $ be $ y = kx + \\frac{p}{2} $. Substituting into the parabola equation $ x^{2} = 2py $ gives $ x^{2} - 2pkx - p^{2} = 0 $, so $ x_{1} + x_{2} = 2pk $, $ x_{1}x_{2} = -p^{2} $. From $ x^{2} = 2py $, we get $ y = \\frac{x^{2}}{2p} $, then $ y' = \\frac{x}{p} $, so $ k_{PA} = \\frac{x_{1}}{p} $, $ k_{PB} = \\frac{x_{2}}{p} $, therefore $ k_{PA} \\cdot k_{PB} = \\frac{x_{1}x_{2}}{p^{2}} = -1 $, i.e., $ PA \\perp PB $. Hence $ \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = 0 $. This problem examines the comprehensive application of the positional relationship between a line and a parabola, tests transformation thinking and computational ability, and is a medium-difficulty problem." }, { "text": "Given that line $l$ intersects the parabola $C$: $y^{2}=4x$ at points $A$ and $B$ (point $A$ is in the first quadrant), $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=-4$, and point $M$ has coordinates $(4,0)$. When the area of triangle $A B M$ is minimized, what is the length of segment $O A$?", "fact_expressions": "l: Line;C: Parabola;A: Point;O: Origin;B: Point;M: Point;Expression(C) = (y^2 = 4*x);Coordinate(M) = (4, 0);DotProduct(VectorOf(O, A), VectorOf(O, B)) = -4;Intersection(l,C)={A,B};Quadrant(A) = 1;WhenMin(Area(TriangleOf(A,B,M)))", "query_expressions": "Length(LineSegmentOf(O,A))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 7]], [[8, 27]], [[30, 33], [40, 44]], [[149, 154]], [[34, 37]], [[108, 111]], [[8, 27]], [[108, 121]], [[53, 106]], [[2, 39]], [[40, 49]], [[122, 146]]]", "query_spans": "[[[147, 159]]]", "process": "According to the problem, let the equation of line $ l $ be $ x = my + n $, $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Combining with the parabola equation gives $ y^{2} - 4my - 4n = 0 $, so $ y_{1} + y_{2} = 4m $, $ y_{1}y_{2} = -4n $, then $ x_{1}x_{2} = \\frac{y_{1}^{2}}{4} \\cdot \\frac{y_{2}^{2}}{4} = \\frac{(y_{1} \\cdot y_{2})^{2}}{16} = n^{2} $, $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = x_{1}x_{2} + y_{1}y_{2} = n^{2} - 4n = -4 $, so $ n^{2} - 4n + 4 = 0 $, solving gives $ n = 2 $, thus line $ l $ passes through the fixed point $ N(2, 0) $, then $ S_{\\Delta ABM} = \\frac{1}{2} |MN| \\cdot |y_{1} - y_{2}| = |y_{1} - y_{2}| = \\sqrt{(y_{1} + y_{2})^{2} - 4y_{1}y_{2}} = \\sqrt{16m^{2} + 32} $. When $ m = 0 $, the area of $ \\triangle ABM $ is minimized, and the equation of the line is $ x = 2 $, the coordinates of point $ A $ are $ (2, 2\\sqrt{2}) $, $ |OA| = \\sqrt{2^{2} + (2\\sqrt{2})^{2}} = 2\\sqrt{3} $." }, { "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>0, b>0)$, point $P$ is a point on the hyperbola $C$ such that $|P F_{1}|=2|P F_{2}|$. If the midpoint $Q$ of segment $P F_{1}$ lies exactly on the line $y=\\frac{b}{a} x$, then the eccentricity of the hyperbola is?", "fact_expressions": "C: Hyperbola;G: Line;a: Number;b: Number;P: Point;F1: Point;F2: Point;Q: Point;Expression(G) = (y = x*(b/a));LeftFocus(C) = F1;RightFocus(C) = F2;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2));MidPoint(LineSegmentOf(P, F1)) = Q;PointOnCurve(Q, G);a > 0;b > 0", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[20, 81], [92, 98], [169, 172]], [[147, 166]], [[28, 81]], [[28, 81]], [[87, 91]], [[2, 9]], [[10, 17]], [[141, 144]], [[147, 166]], [[2, 86]], [[2, 86]], [[20, 81]], [[87, 101]], [[103, 125]], [[127, 144]], [[141, 166]], [[28, 81]], [[28, 81]]]", "query_spans": "[[[169, 178]]]", "process": "According to the problem, point P lies on the right branch of the hyperbola. Given |PF_{1}| = 2|PF_{2}| and |PF_{1}| - |PF_{2}| = 2a, we obtain |PF_{1}| = 4a, |PF_{2}| = 2a. Since Q is the midpoint of PF_{1}, it follows that |OQ| = a, |QF_{1}| = 2a. In \\triangle OQF_{1}, \\tan\\angle QOF_{1} = \\frac{b}{a}, so \\cos\\angle QOF_{1} = \\frac{a}{c} = \\frac{a^{2} + c^{2} - 4a^{2}}{2ac}, solving gives c = \\sqrt{5}a. Therefore, e = \\frac{c}{a} = \\sqrt{5}." }, { "text": "If point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$, satisfies $P F_{1} \\perp P F_{2}$, and $|P F_{1}|=2|P F_{2}|$, \nthen the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P,G);Focus(G)={F1,F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[26, 72], [129, 132]], [[29, 72]], [[29, 72]], [[1, 5]], [[7, 14]], [[15, 22]], [[26, 72]], [[1, 75]], [[6, 72]], [[78, 101]], [[103, 125]]]", "query_spans": "[[[129, 138]]]", "process": "" }, { "text": "Pass a line through point $A(6,1)$ such that it intersects the hyperbola $x^{2}-4 y^{2}=16$ at two points $B$, $C$, and $A$ is the midpoint of segment $BC$. Then the equation of the line (expressed in general form) is?", "fact_expressions": "G: Hyperbola;H: Line;B: Point;C: Point;A: Point;Expression(G) = (x^2 - 4*y^2 = 16);Coordinate(A) = (6, 1);PointOnCurve(A, H);Intersection(H, G) = {B, C};MidPoint(LineSegmentOf(B,C))=A", "query_expressions": "Expression(H)", "answer_expressions": "3*x-2*y-16=0", "fact_spans": "[[[14, 35]], [[11, 13], [65, 67]], [[40, 43]], [[44, 47]], [[1, 10], [50, 53]], [[14, 35]], [[1, 10]], [[0, 13]], [[11, 47]], [[50, 64]]]", "query_spans": "[[[65, 80]]]", "process": "Let the coordinates of points B and C be (x_{1},y_{1}) and (x_{2},y_{2}), respectively. Then the slope of line BC is \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}. By the midpoint formula, we have: x_{1}+x_{2}=12, y_{1}+y_{2}=2. Substituting the coordinates of points B and C, (x_{1},y_{1}) and (x_{2},y_{2}), into the equation x^{2}-4y^{2}=16 gives (x_{1})^{2}-4(y_{1})^{2}=16, (x_{2})^{2}-4(y_{2})^{2}=16. Subtracting these equations yields: (x_{1}+x_{2})(x_{1}-x_{2})=4(y_{1}+y_{2})(y_{1}-y_{2}). Substituting x_{1}+x_{2}=12 and y_{1}+y_{2}=2 and simplifying gives: \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{3}{2}. Since the required line passes through point A(6,1) and has slope \\frac{3}{2}, its equation is: y-1=\\frac{3}{2}(x-6). Simplifying gives: 3x-2y-16=0. Midpoint coordinate problems are common exam topics; students should strengthen practice on this type of problem." }, { "text": "Given the parabola $y^{2}=4x$, denote its focus as $F$. A line $l$ passing through point $F$ intersects the parabola at points $A$ and $B$. Then the minimum value of $|A F|-\\frac{2}{|B F|}$ is?", "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B}", "query_expressions": "Min(Abs(LineSegmentOf(A, F)) - 2/Abs(LineSegmentOf(B, F)))", "answer_expressions": "2*sqrt(2)-2", "fact_spans": "[[[31, 36]], [[2, 16], [37, 40]], [[41, 44]], [[21, 24], [26, 30]], [[45, 48]], [[2, 16]], [[2, 24]], [[25, 36]], [[31, 50]]]", "query_spans": "[[[52, 81]]]", "process": "When the slope of the line exists, solve the system of equations of the line and the parabola; by Vieta's formulas, obtain the relationship between the horizontal coordinates of points A and B. Using the definition of the parabola, derive the expression for |AF| - \\frac{2}{|BF|}, transform it into a function of one variable, and find its extremum. When the slope does not exist, solve using the length of the latus rectum. Detailed solution: Since the parabola is y^{2}=4x, we have F(1,0). When the slope of line l exists, let the equation of line l be y=k(x-1) (k\\neq0). Substituting into y^{2}=4x gives k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Then x_{1}\\cdotx_{2}=1. By the definition of the parabola, |AF|=x_{1}+1, |BF|=x_{2}+1, so |AF|-\\frac{2}{|BF|}=x_{1}+1\\frac{1}{4}. Let x_{2}-1=t (t\\geqslant1), then x_{2}=t+1\\frac{(1+\\sqrt{2})}{+2\\sqrt{2}}=\\frac{2}{1+\\sqrt{2}}=2\\sqrt{2}-2 (equality holds if and only if t=\\sqrt{2}). When the slope of line l does not exist, |AF|=|BF|=2, so |AF|-\\frac{2}{|BF|}=1. In conclusion, the minimum value of |AF|-\\frac{2}{|BF|} is 2\\sqrt{2}-2." }, { "text": "Given fixed points $F_{1}(0,-3)$, $F_{2}(0,3)$, and a moving point $P$ satisfying the condition $|P F_{1}|+|P F_{2}|=t+\\frac{9}{t}$ ($t$ is a constant and $t>0$), what is the trajectory of point $P$?", "fact_expressions": "F1: Point;F2: Point;P: Point;t:Number;Coordinate(F1) = (0, -3);Coordinate(F2) = (0, 3);Abs(LineSegmentOf(P,F1))+Abs(LineSegmentOf(P,F2))=t+9/t;t>0", "query_expressions": "Locus(P)", "answer_expressions": "Line segment F1F2 or ellipse", "fact_spans": "[[[3, 16]], [[19, 31]], [[34, 37], [94, 98]], [[78, 81]], [[3, 16]], [[19, 31]], [[41, 76]], [[86, 91]]]", "query_spans": "[[[94, 103]]]", "process": "" }, { "text": "If the asymptotes of a hyperbola passing through the point $(\\sqrt{2}, 2)$ are $y = \\pm 2x$, then what is the standard equation of the hyperbola?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (sqrt(2), 2);PointOnCurve(H,G) = True;Expression(Asymptote(G)) = (y = pm*2*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/4=1", "fact_spans": "[[[19, 22], [41, 44]], [[2, 18]], [[2, 18]], [[1, 22]], [[19, 38]]]", "query_spans": "[[[41, 51]]]", "process": "Since the asymptotes of the hyperbola are $ y = \\pm 2x $, its equation can be written as $ x^{2} - \\frac{y^{2}}{4} = \\lambda $ ($ \\lambda \\ne 0 $). Since the point $ (\\sqrt{2}, 2) $ lies on the hyperbola, we have $ \\lambda = (\\sqrt{2})^{2} - \\frac{2^{2}}{4} = 1 $. Thus, the required equation is $ x^{2} - \\frac{y^{2}}{4} = 1 $." }, { "text": "On a plane, two fixed points $A$ and $B$ are at a distance of $4$. A moving point $P$ satisfies $PA - PB = 2$. What is the minimum value of the distance from point $P$ to the midpoint of $AB$?", "fact_expressions": "A: Point;B: Point;P: Point;Distance(A, B) = 4;LineSegmentOf(P, A) - LineSegmentOf(P, B) = 2", "query_expressions": "Min(Distance(P, MidPoint(LineSegmentOf(A, B))))", "answer_expressions": "1", "fact_spans": "[[[6, 10]], [[11, 15]], [[26, 29], [42, 46]], [[6, 23]], [[31, 40]]]", "query_spans": "[[[42, 62]]]", "process": "" }, { "text": "Given that point $M$ is a moving point on the parabola $x^{2}=4 y$, find the minimum value of the sum of the distance from point $M$ to point $A(2,0)$ and the distance from point $M$ to the directrix of the parabola.", "fact_expressions": "G: Parabola;A: Point;M: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (2, 0);PointOnCurve(M, G)", "query_expressions": "Min(Distance(M, A) + Distance(M, Directrix(G)))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[7, 21], [53, 56]], [[34, 43]], [[2, 6], [29, 33], [47, 51]], [[7, 21]], [[34, 43]], [[2, 27]]]", "query_spans": "[[[29, 69]]]", "process": "\\because the coordinates of the focus F of the parabola \\( x^{2} = 4y \\) are \\( F(0,1) \\), draw the figure as follows: \\because the directrix of the parabola \\( x^{2} = 4y \\) is \\( y = -1 \\), let \\( d \\) be the distance from point M to this directrix \\( y = -1 \\). By the definition of a parabola, \\( d = |MF| \\), \\therefore \\( |MA| + d = |MA| + |MF| \\geqslant |AF| \\) (equality holds if and only if points F, M, A are collinear and M lies between F and A). \\therefore the minimum value of the sum of the distance from point M to point \\( A(2,0) \\) and the distance from point M to the directrix of the parabola is \\( |AF| \\). \\because \\( F(0,1) \\), \\( A(2,0) \\), \\( |AF| = \\sqrt{2^{2} + 1^{2}} = \\sqrt{5} \\)" }, { "text": "The distance from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to focus $F_{1}$ is $6$, then what is the distance from $P$ to focus $F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;Focus(G) = {F1, F2};Distance(P, F1) = 6", "query_expressions": "Distance(P, F2)", "answer_expressions": "4", "fact_spans": "[[[0, 38]], [[0, 38]], [[41, 44], [64, 67]], [[0, 44]], [[47, 54]], [[70, 77]], [[0, 77]], [[41, 61]]]", "query_spans": "[[[64, 81]]]", "process": "From $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, we get $a^{2}=25$, so $a=5$. By the definition of the ellipse, $|PF_{1}|+|PF_{2}|=6+|PF_{2}|=2a=10$, so $|PF_{2}|=4$." }, { "text": "Given the parabola $E$: $y^{2}=4x$, a line $l$ passing through the point $M(2,0)$ intersects $E$ at points $A$ and $B$, and intersects the directrix of $E$ at point $C$. Point $A$ lies in the first quadrant. If $\\frac{|CB|}{|AB|}=\\frac{2}{3}$, then what is the slope of line $l$?", "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 4*x);M: Point;Coordinate(M) = (2, 0);l: Line;B: Point;A: Point;PointOnCurve(M,l) = True;Intersection(l, E) = {A, B};C: Point;Intersection(Directrix(E) , l) = C;Quadrant(A) = 1 ;Abs(LineSegmentOf(C, B))/Abs(LineSegmentOf(A, B)) = 2/3", "query_expressions": "Slope(l)", "answer_expressions": "2", "fact_spans": "[[[2, 21], [39, 42], [55, 58]], [[2, 21]], [[23, 32]], [[23, 32]], [[33, 38], [115, 120]], [[48, 51]], [[44, 47], [69, 73]], [[22, 38]], [[33, 51]], [[63, 67]], [[33, 67]], [[69, 78]], [[80, 113]]]", "query_spans": "[[[115, 125]]]", "process": "Draw perpendiculars from points A and B to the directrix of E, with feet of perpendiculars H and K, respectively. From the problem, the slope of line l exists and is non-zero, so we can set the equation of line l as y = k(x - 2) (k ≠ 0). Substituting into y² = 4x gives k²x² - 4(k² + 1)x + 4k² = 0. Then Δ_A = 16 + 32k² > 0, x_A x_B = 4 ①. Since |CB| / |AB| = 2/3 and point A lies in the first quadrant, it follows that |CB| / |CA| = 2/5, hence |BK| / |AH| = 2/5. Also, |BK| = x_B + 1, |AH| = x_A + 1, so (x_B + 1)/(x_A + 1) = 2/5, which implies x_B = (2/5)x_A - 3/5 ②. Solving ① and ② simultaneously gives x_A = 4, x_B = 1. Since A and B lie on the parabola y² = 4x, we have y_A = 4, y_B = -2. Therefore, the slope of line l is k = (y_A - y_B)/(x_A - x_B) = (4 + 2)/(4 - 1) = 2." }, { "text": "If point $P$ lies on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ with foci $F_{1}$ and $F_{2}$, such that $P F_{1} \\perp P F_{2}$ and $|P F_{1}|=2 | P F_{2}| $, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: 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);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[28, 74], [133, 136]], [[31, 74]], [[31, 74]], [[1, 5]], [[7, 15]], [[17, 24]], [[28, 74]], [[7, 74]], [[1, 77]], [[80, 103]], [[105, 130]]]", "query_spans": "[[[133, 142]]]", "process": "" }, { "text": "Given that the equations of the asymptotes of a hyperbola are $2x \\pm 3y = 0$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (2*x + pm*3*y = 0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{sqrt(13)/3,sqrt(13)/2}", "fact_spans": "[[[1, 4], [28, 31]], [[1, 26]]]", "query_spans": "[[[28, 37]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$ with left and right foci $F_{1}$, $F_{2}$, and a point $P(2, \\sqrt{2})$ on the hyperbola $C$ such that $\\frac{|P F_{1}|}{|P F_{2}|}=3$. If the line segment $P F_{1}$ intersects the hyperbola $C$ at another point $A$, then the area of $\\Delta P A F_{2}$ is?", "fact_expressions": "C: Hyperbola;a, b: Number;b > a;a > 0;Expression(C) = (x^2/a^2 - y^2/b^2 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;P: Point;Coordinate(P) = (2, sqrt(2));PointOnCurve(P, C);Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2)) = 3;A: Point;Intersection(LineSegmentOf(P, F1), C) = {P, A}", "query_expressions": "Area(TriangleOf(P, A, F2))", "answer_expressions": "9*sqrt(2)/4", "fact_spans": "[[[2, 61], [107, 113], [163, 169]], [[10, 61]], [[10, 61]], [[10, 61]], [[2, 61]], [[68, 75]], [[78, 85]], [[2, 85]], [[2, 85]], [[87, 106]], [[88, 106]], [[88, 116]], [[118, 149]], [[174, 177]], [[151, 177]]]", "query_spans": "[[[179, 202]]]", "process": "" }, { "text": "Let the two foci of the ellipse be $(0, -2)$ and $(0, 2)$, and the distance between the two directrices is $13$. Then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (0,-2);Coordinate(F2) = (0, 2);Focus(G)={F1,F2};L1:Line;L2:Line;Directrix(G)={L1,L2};Distance(L1,L2)=13", "query_expressions": "Expression(G)", "answer_expressions": "y^2/13+x^2/9=1", "fact_spans": "[[[1, 3], [44, 46]], [[10, 19]], [[20, 29]], [[10, 19]], [[20, 29]], [[1, 29]], [], [], [[1, 33]], [[1, 42]]]", "query_spans": "[[[44, 51]]]", "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 $F_{1}(-c, 0)$ and $F_{2}(c, 0)$, respectively. If there exists a point $P$ on the hyperbola such that $\\frac{\\sin \\angle P F_{1} F_{2}}{\\sin \\angle P F_{2} F_{1}}=\\frac{a}{c}$, then the range of values for the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;c:Number;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-c, 0);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);Sin(AngleOf(P, F1, F2))/Sin(AngleOf(P, F2, F1)) = a/c", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,sqrt(2)+1)", "fact_spans": "[[[2, 59], [99, 102], [187, 190]], [[5, 59]], [[5, 59]], [[83, 96]], [[68, 82]], [[83, 96]], [[105, 109]], [[5, 59]], [[5, 59]], [[2, 59]], [[68, 82]], [[83, 96]], [[2, 96]], [[2, 96]], [[99, 109]], [[111, 184]]]", "query_spans": "[[[187, 201]]]", "process": "" }, { "text": "Given the hyperbola $C$: $x^{2}-\\frac{y^{2}}{8}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{2}$ intersects the left and right branches of $C$ at points $A$ and $B$ respectively, and $|A F_{1}|=|B F_{1}|$. Then $|A B|$=?", "fact_expressions": "l: Line;C: Hyperbola;A: Point;F1: Point;B: Point;F2: Point;Expression(C) = (x^2 - y^2/8 = 1);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(F2, l);Intersection(l, LeftPart(C)) = A;Abs(LineSegmentOf(A, F1)) = Abs(LineSegmentOf(B, F1));Intersection(l, RightPart(C)) = B", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "4", "fact_spans": "[[[68, 73]], [[2, 35], [74, 77]], [[86, 89]], [[43, 50]], [[90, 93]], [[51, 58], [60, 67]], [[2, 35]], [[2, 58]], [[2, 58]], [[59, 73]], [[68, 95]], [[97, 118]], [[68, 95]]]", "query_spans": "[[[120, 129]]]", "process": "" }, { "text": "Let the parabola $ C $: $ y^{2} = 2 p x $ have focus $ F $, and let point $ M $ lie on $ C $ such that $ |M F| = 5 $. If the circle with diameter $ M F $ passes through the point $ (0,2) $, and $ p > |M F| $, then the standard equation of the parabola $ C $ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;F: Point;Focus(C) = F;M: Point;PointOnCurve(M, C) = True;Abs(LineSegmentOf(M, F)) = 5;IsDiameter(LineSegmentOf(M, F), G) = True;G: Circle;PointOnCurve(D, G) = True;D: Point;Coordinate(D) = (0,2);p > Abs(LineSegmentOf(M, F))", "query_expressions": "Expression(C)", "answer_expressions": "y^2=16*x", "fact_spans": "[[[1, 22], [35, 38], [84, 90]], [[1, 22]], [[9, 22]], [[26, 29]], [[1, 29]], [[30, 34]], [[30, 39]], [[40, 49]], [[51, 62]], [[61, 62]], [[61, 71]], [[63, 71]], [[63, 71]], [[73, 82]]]", "query_spans": "[[[84, 97]]]", "process": "Let point M(x_{0},y_{0}), then |MF|=x_{0}+\\frac{p}{2}=5,\\therefore x_{0}=5-\\frac{p}{2}, and point F is (\\frac{p}{2},0), so the equation of the circle with MF as diameter is (x-x_{0})(x-\\frac{p}{2})+(y-y_{0})y=0. Substituting x=0, y=2 gives px_{0}+8-4y_{0}=0, that is \\frac{y_{0}^{2}}{2}-4y_{0}+8=0, so y_{0}=4. From y_{0}^{2}=2px_{0}, we get 16=2p(5-\\frac{p}{2}). Solving this yields p=2 or p=8. Since p>|MF|, the equation of C is y^{2}=16x." }, { "text": "In the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, what is the slope of the line containing the chord for which point $M(-1, \\frac{1}{2})$ is the midpoint?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);M: Point;Coordinate(M) = (-1, 1/2);H: LineSegment;IsChordOf(H, G) = True;MidPoint(H) = M", "query_expressions": "Slope(OverlappingLine(H))", "answer_expressions": "1/2", "fact_spans": "[[[0, 27]], [[0, 27]], [[30, 51]], [[30, 51]], [], [[0, 56]], [[0, 56]]]", "query_spans": "[[[0, 65]]]", "process": "Analysis: Using the \"point difference method,\" let the two endpoints of the chord be A(x_{1},y_{1}) and B(x_{2},y_{2}). Then \\frac{x_{1}^{2}}{4} + y_{1}^{2} = 1, \\frac{x_{2}^{2}}{4} + y_{2}^{2} = 1. Subtracting these two equations gives \\frac{x_{1}^{2} - x_{2}^{2}}{4} + (y_{1}^{2} - y_{2}^{2}) = 0. Rearranging yields \\frac{(y_{1} - y_{2})}{(x_{1} - x_{2})} \\cdot \\frac{(y_{1} + y_{2})}{(x_{1} + x_{2})} = -\\frac{1}{4}, that is, k_{AB} \\cdot k_{OM} = -\\frac{1}{4}. Given k_{OM} = -\\frac{1}{2}, we have k_{AB} \\cdot (-\\frac{1}{2}) = -\\frac{1}{4}, solving gives k_{AB} = \\frac{1}{2}. Thus, the slope of the line containing the chord with midpoint M is \\frac{1}{2}." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ passes through the point $(-2,0)$. A line passing through the left focus $F_{1}$ intersects the left branch of the hyperbola at points $A$ and $B$, the right focus is $F_{2}$. If $\\angle A F_{2} B=45^{\\circ}$ and $|A F_{2}|=8$, then the area of $\\triangle A B F_{2}$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Line;H: Point;A: Point;B: Point;F2: Point;F1:Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (-2, 0);PointOnCurve(H, C);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(F1, G);Intersection(G, LeftPart(C)) = {A, B};AngleOf(A, F2, B) = ApplyUnit(45, degree);Abs(LineSegmentOf(A, F2)) = 8", "query_expressions": "Area(TriangleOf(A, B, F2))", "answer_expressions": "16", "fact_spans": "[[[2, 63], [89, 92]], [[10, 63]], [[10, 63]], [[86, 88]], [[64, 73]], [[97, 100]], [[101, 104]], [[111, 118]], [[78, 85]], [[9, 63]], [[9, 63]], [[2, 63]], [[64, 73]], [[2, 73]], [[2, 85]], [[89, 118]], [[74, 88]], [[86, 106]], [[120, 149]], [[151, 164]]]", "query_spans": "[[[166, 192]]]", "process": "Given the hyperbola C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) passes through the point (-2,0), we get a=2. Since a line passing through the left focus F_{1} intersects the left branch of the hyperbola at points A and B, and the right focus is F_{2}, with |AF_{2}|=8, we obtain |AF_{1}|=4. Let |BF_{1}|=x, then |BF_{2}|=x+4, and we have |AB|=|BF_{2}|=x+4. Also, since \\angle AF_{2}B=45^{\\circ}, triangle ABF_{2} is an isosceles right triangle with the right angle at B. Thus, |AB|=|BF_{2}|=4\\sqrt{2}, and we get S_{\\triangle ABF_{2}}=\\frac{1}{2}\\times4\\sqrt{2}\\times4\\sqrt{2}=16. Hence, the answer is: 16. This problem mainly examines the definition and simple properties of hyperbolas; recognizing that triangle ABF_{2} is an isosceles right triangle with the right angle at B is key to solving the problem." }, { "text": "Given that points $F$ and $A$ are the left focus and the right vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line perpendicular to the $x$-axis is drawn through $F$, intersecting the ellipse at point $P$. The inradius of triangle $\\triangle A F P$ is $\\frac{\\sqrt{a^{2}-b^{2}}}{2}$. Find the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;F: Point;P: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;RightVertex(G) = A;Z: Line;PointOnCurve(F, Z);IsPerpendicular(Z, xAxis);Intersection(Z, G) = P;Radius(InscribedCircle(TriangleOf(A, F, P))) = sqrt(a^2 - b^2)/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(7) - 1)/3", "fact_spans": "[[[13, 65], [87, 89], [152, 154]], [[15, 65]], [[15, 65]], [[7, 10]], [[2, 6], [75, 78]], [[90, 94]], [[15, 65]], [[15, 65]], [[13, 65]], [[2, 73]], [[2, 73]], [], [[74, 86]], [[74, 86]], [[74, 94]], [[96, 150]]]", "query_spans": "[[[152, 160]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has an asymptote with equation $y=\\frac{\\sqrt{5}}{2} x$, and shares a common focus with the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$. Then the equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (y = (sqrt(5)/2)*x);G: Ellipse;Expression(G) = (x^2/12 + y^2/3 = 1);Focus(C) = Focus(G)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4-y^2/5=1", "fact_spans": "[[[2, 63], [146, 152]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 96]], [[100, 138]], [[100, 138]], [[2, 143]]]", "query_spans": "[[[146, 157]]]", "process": "The hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) has asymptotes given by $ y = \\pm \\frac{b}{a}x $. Given one asymptote equation $ y = \\frac{\\sqrt{5}}{2}x $, we obtain $ \\frac{b}{a} = \\frac{\\sqrt{5}}{2} $ ①. The ellipse $ \\frac{x^{2}}{12} + \\frac{y^{2}}{3} = 1 $ has foci at $ (-3,0) $, $ (3,0) $, from which we get $ a^{2} + b^{2} = 9 $ ②. From ① and ②, we obtain $ a = 2 $, $ b = \\sqrt{5} $, so the equation of the hyperbola is $ \\frac{x^{2}}{4} - \\frac{y^{2}}{5} = 1 $." }, { "text": "In $Rt \\triangle ABC$, $ AB=AC=1 $. If an ellipse passes through points $A$ and $B$, has one focus at $C$, and the other focus on $AB$, then what is the eccentricity of this ellipse?", "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;LineSegmentOf(A,B)=LineSegmentOf(A,C);LineSegmentOf(A,C)=1;PointOnCurve(A, G);PointOnCurve(B,G);OneOf(Focus(G))=C;F:Point;OneOf(Focus(G))=F;Negation(C=F);PointOnCurve(F, LineSegmentOf(A, B))", "query_expressions": "Eccentricity(G)", "answer_expressions": "", "fact_spans": "[[[34, 36], [77, 79], [50, 51]], [[38, 41]], [[44, 47]], [[57, 60]], [[21, 31]], [[21, 31]], [[34, 49]], [[34, 49]], [[50, 60]], [], [[50, 66]], [[50, 66]], [[50, 73]]]", "query_spans": "[[[77, 85]]]", "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}}{9}=1$ $(a>3)$, respectively, and $P$ is a point on the ellipse $C$ such that $\\angle F_{1} P F_{2}=120^{\\circ}$, then $|P F_{1}| \\cdot|P F_{2}|$=?", "fact_expressions": "C: Ellipse;a: Number;F1: Point;P: Point;F2: Point;a>3;Expression(C) = (y^2/9 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(120, degree)", "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))", "answer_expressions": "36", "fact_spans": "[[[20, 71], [82, 87]], [[27, 71]], [[2, 9]], [[78, 81]], [[10, 17]], [[27, 71]], [[20, 71]], [[2, 77]], [[2, 77]], [[78, 90]], [[92, 126]]]", "query_spans": "[[[128, 156]]]", "process": "By the definition of an ellipse, |PF₁| + |PF₂| = 2a, and |F₁F₂| = 2c = 2√(a²−9). Applying the law of cosines: |F₁F₂|² = |PF₁|² + |PF₂|² − 2|PF₁||PF₂|cos120°. Thus, 4(a²−9) = 4a² − 2|PF₁||PF₂| + |PF₁||PF₂| = 4a² − |PF₁||PF₂|. Solving gives |PF₁||PF₂| = 36. Therefore, fill in 36." }, { "text": "Given points $A(-1,-3)$, $B(2,0)$, and $P(x, y)$ with $-10, n>0)$, then what is the range of the slope of line $PA$?", "fact_expressions": "G: Ellipse;n: Number;m: Number;P: Point;A: Point;B: Point;m>0;n>0;Expression(G) = (y^2/n + x^2/m = 1);Coordinate(A) = (-1, -3);Coordinate(B) = (2, 0);Coordinate(P) = (x1, y1);x1:Number;y1:Number;-10, b>0)$ has eccentricity $\\sqrt{3}$, what are the equations of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Eccentricity(G) = sqrt(3)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "sqrt(2)*x+pm*2*y=0", "fact_spans": "[[[2, 58], [76, 79]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 73]]]", "query_spans": "[[[76, 87]]]", "process": "According to the problem, the eccentricity of the hyperbola is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{3} $, solving yields $ \\frac{b}{a} = \\sqrt{2} $. Since the foci of the hyperbola lie on the vertical axis, the asymptotes of the hyperbola are $ y = \\pm\\frac{a}{b}x \\Rightarrow y = \\pm\\frac{\\sqrt{2}}{2}x $, that is, $ \\sqrt{2}x \\pm 2y = 0 $." }, { "text": "Given the line $l$: $k x - y + 1 = 0$ intersects the parabola $C$: $y^{2} = 4 x$ at points $A$ and $B$, $O$ is the origin, and $P$ is the intersection point of the directrix of parabola $C$ with the $x$-axis. If $\\overrightarrow{O A} \\cdot \\overrightarrow{O B} < 0$, then what is the equation of the line when $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ takes its minimum value?", "fact_expressions": "l: Line;Expression(l) = (k*x - y + 1 = 0);k: Number;C: Parabola;Expression(C) = (y^2 = 4*x);A: Point;B: Point;Intersection(l, C) = {A, B};O: Origin;P: Point;Intersection(Directrix(C), xAxis) = P;DotProduct(VectorOf(O, A), VectorOf(O, B)) < 0;WhenMin(DotProduct(VectorOf(P, A), VectorOf(P, B)))", "query_expressions": "Expression(l)", "answer_expressions": "5*x+y-1=0", "fact_spans": "[[[2, 20], [192, 194]], [[2, 20]], [[9, 20]], [[21, 40], [61, 67]], [[21, 40]], [[42, 45]], [[46, 49]], [[2, 51]], [[52, 55]], [[79, 82]], [[61, 82]], [[84, 135]], [[137, 191]]]", "query_spans": "[[[192, 199]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, its two asymptotes intersect the directrix of the parabola $y^{2}=4x$ at points $A$ and $B$, respectively. Let $O$ be the origin. If $S_{\\triangle A O B}=2 \\sqrt{3}$, then the eccentricity $e$ 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;Z1: Line;Z2: Line;Asymptote(G) = {Z1, Z2};H: Parabola;Expression(H) = (y^2 = 4*x);A: Point;Intersection(Z1, Directrix(H)) = A;B: Point;Intersection(Z2, Directrix(H)) = B;O: Origin;Area(TriangleOf(A, O, B)) = 2*sqrt(3);e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(13)", "fact_spans": "[[[2, 58], [142, 145]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [], [], [[2, 64]], [[65, 79]], [[65, 79]], [[86, 89]], [[2, 97]], [[92, 95]], [[2, 97]], [[98, 101]], [[108, 140]], [[149, 152]], [[142, 152]]]", "query_spans": "[[[149, 154]]]", "process": "Since the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are $y=\\pm\\frac{b}{a}x$, and the directrix of the parabola $y^{2}=4x$ is $x=-1$, therefore $A(-1,\\frac{b}{a})$, $B(-1,-\\frac{b}{a})$. Thus $S_{\\triangle AOB}=\\frac{1}{2}\\times1\\times2\\frac{b}{a}=2\\sqrt{3}$, $\\therefore b=2\\sqrt{3}a$, $\\therefore c=\\sqrt{13}a$, $e=\\sqrt{13}$." }, { "text": "Given that the directrix of the parabola $y^{2}=2 a x$ is $x=-2$, then $a$=?", "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y^2 = 2*(a*x));Expression(Directrix(G)) = (x = -2)", "query_expressions": "a", "answer_expressions": "4", "fact_spans": "[[[2, 18]], [[32, 35]], [[2, 18]], [[2, 30]]]", "query_spans": "[[[32, 37]]]", "process": "Since the equation of the directrix is $ x = -\\frac{a}{2} = -2 $, we have $ a = 4 $." }, { "text": "If a moving point $P(x_{0}, y_{0})$ moves on the circle $C$: $x^{2}+y^{2}=1$, then the equation of the trajectory of the moving point $Q(x_{0} y_{0}, x_{0}+y_{0})$ is?", "fact_expressions": "C: Circle;P: Point;Q: Point;x0:Number;y0:Number;Expression(C) = (x^2 + y^2 = 1);Coordinate(P) = (x0, y0);Coordinate(Q) = (x0*y0, x0 + y0);PointOnCurve(P, C)", "query_expressions": "LocusEquation(Q)", "answer_expressions": "(y^2=2*x+1)&(-1/2<=x<=1/2)", "fact_spans": "[[[21, 42]], [[3, 20]], [[49, 78]], [[3, 20]], [[3, 20]], [[21, 42]], [[3, 20]], [[49, 78]], [[3, 43]]]", "query_spans": "[[[49, 85]]]", "process": "Let Q(x,y), then x=x_{0}y_{0}, y=x_{0}+y_{0}. Since the moving point P(x_{0},y_{0}) moves on the circle C: x^{2}+y^{2}=1, \\therefore x_{0}^{2}+y_{0}^{2}=1, 2x+1\\cdot x_{0}^{2}+y_{0}^{2}=1\\geqslant 2|x_{0}y_{0}|=|2x|. \\therefore -\\frac{1}{2}\\leqslant x\\leqslant \\frac{1}{2}, \\therefore the required trajectory equation is: y^{2}=2x+1 \\left(-\\frac{1}{2}\\leqslant x\\leqslant \\frac{1}{2}\\right)" }, { "text": "The focus of the parabola $y^{2}=2 p x$ is $F$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$ be two moving points on the parabola. If $x_{1}+x_{2}+p=\\frac{2 \\sqrt{3}}{3}|A B|$, then the maximum value of $\\angle A F B$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;F: Point;Focus(G) = F;A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(A, G);PointOnCurve(B, G);p + x1 + x2 = ((2*sqrt(3))/3)*Abs(LineSegmentOf(A, B))", "query_expressions": "Max(AngleOf(A, F, B))", "answer_expressions": "2*pi/3", "fact_spans": "[[[0, 16], [62, 65]], [[0, 16]], [[3, 16]], [[20, 23]], [[0, 23]], [[25, 42]], [[44, 61]], [[25, 42]], [[44, 61]], [[25, 42]], [[44, 61]], [[25, 42]], [[44, 61]], [[25, 71]], [[25, 71]], [[73, 114]]]", "query_spans": "[[[116, 136]]]", "process": "Let A(x₁,y₁) and B(x₂,y₂) be two moving points on the parabola y² = 2px. Then |AF| + |BF| = x₁ + x₂ + p. Also, x₁ + x₂ + p = (2√3)/3 |AB|, so |AF| + |BF| = (2√3)/3 |AB|. In triangle AFB, by the cosine law: \ncos∠AFB = (|AF|² + |BF|² − |AB|²)/(2|AF||BF|) = ((|AF| + |BF|)² − 2|AF||BF| − |AB|²)/(2|AF||BF|) = 4/(2|AF||BF|) − 1 = 1/(2|AF||BF|) − 1. \nAlso, |AF| + |BF| = (2√3)/3 |AB| ≥ 2√(|AF||BF|), which implies |AF||BF| ≤ (1/3)|AB|², so cos∠AFB ≥ (1/(2·(1/3)|AB|²)) − 1 = (3/(2|AB|²)) − 1 ≥ −1/2. Therefore, the maximum value of ∠AFB is 2π/3." }, { "text": "Through a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a perpendicular line to its asymptote, with foot of perpendicular at $M$. Extend $FM$ to intersect the $y$-axis at point $E$. If $FM = ME$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;M: Point;E: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(G))=F;L:Line;PointOnCurve(F,L);IsPerpendicular(L,Asymptote(G));FootPoint(L,Asymptote(G))=M;Intersection(OverlappingLine(LineSegmentOf(F, M)), yAxis) = E;LineSegmentOf(F, M) = LineSegmentOf(M, E)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 57], [66, 67], [111, 114]], [[4, 57]], [[4, 57]], [[62, 65]], [[78, 81]], [[94, 98]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [], [[0, 74]], [[0, 74]], [[0, 81]], [[82, 98]], [[100, 109]]]", "query_spans": "[[[111, 120]]]", "process": "" }, { "text": "The center of the hyperbola is at the origin, the foci lie on the $y$-axis, the focal distance is $16$, and one asymptote has the equation $y=\\frac{3}{\\sqrt{7}} x$. Then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), yAxis);FocalLength(G) = 16;Expression(OneOf(Asymptote(G))) =(y=(3/sqrt(7))*x)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/36-x^2/28=1", "fact_spans": "[[[0, 3], [61, 64]], [[7, 9]], [[0, 9]], [[0, 18]], [[0, 26]], [[0, 59]]]", "query_spans": "[[[61, 68]]]", "process": "Test Analysis: According to the problem, assume the required hyperbola equation is \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1(a>0,b>0). From \\begin{cases}2c=16\\\\\\frac{a}{b}=\\frac{3}{\\sqrt{7}}\\\\c^{2}=a^{2}+b^{2}\\end{cases}, solve to get \\begin{cases}a=6\\\\b=2\\sqrt{7},\\\\c=8\\end{cases}. Therefore, the required hyperbola equation is \\frac{y^{2}}{36}-\\frac{x^{2}}{28}=1" }, { "text": "$P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $F_{1}$, $F_{2}$ are the left and right foci. If $\\angle F_{1} P F_{2}=60^{\\circ}$, then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "3*sqrt(3)", "fact_spans": "[[[4, 42]], [[4, 42]], [[0, 3]], [[0, 45]], [[46, 53]], [[54, 61]], [[4, 66]], [[4, 66]], [[68, 101]]]", "query_spans": "[[[103, 130]]]", "process": "Test Analysis: From the ellipse equation $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, we know $a=5$, $b=3$, $c=4$. Since point $P$ lies on the ellipse and $F_{1}$, $F_{2}$ are the left and right foci of the ellipse, we have $|AF_{1}|+|AF_{2}|=10$, $|F_{1}F_{2}|=8$. Let $|AF_{1}|=m$, $|AF_{2}|=n$, then $\\begin{cases}m+n=10\\\\m^{2}+n^{2}-mn=64\\end{cases}$, solving gives $mn=12$. Therefore, the area of $\\triangle F_{1}PF_{2}$ is $\\frac{1}{2}mn\\sin60^{\\circ}=3\\sqrt{3}$. So the answer should be: $3\\sqrt{3}$" }, { "text": "Given that the left focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ is $F_{1}$, the right focus is $F_{2}$, and a line perpendicular to the $x$-axis is drawn through $F_{1}$ intersecting the ellipse at points $A$ and $B$, then the area of $\\triangle A B F_{2}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/5 + y^2/4 = 1);F1: Point;LeftFocus(G) = F1;F2: Point;RightFocus(G) = F2;L: Line;PointOnCurve(F1, L)= True;IsPerpendicular(L, xAxis) = True;Intersection(L, G) = {A, B};A: Point;B: Point", "query_expressions": "Area(TriangleOf(A, B, F2))", "answer_expressions": "8*sqrt(5)/5", "fact_spans": "[[[2, 39], [81, 83]], [[2, 39]], [[44, 51], [65, 72]], [[2, 51]], [[56, 63]], [[2, 63]], [], [[64, 80]], [[64, 80]], [[64, 95]], [[86, 89]], [[90, 93]]]", "query_spans": "[[[97, 123]]]", "process": "From the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, we have $a^{2}=5$, $b^{2}=4$, so $c=\\sqrt{a^{2}-b^{2}}=1$, thus the left and right foci of the ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$, and $|F_{1}F_{2}|=2$. Draw a line perpendicular to the x-axis through $F_{1}$ intersecting the ellipse at points $A$, $B$. It is clear that $x=-1$ intersects the ellipse at points $A$, $B$. Substitute $x=-1$ into the ellipse equation $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$, solving gives $y=\\pm\\frac{4\\sqrt{5}}{5}$, so $|AB|=\\frac{8\\sqrt{5}}{5}$, hence $s_{\\triangle ABF_{2}}=\\frac{1}{2}|AB||F_{1}F_{2}|=\\frac{8\\sqrt{5}}{5}$." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at point $F$. A line perpendicular to the $x$-axis is drawn through $F$, intersecting the ellipse at points $A$ and $B$. If $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$, then the eccentricity of the ellipse is equal to?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;O: Origin;A: Point;B: Point;F: Point;a > b;b > 0;RightFocus(G) = F;L:Line;IsPerpendicular(L,xAxis);PointOnCurve(F,L);Intersection(L, G) = {A, B};DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0", "query_expressions": "Eccentricity(G)", "answer_expressions": "(-1+sqrt(5))/2", "fact_spans": "[[[2, 54], [77, 79], [144, 146]], [[2, 54]], [[4, 54]], [[4, 54]], [[91, 142]], [[80, 83]], [[84, 87]], [[59, 62], [64, 67]], [[4, 54]], [[4, 54]], [[2, 62]], [], [[68, 76]], [[63, 76]], [[74, 89]], [[91, 142]]]", "query_spans": "[[[144, 153]]]", "process": "From the given conditions, the coordinates of points A and B can be determined. Then, since $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=0$, it follows that $\\triangle OAB$ is an isosceles right triangle, leading to the equation $a^{2}-c^{2}=ac$, from which the eccentricity can be found. $\\because$ the ellipse $C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has its right focus at $F$. Drawing a vertical line through $F$ intersecting the ellipse $C$ at points $A$ and $B$, substitute $x=c$ into $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, yielding: $y=\\pm\\frac{b^{2}}{a}$. If $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=0$, then $\\triangle OAB$ is an isosceles right triangle ($O$ being the origin), so we obtain: $\\frac{b^{2}}{a}=c$, i.e., $a^{2}-c^{2}=ac$, or $e^{2}+e-1=0$ with $e\\in(0,1)$. Solving gives: $e=\\frac{\\sqrt{5}-1}{2}$" }, { "text": "Hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, $P$ is a point on hyperbola $C$. If the product of the distances from point $P$ to the two asymptotes of hyperbola $C$ is $1$, what is the range of values for the semi-focal length $c$ 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, C) = True;L1: Line;L2: Line;Asymptote(C) = {L1,L2};Distance(P,L1) * Distance(P,L2) = 1;c: Number;HalfFocalLength(C) = c", "query_expressions": "Range(c)", "answer_expressions": "[2,+oo)", "fact_spans": "[[[0, 61], [67, 73], [84, 90], [107, 110]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[63, 66], [79, 83]], [[63, 77]], [], [], [[84, 96]], [[79, 105]], [[114, 117]], [[107, 117]]]", "query_spans": "[[[114, 123]]]", "process": "By the given condition, the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, its asymptotes are $ bx \\pm ay = 0 $. Let $ P(x, y) $, then the distances from point $ P $ to the two asymptotes are $ d_{1} = \\frac{|bx - ay|}{\\sqrt{a^{2} + b^{2}}} $, $ d_{2} = \\frac{|bx + ay|}{\\sqrt{a^{2} + b^{2}}} $. Since the product of the distances from point $ P $ to the two asymptotes of hyperbola $ C $ is 1, we have $ d_{1}d_{2} = \\frac{|bx - ay|}{\\sqrt{a^{2} + b^{2}}} \\cdot \\frac{|bx + ay|}{\\sqrt{a^{2} + b^{2}}} = \\frac{|b^{2}x^{2} - a^{2}y^{2}|}{a^{2} + b^{2}} = 1 $. From $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $, we get $ b^{2}x^{2} - a^{2}y^{2} = a^{2}b^{2} $, so $ \\frac{a^{2}b^{2}}{a^{2} + b^{2}} = 1 $, that is, $ a^{2} + b^{2} = a^{2}b^{2} \\leqslant \\frac{(a^{2} + b^{2})^{2}}{4} $, thus $ c^{2} \\leqslant \\frac{c^{4}}{4} $, so $ c \\geqslant 2 $, with equality if and only if $ a = b $. Therefore, the range of the semi-focal length $ c $ of the hyperbola is $ [2, +\\infty) $." }, { "text": "The line $l$: $y=2x-4$ passes through the focus $F$ of the parabola $C$: $y^{2}=2px$, and intersects $C$ at points $A$ and $B$. Then $|AB|$=?", "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;B: Point;Expression(C) = (y^2 = 2*p*x);Expression(l)=(y=2*x-4);Focus(C)=F;PointOnCurve(F,l);Intersection(l,C)={A,B};F:Point", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "10", "fact_spans": "[[[0, 16]], [[17, 38], [46, 49]], [[24, 38]], [[51, 54]], [[55, 58]], [[17, 38]], [[0, 16]], [[17, 44]], [[0, 44]], [[0, 60]], [[41, 44]]]", "query_spans": "[[[62, 71]]]", "process": "Since the line $ l: y = 2x - 4 $ passes through the focus $ F $ of the parabola $ C: y^2 = 2px $, we have $ F(2, 0) $, so $ p = 4 $, hence the parabola $ C: y^2 = 8x $. Let $ A(x_1, y_1) $, $ B(x_2, y_2) $. From \n\\[\n\\begin{cases}\ny = 2x - 4 \\\\\ny^2 = 8x\n\\end{cases}\n\\]\nwe obtain $ x^2 - 6x + 4 = 0 $, thus $ |AB| = x_1 + x_2 + p = 10 $." }, { "text": "The standard equation of a hyperbola that has the same foci as the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and has eccentricity $2$ is?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);G: Hyperbola;Focus(H) = Focus(G);Eccentricity(G) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2/12=1", "fact_spans": "[[[1, 39]], [[1, 39]], [[54, 57]], [[0, 57]], [[46, 57]]]", "query_spans": "[[[54, 63]]]", "process": "" }, { "text": "$P$ is a point on the hyperbola $C_{1}$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, $F$ is the right focus of the hyperbola, and $PF=4$. Then the distance from point $P$ to the left directrix of the hyperbola is?", "fact_expressions": "C1: Hyperbola;P: Point;F: Point;Expression(C1) = (x^2/4 - y^2/5 = 1);PointOnCurve(P, C1);RightFocus(C1) = F;LineSegmentOf(P, F) = 4", "query_expressions": "Distance(P, LeftDirectrix(C1))", "answer_expressions": "16/3", "fact_spans": "[[[4, 51], [59, 62], [82, 85]], [[0, 3], [77, 81]], [[55, 58]], [[4, 51]], [[0, 54]], [[55, 66]], [[68, 75]]]", "query_spans": "[[[77, 93]]]", "process": "a^{2}=4, b^{2}=5, \\therefore c^{2}=9. Let the left focus of the ellipse be F_{1}, ||PF_{1}|-|PF||=4 \\Rightarrow ||PF_{1}|-4|=4, solving gives |PF_{1}|=8. Let the distance from point P to the left directrix of the hyperbola be d, then \\frac{|PF_{1}|}{d}=\\frac{c}{a} \\Rightarrow \\frac{8}{d}=\\frac{3}{2}, \\therefore d=\\frac{16}{3}. Hence fill in: \\frac{16}{3}. This problem involves two definitions, intended to examine transformation and computational ability, belonging to a simple type." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, where $\\frac{b}{a}=2$, then the eccentricity $e$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;b/a = 2;e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 58]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[59, 74]], [[79, 82]], [[2, 82]]]", "query_spans": "[[[79, 84]]]", "process": "Problem Analysis: Since $\\frac{b}{a}=2$, that is, $b=2a$, $c=\\sqrt{a^{2}+b^{2}}=\\sqrt{a^{2}+4a^{2}}=\\sqrt{5}a$, then $e=\\frac{c}{a}=\\sqrt{5}$" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the left and right vertices are denoted as $A$ and $B$ respectively. Point $P$ lies on the curve $C$. If in $\\triangle P A B$, $\\angle P B A = \\angle P A B + \\frac{\\pi}{2}$, then what is the equation of the asymptotes of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;A: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(P, C);AngleOf(P, B, A) = pi/2 + AngleOf(P, A, B)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y = pm*x", "fact_spans": "[[[2, 63], [85, 90], [155, 161]], [[9, 63]], [[9, 63]], [[80, 84]], [[72, 75]], [[76, 79]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 79]], [[2, 79]], [[80, 91]], [[112, 153]]]", "query_spans": "[[[155, 169]]]", "process": "As shown in the figure, draw BM perpendicular to the x-axis from point B. Since ∠PBA = ∠PAB + π/2, it follows that ∠PAB = ∠PBM, so ∠PAB + ∠PBx = π/2. Thus, k_{PA} ⋅ k_{PB} = 1. Let P(x, y), and given A(-a, 0), B(a, 0), then (y/(x+a)) ⋅ (y/(x−a)) = 1, so x² − y² = a², hence a = b. Therefore, the asymptotes of hyperbola C are y = ±x." }, { "text": "Given that the hyperbola $C$ passing through the point $P(-2 , 0)$ has the same foci as the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, what is the equation of the asymptotes of the hyperbola $C$?", "fact_expressions": "P: Point;Coordinate(P) = (-2, 0);PointOnCurve(P, C);C: Hyperbola;G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(C) = Focus(G)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "sqrt(3)*x + pm*y=0", "fact_spans": "[[[3, 15]], [[3, 15]], [[2, 22]], [[16, 22], [69, 75]], [[23, 61]], [[23, 61]], [[16, 67]]]", "query_spans": "[[[69, 83]]]", "process": "" }, { "text": "Given the parabola $C$: $x^{2}=2 y$, then the equation of the directrix of $C$ is?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*y)", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "y = -1/2", "fact_spans": "[[[2, 21], [23, 26]], [[2, 21]]]", "query_spans": "[[[23, 33]]]", "process": "Since the parabola C: x^{2}=2y, we have 2p=2, \\therefore p=1, so the directrix equation of C is y=-\\frac{1}{2}." }, { "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}$ and $F_{2}$ respectively, and the ellipse $C$ shares foci with the hyperbola $C^{\\prime}$: $\\frac{2 x^{2}}{a^{2}}-y^{2}=1$. If a point $M$, an intersection of ellipse $C$ and hyperbola $C^{\\prime}$, satisfies $|M F_{1}| \\cdot|M F_{2}|=2$, then what is the area of $\\Delta M F_{1} F_{2}$?", "fact_expressions": "C1: Hyperbola;C: Ellipse;b: Number;a: Number;M: Point;F1: Point;F2: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Expression(C1)=(2*x^2/a^2-y^2=1);Focus(C)=Focus(C1);OneOf(Intersection(C,C1))=M;Abs(LineSegmentOf(M,F1))*Abs(LineSegmentOf(M,F2))=2", "query_expressions": "Area(TriangleOf(M, F1, F2))", "answer_expressions": "1", "fact_spans": "[[[91, 139], [150, 165]], [[2, 59], [85, 90], [144, 149]], [[9, 59]], [[9, 59]], [[170, 173]], [[68, 75]], [[76, 83]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[91, 139]], [[85, 142]], [[144, 173]], [[175, 203]]]", "query_spans": "[[[205, 232]]]", "process": "From the definitions of the ellipse and hyperbola, we obtain \n\\begin{cases}|MF_{1}|+|MF_{2}|=2a\\\\|MF_{1}|-|MF_{2}|=\\sqrt{2}a\\end{cases}, \nsolving gives \n\\begin{cases}|MF_{1}|=\\frac{2+\\sqrt{2}}{2}a\\\\|MF_{2}|=\\frac{2-\\sqrt{2}}{2}a\\end{cases}. \nSubstituting into |MF_{1}|\\cdot|MF_{2}|=2, solve for the value of a, and then find the lengths of |MF_{1}|, |MF_{2}|, and |F_{1}F_{2}|. By the Pythagorean theorem, \\triangle MF_{1}F_{2} is a right triangle. Combining with the area formula, the solution can be obtained. \n\n[Detailed Solution] According to the problem, rewrite the hyperbola C: \\frac{2x^{2}}{a^{2}}-y^{2}=1 into standard form as \\frac{x^{2}}{2}-y^{2}=1. Without loss of generality, assume point M lies on the right branch of the hyperbola. Then, from the definitions of the ellipse and hyperbola, we have \n\\begin{cases}|MF_{1}|+|MF_{2}|=2a\\\\|MF_{1}|-|MF_{2}|=2\\cdot\\frac{\\sqrt{2}}{2}a=\\sqrt{2}a\\end{cases}, \nsolving gives \n\\begin{cases}|MF_{1}|=\\frac{2+\\sqrt{2}}{2}a\\\\|MF_{2}|=\\frac{2-\\sqrt{2}}{2}a\\end{cases}. \nSince |MF_{1}|\\cdot|MF_{2}|=2, substituting yields \n\\frac{2+\\sqrt{2}}{2}a\\cdot\\frac{2-\\sqrt{2}}{2}a=2, \nsolving gives a=2 or a=-2 (discard negative). \nThus, |MF_{1}|=2+\\sqrt{2}, |MF_{2}|=2-\\sqrt{2}, and the focal distance of the hyperbola |F_{1}F_{2}|=2\\sqrt{\\frac{a^{2}}{2}+1}=2\\sqrt{3}. Clearly, |MF_{1}|^{2}+|MF_{2}|^{2}=|F_{1}F_{2}|^{2}, so \\triangle MF_{1}F_{2} is a right triangle. Therefore, the area of \\triangle MF_{1}F_{2} is: \nS_{\\triangle MF_{1}F_{2}}=\\frac{1}{2}\\cdot|MF_{1}|\\cdot|MF_{2}|=\\frac{1}{2}\\times(2+\\sqrt{2})\\times(2-\\sqrt{2})=1." }, { "text": "The line $y=\\sqrt{3}(x-1)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$, and $O$ is the origin. Then the area of triangle $AOB$ is?", "fact_expressions": "H: Line;Expression(H) = (y = sqrt(3)*(x - 1));G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(H, G) = {A, B};O: Origin", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[0, 19]], [[0, 19]], [[20, 34]], [[20, 34]], [[37, 40]], [[41, 44]], [[0, 46]], [[47, 50]]]", "query_spans": "[[[55, 69]]]", "process": "Solve the system \\begin{cases}y=\\sqrt{3}(x-1)\\\\v2=4x\\end{cases}, eliminate x to obtain y^{2}-\\frac{4}{\\sqrt{3}}y-4=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), y_{1}+y_{2}=\\frac{4}{\\sqrt{3}}, y_{1}y_{2}=-s_{AAOB}=\\frac{1}{2}|y_{1}-y_{2}|=\\frac{1}{2}\\sqrt{(y_{1}+y_{2})^{2}-4y_{1}y_{2}}=\\frac{1}{2}\\sqrt{\\frac{16}{3}+16}=\\frac{4\\sqrt{3}}{3}" }, { "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, find the minimum value of the sum of the distances from point $P$ to the line $l$: $2x - y + 3 = 0$ and to the $y$-axis.", "fact_expressions": "l: Line;G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);Expression(l) = (2*x - y + 3 = 0);PointOnCurve(P, G)", "query_expressions": "Min(Distance(P, l) + Distance(P, yAxis))", "answer_expressions": "sqrt(5) - 1", "fact_spans": "[[[31, 48]], [[6, 20]], [[2, 5], [26, 30]], [[6, 20]], [[31, 48]], [[2, 24]]]", "query_spans": "[[[26, 64]]]", "process": "By the given condition, the focus of the parabola is $ F(1,0) $. Let $ d $ be the distance from point $ P $ to the line $ l $. According to the definition of the parabola, the distance from point $ P $ to the $ y $-axis is $ |PF| - 1 $. Therefore, the sum of the distance from point $ P $ to the line $ l $ and to the $ y $-axis is $ d + |PF| - 1 $. Hence, the minimum value of $ d + |PF| $ is the distance from point $ F $ to the line $ l $. Thus, the minimum value of $ d + |PF| $ is $ \\frac{|2+3|}{\\sqrt{2^{2}+(-1)^{2}}} = \\sqrt{5} $, so the minimum value of $ d + |PF| - 1 $ is $ \\sqrt{5} - 1 $." }, { "text": "Given that points $P$ and $Q$ have the same horizontal coordinate, the vertical coordinate of $P$ is twice that of $Q$, and the trajectories of $P$ and $Q$ are hyperbolas $C_{1}$ and $C_{2}$, respectively. If the asymptotes of $C_{1}$ are given by $y = \\pm \\sqrt{3} x$, then the equations of the asymptotes of $C_{2}$ are?", "fact_expressions": "C1:Hyperbola;C2:Hyperbola;P:Point;Q:Point;XCoordinate(P)=XCoordinate(Q);YCoordinate(P)=2*YCoordinate(Q);Locus(P)=C1;Locus(Q)=C2;Expression(Asymptote(C1)) = (y = pm*(sqrt(3)*x))", "query_expressions": "Expression(Asymptote(C2))", "answer_expressions": "y = pm*(sqrt(3)/2)*x", "fact_spans": "[[[51, 61], [72, 79]], [[62, 69], [106, 113]], [[2, 6], [17, 20], [38, 41]], [[7, 10], [25, 28], [42, 45]], [[2, 16]], [[17, 37]], [[38, 69]], [[38, 69]], [[72, 104]]]", "query_spans": "[[[106, 121]]]", "process": "From the given condition: $ C_{1}: 3x^{2} - y^{2} = \\lambda $, ($ \\lambda \\neq 0 $). Let $ Q(x, y) $, then $ P(x, 2y) $, so $ 3x^{2} - 4y^{2} = \\lambda $, thus the asymptotes of $ C_{2} $ are $ y = \\pm \\frac{\\sqrt{3}}{2}x $." }, { "text": "Let the circle centered at the focus $ F $ of the parabola $ C: y^2 = 4x $ intersect $ C $ at points $ A $ and $ B $, and intersect the directrix of $ C $ at points $ D $ and $ E $. Given that $ |AB| = 8 $, then $ |DE| = $?", "fact_expressions": "C: Parabola;G: Circle;A: Point;B: Point;D: Point;E: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Center(G) = F;Intersection(G, C) = {A, B};Intersection(G, Directrix(C)) = {D, E};Abs(LineSegmentOf(A, B)) = 8", "query_expressions": "Abs(LineSegmentOf(D, E))", "answer_expressions": "2*sqrt(21)", "fact_spans": "[[[1, 20], [32, 35], [47, 50]], [[30, 31]], [[36, 39]], [[40, 43]], [[54, 57]], [[58, 61]], [[23, 26]], [[1, 20]], [[1, 26]], [[0, 31]], [[30, 45]], [[30, 63]], [[66, 75]]]", "query_spans": "[[[77, 86]]]", "process": "From the parabola equation we know: $\\frac{p}{2}=1$, $\\therefore F(1,0)$. Without loss of generality, assume point $A$ is in the first quadrant. As shown in the figure, from $|AB|=8$, $y^{2}=4x$ we get: $A(4,4)$, $\\therefore$ the radius of the circle $r=\\sqrt{3^{2}+4^{2}}=5$, $\\therefore |DE|=2\\sqrt{r^{2}-p^{2}}=2\\sqrt{25-4}=2\\sqrt{21}$" }, { "text": "Given line $l_{1}$: $x-y-5=0$ and line $l_{2}$: $y=-4$, find the minimum value of the sum of distances from a moving point $P$ on the parabola $x^{2}=16 y$ to line $l_{1}$ and line $l_{2}$.", "fact_expressions": "G: Parabola;l1: Line;l2: Line;P:Point;Expression(G) = (x^2 = 16*y);Expression(l1) = (x - y - 5 = 0);Expression(l2) = (y=-4);PointOnCurve(P,G)", "query_expressions": "Min(Distance(P,l1)+Distance(P,l2))", "answer_expressions": "(9/2)*sqrt(2)", "fact_spans": "[[[41, 56]], [[2, 22], [64, 73]], [[23, 40], [74, 83]], [[60, 63]], [[41, 56]], [[2, 22]], [[23, 40]], [[41, 63]]]", "query_spans": "[[[60, 94]]]", "process": "Let the focus of the parabola $ x^{2} = 16y $ be $ F $, then $ F(0,4) $, and the line $ l_{2}: y = -4 $ is its directrix, so the distance from $ P $ to the line $ l_{2} $ is $ d_{2} = |PF| $. Let $ d_{1} $ be the distance from $ P $ to the line $ l_{1} $. As shown in the figure, the minimum value of the sum of the distances from the moving point $ P $ to the lines $ l_{1} $ and $ l_{2} $ is the distance from the point $ F(0,4) $ to the line $ l_{1}: x - y - 5 = 0 $, which is $ \\frac{|0 - 4 - 5|}{\\sqrt{5}} = \\frac{9\\sqrt{2}}{2} $." }, { "text": "Given that the coordinates of one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{16}=1$ $(a>0)$ are $(5,0)$, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/16 + x^2/a^2 = 1);Coordinate(OneOf(Focus(G))) = (5, 0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "5/3", "fact_spans": "[[[2, 50], [69, 72]], [[5, 50]], [[5, 50]], [[2, 50]], [[2, 66]]]", "query_spans": "[[[69, 78]]]", "process": "The coordinates of the foci of the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{16}=1 (a>0) are (5,0), so a^{2}=5^{2}-16=9, a=3, thus the eccentricity e=\\frac{c}{a}=\\frac{5}{3}. The answer is: 5" }, { "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has a distance of $5$ to one focus, then what is the distance from $P$ to the other focus?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 5", "query_expressions": "Distance(P, F2)", "answer_expressions": "5", "fact_spans": "[[[1, 39]], [[42, 45], [59, 62]], [[1, 39]], [[1, 45]], [], [], [[1, 50]], [[1, 68]], [[1, 68]], [[1, 57]]]", "query_spans": "[[[1, 73]]]", "process": "By the definition of an ellipse, a=5, and the sum of the distances from P to the two foci is 2a=10. Therefore, the distance to the other focus is 5. Answer: 5" }, { "text": "The length of the minor axis of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1)", "query_expressions": "Length(MinorAxis(G))", "answer_expressions": "6", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 44]]]", "process": "Using the standard equation of the ellipse, solve directly to obtain the answer. \\because the equation of the ellipse is: \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1 \\therefore the length of the minor axis 2b=6." }, { "text": "Given the parabola $y^{2}=2 p x$ ($p>0$), a perpendicular is drawn from the origin to a certain line passing through the focus, and the foot of the perpendicular has coordinates $(2,1)$. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;p: Number;H: Line;O:Origin;p>0;Expression(G) = (y^2 = 2*(p*x));L:Line;PointOnCurve(Focus(G),H);PointOnCurve(O,L);IsPerpendicular(H,L);Coordinate(FootPoint(H,L))=(2,1)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=10*x", "fact_spans": "[[[1, 22], [54, 57]], [[4, 22]], [[33, 35]], [[24, 26]], [[4, 22]], [[1, 22]], [], [[27, 35]], [[23, 38]], [[23, 38]], [[23, 51]]]", "query_spans": "[[[53, 61]]]", "process": "" }, { "text": "Given that the foci of the hyperbola lie on the $y$-axis, the focal distance is $4$, and one asymptote has the equation $y=\\sqrt{3}x$, then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G),yAxis);FocalLength(G) = 4;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/3-x^2=1", "fact_spans": "[[[2, 5], [47, 50]], [[2, 14]], [[2, 21]], [[2, 45]]]", "query_spans": "[[[47, 57]]]", "process": "Using the method of undetermined coefficients, the result can be obtained. [Detailed solution] According to the given conditions, assume the standard equation of the hyperbola is \\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1. Then \\begin{cases}2c=4\\\\\\frac{a}{b}=\\sqrt{3}\\\\c=a^{2}+b^{2}\\end{cases}, solving yields a^{2}=3,b^{2}=1, so the standard equation of the hyperbola is \\frac{y^{2}}{3}-x^{2}=1" }, { "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a}-\\frac{y^{2}}{b}=1$ is $\\sqrt{5}$, then $\\frac{a}{b}=$?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;Expression(G) = (-y^2/b + x^2/a = 1);Eccentricity(G) = sqrt(5)", "query_expressions": "a/b", "answer_expressions": "{4,1/4}", "fact_spans": "[[[1, 39]], [[56, 69]], [[56, 69]], [[1, 39]], [[1, 54]]]", "query_spans": "[[[56, 71]]]", "process": "" }, { "text": "The line $l$ passing through the point $(0, -3)$ has only one common point with the parabola $y^2 = 4x$. Then the equation of line $l$ is?", "fact_expressions": "H: Point;Coordinate(H) = (0, -3);l: Line;PointOnCurve(H, l);G: Parabola;Expression(G) = (y^2 = 4*x);NumIntersection(l, G) = 1", "query_expressions": "Expression(l)", "answer_expressions": "{x=0, y=-3, x+3*y+9=0}", "fact_spans": "[[[1, 10]], [[1, 10]], [[11, 16], [40, 45]], [[0, 16]], [[17, 31]], [[17, 31]], [[11, 38]]]", "query_spans": "[[[40, 50]]]", "process": "Discuss the cases when the slope of a line does not exist, is zero, and is non-zero to obtain the answer: According to the problem, when the slope of line $ l $ does not exist, the equation of line $ l $ is $ x = 0 $, which satisfies the condition. When the slope of line $ l $ exists, let the equation of line $ l $ be $ y = kx - 3 $, substitute into $ y^2 = 4x $. We get $ k^2x^2 - (6k + 4)x + 9 = 0 $. Thus, when $ k = 0 $, the equation of line $ l $ is $ y = -3 $, which satisfies the condition; when $ k \\neq 0 $, from $ \\Delta = (6k + 4)^2 - 36k^2 = 0 $, solving gives $ k = -\\frac{1}{3} $, at this time the equation of line $ l $ is $ x + 3y + 9 = 0 $. In conclusion, the equation of line $ l $ is $ x = 0 $ or $ y = -3 $ or $ x + 3y + 9 = 0 $." }, { "text": "The equation of a hyperbola with asymptotes given by $\\frac{x}{a} \\pm \\frac{y}{b}=0$ is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (x/a + pm*y/b = 0);a: Number;b: Number", "query_expressions": "Expression(G)", "answer_expressions": "(x^2/a^2 - y^2/b^2 = k)&Negation(k = 0)", "fact_spans": "[[[39, 42]], [[0, 42]], [[3, 34]], [[3, 34]]]", "query_spans": "[[[39, 46]]]", "process": "" }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{m}=1$ with foci on the $x$-axis is $\\frac{1}{2}$, then $m$=?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/2 + y^2/m = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "3/2", "fact_spans": "[[[10, 47]], [[67, 70]], [[10, 47]], [[1, 47]], [[10, 65]]]", "query_spans": "[[[67, 72]]]", "process": "" }, { "text": "Given that $A$, $B$, $P$ are three distinct points on the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$, and satisfy $\\overrightarrow{P A}+\\overrightarrow{P B}=2 \\overrightarrow{P O}$ ($O$ being the origin), let $m$, $n$ denote the slopes of lines $PA$, $PB$ respectively. Then the minimum value of $m^{2}+\\frac{n^{2}}{9}$ is?", "fact_expressions": "G: Hyperbola;P: Point;A: Point;B: Point;O: Origin;m: Number;n: Number;Expression(G) = (x^2 - y^2/4 = 1);PointOnCurve(A, G);PointOnCurve(B, G);PointOnCurve(P, G);Negation(A = B);Negation(A = P);Negation(P = B);VectorOf(P, A) + VectorOf(P, B) = 2*VectorOf(P, O);Slope(LineSegmentOf(P, A)) = m;Slope(LineSegmentOf(P, B)) = n", "query_expressions": "Min(m^2 + n^2/9)", "answer_expressions": "8/3", "fact_spans": "[[[18, 46]], [[13, 17]], [[2, 5]], [[8, 11]], [[122, 125]], [[150, 153]], [[155, 158]], [[18, 46]], [[2, 51]], [[2, 51]], [[2, 51]], [[2, 51]], [[2, 51]], [[2, 51]], [[55, 121]], [[132, 158]], [[132, 158]]]", "query_spans": "[[[160, 189]]]", "process": "\\because satisfying \\overrightarrow{PA}+\\overrightarrow{PB}=2\\overrightarrow{PO} (O is the coordinate origin), \\therefore A,B are symmetric about the origin, let A(x_{1},y_{1}), B(-x_{1},-y_{1}), P(x_{0},y_{0}), then \\frac{y_{0}^{2}}{4}=x_{0}^{2}-1, \\frac{y_{1}^{2}}{4}=x_{1}^{2}-1; the slopes of lines PA, PB are denoted by m,n, satisfying mn=\\frac{y_{0}^{2}-y_{1}^{2}}{x_{0}^{2}-x_{1}^{2}}=4, then m^{2}+\\frac{n^{2}}{9}\\geqslant2\\cdot m\\cdot\\frac{n}{3}=\\frac{2mn}{3}=\\frac{8}{3}, thus the minimum value of m^{2}+\\frac{n^{2}}{9} is \\frac{8}{3}" }, { "text": "Given that $P$ is an intersection point of the ellipse $\\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1$ $(a_{1}>b_{1}>0)$ and the hyperbola $\\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1$ $(a_{2}>0, b_{2}>0)$, $F_{1}$, $F_{2}$ are the common foci of the ellipse and the hyperbola, $e_{1}$, $e_{2}$ are the eccentricities of the ellipse and the hyperbola respectively, and if $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$, then the minimum value of $e_{1} \\cdot e_{2}$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;a2: Number;b2: Number;a1: Number;b1: Number;F1: Point;P: Point;F2: Point;a2>0;b2>0;Expression(G) = (-y^2/b2^2 + x^2/a2^2 = 1);a1>b1;b1>0;Expression(H) = (y^2/b1^2 + x^2/a1^2 = 1);OneOf(Intersection(H, G)) = P;Focus(G) = {F1, F2};Focus(H) = {F1, F2};Focus(G) = Focus(H);e1: Number;e2: Number;Eccentricity(H) = e1;Eccentricity(G) = e2;AngleOf(F1, P, F2) = pi/3", "query_expressions": "Min(e1*e2)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[77, 151], [176, 179], [207, 210]], [[6, 76], [173, 175], [204, 206]], [[80, 151]], [[80, 151]], [[8, 76]], [[8, 76]], [[157, 164]], [[2, 5]], [[165, 172]], [[80, 151]], [[80, 151]], [[77, 151]], [[8, 76]], [[8, 76]], [[6, 76]], [[2, 156]], [[157, 184]], [[157, 184]], [[173, 184]], [[185, 192]], [[194, 201]], [[185, 214]], [[185, 214]], [[216, 252]]]", "query_spans": "[[[254, 279]]]", "process": "By the symmetry of the ellipse and hyperbola, assume without loss of generality that point P lies in the first quadrant, so |PF_{1}| > |PF_{2}|. Since the ellipse and hyperbola share common foci, let the semi-focal length of the ellipse and hyperbola be c. By the definitions of the ellipse and hyperbola, we have: |PF_{1}| + |PF_{2}| = 2a_{1}, |PF_{1}| - |PF_{2}| = 2a_{2}. Solving gives |PF_{1}| = a_{1} + a_{2}, |PF_{2}| = a_{1} - a_{2}. In triangle F_{1}PF_{2}, by the law of cosines, we obtain: |F_{1}F_{2}|^{2} = |PF_{1}|^{2} + |PF_{2}|^{2} - 2|PF_{1}||PF_{2}|\\cos\\frac{\\pi}{3}, that is, 4c^{2} = (a_{1}+a_{2})^{2} + (a_{1}-a_{2})^{2} - (a_{1}+a_{2})(a_{1}-a_{2}). Simplifying yields 4c^{2} = a_{1}^{2} + 3a_{2}^{2}, so \\frac{1}{e_{1}^{2}} + 3\\frac{1}{e_{2}^{2}} = 4. Also, \\frac{1}{e_{1}^{2}} + 3\\frac{1}{e_{2}^{2}} \\geqslant \\frac{2\\sqrt{3}}{e_{1}e_{2}}, therefore e_{1}e_{2} \\geqslant \\frac{\\sqrt{3}}{2}." }, { "text": "Given that the asymptotes of a hyperbola are $y=\\pm \\sqrt{2} x$, and it passes through the point $(1,2)$, then the focal distance of this hyperbola is?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, 2);Expression(Asymptote(G)) = (y = pm*sqrt(2)*x);PointOnCurve(H,G)", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 5], [44, 47]], [[33, 41]], [[33, 41]], [[2, 30]], [[2, 41]]]", "query_spans": "[[[44, 52]]]", "process": "The asymptotes of the hyperbola are given by $ y = \\pm\\sqrt{2}x $. Thus, the equation of the hyperbola can be written as $ y^{2} - 2x^{2} = t $ ($ t \\neq 0 $). Substituting the point $ (1, 2) $ gives $ t = 4 - 2 = 2 $. Therefore, the equation of the hyperbola is $ \\frac{y^{2}}{2} - x^{2} = 1 $. It follows that $ a = \\sqrt{2} $, $ b = 1 $, $ c = \\sqrt{3} $, and the focal distance is $ 2c = 2\\sqrt{3} $." }, { "text": "Given that $A$ and $B$ are the common vertices of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ and the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. $P$ is a moving point on the hyperbola, and $M$ is a moving point on the ellipse ($P$, $M$ are distinct from $A$, $B$), satisfying $AP+BP= \\lambda(AM+BM)$, where $\\lambda \\in \\mathbb{R}$. Let the slopes of lines $AP$, $BP$, $AM$, $BM$ be denoted by $k_{1}$, $k_{2}$, $k_{3}$, $k_{4}$ respectively. If $k_{1}+k_{2}=5$, then $k_{3}+k_{4}=$?", "fact_expressions": "G: Hyperbola;H: Ellipse;b: Number;a: Number;A: Point;P: Point;B: Point;M: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a1:Number;b1:Number;a1 > b1;b1 > 0;Expression(H) = (y^2/b1^2 + x^2/a1^2 = 1);Vertex(G)={A,B};In(A,Vertex(H));In(B,Vertex(H));PointOnCurve(P,G);PointOnCurve(M,H);Negation(P=A);Negation(P=B);Negation(M=A);Negation(M=B);LineSegmentOf(A, P) + LineSegmentOf(B, P) = lambda*(LineSegmentOf(A, M) + LineSegmentOf(B, M));lambda:Real;Slope(LineOf(A,P))=k1;Slope(LineOf(B,P))=k2;Slope(LineOf(A,M))=k3;Slope(LineOf(B,M))=k4;k1:Number;k2:Number;k3:Number;k4:Number;k1+k2=5", "query_expressions": "k3 + k4", "answer_expressions": "-5", "fact_spans": "[[[63, 120], [130, 133]], [[10, 62], [142, 144]], [[66, 120]], [[66, 120]], [[2, 5], [159, 162]], [[126, 129], [149, 152]], [[6, 9], [165, 168]], [[138, 141], [153, 156]], [[66, 120]], [[66, 120]], [[63, 120]], [[12, 62]], [[12, 62]], [[12, 62]], [[12, 62]], [[10, 62]], [[2, 125]], [[2, 125]], [[2, 125]], [[126, 137]], [[138, 148]], [[149, 168]], [[149, 168]], [[149, 168]], [[149, 168]], [[173, 197]], [[200, 215]], [[217, 289]], [[217, 289]], [[217, 289]], [[217, 289]], [[255, 262]], [[263, 270]], [[273, 280]], [[282, 289]], [[291, 306]]]", "query_spans": "[[[308, 323]]]", "process": "" }, { "text": "Ellipse $r$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, with focal distance $2c$. If a line $y=\\sqrt{3}(x+c)$ intersects the ellipse $r$ at a point $M$ such that $\\angle M F_{1} F_{2}=2 \\angle M F_{2} F_{1}$, then the eccentricity of the ellipse equals?", "fact_expressions": "r: Ellipse;b: Number;a: Number;c: Number;M: Point;F1: Point;F2: Point;G:Line;a > b;b > 0;Expression(r) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = sqrt(3)*(c + x));LeftFocus(r)=F1;RightFocus(r)=F2;FocalLength(r)=2*c;OneOf(Intersection(G,r))=M;AngleOf(M,F1,F2)=2*AngleOf(M,F2,F1)", "query_expressions": "Eccentricity(r)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[0, 57], [111, 116], [175, 177]], [[7, 57]], [[7, 57]], [[84, 89]], [[121, 124]], [[66, 73]], [[74, 81]], [[91, 110]], [[7, 57]], [[7, 57]], [[0, 57]], [[91, 110]], [[0, 81]], [[0, 81]], [[0, 89]], [[91, 124]], [[126, 171]]]", "query_spans": "[[[175, 184]]]", "process": "Notice that the line passes through the point (-c,0), which is the left focus F_{1}, and has a slope of \\sqrt{3}, so the inclination angle is 60^{\\circ}, that is, \\angle MF_{1}F_{2}=60^{\\circ}. Also, \\angle MF_{1}F_{2}=2\\angle MF_{2}F_{1}, hence \\angle MF_{2}F_{1}=30^{\\circ}, then \\angle F_{2}MF_{1}=90^{\\circ}, MF_{1}=F_{1}F_{2}\\cdot\\cos60^{\\circ}=2c\\cdot\\frac{1}{2}=c, MF_{2}=F_{1}F_{2}\\cdot\\sin60^{\\circ}=2c\\cdot\\frac{\\sqrt{3}}{2}=\\sqrt{3}c, e=\\frac{2c}{2a}=\\frac{2c}{MF_{1}+MF_{2}}=\\frac{2c}{\\sqrt{3}c+c}=\\sqrt{3}-1. [Knowledge Point] Examines the calculation of eccentricity, requiring students to have sharp observation skills, such as recognizing characteristics of lines. Considered a difficult problem." }, { "text": "Given that $O$ is the coordinate origin, point $P$ lies on the parabola $y^{2}=16x$, point $F$ is the focus of the parabola, and the area of $\\Delta OPF$ is $32$, then $|PF|$=?", "fact_expressions": "O: Origin;G: Parabola;Expression(G) = (y^2 = 16*x);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;Area(TriangleOf(O, P, F)) = 32", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "20", "fact_spans": "[[[2, 5]], [[16, 31], [38, 41]], [[16, 31]], [[11, 15]], [[11, 32]], [[33, 37]], [[33, 44]], [[46, 68]]]", "query_spans": "[[[70, 79]]]", "process": "From the parabola equation, the coordinates of the focus can be determined. Using the triangle area formula, the y-coordinate of point P can be found, and then the x-coordinate of point P is obtained. The result is calculated using the focal radius formula of the parabola. From the given condition: F(4,0), ∴ |OF| = 4. Let P(x,y), then S_{AOPF} = \\frac{1}{2}|OF|⋅|y| = 2|y| = 32, solving gives: |y| = 16, ∴ x = \\frac{y^{2}}{16} = 16, ∴ |PF| = 16 + 4 = 20" }, { "text": "Given the parabola $C$: $y^{2}=8x$ with focus $F$, and the intersection point of the directrix and the $x$-axis denoted as $K$. Point $A$ lies on $C$ such that $|AK|=\\sqrt{2}|AF|$. Then the area of $\\Delta AFK$ is?", "fact_expressions": "C: Parabola;A: Point;F: Point;K: Point;Expression(C) = (y^2 = 8*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = K;PointOnCurve(A, C);Abs(LineSegmentOf(A, K)) = sqrt(2)*Abs(LineSegmentOf(A, F))", "query_expressions": "Area(TriangleOf(A, F, K))", "answer_expressions": "8", "fact_spans": "[[[2, 21], [49, 52]], [[44, 48]], [[25, 28]], [[40, 43]], [[2, 21]], [[2, 28]], [[2, 43]], [[44, 53]], [[54, 75]]]", "query_spans": "[[[77, 96]]]", "process": "The parabola C: y^{2}=8x has focus F(2,0), and the intersection point of the directrix with the x-axis is K(-2,0). Let point A have coordinates (\\frac{y^{2}}{8},y). Then (\\frac{y^{2}}{8}+2)^{2}+y^{2}=2\\times((\\frac{y^{2}}{8}-2)^{2}+y^{2}). Solving gives y^{2}=16. Therefore, the area of \\DeltaAFK is \\frac{1}{2}\\times4\\times4=8." }, { "text": "Given that the minimum sum of distances from a point on the parabola $C$: $y^{2}=2 p x$ ($p>0$) to the focus $F$ and the point $(4,0)$ is $5$, find the equation of this parabola?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;K: Point;PointOnCurve(K, C);F: Point;Focus(C) = F;G: Point;Coordinate(G) = (4, 0);Min(Distance(K, F) + Distance(K, G)) = 5", "query_expressions": "Expression(C)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[2, 28], [62, 65]], [[2, 28]], [[10, 28]], [[10, 28]], [], [[2, 31]], [[34, 37]], [[2, 37]], [[38, 46]], [[38, 46]], [[2, 59]]]", "query_spans": "[[[62, 69]]]", "process": "According to the definition of a parabola, the sum of the distance from point P to the focus and to the point (4,0) equals the sum of the distance from point P to the directrix and to the point (4,0). The minimum value is the distance from the point (4,0) to the directrix, which is \\frac{p}{2}+4=5, so p=2. Therefore, the equation of the parabola is y^{2}=4x. This question mainly examines the equation and geometric properties of a parabola, as well as mathematical operation ability and the combination of number and shape thinking." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of ellipse $C$, point $P$ lies on the ellipse and satisfies $P F_{1}=2 P F_{2}$, $\\angle P F_{1} F_{2}=30^{\\circ}$, find the eccentricity of the ellipse.", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2);AngleOf(P, F1, F2) = ApplyUnit(30, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[18, 23], [35, 37], [99, 101]], [[30, 34]], [[2, 9]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 38]], [[42, 61]], [[63, 97]]]", "query_spans": "[[[99, 107]]]", "process": "In $\\triangle PF_{1}F_{2}$, by the law of sines, $\\sin\\angle PF_{2}F_{1}=1$, so $\\angle PF_{2}F_{1}=\\frac{\\pi}{2}$. Let $PF_{2}=1$, then $PF_{1}=2$, $F_{2}F_{1}=\\sqrt{3}$, therefore the eccentricity $e=\\frac{2c}{2a}=\\frac{\\sqrt{3}}{2}$." }, { "text": "The standard equation of a parabola with the directrix $x=1$ is?", "fact_expressions": "G: Parabola;Expression(Directrix(G))= (x = 1)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=-4*x", "fact_spans": "[[[10, 13]], [[0, 13]]]", "query_spans": "[[[10, 20]]]", "process": "The standard equation of a parabola with directrix $ x = 1 $ is $ y^{2} = -2px $, $ \\frac{p}{2} = 1 $, $ \\therefore y^{2} = -4x $." }, { "text": "It is known that $F$ is the focus of the parabola $C$: $y^{2}=4x$. A line passing through $F$ with slope $1$ intersects the parabola $C$ at points $A$ and $B$. Then the value of $|AB|$ is equal to?", "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) = 1;Intersection(G, C) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[6, 24], [43, 49]], [[40, 42]], [[50, 53]], [[54, 57]], [[2, 5], [29, 32]], [[6, 24]], [[2, 27]], [[28, 42]], [[33, 42]], [[40, 59]]]", "query_spans": "[[[61, 73]]]", "process": "" }, { "text": "If a point $M$ on the parabola $y^{2}=2 p x(p>0)$ has distances of $10$ and $6$ to the directrix and the axis of symmetry, respectively, then its standard equation is?", "fact_expressions": "G: Parabola;p: Number;M: Point;p>0;Expression(G) = (y**2 = 2*(p*x));PointOnCurve(M, G) = True;Distance(M, Directrix(G)) = 10;Distance(M, SymmetryAxis(G)) = 6", "query_expressions": "Expression(G)", "answer_expressions": "{y^2 = 4*x, y^2 = 36*x}", "fact_spans": "[[[1, 22], [51, 52]], [[4, 22]], [[25, 28]], [[4, 22]], [[1, 22]], [[1, 28]], [[1, 49]], [[1, 49]]]", "query_spans": "[[[51, 58]]]", "process": "The distance from point M to the axis of symmetry is 6, so assume the coordinates of point M are (x,6). Also, since the distance from point M to the directrix is 10, we have 6^{2}=2px and x+\\frac{p}{2}=10. Solving gives \\begin{cases}x=9,\\\\p=2\\end{cases} or \\begin{cases}x=1,\\\\p=18.\\end{cases} Therefore, when the x-coordinate of point M is 9, the equation of the parabola is y^2=4x; when the x-coordinate of point M is 1, the equation of the parabola is y^{2}=36x." }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$, let chord $AB$ pass through the focus, with endpoints $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": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;Endpoint(LineSegmentOf(A, B)) = {A, B};Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);IsChordOf(LineSegmentOf(A,B),G) = True;PointOnCurve(Focus(G),LineSegmentOf(A,B)) = True;x1: Number;y1: Number;x2: Number;y2: Number", "query_expressions": "(y1*y2)/(x1*x2)", "answer_expressions": "-4", "fact_spans": "[[[2, 23]], [[2, 23]], [[5, 23]], [[5, 23]], [[43, 61]], [[64, 82]], [[29, 82]], [[43, 61]], [[64, 82]], [[2, 34]], [[2, 34]], [[43, 61]], [[43, 61]], [[64, 82]], [[64, 82]]]", "query_spans": "[[[84, 121]]]", "process": "From the parabola $ y^{2} = 2px $ ($ p > 0 $), we obtain its focus coordinates as $ \\left( \\frac{p}{2}, 0 \\right) $. Let the equation of the line containing chord AB passing through the focus be $ x = my + \\frac{p}{2} $. From \n$$\n\\begin{cases}\nx = my + \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n$$\neliminating $ x $ and simplifying, we get $ y^{2} - 2pmy - p^{2} = 0 $. Thus, $ y_{1}y_{2} = -p^{2} $. Since points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $ lie on the parabola, \nso $ x_{1} = \\frac{y_{1}^{2}}{2p} $, $ x_{2} = \\frac{y_{2}^{2}}{2p} $, hence $ x_{1}x_{2} = \\frac{y_{1}^{2}}{2p} \\cdot \\frac{y_{2}^{2}}{2p} = \\frac{(y_{1}y_{2})^{2}}{4p^{2}} = \\frac{(-p^{2})^{2}}{4p^{2}} = \\frac{p^{4}}{4p^{2}} = \\frac{p^{2}}{4} $." }, { "text": "Given that a line with slope $1$ passes through the focus of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ and intersects the ellipse at points $A$ and $B$, what is the length of the segment $AB$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);H: Line;Slope(H) = 1;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "(2*sqrt(6))/5", "fact_spans": "[[[12, 39], [45, 47]], [[12, 39]], [[9, 11]], [[2, 11]], [[9, 42]], [[49, 52]], [[53, 56]], [[9, 58]]]", "query_spans": "[[[60, 71]]]", "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}$. A line passing through the right focus $F$ with slope $k$ $(k>0)$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{A F}=3 \\overrightarrow{F B}$, then $k=$?", "fact_expressions": "C: Ellipse;a: Number;b: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;k:Number;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(3)/2;RightFocus(C)=F;PointOnCurve(F, G);Slope(G)=k;k>0;Intersection(G, C) = {A,B};VectorOf(A, F) = 3*VectorOf(F, B)", "query_expressions": "k", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 59], [108, 111]], [[8, 59]], [[8, 59]], [[105, 107]], [[114, 117]], [[89, 92]], [[118, 121]], [[8, 59]], [[8, 59]], [[173, 176], [96, 104]], [[2, 59]], [[2, 84]], [[2, 92]], [[85, 107]], [[93, 107]], [[96, 104]], [[105, 123]], [[125, 170]]]", "query_spans": "[[[173, 178]]]", "process": "" }, { "text": "Given an ellipse with foci on the $x$-axis, $\\frac{x^{2}}{m}+\\frac{y^{2}}{9}=1$, has eccentricity $e=\\frac{1}{2}$. Find the real number $m=?$", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/9 + x^2/m = 1);e: Number;Eccentricity(G) = e;e = 1/2;m: Real;PointOnCurve(Focus(G), xAxis)", "query_expressions": "m", "answer_expressions": "12", "fact_spans": "[[[11, 48]], [[11, 48]], [[52, 67]], [[11, 67]], [[52, 67]], [[69, 74]], [[2, 48]]]", "query_spans": "[[[69, 76]]]", "process": "Problem Analysis: Directly use the given conditions to find the geometric quantities a, b, c of the ellipse, and then use the eccentricity formula to calculate and solve. For the ellipse with foci on the x-axis, \\frac{x^{2}}{m}+\\frac{y^{2}}{9}=1, we have a=\\sqrt{m}, b=3, c=\\sqrt{m-9}. \\because the eccentricity is e=\\frac{1}{2}, \\therefore \\frac{c}{a}=\\frac{\\sqrt{m-9}}{\\sqrt{m}}=\\frac{1}{2}. Solving gives m=12." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, with right focus $F$. Point $P$ is a moving point on the line $x=2$ in the first quadrant. The line $PF$ intersects one asymptote of the hyperbola $C$ at point $Q$ in the first quadrant. If $\\overrightarrow{F P}=2 \\overrightarrow{P Q}$, then $|F Q|$=?", "fact_expressions": "C: Hyperbola;G: Line;F: Point;P: Point;Q: Point;Expression(C) = (x^2/4 - y^2/12 = 1);Expression(G) = (x = 2);RightFocus(C) = F;Quadrant(P) = 1;PointOnCurve(P, G);Quadrant(Q) = 1;Intersection(LineOf(P, F), OneOf(Asymptote(C))) = Q;VectorOf(F, P) = 2*VectorOf(P, Q)", "query_expressions": "Abs(LineSegmentOf(F, Q))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 46], [85, 91]], [[60, 67]], [[51, 54]], [[55, 59]], [[107, 110]], [[2, 46]], [[60, 67]], [[2, 54]], [[55, 76]], [[55, 76]], [[77, 110]], [[77, 110]], [[112, 157]]]", "query_spans": "[[[159, 168]]]", "process": "In the hyperbola $ C: \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 $, $ a^{2} = 4 $, $ b^{2} = 12 $, then $ c^{2} = a^{2} + b^{2} = 16 $, that is, $ c = 4 $, so the right focus is $ F(4,0) $. Let $ P(2,y_{0}) $, $ y_{0} > 0 $, $ Q(x_{1},y_{1}) $. The asymptotes of the hyperbola $ C: \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 $ are given by: $ y = \\pm\\sqrt{3}x $. From $ \\overrightarrow{FP} = 2\\overrightarrow{PQ} $, it follows that point $ Q $ lies on the line $ y = \\sqrt{3}x $. $ \\overrightarrow{FP} = (-2,y_{0}) $, $ \\overrightarrow{PQ} = (x_{1}-2,y_{1}-y_{0}) $. Therefore, $ -2 = 2(x_{1}-2) $, solving gives $ x_{1} = 1 $, then $ y_{1} = \\sqrt{3} \\times 1 = \\sqrt{3} $." }, { "text": "Given that point $P(x_1, y_1)$ moves on the ellipse $\\frac{x^{2}}{3}+\\frac{2 y^{2}}{3}=1$, then the minimum value of $\\frac{1}{x^{2}}+\\frac{2}{1+y^{2}}$ is?", "fact_expressions": "G: Ellipse;P: Point;x1: Number;y1: Number;Expression(G) = (x^2/3 + (2*y^2)/3 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G)", "query_expressions": "Min(2/(y^2 + 1) + 1/(x^2))", "answer_expressions": "9/5", "fact_spans": "[[[15, 54]], [[2, 14]], [[3, 14]], [[3, 14]], [[15, 54]], [[2, 14]], [[2, 55]]]", "query_spans": "[[[59, 99]]]", "process": "Analysis: From the given condition, we have $x^{2}+2y^{2}=3$, that is, $x^{2}=3-2y^{2}$, then $\\frac{1}{2}+\\frac{2}{1+y^{2}}=\\frac{1}{3-2y^{2}}+\\frac{2}{1+y^{2}}=\\frac{1}{5}\\times\\left(\\frac{1}{3-2y^{2}}+\\frac{4}{2+2y^{2}}\\right)(3-2y^{2}+2+2y^{2})$. Using the basic inequality, the minimum value can be found. Point $P(x,y)$ moves on the ellipse $x^{2}+2y^{2}=3$, $\\therefore x^{2}+2y^{2}=3$, that is, $x^{2}=3-2y^{2}$. Then $\\frac{1}{x^{2}}+\\frac{2}{1+y^{2}}=\\frac{1}{5}\\times\\left(\\frac{1}{3-2y^{2}}+\\frac{4}{2+2y^{2}}\\right)(3-2y^{2}+2+2y^{2})=\\frac{1}{5}\\left(5+\\frac{2+2y^{2}}{3-2y^{2}}+\\frac{4(3-2y^{2})}{2+2y^{2}}\\right)\\geqslant\\frac{1}{5}(5+2\\sqrt{4})=\\frac{9}{5}$, that is, the required minimum value is $\\frac{9}{5}$." }, { "text": "Let the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0)$ have its right vertex and upper vertex denoted as $A$ and $B$, respectively. The line $AB$ intersects the line $y=-x$ at point $P$. If point $P$ lies on the parabola $y^{2}=-a x$, then the eccentricity of ellipse $M$ is equal to?", "fact_expressions": "G: Parabola;a: Number;M: Ellipse;b: Number;H: Line;B: Point;A: Point;P: Point;Expression(G) = (y^2 = -a*x);a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = -x);RightVertex(M) = A;UpperVertex(M) = B;Intersection(LineOf(A,B), H) = P;PointOnCurve(P, G)", "query_expressions": "Eccentricity(M)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[107, 122]], [[7, 59]], [[1, 59], [125, 130]], [[7, 59]], [[85, 93]], [[73, 76]], [[69, 72]], [[96, 100], [102, 106]], [[107, 122]], [[7, 59]], [[7, 59]], [[1, 59]], [[85, 93]], [[1, 76]], [[1, 76]], [[77, 100]], [[102, 123]]]", "query_spans": "[[[125, 137]]]", "process": "" }, { "text": "Given points $A(0,1)$, $C(0,5)$, the locus of moving point $M$ is $x^{2}=4 y$, and moving point $N$ satisfies $|N C|=2$, then the minimum value of $|M N|+\\frac{1}{2}|N A|$ is?", "fact_expressions": "A: Point;C: Point;N: Point;M: Point;Coordinate(A) = (0, 1);Coordinate(C) = (0, 5);E: Curve;Expression(E) = (x^2 = 4*y);Locus(M) = E;Abs(LineSegmentOf(N, C)) = 2", "query_expressions": "Min(Abs(LineSegmentOf(M, N)) + Abs(LineSegmentOf(N, A))/2)", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 11]], [[13, 21]], [[47, 50]], [[25, 28]], [[2, 11]], [[13, 21]], [[32, 43]], [[32, 43]], [[25, 43]], [[52, 61]]]", "query_spans": "[[[64, 94]]]", "process": "According to the problem, draw the graph: the moving point N satisfies |NC| = 2. Let N(x, y), then the trajectory of N is the circle x^{2} + y^{2} - 10y + 21 = 0. Let Q(m, n), and |QN| = \\frac{1}{2}|NA|, we obtain \\sqrt{(x - m)^{2} + (y - n)^{2}} = \\frac{1}{2}\\sqrt{x^{2} + (y - 1)^{2}}. Simplifying yields 3x^{2} + 3y^{2} - 8mx + (2 - 8n)y + 4m^{2} + 4n^{2} - 1 = 0. Since the equation of N is also 3x^{2} + 3y^{2} - 30y + 63 = 0, we get m = 0, n = 4, so Q(0, 4). Then the minimum value of |MN| + \\frac{1}{2}|NA| is the minimum value of |MN| + |QM|. The minimum occurs when Q, N, M are collinear and lie on the normal line of the parabola. Let M(s, t). Since the derivative of y = \\frac{1}{4}x^{2} is y' = \\frac{1}{2}x, we have \\frac{1}{2}s \\cdot \\frac{t - 4}{s} = -1, solving gives t = 2, s = \\pm 2\\sqrt{2}, so M(\\pm 2\\sqrt{2}, 2), and thus |QM| = \\sqrt{4 + 8} = 2\\sqrt{3}." }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and a line $l$ passing through $F$ intersects the parabola at points $A$ and $B$. Let $P(-\\frac{5}{3}, 0)$. If $PB \\perp AB$, then $|AF|$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l) = True;Intersection(l,G) = {A,B};B: Point;A: Point;P: Point;Coordinate(P) = (-5/3, 0);IsPerpendicular(LineSegmentOf(P, B), LineSegmentOf(A, B))", "query_expressions": "Abs(LineSegmentOf(A, F))", "answer_expressions": "4", "fact_spans": "[[[2, 16], [35, 38]], [[2, 16]], [[20, 23], [25, 28]], [[2, 23]], [[29, 34]], [[24, 34]], [[29, 50]], [[45, 48]], [[41, 44]], [[51, 71]], [[51, 71]], [[73, 88]]]", "query_spans": "[[[90, 99]]]", "process": "" }, { "text": "The circle passing through the two intersection points of the line $y=2$ and the parabola $x^{2}=8 y$, and tangent to the directrix of the parabola is?", "fact_expressions": "G: Parabola;H: Circle;L: Line;P:Point;D:Point;Expression(G) = (x^2 = 8*y);Expression(L) = (y = 2);Intersection(G,L)={P,D};PointOnCurve(P,H);PointOnCurve(D,H);IsTangent(Directrix(G),H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2+(y-2)^2=16", "fact_spans": "[[[9, 23], [32, 35]], [[40, 41]], [[1, 8]], [], [], [[9, 23]], [[1, 8]], [[1, 28]], [[0, 41]], [[0, 41]], [[31, 41]]]", "query_spans": "[[[40, 44]]]", "process": "The directrix of the parabola is given by y = -2. Substituting y = 2 into the equation of the parabola, we obtain x = \\pm4, so the coordinates of the two intersection points are known. Since the circle is tangent to y = -2, by symmetry of the circle, the center lies on the y-axis. Let the center be (0, b). Then \\frac{2}{2} + (b - 2)^{2} = |b - (-2)|, solving gives b = 2. Hence, the center is (0, 2) and the radius is 4. Therefore, the equation of the circle is x^{2} + (y - 2)^{2} = 16." }, { "text": "It is known that point $F$ is the left focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A$ is the right vertex, and point $P$ is a point on the ellipse such that $P F \\perp x$-axis. If $|P F|=\\frac{1}{4}|A F|$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F: Point;A:Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G)=F;RightVertex(G)=A;PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P,F),xAxis);Abs(LineSegmentOf(P,F))=(1/4)*Abs(LineSegmentOf(A,F))", "query_expressions": "Eccentricity(G)", "answer_expressions": "3/4", "fact_spans": "[[[7, 59], [77, 79], [126, 128]], [[9, 59]], [[9, 59]], [[72, 76]], [[2, 6]], [[64, 67]], [[9, 59]], [[9, 59]], [[7, 59]], [[2, 63]], [[7, 71]], [[72, 82]], [[83, 97]], [[99, 123]]]", "query_spans": "[[[126, 134]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, respectively, $P$ a point on the ellipse, and $M$ the midpoint of $F_{1} P$. If $|O M|=1$, then what is the distance from point $P$ to the left focus of the ellipse?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);M: Point;MidPoint(LineSegmentOf(F1, P)) = M;Abs(LineSegmentOf(O, M)) = 1;O: Origin", "query_expressions": "Distance(P, LeftFocus(G))", "answer_expressions": "4", "fact_spans": "[[[19, 56], [67, 69], [106, 108]], [[19, 56]], [[1, 8]], [[9, 16]], [[1, 62]], [[1, 62]], [[63, 66], [101, 105]], [[63, 72]], [[73, 76]], [[73, 89]], [[90, 99]], [[90, 99]]]", "query_spans": "[[[101, 116]]]", "process": "As shown in the figure, OM is the midline of AF₁F₂P. From |OM| = 1 we get: PF₂ = 2. From the ellipse \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1 we get: a^{2} = 9, that is: a = 3. Also PF₂ + PF₁ = 2a = 6, solving gives: PF = 4." }, { "text": "Given the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{m}=1$, with the major axis on the $y$-axis, if the focal distance is $4$, then what is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/8 + y^2/m = 1);OverlappingLine(MajorAxis(G),yAxis);FocalLength(G) = 4", "query_expressions": "m", "answer_expressions": "12", "fact_spans": "[[[2, 39]], [[58, 61]], [[2, 39]], [[2, 48]], [[2, 56]]]", "query_spans": "[[[58, 64]]]", "process": "" }, { "text": "A point $A(m, 2 \\sqrt{3})$ on the parabola $y^{2}=2 p x(p>0)$ is at a distance of $4$ from the focus. Find the value of the real number $p$.", "fact_expressions": "G: Parabola;p: Real;A: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (m, 2*sqrt(3));PointOnCurve(A, G);Distance(A, Focus(G)) = 4", "query_expressions": "p", "answer_expressions": "{2,6}", "fact_spans": "[[[0, 21]], [[54, 59]], [[24, 42]], [[3, 21]], [[0, 21]], [[24, 42]], [[0, 42]], [[0, 52]]]", "query_spans": "[[[54, 63]]]", "process": "Since the distance from point A(m, 2\\sqrt{3}) on the parabola y^{2} = 2px (p > 0) to the focus is 4, we have \\begin{cases} m + \\frac{p}{2} = 4 \\\\ 12 = 2pm \\end{cases}, solving yields \\begin{cases} p = 2 \\\\ m = 3 \\end{cases} or \\begin{cases} p = 6 \\\\ m = 1 \\end{cases}, therefore the value of p is 2 or 6." }, { "text": "The equations of the asymptotes of the hyperbola $3 x^{2}-y^{2}=3$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (3*x^2 - y^2 = 3)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(3)*x", "fact_spans": "[[[0, 20]], [[0, 20]]]", "query_spans": "[[[0, 28]]]", "process": "From the hyperbola equation $ 3x^{2} - y^{2} = -3 $, we have $ \\frac{y^{2}}{3} - x^{2} = 1 $, so $ a = \\sqrt{3} $, $ b = 1 $, therefore the asymptotes of the hyperbola are $ y = \\pm\\frac{a}{b}x = \\pm\\sqrt{3}x $, that is, $ \\sqrt{3}x \\pm y = 0 $." }, { "text": "Given that $F$ is the focus of the parabola $y^{2}=4x$, a line passing through point $F$ intersects the parabola at points $A$ and $B$. If point $A$ lies in the first quadrant and $|FA|=3|FB|$, then what are the coordinates of the midpoint of segment $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);Intersection(H, G) = {A, B};Quadrant(A)=1;Abs(LineSegmentOf(F, A)) = 3*Abs(LineSegmentOf(F, B))", "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A,B)))", "answer_expressions": "(5/3,2*sqrt(3)/3)", "fact_spans": "[[[6, 20], [34, 37]], [[30, 32]], [[38, 41], [49, 53]], [[42, 45]], [[2, 5], [25, 29]], [[6, 20]], [[2, 23]], [[24, 32]], [[30, 47]], [[49, 58]], [[60, 74]]]", "query_spans": "[[[76, 90]]]", "process": "According to the problem and the geometric properties of the parabola, the geometric relationship is shown in the following figure: as shown, draw perpendiculars from A and B to the directrix, with feet C and D respectively; draw a perpendicular from B to AC, with foot E. Let BF = m, then AF = 3m, AB = 4m. By the definition of the parabola, AC = 3m, BD = m. Hence in right triangle ABE, AE = 2m = \\frac{1}{2}AB, so \\angle EAB = 60^{\\circ}, that is, the inclination angle of line AB is 60^{\\circ}. Therefore, we can assume the equation of line AB is y = \\sqrt{3}(x - 1). Substituting into y^{2} = 4x, we obtain 3x^{2} - 10x + 3 = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), and let M(x_{0}, y_{0}) be the midpoint of AB. Then x_{1} + x_{2} = \\frac{10}{3}, so x_{0} = \\frac{x_{1} + x_{2}}{2} = \\frac{5}{3}. Thus y_{0} = \\sqrt{3}(x_{0} - 1) = \\sqrt{3}(\\frac{5}{3} - 1) = \\frac{2\\sqrt{3}}{3}. Hence the coordinates of the midpoint of AB are (\\frac{5}{3}, \\frac{2\\sqrt{3}}{3})." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0 , b>0)$ has its right focus at $F$, and its two asymptotes are $l_{1}$: $y=\\frac{b}{a} x$, $l_{2}$: $y=-\\frac{b}{a} x$. A perpendicular is drawn from $F$ to $l_{1}$, with foot of perpendicular at $M$, and this perpendicular intersects $l_{2}$ at point $N$, where $O$ is the origin. If $|O F|=|F N|$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;O: Origin;F: Point;N: Point;M:Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;l1:Line;l2:Line;Expression(l1)=(y=(b/a)*x);Expression(l2)=(y=-(b/a)*x);L:Line;PointOnCurve(F,L);IsPerpendicular(l1,L);FootPoint(l1,L)=M;Intersection(L,l2)=N;Abs(LineSegmentOf(O, F)) = Abs(LineSegmentOf(F, N));Asymptote(C)={l1,l2}", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 66], [206, 212]], [[10, 66]], [[10, 66]], [[180, 183]], [[71, 74], [141, 144]], [[175, 179]], [[159, 162]], [[10, 66]], [[10, 66]], [[2, 66]], [[2, 74]], [[82, 108], [145, 152]], [[111, 138], [167, 174]], [[82, 108]], [[111, 138]], [], [[140, 155]], [[140, 155]], [[139, 162]], [[140, 179]], [[191, 204]], [[2, 138]]]", "query_spans": "[[[206, 218]]]", "process": "Since $ l_{1}: y = \\frac{b}{a}x' $, we can assume the equation of line $ FM $ is $ y = -\\frac{a}{b}(x - c) $. Solving the system \n\\[\n\\begin{cases}\ny = -\\frac{a}{b}(x - c) \\\\\ny = -\\frac{b}{a}x\n\\end{cases}\n\\]\ngives the coordinates of point $ N $ as $ \\left( \\frac{a^{2}c}{a^{2} - b^{2}}, -\\frac{abc}{a^{2} - b^{2}} \\right) $, and since $ F(c, 0) $, \n$ |FN| = \\sqrt{ \\left( c - \\frac{a^{2}c}{a^{2} - b^{2}} \\right)^{2} + \\left( 0 + \\frac{abc}{a^{2} - b^{2}} \\right)^{2} } = |OF| = c $. Simplifying yields: $ 4a^{2} = 3c^{2} $, so $ e = \\frac{c}{a} = \\sqrt{\\frac{4}{3}} = \\frac{2\\sqrt{3}}{3} $." }, { "text": "The hyperbola has its center at the origin, foci on the $y$-axis, a focal distance of $16$, and one asymptote given by $y=\\frac{3}{\\sqrt{7}} x$. Then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;PointOnCurve(Focus(G), yAxis) = True;FocalLength(G) = 16;Expression(OneOf(Asymptote(G))) = (y = x*3/sqrt(7))", "query_expressions": "Expression(G)", "answer_expressions": "y^2/36-x^2/28=1", "fact_spans": "[[[0, 3], [62, 65]], [[7, 9]], [[0, 9]], [[0, 18]], [[0, 26]], [[0, 60]]]", "query_spans": "[[[62, 69]]]", "process": "Test analysis: The focal length is 16, so c = 8. One asymptote equation is y = \\frac{3}{\\sqrt{7}}x, that is, \\frac{3}{\\sqrt{7}} = \\frac{a}{b}, solving gives a = 6, b = 2\\sqrt{7}. The hyperbola equation is: \\frac{y^{2}}{36} - \\frac{x^{2}}{28} = 1" }, { "text": "If the focal distance of the hyperbola $C$: $\\frac{x^{2}}{m}-y^{2}=1$ is $2 \\sqrt{2}$, then what is the equation of the asymptotes of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;m: Number;Expression(C) = (-y^2 + x^2/m = 1);FocalLength(C) = 2*sqrt(2)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*x", "fact_spans": "[[[1, 34], [52, 58]], [[9, 34]], [[1, 34]], [[1, 50]]]", "query_spans": "[[[52, 66]]]", "process": "From the problem, we have: $2\\sqrt{1+m}=2\\sqrt{2}$, solving gives: $m=1$, then the asymptotes of the hyperbola are: $y=\\pm x$" }, { "text": "Given point $M(1,-1)$ and parabola $C$: $y=\\frac{1}{4} x^{2}$, a line passing through the focus of $C$ with slope $k$ intersects $C$ at points $A$ and $B$. If $\\overrightarrow{A M} \\cdot \\overrightarrow{B M}=0$, then $k=?$", "fact_expressions": "C: Parabola;G: Line;M: Point;A: Point;B: Point;Expression(C) = (y = x^2/4);Coordinate(M) = (1, -1);PointOnCurve(Focus(C), G);Slope(G) = k;Intersection(G, C) = {A, B};DotProduct(VectorOf(A, M), VectorOf(B, M)) = 0;k:Number", "query_expressions": "k", "answer_expressions": "1/2", "fact_spans": "[[[13, 42], [44, 47], [61, 64]], [[58, 60]], [[2, 12]], [[66, 69]], [[70, 73]], [[13, 42]], [[2, 12]], [[43, 60]], [[51, 60]], [[58, 75]], [[77, 128]], [[130, 133], [54, 57]]]", "query_spans": "[[[130, 135]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), and the equation of line AB be y = kx + 1. By solving the system of equations consisting of the line and the parabola, eliminate y, use the relationship between roots and coefficients to express x_{1}+x_{2}, x_{1}x_{2}, thereby express y_{1}+y_{2}, y_{1}y_{2}, and then express \\overrightarrow{AM}\\cdot\\overrightarrow{BM}; set up an equation to solve for the value of the slope. [Solution] The standard form of the parabola is x^{2} = 4y, with focus coordinates (0,1). Let A(x_{1},y_{1}), B(x_{2},y_{2}), and the equation of line AB be y = kx + 1. Substituting into the parabola equation gives x^{2} - 4kx - 4 = 0, so x_{1}+x_{2} = 4k, x_{1}x_{2} = -4. y_{1}+y_{2} = k(x_{1}+x_{2}) + 2 = 4k^{2} + 2, y_{1}y_{2} = \\frac{1}{16}x_{1}^{2}x_{2}^{2} = 1. Therefore, \\overrightarrow{AM}\\cdot\\overrightarrow{BM} = (1-x_{1}, -1-y_{1}) \\cdot (1-x_{2}, -1-y_{2}) = 1 + x_{1}x_{2} + y_{1}y_{2} - (x_{1}+x_{2}) + (y_{1}+y_{2}). Substituting gives 4k^{2} - 4k + 1 = 0 \\Rightarrow k = \\frac{1}{2}." }, { "text": "Given a point $M(x_{1}, y_{1})$ on the parabola $x^{2} = 4y$ such that the distance from $M$ to its focus is $3$, what is the distance from point $M$ to the origin?", "fact_expressions": "G: Parabola;M: Point;x1:Number;y1:Number;Expression(G) = (x^2 = 4*y);Coordinate(M) = (x1, y1);PointOnCurve(M, G);Distance(M, Focus(G)) = 3;O:Origin", "query_expressions": "Distance(M,O)", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 16], [37, 38]], [[19, 36], [49, 53]], [[19, 36]], [[19, 36]], [[2, 16]], [[19, 36]], [[2, 36]], [[19, 47]], [[54, 56]]]", "query_spans": "[[[49, 60]]]", "process": "According to the parabola equation, the focus coordinates are (0,1) and the directrix equation is y = -1. By the definition of a parabola, we obtain y_{1} = 2. Substituting into the parabola equation gives x = \\pm2\\sqrt{2}. Therefore, the coordinates of point M are (\\pm2\\sqrt{2}, 2), and the distance from point M to the origin is |OM| = \\sqrt{2^{2} + (\\pm2\\sqrt{2})^{2}} = 2\\sqrt{3}." }, { "text": "The equation of the directrix of the parabola $y^{2}=-8 x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -8*x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=2", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "y^{2}=-8x, where 2p=8, therefore \\frac{p}{2}=2, thus the equation of the directrix is x=2" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, what is the distance from a focus of $C$ to one of its asymptotes?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/6 - y^2/3 = 1)", "query_expressions": "Distance(Focus(C), Asymptote(C))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 45], [47, 50], [54, 55]], [[2, 45]]]", "query_spans": "[[[47, 63]]]", "process": "It is easy to obtain that for the hyperbola $ C: \\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1 $, $ b=\\sqrt{3} $, hence the distance from the foci of $ C $ to its asymptotes" }, { "text": "Given points $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ lie on the parabola $C$: $y^{2}=4x$, the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $D(m, 0)$. If $x_{1}+x_{2}=2$, then $m=$?", "fact_expressions": "C: Parabola;A: Point;B: Point;D: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Coordinate(D) = (m, 0);x1:Number;x2:Number;y1:Number;y2:Number;m:Number;PointOnCurve(A, C);PointOnCurve(B, C);Intersection(PerpendicularBisector(LineSegmentOf(A,B)), xAxis) = D;x1 + x2 = 2", "query_expressions": "m", "answer_expressions": "3", "fact_spans": "[[[39, 58]], [[2, 19]], [[21, 38]], [[83, 93]], [[39, 58]], [[2, 19]], [[21, 38]], [[83, 93]], [[2, 19]], [[21, 38]], [[2, 19]], [[21, 38]], [[112, 115]], [[2, 61]], [[2, 61]], [[62, 93]], [[95, 110]]]", "query_spans": "[[[112, 117]]]", "process": "Given that A(x₁, y₁) and B(x₂, y₂) are two points on the parabola C: y² = 4x, then \n\\begin{cases} y₁² = 4x₁ \\\\ y₂² = 4x₂ \\end{cases}, \nsubtracting these two equations gives (y₁ + y₂)(y₁ - y₂) = 4(x₁ - x₂). Since the perpendicular bisector of segment AB intersects the x-axis, the slope of line AB exists and is not zero. Therefore, \n\\frac{y₁ - y₂}{x₁ - x₂} = \\frac{4}{y₁ + y₂}. \nThe midpoint of segment AB has coordinates (\\frac{x₁ + x₂}{2}, \\frac{y₁ + y₂}{2}), so the slope of the perpendicular bisector of segment AB is -. Therefore, \n\\frac{y₁ - y₂}{x₁ - x₂} \\cdot \\frac{y₁ + y₂}{1 - m} = \\frac{y}{y₁} \\cdot \\frac{4}{y₁ + y₂}, \nsolving gives m = 3." }, { "text": "Given fixed point $A(4,0)$ and a moving point $B$ on the circle $x^{2}+y^{2}=4$, a moving point $P(x, y)$ satisfies $\\overrightarrow{O A}+\\overrightarrow{O B}=2 \\overrightarrow{O P}$. Then the trajectory equation of point $P$ is?", "fact_expressions": "G: Circle;A: Point;P: Point;O: Origin;B: Point;x1:Number;y1:Number;Expression(G) = (x^2 + y^2 = 4);Coordinate(A) = (4, 0);Coordinate(P) = (x1, y1);PointOnCurve(B, G);VectorOf(O, A) + VectorOf(O, B) = 2*VectorOf(O, P)", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x-2)^2+y^2=1", "fact_spans": "[[[13, 29]], [[4, 12]], [[39, 48], [118, 122]], [[50, 116]], [[33, 36]], [[39, 48]], [[39, 48]], [[13, 29]], [[4, 12]], [[39, 48]], [[13, 36]], [[50, 116]]]", "query_spans": "[[[118, 129]]]", "process": "Since fixed point A(4,0) and moving point B on the circle x^{2}+y^{2}=4, let moving point P(x,y) and point B(m,n) on the circle, then \\because\\overrightarrow{OA}+\\overrightarrow{OB}=2\\overrightarrow{OP}, \\therefore(m+4,n)=(2x,2y), \\therefore m=2x-4, n=2y, \\therefore m^{2}+n^{2}=4, i.e. (x-2)^{2}+y^{2}=1" }, { "text": "The distance from a focus of the hyperbola $\\frac{x^{2}}{4}+\\frac{y^{2}}{k}=1$ to its asymptote is $3$. Find the value of the real number $k$.", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (x^2/4 + y^2/k = 1);Distance(OneOf(Focus(G)),Asymptote(G))=3", "query_expressions": "k", "answer_expressions": "-9", "fact_spans": "[[[0, 38], [44, 45]], [[56, 61]], [[0, 38]], [[0, 54]]]", "query_spans": "[[[56, 65]]]", "process": "The equation of the hyperbola is $\\frac{x^2}{4}-\\frac{y^{2}}{-k}=1$. Take one asymptote of the hyperbola as $y=\\frac{\\sqrt{-k}}{2}x$, that is, $-kx-2y=0$; one focus of the hyperbola is $(\\sqrt{4-k},0)$. According to the problem, $\\frac{|-k\\sqrt{4-k}|}{\\sqrt{-k+4}}=3$, solving gives $k=-9$ or $k=4$ (discarded). Answer: $-9$" }, { "text": "Given that the line $x=m$ intersects the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) at points $A$ and $B$, respectively. If the area of $\\triangle A O B$ ($O$ is the origin) is $\\sqrt{2}$, and the eccentricity of the hyperbola $C$ is $\\sqrt{3}$, then $m=?$", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;m: Number;A: Point;O: Origin;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x = m);l1: Line;l2: Line;Asymptote(C) = {l1, l2} ;Intersection(G, l1) = A;Intersection(G, l2) = B;Area(TriangleOf(A, O, B)) = sqrt(2);Eccentricity(C) = sqrt(3)", "query_expressions": "m", "answer_expressions": "pm*1", "fact_spans": "[[[10, 71], [136, 142]], [[18, 71]], [[18, 71]], [[2, 9]], [[159, 162]], [[81, 84]], [[110, 113]], [[85, 88]], [[18, 71]], [[18, 71]], [[10, 71]], [[2, 9]], [], [], [[10, 77]], [[2, 90]], [[2, 90]], [[92, 134]], [[136, 157]]]", "query_spans": "[[[159, 164]]]", "process": "The asymptotes of the hyperbola are given by $ y = \\pm\\frac{b}{a}x $. By solving the system of equations, the coordinates of points $ A $ and $ B $ are obtained, and $ |AB| = \\left|\\frac{2b}{a}m\\right| $. Given that the eccentricity of the hyperbola is $ \\sqrt{3} $, it follows that $ \\frac{b}{a} = \\sqrt{2} $, so $ |AB| = 2\\sqrt{2}|m| $. Using the area formula, the solution can be found. [Detailed Solution] For the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $), the equations of the asymptotes are $ y = \\pm\\frac{b}{a}x $. Solving the system \n$$\n\\begin{cases}\nx = m \\\\\ny = \\frac{b}{a}x\n\\end{cases},\n$$\nwe get \n$$\n\\begin{cases}\nx = m \\\\\ny = \\frac{bm}{a}\n\\end{cases};\n$$\nsolving the system \n$$\n\\begin{cases}\nx = m \\\\\ny = -\\frac{b}{a}x\n\\end{cases},\n$$\nwe get \n$$\n\\begin{cases}\nx = m \\\\\ny = -\\frac{bm}{a}\n\\end{cases},\n$$\nthus $ |AB| = \\left|\\frac{2b}{a}m\\right| $. Since the eccentricity of the hyperbola is $ \\sqrt{3} $, we have $ \\frac{b^{2}}{a^{2}} = \\frac{c^{2}-a^{2}}{a^{2}} = e^{2} - 1 = 2 $, hence $ \\frac{b}{a} = \\sqrt{2} $, so $ |AB| = |2\\sqrt{2}m| $. Therefore, $ S_{\\triangle AOB} = \\frac{1}{2} \\times |2\\sqrt{2}m| \\times |m| = \\sqrt{2} $, solving gives $ m = \\pm1 $." }, { "text": "The distance from the focus to the directrix of the parabola $x^{2}=4 a y(a>0)$ is?", "fact_expressions": "G: Parabola;a: Number;a>0;Expression(G) = (x^2 = 4*(a*y))", "query_expressions": "Distance(Focus(G),Directrix(G))", "answer_expressions": "2*a", "fact_spans": "[[[0, 21], [25, 26]], [[3, 21]], [[3, 21]], [[0, 21]]]", "query_spans": "[[[0, 33]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a}+\\frac{y^{2}}{9}=1$ has a focal distance of $8$, then $a$=?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/9 + x^2/a = 1);a: Number;FocalLength(C) = 8", "query_expressions": "a", "answer_expressions": "25", "fact_spans": "[[[2, 44]], [[2, 44]], [[53, 56]], [[2, 51]]]", "query_spans": "[[[53, 58]]]", "process": "According to the problem, 2c=8, so c=4; if the foci of the ellipse lie on the x-axis, that is, when a>9, then a=9+16=25; if the foci of the ellipse lie on the y-axis, that is, when 0 0; \nlet \n{ x₀ = √2 cos( \n{ y₀ = 1 + sinθ \nx₀/y₀ √2 + sinθ + 3sinθ − 9| = |√17 cos(θ+φ) − 9| / √13, \nwhere θ ≠ 2kπ − π/2, k ∈ Z, then the distance from point M(x₀,y₀) to the line 2x+3y−12=0 is \nso d ≤ (9+√17)/√13; \nbecause by the trapezoid midline property, the sum of distances from A and B to the line 2x+3y−12=0 is twice the distance from point M(x₀,y₀) to the line 2x+3y−12=0. \nThus |2x₁+3y₁−12|+|2x₂+3y₂−12| = 2√13 d ≤ 18+2√17. \nIn summary, the maximum value of |2x₁+3y₁−12|+|2x₂+3y₂−12| is 18+2√17. \nThis problem mainly examines maximum value problems in ellipses, has strong comprehensiveness, high difficulty, and emphasizes the awareness of substitution and core competence in mathematical operations." }, { "text": "In $\\triangle A B C$, points $A(-3,0)$, $B(3,0)$, and point $C$ lies on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. Then the perimeter of $\\triangle A B C$ is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;C: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(A) = (-3, 0);Coordinate(B) = (3, 0);PointOnCurve(C, G)", "query_expressions": "Perimeter(TriangleOf(A, B, C))", "answer_expressions": "16", "fact_spans": "[[[47, 86]], [[20, 30]], [[33, 41]], [[42, 46]], [[47, 86]], [[20, 30]], [[33, 41]], [[42, 87]]]", "query_spans": "[[[89, 111]]]", "process": "From the equation of the ellipse, we know that a=5, b=4, then c=\\sqrt{a^{2}-b^{2}}=3, so A and B are the two foci of the ellipse. Therefore, the perimeter of \\triangle ABC is l=CA+CB+AB=2a+2c=16." }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and point $M$ is a point on the parabola distinct from the vertex, $\\overrightarrow{O M}=2 \\overrightarrow{O N}$ (where point $O$ is the origin), a line perpendicular to line $OM$ is drawn through point $N$ and intersects the $x$-axis at point $P$. Then $2|O P|-|M F|$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;M: Point;PointOnCurve(M, G);Negation(Vertex(G) = M);O: Origin;VectorOf(O, M) = 2*VectorOf(O, N);N: Point;P: Point;H: Line;PointOnCurve(N,H);IsPerpendicular(LineOf(O,M),H);Intersection(H,xAxis) = P", "query_expressions": "-Abs(LineSegmentOf(M, F)) + 2*Abs(LineSegmentOf(O, P))", "answer_expressions": "3", "fact_spans": "[[[2, 16], [29, 32]], [[2, 16]], [[20, 23]], [[2, 23]], [[24, 28]], [[24, 40]], [[24, 40]], [[87, 91]], [[41, 86]], [[99, 103]], [[121, 125]], [], [[98, 114]], [[98, 114]], [[98, 125]]]", "query_spans": "[[[127, 143]]]", "process": "According to the problem, let M(\\frac{y_{0}^{2}}{4}, y_{0}), and since \\overrightarrow{OM} = 2\\overrightarrow{ON}, it follows that N is the midpoint of OM and N(\\frac{y_{0}^{2}}{8}, \\frac{y_{0}}{2}). Then k_{OM} = \\frac{4}{y_{0}}. It is easy to obtain the equation of the line NP perpendicular to OM as y - \\frac{y_{0}}{2} = -\\frac{y_{0}}{4}(x - \\frac{y_{0}^{2}}{8}). Setting y = 0, we get x = \\frac{y_{0}^{2}}{8} + 2, so P(\\frac{y_{0}^{2}}{8} + 2, 0). By the definition of the parabola, MF = \\frac{y_{0}^{2}}{4} + 1. Hence, 2|OP| - |MF| = 2(\\frac{y_{0}^{2}}{8} + 2) - (\\frac{y_{0}^{2}}{4} + 1) = 3." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$, $F_{1}$, $F_{2}$ are its two foci, and $P$ is an arbitrary point on $C$. Then the maximum value of $P F_{1} \\cdot P F_{2}$ is?", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2 = 1);Focus(C) = {F1, F2};PointOnCurve(P, C)", "query_expressions": "Max(LineSegmentOf(P, F1)*LineSegmentOf(P, F2))", "answer_expressions": "1", "fact_spans": "[[[2, 34], [55, 56], [65, 68]], [[61, 64]], [[37, 44]], [[47, 54]], [[2, 34]], [[36, 60]], [[61, 73]]]", "query_spans": "[[[75, 104]]]", "process": "According to the problem, F_{1}(-\\sqrt{3},0), F_{2}(\\sqrt{3},0). Let P(x,y), then \\frac{x^{2}}{4}+y^{2}=1, that is, y^{2}=1-\\frac{x^{2}}{4}. \\therefore PF_{1}\\cdot PF_{2}=(-\\sqrt{3}-x,-y)\\cdot(\\sqrt{3}-x,-y)=x^{2}-3+y^{2}=x^{2}-3+1-\\frac{x^{2}}{4}=\\frac{3}{4}x^{2}-2\\leqslant\\frac{3}{4}\\times4-2=1" }, { "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. If $|AF|=3$, then $|BF|=$?", "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};Abs(LineSegmentOf(A, F)) = 3", "query_expressions": "Abs(LineSegmentOf(B, F))", "answer_expressions": "3/2", "fact_spans": "[[[1, 15], [26, 29]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[30, 33]], [[34, 37]], [[22, 39]], [[41, 49]]]", "query_spans": "[[[51, 60]]]", "process": "" }, { "text": "Through the right focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, draw a perpendicular line to the $x$-axis, intersecting the ellipse $C$ at points $A$ and $B$. The line $l$ passes through the left focus and the upper vertex of $C$. If the circle with $AB$ as diameter has common points with $l$, then what is the range of values for the eccentricity of the ellipse $C$?", "fact_expressions": "l: Line;C: Ellipse;b: Number;a: Number;G: Circle;A: Point;B: Point;a > b;b > 0;l1:Line;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(RightFocus(C),l1);IsPerpendicular(l1,xAxis);Intersection(l1,C)={A,B};PointOnCurve(LeftFocus(C),l);PointOnCurve(UpperVertex(C),l);IsDiameter(LineSegmentOf(A,B),G);IsIntersect(l,G)", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(0,sqrt(5)/5]", "fact_spans": "[[[87, 92], [118, 121]], [[1, 57], [71, 76], [93, 96], [128, 133]], [[7, 57]], [[7, 57]], [[116, 117]], [[77, 80]], [[81, 84]], [[7, 57]], [[7, 57]], [], [[1, 57]], [[0, 69]], [[0, 69]], [[0, 86]], [[87, 100]], [[87, 104]], [[106, 117]], [[116, 126]]]", "query_spans": "[[[128, 144]]]", "process": "Find the equation of line l, use the relationship between the distance from a point to a line and the semi-latus rectum, set up an inequality, and solve it. The equation of line l is: \\frac{x}{-c}+\\frac{y}{b}=1. The right focus of the ellipse (c,0) passes through the right focus of the ellipse C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). A perpendicular line to the x-axis is drawn through the right focus of ellipse C, intersecting C at points A and B. Line l passes through the left focus and the upper vertex of C. If the circle with AB as diameter has a common point with line l, then: \\frac{|-1-1|}{\\sqrt{\\frac{1}{c^{2}}+\\frac{1}{b^{2}}}}\\leqslant\\frac{b^{2}}{a}, which gives: b\\geqslant2c, that is, a^{2}-c^{2}\\geqslant4c^{2}, i.e., e^{2}\\leqslant\\frac{1}{5}, where e\\in(0,1). Solving yields: 00 , b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A point $P$ on the right branch has a perpendicular drawn to an asymptote of the hyperbola $C$, with foot of the perpendicular at $H$. If the minimum value of $|P H|+|P F_{1}|$ is $4 a$, then what is the eccentricity of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;P: Point;H: 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,RightPart(C));L:Line;PointOnCurve(P,L);IsPerpendicular(L,OneOf(Asymptote(C)));FootPoint(L,OneOf(Asymptote(C)))=H;Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, H))) = 4*a", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 64], [99, 105], [153, 159]], [[146, 151]], [[10, 64]], [[95, 98]], [[118, 121]], [[73, 80]], [[81, 88]], [[10, 64]], [[10, 64]], [[2, 64]], [[2, 88]], [[2, 88]], [[2, 98]], [], [[89, 114]], [[89, 114]], [[89, 121]], [[124, 151]]]", "query_spans": "[[[153, 165]]]", "process": "Using the definition of the hyperbola |PF_{1}| = |PF_{2}| + 2a, it follows that |PH| + |PF_{1}| = |PH| + |PF_{2}| + 2a. Using the point-to-line distance formula, we obtain |PH| + |PF_{2}| = b. According to the problem, b + 2a = 4a, thus the eccentricity can be found. From the definition of the hyperbola, |PF_{1}| - |PF_{2}| = 2a, then |PF_{1}| = |PF_{2}| + 2a. Therefore, |PH| + |PF_{1}| = |PH| + |PF_{2}| + 2a. Hence, drawing a perpendicular from F_{2} to an asymptote of the hyperbola with foot H, intersecting the right branch at point P, |PH| + |PF_{2}| + 2a is minimized, and the minimum value is 4a. It is easy to find that the distance from the focus to the asymptote is b, i.e., |PH| + |PF_{2}| = b. Thus, b + 2a = 4a, so b = 2a, c^{2} = 5a^{2}, and the eccentricity e = \\sqrt{5}." }, { "text": "The standard equation of a parabola with focus $(0,-5)$ is?", "fact_expressions": "G: Parabola;Coordinate(Focus(G)) = (0, -5)", "query_expressions": "Expression(G)", "answer_expressions": "x^2=-20*y", "fact_spans": "[[[12, 15]], [[0, 15]]]", "query_spans": "[[[12, 21]]]", "process": "The focus of the parabola is at (0, -5) on the y-axis. Let the standard equation of the parabola be x^{2} = -2py. Then \\frac{p}{2} = 5, solving gives p = 10. Hence, the standard equation of the parabola is x^{2} = -20y." }, { "text": "Given an ellipse $\\Gamma$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with eccentricity $\\frac{\\sqrt{2}}{2}$, the three vertices of triangle $A B C$ lie on the ellipse $\\Gamma$. Let the midpoints of its three sides $A B$, $B C$, $A C$ be $D$, $E$, $F$ respectively, and let the slopes of the lines containing the three sides be $k_{1}$, $k_{2}$, $k_{3}$ respectively, where $k_{1}$, $k_{2}$, $k_{3}$ are all non-zero. Let $O$ be the origin. If the sum of the slopes of the lines $O D$, $O E$, $O F$ is $1$, then $\\frac{1}{k_{1}}+\\frac{1}{k_{2}}+\\frac{1}{k_{3}}$=?", "fact_expressions": "Gamma: Ellipse;Expression(Gamma) = (x^2/a^2 + y^2/b^2 = 1);a: Number;b: Number;a > b;b > 0;Eccentricity(Gamma) = sqrt(2)/2;A: Point;B: Point;C: Point;PointOnCurve(Vertex(TriangleOf(A, B, C)), Gamma) = True;MidPoint(LineSegmentOf(A, B)) = D;MidPoint(LineSegmentOf(B, C)) = E;MidPoint(LineSegmentOf(A, C)) = F;D: Point;E: Point;F: Point;k1: Number;k2: Number;k3: Number;Slope(OverlappingLine(LineSegmentOf(A, B))) = k1;Slope(OverlappingLine(LineSegmentOf(B, C))) = k2;Slope(OverlappingLine(LineSegmentOf(A, C))) = k3;Negation(k1 = 0);Negation(k2 = 0);Negation(k3 = 0);O: Origin;Slope(LineOf(O, D)) + Slope(LineOf(O, E)) + Slope(LineOf(O, F)) = 1", "query_expressions": "1/k2 + 1/k1 + 1/k3", "answer_expressions": "-2", "fact_spans": "[[[2, 64], [107, 117]], [[2, 64]], [[14, 64]], [[14, 64]], [[14, 64]], [[14, 64]], [[2, 89]], [[93, 100]], [[93, 100]], [[93, 100]], [[90, 118]], [[125, 163]], [[125, 163]], [[125, 163]], [[152, 155]], [[156, 159]], [[160, 163]], [[177, 184], [206, 213]], [[187, 194], [215, 224]], [[197, 204], [226, 233]], [[165, 204]], [[165, 204]], [[165, 204]], [[206, 239]], [[206, 239]], [[206, 239]], [[242, 245]], [[252, 284]]]", "query_spans": "[[[286, 337]]]", "process": "" }, { "text": "The standard equation of the hyperbola with asymptotes $2 x \\pm 3 y=0$ and passing through the point $(1,2)$ is?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (1, 2);Expression(Asymptote(G)) = (2*x+pm*3*y=0);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/(32/9)-x^2/8=1", "fact_spans": "[[[33, 36]], [[24, 32]], [[24, 32]], [[0, 36]], [[22, 36]]]", "query_spans": "[[[33, 42]]]", "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 by $F_{1}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the right branch of $C$ at points $A$ and $B$, with $AF_{1} \\perp AB$ and $4|AF_{1}|=3|AB|$. Find the eccentricity of $C$.", "fact_expressions": "C: Hyperbola;b: Number;a: Number;H: Line;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;PointOnCurve(F2, H);Intersection(H, RightPart(C)) = {A,B};IsPerpendicular(LineSegmentOf(A,F1),LineSegmentOf(A,B));4*Abs(LineSegmentOf(A,F1))=3*Abs(LineSegmentOf(A,B))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[2, 58], [95, 98], [151, 154]], [[5, 58]], [[5, 58]], [[92, 94]], [[102, 105]], [[106, 110]], [[67, 74]], [[75, 82], [84, 91]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 82]], [[2, 82]], [[83, 94]], [[92, 112]], [[113, 130]], [[131, 148]]]", "query_spans": "[[[151, 160]]]", "process": "Let |AF₁| = 3t, t > 0. From 4|AF₁| = 3|AB|, we get |AB| = 4t. By the definition of the hyperbola, |AF₂| = |AF₁| - 2a = 3t - 2a. |BF₂| = |AB| - |AF₂| = 4t - (3t - 2a) = t + 2a. By the definition of the hyperbola, |BF| = |BF₂| + 2a = t + 4a. In right triangle ABF₁, we have |BF₁| = √(|AB|² + |AF₁|²) = 5t = t + 4a, so t = a. In right triangle AF₁F₂, we have |AF₁|² + |AF₂|² = |F₁F₂|², which gives 9a² + a² = 4c², so c = (√10)/2 a, thus e = c/a = (√10)/2." }, { "text": "The left and right foci of the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ ($a>b>0$) are $F_{1}$ and $F_{2}$ respectively, and $P$ is any point on the ellipse $M$. Given that the maximum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is $3 c^{2}$, where $c^{2}=a^{2}-b^{2}$, what is the eccentricity of the ellipse $M$?", "fact_expressions": "M: Ellipse;b: Number;a: Number;c: Number;P: Point;F1: Point;F2: Point;a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(M) = F1;RightFocus(M) = F2;PointOnCurve(P,M);Max(DotProduct(VectorOf(P, F1), VectorOf(P,F2))) = 3*c^2;c^2 = a^2-b^2", "query_expressions": "Eccentricity(M)", "answer_expressions": "1/2", "fact_spans": "[[[0, 56], [85, 90], [191, 196]], [[6, 56]], [[6, 56]], [[158, 167]], [[81, 84]], [[65, 72]], [[73, 80]], [[6, 56]], [[6, 56]], [[0, 56]], [[0, 80]], [[0, 80]], [[81, 94]], [[96, 167]], [[170, 189]]]", "query_spans": "[[[191, 202]]]", "process": "" }, { "text": "Given the ellipse $x^{2}+4 y^{2}=16$, the line $A B$ passes through the point $P(2,-1)$ and intersects the ellipse at points $A$ and $B$. If the slope of the line $A B$ is $\\frac{1}{2}$, then the value of $| AB| $ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 16);A: Point;B: Point;P: Point;Coordinate(P) = (2, -1);PointOnCurve(P, LineOf(A,B));Intersection(LineOf(A,B),G)={A,B};Slope(LineOf(A,B)) = 1/2", "query_expressions": "Abs(LineSegmentOf(A,B))", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 22], [45, 47]], [[2, 22]], [[49, 52]], [[53, 56]], [[31, 42]], [[31, 42]], [[23, 42]], [[23, 58]], [[60, 84]]]", "query_spans": "[[[86, 98]]]", "process": "" }, { "text": "Given that $O$ is the coordinate origin, the tangent line $\\frac{x_{0} x}{4}+y_{0} y=1$ at a point $P(x_{0}, y_{0})$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. Then when $|A B|$ is minimized, $|O P|=$?", "fact_expressions": "G: Ellipse;P: Point;O: Origin;B:Point;A:Point;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(P) = (x0,y0);PointOnCurve(P,G);WhenMin(Abs(LineSegmentOf(A,B)));x0:Number;y0:Number;l1:Line;TangentOfPoint(P,G)=l1;Expression(l1)=(x0*x/4+y0=1);Intersection(l2,yAxis)=A;Intersection(l1,yAxis)=B", "query_expressions": "Abs(LineSegmentOf(O,P))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[12, 39]], [[42, 59]], [[2, 5]], [[106, 109]], [[102, 105]], [[12, 39]], [[42, 59]], [[12, 59]], [[113, 124]], [[42, 59]], [[42, 59]], [61, 88], [11, 88], [61, 88], [61, 108], [61, 108]]", "query_spans": "[[[125, 134]]]", "process": "Let the point of tangency be $(2\\cos\\theta,\\sin\\theta)$ according to the problem. Then the tangent line equation to the ellipse at this point is $\\frac{\\cos\\theta\\cdot x}{2} + \\sin\\theta\\cdot y = 1$. Letting $x=0$, we get $y = \\frac{1}{\\sin\\theta}$; letting $y=0$, we get $x = \\frac{2}{\\cos\\theta}$. Therefore, $|AB|^{2} = x^{2} + y^{2} = \\frac{1}{\\sin^{2}\\theta} + \\frac{4}{\\cos^{2}\\theta} = \\left(\\frac{1}{\\sin^{2}\\theta} + \\frac{4}{\\cos^{2}\\theta}\\right)(\\sin^{2}\\theta + \\cos^{2}\\theta) = 1 + 4 + \\frac{\\cos^{2}\\theta}{\\sin^{2}\\theta} + \\frac{4\\sin^{2}\\theta}{\\cos^{2}\\theta} \\geqslant 5 + 2\\sqrt{4} = 9$. The equality holds if and only if $2\\sin^{2}\\theta = \\cos^{2}\\theta$, that is, $\\tan\\theta = \\frac{\\sqrt{2}}{2}$. At this time, $\\cos^{2}\\theta = \\frac{1}{\\tan^{2}\\theta + 1} = \\frac{2}{3}$. Therefore, $|AB|^{2}_{\\min} = 9$, so $|AB|_{\\min} = 3$. At this time, $OP = \\sqrt{3\\cos^{2}\\theta + 1} = \\sqrt{3}$." }, { "text": "The focus of the parabola $y^{2}=2 x$ is $F$. If $P(2, y)$ lies on the parabola, then $|PF|=$?", "fact_expressions": "G: Parabola;F: Point;P: Point;y1: Number;Expression(G) = (y^2 = 2*x);Coordinate(P) = (2, y1);Focus(G) = F;PointOnCurve(P, G)", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "5/2", "fact_spans": "[[[0, 14], [33, 36]], [[18, 21]], [[23, 32]], [[23, 32]], [[0, 14]], [[23, 32]], [[0, 21]], [[23, 37]]]", "query_spans": "[[[39, 47]]]", "process": "" }, { "text": "Given $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 line $l$ is drawn through $F_{1}$ intersecting the left and right branches of $C$ at points $P$ and $Q$ respectively. If $\\angle Q P F_{2}=60^{\\circ}$ and $|P Q|=|P F_{2}|$, then the equations of the asymptotes of the hyperbola are?", "fact_expressions": "F2: Point;F1: Point;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;PointOnCurve(F1, l);l: Line;Intersection(l, LeftPart(C)) = P;Intersection(l, RightPart(C)) = Q;Q: Point;P: Point;AngleOf(Q, P, F2) = ApplyUnit(60, degree);Abs(LineSegmentOf(P, Q)) = Abs(LineSegmentOf(P, F2))", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*sqrt(6)*x", "fact_spans": "[[[10, 17]], [[2, 9], [87, 94]], [[18, 79], [101, 104], [173, 176]], [[18, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[26, 79]], [[2, 85]], [[2, 85]], [[86, 100]], [[95, 100]], [[95, 121]], [[95, 121]], [[116, 119]], [[112, 115]], [[123, 152]], [[154, 171]]]", "query_spans": "[[[173, 185]]]", "process": "By the definition of a hyperbola, |QF₁| - |QF₂| = |PF₁| = 2a, |PF₂| - |PF₁| = 2a ∴ |PF₂| = 4a, ∵ ∠QPF₂ = 60°, ∴ ∠F₁PF₂ = 120°. In △F₁PF₂, by the law of cosines |F₁F₂|² = |PF₁|² + |PF₂|² - 2|PF₁|⋅|PF₂|cos120°, we get 4c² = 4a² + 3|PF₁|⋅|PF₂| = 28a², that is, c²/a² = 7 ∴ b²/a² = 6, so the asymptotes of the hyperbola are y = ±√6x" }, { "text": "$P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, $F_{1}$ and $F_{2}$ are its two foci, $O$ is the coordinate origin, and a moving point $Q$ satisfies $\\overrightarrow{O Q}=\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}$. Then the trajectory equation of the moving point $Q$ is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;O: Origin;Q: Point;P: Point;F1: Point;F2: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);Focus(G)={F1,F2};VectorOf(O, Q) = VectorOf(P, F1) + VectorOf(P, F2)", "query_expressions": "LocusEquation(Q)", "answer_expressions": "x^2/(4*a^2)+y^2/(4*b^2)=1", "fact_spans": "[[[4, 49], [72, 73]], [[6, 49]], [[6, 49]], [[79, 82]], [[92, 95], [173, 176]], [[0, 3]], [[56, 63]], [[64, 71]], [[4, 49]], [[0, 55]], [[56, 78]], [[97, 169]]]", "query_spans": "[[[173, 183]]]", "process": "" }, { "text": "Given the circle $F$: $(x+3)^{2}+y^{2}=1$, and the line $l$: $x=2$, find the equation of the locus of the center $M^{\\prime}$ of a circle that is tangent to the line $l$ and externally tangent to the circle $F$.", "fact_expressions": "F: Circle;Expression(F) = (y^2 + (x + 3)^2 = 1);l: Line;Expression(l) = (x = 2);M: Circle;IsTangent(l, M);IsOutTangent(F, M);M1: Point;Center(M) = M1", "query_expressions": "LocusEquation(M1)", "answer_expressions": "y^2=-12*x", "fact_spans": "[[[2, 27], [52, 56]], [[2, 27]], [[28, 40], [43, 48]], [[28, 40]], [[59, 60]], [[42, 60]], [[51, 60]], [[63, 75]], [[59, 75]]]", "query_spans": "[[[63, 82]]]", "process": "Since the distance from point M to fixed point F is equal to the distance from M to the line x=2 plus 1, the trajectory is a parabola with focus F(-3,0) and directrix x=3, and its equation is y^{2}=-12x." }, { "text": "The equations of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/9 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(3/2)*x", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 46]]]", "process": "From the equation of the hyperbola, we obtain a=2, b=3. Then, based on the geometric properties of the curve, we can find the equations of the asymptotes and obtain the answer. [Detailed solution] From the equation of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1, we get a=2, b=3. Since the foci lie on the x-axis, the equations of the asymptotes are y=\\pm\\frac{b}{a}x=\\pm\\frac{3}{2}x" }, { "text": "Given the parabola $C$: $y^{2}=4x$, let point $P$ be any point on the line $x=-2$. From point $P$, draw two tangent lines to the parabola $C$, with points of tangency $A$ and $B$. What is the maximum distance from point $M(0,1)$ to the line $AB$?", "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;M: Point;P: Point;l1: Line;l2: Line;Expression(C) = (y^2 = 4*x);Expression(G) = (x=-2);Coordinate(M) = (0, 1);PointOnCurve(P,G);TangentOfPoint(P, C) = {l1, l2};TangentPoint(l1,C)=A;TangentPoint(l2,C)=B", "query_expressions": "Min(Distance(M,LineOf(A,B)))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 21], [45, 51]], [[27, 33]], [[62, 65]], [[66, 69]], [[71, 80]], [[40, 44]], [], [], [[2, 21]], [[27, 33]], [[71, 80]], [[22, 38]], [[39, 56]], [[39, 69]], [[39, 69]]]", "query_spans": "[[[71, 97]]]", "process": "Let P(-2, m), A(x_{1}, y_{1}), B(x_{2}, y_{2}). It is clear that the slope at point A is not 0; let the slope be k. Thus, the tangent line equation at point A is y - y_{1} = k(x - x_{1}). From \n\\begin{cases} y - y_{1} = k(x - x_{1}) \\\\ y^{2} = 4x \\end{cases}, \neliminating x gives y^{2} - \\frac{4}{k}y + \\frac{4y_{1}}{k} - 4x_{1} = 0. From \\triangle = 0, we get k = \\frac{2}{y_{1}}. Therefore, the tangent line equation at point A is 2x - y_{1}y + 2x_{1} = 0. Since the tangent passes through point P(-2, m), we have -4 - y_{1}m + 2x_{1} = 0. Similarly, the tangent line equation at point B is -4 - y_{2}m + 2x_{2} = 0. Hence, the equation of line AB is -4 - ym + 2x = 0, so line AB passes through the fixed point (2, 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 (2, 0), which is \\sqrt{5}." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{5}+\\frac{y^{2}}{9}=1$, and let $P$ be a point on $C$ such that $|P F_{1}|=|F_{1} F_{2}|$. Then the inradius $r$ of $\\Delta P F_{1} F_{2}$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/5 + y^2/9 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(F1, F2));r: Number;Radius(InscribedCircle(TriangleOf(P, F1, F2))) = r", "query_expressions": "r", "answer_expressions": "sqrt(15)/5", "fact_spans": "[[[17, 59], [69, 72]], [[17, 59]], [[1, 8]], [[9, 16]], [[1, 64]], [[65, 68]], [[65, 75]], [[77, 102]], [[133, 136]], [[104, 136]]]", "query_spans": "[[[133, 138]]]", "process": "Ellipse $ C: \\frac{x^{2}}{5} + \\frac{y^{2}}{9} = 1 $, $ a^{2} = 9 $, $ b^{2} = 5 $, then $ c^{2} = a^{2} - b^{2} = 4 $, $ \\therefore a = 3 $, $ c = 2 $, $ \\therefore |F_{1}F_{2}| = 2c = 4 $, $ \\because |PF_{1}| + |PF_{2}| = 2a $, $ |PF_{1}| = |F_{1}F_{2}| $, $ \\therefore |PF_{2}| = 2 $, $ \\because S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2} \\times 2 \\times \\sqrt{4^{2} - 1^{2}} = \\sqrt{15} $, $ \\therefore S_{\\Delta PF_{1}F_{2}} = \\frac{1}{2} \\cdot (|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\cdot r = \\sqrt{15} $, solving gives $ r = \\frac{\\sqrt{15}}{5} $. Hence, fill in $ \\frac{\\sqrt{15}}{5} $. This question examines the geometric properties of ellipses, involving the application of the definition of an ellipse, the properties of the incircle of a triangle, and the method of equal area transformation. Note the values corresponding to $ a^{2} $, $ b^{2} $ in the standard equation of an ellipse." }, { "text": "The vertex of the parabola is at the origin, and the focus lies on the $x$-axis. The line $2x - y = 0$ intersects the parabola at points $A$ and $B$, and $P(1, 2)$ is the midpoint of segment $AB$. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;PointOnCurve(Focus(G), xAxis);H: Line;Expression(H) = (2*x - y = 0);A: Point;B: Point;Intersection(H, G) = {A, B};P: Point;Coordinate(P) = (1, 2);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(G)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[0, 3], [32, 35], [69, 72]], [[7, 9]], [[0, 9]], [[0, 18]], [[20, 31]], [[20, 31]], [[37, 40]], [[41, 44]], [[20, 46]], [[47, 57]], [[47, 57]], [[47, 67]]]", "query_spans": "[[[69, 77]]]", "process": "Test analysis: Since the line 2x - y = 0 passes through the origin, one of the points A and B is the origin. Without loss of generality, let A(0,0). Given that P(1,2) is the midpoint of AB, then B(2,4). Let the equation of the parabola be y^{2} = mx. Then 4^{2} = m \\times 2, so m = 8. Thus, the 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)$, the right focus is $F$, the left vertex is $A$, and $O$ is the origin. A circle is drawn with $OF$ as diameter, intersecting an asymptote of the hyperbola at point $P$, and $|P A|=|P F|$. Then the eccentricity $e$ 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;A: Point;LeftVertex(C) = A;O: Origin;G: Circle;IsDiameter(LineSegmentOf(O,F),G);Intersection(G,OneOf(Asymptote(C))) = P;P: Point;e: Number;Abs(LineSegmentOf(P, A)) = Abs(LineSegmentOf(P, F));Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "2", "fact_spans": "[[[2, 63], [101, 104], [133, 136]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[76, 79]], [[2, 79]], [[80, 83]], [[99, 100]], [[89, 100]], [[99, 115]], [[111, 115]], [[140, 143]], [[118, 131]], [[133, 143]]]", "query_spans": "[[[140, 145]]]", "process": "From the given conditions, F(c,0). Then the equation of the circle with OF as diameter is (x-\\frac{c}{2})^{2}+y^{2}=\\frac{c^{2}}{4}. Take one asymptote y=\\frac{b}{a}x and solve the system of equations of the line and the circle to find the coordinates of P. Then use |PA|=|PF| to obtain the relationship between a and c, and find the eccentricity of the hyperbola. [Solution] From the given conditions, A(-a,0), F(c,0), the equations of the asymptotes of the hyperbola are y=\\pm\\frac{b}{a}x. Take y=\\frac{b}{a}x. The equation of the circle with OF as diameter is (x-\\frac{c}{2})^{2}+y^{2}=\\frac{c^{2}}{4}. Solve the system \\begin{cases}y=\\frac{b}{a}x\\\\(x-\\frac{c}{2})^{2}+y^{2}=\\frac{c^{2}}{4}\\end{cases}, yielding \\begin{cases}x=\\frac{a2}{c}\\\\y=\\frac{ab}{c}\\end{cases} or \\begin{cases}x=0\\\\y=0\\end{cases} (discarded). Thus P(\\frac{a^{2}}{c},\\frac{ab}{c}). We get \\frac{c-a}{2}=\\frac{a^{2}}{c}, i.e., c^{2}-ac=2a^{2}, e^{2}-e-2=0, so (e-2)(e+1)=0, solving gives e=2 or e=-1 (discarded). Therefore, the eccentricity of the hyperbola is e=2." }, { "text": "Given that points $F$ and $B$ are the focus and the endpoint of the imaginary axis of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively, if the midpoint of segment $FB$ lies on the hyperbola $C$, then the equation of the asymptotes of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;B: Point;F: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);OneOf(Focus(C))=F;OneOf(Endpoint(ImageinaryAxis(C)))=B;PointOnCurve(MidPoint(LineSegmentOf(F,B)), C)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*2*x", "fact_spans": "[[[13, 74], [95, 101], [104, 110]], [[21, 74]], [[21, 74]], [[7, 10]], [[2, 6]], [[21, 74]], [[21, 74]], [[13, 74]], [[2, 82]], [[2, 82]], [[84, 102]]]", "query_spans": "[[[104, 118]]]", "process": "From the given conditions, point B(0,b) and point F(c,0) are known, so the midpoint of segment FB is (\\frac{c}{2},\\frac{b}{2}). Since the midpoint M of segment FB lies on hyperbola C, it follows that \\frac{(\\frac{c}{2})^{2}}{a^{2}}-\\frac{(\\frac{b}{2})^{2}}{b^{2}}. Solving gives (\\frac{c}{a})^{2}=5, \\therefore\\frac{c}{a}=\\sqrt{5}, \\therefore\\frac{b}{a}=\\sqrt{\\frac{b^{2}}{a^{2}}}=\\sqrt{\\frac{c^{2}-a^{2}}{a^{2}}}=2. The asymptotes of hyperbola C are y=\\pm2x." }, { "text": "Given that the focus of the parabola $x^{2}=a y$ is exactly the lower focus of the hyperbola $\\frac{y^{2}}{3}-x^{2}=1$, then $a$=?", "fact_expressions": "H: Parabola;Expression(H) = (x^2 = a*y);a: Number;G: Hyperbola;Expression(G) = (-x^2 + y^2/3 = 1);Focus(H) = LowerFocus(G)", "query_expressions": "a", "answer_expressions": "-8", "fact_spans": "[[[2, 16]], [[2, 16]], [[56, 59]], [[22, 50]], [[22, 50]], [[2, 54]]]", "query_spans": "[[[56, 61]]]", "process": "The focus of the parabola \\( x^{2} = ay \\) is \\( \\left(0, \\frac{a}{4}\\right) \\). The lower focus of the hyperbola \\( \\frac{y^{2}}{3} - x^{2} = 1 \\) is \\( (0, -2) \\). \\( \\therefore \\frac{a}{4} = -2 \\), \\( \\therefore a = -8 \\)" }, { "text": "Given that ellipse $C$ and hyperbola $Q$ have the same foci $F_{1}$, $F_{2}$, with eccentricities $e_{1}$, $e_{2}$ respectively. Let point $M$ be a common point of $C$ and $Q$. If $\\angle F_{2} M F_{1}=60^{\\circ}$, then the minimum value of $\\frac{e_{1} e_{2}}{e_{1}+e_{2}}$ is?", "fact_expressions": "Q: Hyperbola;C: Ellipse;F2: Point;M: Point;F1: Point;e1:Number;e2:Number;Focus(C) = {F1, F2};Focus(Q) = {F1, F2};Focus(C)=Focus(Q);OneOf(Intersection(C,Q))=M;Eccentricity(C)=e1;Eccentricity(Q)=e2;AngleOf(F2, M, F1) = ApplyUnit(60, degree)", "query_expressions": "Min((e1*e2)/(e1 + e2))", "answer_expressions": "sqrt(3)/4", "fact_spans": "[[[8, 14], [72, 75]], [[2, 7], [68, 71]], [[27, 34]], [[63, 67]], [[19, 26]], [[45, 52]], [[54, 61]], [[2, 34]], [[8, 34]], [[2, 34]], [[63, 81]], [[36, 61]], [[36, 61]], [[83, 117]]]", "query_spans": "[[[118, 157]]]", "process": "Let the ellipse equation be \\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1, and the hyperbola equation be \\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1. From the definitions of the ellipse and hyperbola, we have: |MF_{1}|+|MF_{2}|=2a_{1}, |MF_{1}|-|MF_{2}|=2a_{2}. Solving gives |MF_{1}|=a_{1}+a_{2}, |MF_{2}|=a_{1}-a_{2}. Using the law of cosines, simplify the expression 4=\\frac{1}{e^{2}}+\\frac{3}{e^{2}}, and using the Cauchy-Schwarz inequality, the answer can be obtained. Let the ellipse equation be \\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1, and the hyperbola equation be \\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1. From the definitions of the ellipse and hyperbola, we have: |MF_{1}|+|MF_{2}|=2a_{1}, |MF_{1}|-|MF_{2}|=2a_{2}, thus we obtain: |MF_{1}|=a_{1}+a_{2}, |MF_{2}|=a_{1}-a_{2}. In \\triangle F_{1}MF_{2}, by the law of cosines: |F_{1}F_{2}|^{2}=|MF_{1}|^{2}+|MF_{2}|^{2}-2|MF_{1}||MF_{2}|\\cos\\angle F_{2}MF_{1}. \\therefore (2c)^{2}=(a_{1}+a_{2})^{2}+(a_{1}-a_{2})^{2}+2(a_{1}+a_{2})(a_{1}-a_{2})\\cos60^{\\circ}, which gives 4c^{2}=a_{1}^{2}+3a_{2}^{2}, \\therefore 4=\\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}}. Applying the Cauchy-Schwarz inequality: (1+\\frac{1}{3})(\\frac{1}{e_{1}^{2}}+\\frac{3}{e_{2}^{2}})\\geqslant(1\\times\\frac{1}{e_{1}}+\\frac{1}{\\sqrt{3}}\\times\\frac{\\sqrt{3}}{e_{2}})^{2}=(\\frac{1}{e_{1}}+\\frac{1}{e_{2}})^{2}, so (\\frac{1}{e_{1}}+\\frac{1}{e_{2}})^{2}\\leqslant\\frac{4}{3}\\times4=\\frac{16}{3}, hence \\frac{1}{e_{1}}+\\frac{1}{e_{2}}\\leqslant\\frac{4\\sqrt{3}}{3}, yielding \\frac{1}{e_{1}+\\frac{1}{e_{2}}}\\geqslant\\frac{\\sqrt{3}}{4}, therefore \\frac{e_{1}e_{2}}{e_{1}+e_{2}}\\geqslant\\frac{\\sqrt{3}}{4}, with equality when e_{1}=\\frac{\\sqrt{3}}{2}, e_{2}=\\sqrt{3}. The minimum value of \\frac{e_{1}e_{2}}{e_{1}+e_{2}} is \\frac{\\sqrt{3}}{4}." }, { "text": "The focal distance of the hyperbola $2 x^{2}-y^{2}=6$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (2*x^2 - y^2 = 6)", "query_expressions": "FocalLength(G)", "answer_expressions": "6", "fact_spans": "[[[0, 20]], [[0, 20]]]", "query_spans": "[[[0, 25]]]", "process": "The hyperbola 2x^{2}\\cdoty^{2}=6 is equivalent to \\frac{x^{2}}{3}-\\frac{y^{2}}{6}=1, yielding a=\\sqrt{3}, b=\\sqrt{6}, c=\\sqrt{a^{2}+b^{2}}=3, so the focal distance is 2c=6." }, { "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$, and 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);b: Number;a: 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, 58], [112, 115]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 95]], [[2, 110]]]", "query_spans": "[[[112, 119]]]", "process": "Given $\\frac{b}{a} = \\frac{\\sqrt{3}}{3}$, that is, $a = \\sqrt{3}b$, take the hyperbola's vertex $(a,0)$ and the asymptote $y = \\frac{\\sqrt{3}}{3}x$, then the distance from the vertex to the asymptote $\\sqrt{3}x - 3y = 0$ is $\\frac{\\sqrt{3}a}{\\sqrt{(\\sqrt{3})^{2} + 3^{2}}} = \\frac{a}{2}$. From the given condition, $a = 2$, so $b = \\frac{2}{\\sqrt{3}}$, then the required equation of the hyperbola is $\\frac{x^{2}}{4} - \\frac{3y^{2}}{4} = 1$." }, { "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=2$, then $a$=?", "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 2)", "query_expressions": "a", "answer_expressions": "-1/8", "fact_spans": "[[[0, 14]], [[27, 30]], [[0, 14]], [[0, 25]]]", "query_spans": "[[[27, 32]]]", "process": "" }, { "text": "Hyperbola $C$: $x^{2}-y^{2}=1$. If the right vertex of hyperbola $C$ is $A$, and a line $l$ passing through $A$ intersects the two asymptotes of hyperbola $C$ at points $P$ and $Q$, such that $\\overrightarrow{P A}=2 \\overrightarrow{A Q}$, then the slope of line $l$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2 = 1);A: Point;RightVertex(C) = A;PointOnCurve(A, l);l: Line;Intersection(l, Asymptote(C)) = {P, Q};P: Point;Q: Point;VectorOf(P, A) = 2*VectorOf(A, Q)", "query_expressions": "Slope(l)", "answer_expressions": "pm*3", "fact_spans": "[[[0, 22], [24, 30], [50, 56]], [[0, 22]], [[35, 38], [40, 43]], [[24, 38]], [[39, 49]], [[44, 49], [122, 127]], [[44, 73]], [[64, 67]], [[68, 71]], [[75, 120]]]", "query_spans": "[[[122, 132]]]", "process": "" }, { "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$, if there exists a point $P$ on the hyperbola such that $\\angle F_{1} P F_{2}=\\frac{2 \\pi}{3}$ and $|F_{1} P|=3|P F_{2}|$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G);AngleOf(F1,P,F2)=2*pi/3;Abs(LineSegmentOf(F1, P)) = 3*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(13)/2", "fact_spans": "[[[18, 64], [73, 76], [149, 152]], [[21, 64]], [[21, 64]], [[2, 9]], [[79, 83]], [[10, 17]], [[18, 64]], [[2, 71]], [[2, 71]], [[73, 83]], [[85, 124]], [[125, 147]]]", "query_spans": "[[[149, 158]]]", "process": "\\because|F_{1}P|=3|PF_{2}|,||PF_{1}|-|PF_{2}||=2a\\therefore|PF_{2}|=a,|PF_{1}|=3a\\because\\angleF_{1}PF_{2}=\\frac{2\\pi}{3}=\\frac{|PF|^{2}+|PF_{2}^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}|\\cdot|PF_{2}|}=-\\frac{1}{2}, i.e., \\frac{10a^{2}-4c^{2}}{6a^{2}}=-\\frac{1}{2}\\thereforee^{2}=\\frac{13}{4}, i.e., e=\\frac{\\sqrt{13}}{2}." }, { "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{5}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on $\\Gamma$. If $|P F_{1}| \\cdot|P F_{2}|=3$, then what is the value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$?", "fact_expressions": "Gamma: Ellipse;Expression(Gamma) = (x^2/5 + y^2 = 1);F1: Point;F2: Point;LeftFocus(Gamma) = F1;RightFocus(Gamma) = F2;P: Point;PointOnCurve(P, Gamma);Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)) = 3", "query_expressions": "DotProduct(VectorOf(P, F1), VectorOf(P, F2))", "answer_expressions": "-1", "fact_spans": "[[[2, 39], [68, 76]], [[2, 39]], [[48, 55]], [[56, 63]], [[2, 63]], [[2, 63]], [[64, 67]], [[64, 79]], [[81, 109]]]", "query_spans": "[[[111, 172]]]", "process": "According to the definition of the ellipse, write |PF_{1}|+|PF_{2}| and |F_{1}F_{2}|, then substitute to solve for \\cos\\angle F_{1}PF_{2}, thus obtaining \\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF}_{2}. From the problem, a=\\sqrt{5}, c=2. By the definition of the ellipse, |PF_{1}|+|PF_{2}|=2\\sqrt{5}, |F_{1}F_{2}|=4. Then \\cos\\angle F_{1}PF_{2}=\\frac{20-6-16}{2\\times3}=-\\frac{1}{3}, so \\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=|PF_{1}|\\cdot|PF_{2}|\\cos\\angle F_{1}PF=-1" }, { "text": "Let points $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, respectively, and let point $N$ be the upper vertex of the ellipse $C$. If a moving point $M$ satisfies: $|\\overrightarrow{M N}|^{2}=2 \\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}$, then the maximum value of $|\\overrightarrow{M F_{1}}+2 \\overrightarrow{M F_{2}}|$ is?", "fact_expressions": "C: Ellipse;M: Point;N: Point;F1: Point;F2: Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;UpperVertex(C)=N;Abs(VectorOf(M, N))^2 = 2*DotProduct(VectorOf(M, F1), VectorOf(M, F2))", "query_expressions": "Max(Abs(VectorOf(M, F1) + 2*VectorOf(M, F2)))", "answer_expressions": "6 +sqrt(10)", "fact_spans": "[[[19, 51], [64, 69]], [[77, 80]], [[59, 63]], [[0, 8]], [[9, 16]], [[19, 51]], [[0, 58]], [[0, 58]], [[59, 73]], [[83, 169]]]", "query_spans": "[[[171, 232]]]", "process": "Let M(x_{0},y_{0}), from \\frac{x^{2}}{2}+y^{2}=1, we get N(0,1), F_{1}(-1,0), F_{2}(1,0). Then from |MN|^{2}=2\\overrightarrow{MF_{1}}\\cdot\\overrightarrow{MF_{2}}, we obtain x_{0}^{2}+(y_{0}-1)^{2}=2x_{0}^{2}-2+2y_{0}^{2}, which simplifies to x_{0}^{2}+(y_{0}+1)^{2}=4. We can set \\begin{cases}x_{0}=2\\cos\\alpha\\\\y_{0}=2\\sin\\alpha\\end{cases}. Then \\overrightarrow{MF_{1}}=(-2\\cos\\alpha-1,-2\\sin\\alpha+1), 2\\overrightarrow{MF_{2}}=(-4\\cos\\alpha+2,-4\\sin\\alpha+2), \\overrightarrow{MF_{1}}+2\\overrightarrow{MF_{2}}=(-6\\cos\\alpha+1,-6\\sin\\alpha+3), |\\overrightarrow{MF_{1}}+2\\overrightarrow{MF_{2}}|=\\sqrt{(1-6\\cos\\alpha)^{2}+(3-6\\sin\\alpha)^{2}}. The maximum value of |\\overrightarrow{MF_{1}}+2\\overrightarrow{MF_{2}}| is 6+\\sqrt{10}, so the answer is 36+\\sqrt{10}. Method point: This problem mainly examines simple properties of ellipses, the dot product formula of planar vectors, and trigonometric functions for finding maximum values, belonging to a difficult problem. Common methods for finding maximum/minimum values include \\textcircled{1} completing the square: if the function is quadratic in one variable, completing the square is often used; \\textcircled{2} graphical method; \\textcircled{3} inequality method; \\textcircled{4} monotonicity method; \\textcircled{5} substitution method: commonly uses algebraic or trigonometric substitutions; when using substitution to find range, carefully analyze the range change of the substitution parameter; after trigonometric substitution, auxiliary angle formulas combined with trigonometric function monotonicity are often used for solving." }, { "text": "Given the curve $C$: $\\frac{x^{2}}{a}+\\frac{y^{2}}{9}=1$ has a focal distance of $8$, then $a$=?", "fact_expressions": "C: Curve;a: Number;Expression(C) = (y^2/9 + x^2/a = 1);FocalLength(C) = 8", "query_expressions": "a", "answer_expressions": "{25, -7}", "fact_spans": "[[[2, 44]], [[53, 56]], [[2, 44]], [[2, 51]]]", "query_spans": "[[[53, 58]]]", "process": "From the given condition, the semi-focal distance is c=4. When a>0, the curve C is an ellipse, and since 9<4^{2}, it follows that a=9+4^{2}=25; when a<0, the curve C is a hyperbola, so -a+9=4^{2}, hence a=-. Therefore, the value of a is 25 or --" }, { "text": "If the curve $x=\\sqrt{a-y^{2}}$ and the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$ have two distinct intersection points, then the range of values for $a$ is?", "fact_expressions": "H: Curve;Expression(H) = (x = sqrt(a - y^2));a: Number;G: Ellipse;Expression(G) = (x^2/9 + y^2/7 = 1);NumIntersection(H, G) = 2", "query_expressions": "Range(a)", "answer_expressions": "[7,9)", "fact_spans": "[[[1, 21]], [[1, 21]], [[69, 72]], [[22, 59]], [[22, 59]], [[1, 67]]]", "query_spans": "[[[69, 79]]]", "process": "In the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$, $a_{1}=9$, $b_{1}^{2}=7$, so $a_{1}=3$, $b_{1}=\\sqrt{7}$, then the upper vertex of the ellipse is $(0,\\sqrt{7})$, the right vertex is $(3,0)$. The curve $x=\\sqrt{a-y^{2}}$ $(x\\geqslant0)$, that is, $x^{2}+y^{2}=a$ $(x\\geqslant0)$, represents the right half-circle with the origin as the center and $\\sqrt{a}$ as the radius, including the $y$-axis. According to the condition, we have $\\sqrt{7}\\leqslant\\sqrt{a}<3$, so $7\\leqslant a<9$, that is, $a\\in[7,9)$;" }, { "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{45}+\\frac{y^{2}}{20}=1$ such that the lines connecting $P$ to the two foci are perpendicular to each other, if point $P$ lies in the second quadrant, then the coordinates of this point are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/45 + y^2/20 = 1);P:Point;PointOnCurve(P,G);F1:Point;F2:Point;Focus(G)={F1,F2};IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2));Quadrant(P)=2", "query_expressions": "Coordinate(P)", "answer_expressions": "(-3,4)", "fact_spans": "[[[2, 41]], [[2, 41]], [[44, 47], [61, 65], [73, 74]], [[2, 47]], [], [], [[2, 52]], [[2, 59]], [[61, 70]]]", "query_spans": "[[[73, 79]]]", "process": "\\because \\text{ellipse} \\frac{x^2}{45} + \\frac{y^2}{20} = 1, \\therefore a^2 = 45, b^2 = 20, c^2 = a^2 - b^2 = 25, \\text{circle centered at } O \\text{ with radius } 5 \\text{ has equation } x^2 + y^2 = 25 \\quad (1) \\quad \\frac{x^2}{45} + \\frac{y^2}{20} = 1 \\text{ from } (1)(2) \\text{ we get: } x^2 = 9, y^2 = 16 \\text{ solving gives: } x_1 = -3, x_2 = 3, y_1 = 4, y_2 = -4, \\text{ a point } P \\text{ on the ellipse } \\frac{x^2}{45} + \\frac{y^2}{20} = 1 \\text{ such that the lines connecting } P \\text{ to the two foci are perpendicular, i.e., the intersection points of the ellipse and the circle. If point } P \\text{ is in the second quadrant, then the intersection point is } (-3, 4)." }, { "text": "If the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{9}=1$ $(m>9)$ and the hyperbola $\\frac{x^{2}}{n}-\\frac{y^{2}}{9}=1$ $(n>0)$ have the same foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves, then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Ellipse;Expression(G)=(x^2/m+y^2/9=1);m:Number;m>9;C:Hyperbola;Expression(C)=(x^2/n-y^2/9=1);n>0;n:Number;F1:Point;F2:Point;P:Point;Focus(G)={F1,F2};Focus(C)={F1,F2};OneOf(Intersection(G,C))=P", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "9", "fact_spans": "[[[1, 45]], [[1, 45]], [[3, 45]], [[3, 45]], [[46, 91]], [[46, 91]], [[49, 91]], [[49, 91]], [[97, 104]], [[107, 114]], [[117, 120]], [[1, 114]], [[46, 114]], [[1, 129]]]", "query_spans": "[[[131, 158]]]", "process": "" }, { "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 a point $B$, such that $\\overrightarrow{F A}=3 \\overrightarrow{F B}$. Then $|A B|$=?", "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": "8/3", "fact_spans": "[[[2, 21], [53, 56]], [[40, 42]], [[35, 39], [24, 27]], [[47, 51]], [[62, 65]], [[30, 33], [43, 46]], [[2, 21]], [[2, 27]], [[2, 33]], [[34, 42]], [[40, 51]], [[40, 65]], [[67, 112]]]", "query_spans": "[[[114, 123]]]", "process": "From the equation of the parabola C: y^{2}=4x, it is easy to know that the focus is F(1,0) and the directrix is l: x=-1. Draw BM\\botl from point B, with M as the foot of the perpendicular. Let the directrix l intersect the x-axis at point N, as shown in the figure below. Without loss of generality, let |FB|=t, then |FA|=3|FB|=3t, thus |AB|=2t. Also, by the definition of the parabola, BM=FB=t. Since \\frac{|BM|}{|NF|}=\\frac{|AB|}{|FA|}, and |NF|=2, we have \\frac{t}{2}=\\frac{2t}{3t}, solving gives t=\\frac{4}{3}. Therefore, |AB|=2t=\\frac{8}{3}. The answer is: \\underline{8}" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus of the ellipse is $F$. The line $l$: $y=\\sqrt{3} x$ intersects the ellipse $C$ at points $A$ and $B$. If $A F \\perp B F$, then the eccentricity of the ellipse $C$ is?", "fact_expressions": "l: Line;C: Ellipse;a: Number;b: Number;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;Expression(l)=(y = sqrt(3)*x);Intersection(l, C) = {A, B};IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[68, 89]], [[2, 59], [90, 95], [126, 131]], [[8, 59]], [[8, 59]], [[98, 101]], [[64, 67]], [[102, 105]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[68, 89]], [[68, 107]], [[109, 124]]]", "query_spans": "[[[126, 137]]]", "process": "Draw the graph, let the left focus be $ F_{1} $, connect $ AF_{1} $, $ BF_{1} $. According to the symmetry axis of the ellipse and $ AF \\perp BF $, determine that quadrilateral $ AF_{1}BF $ is a rectangle. Use the inclination angle of the line $ y = \\sqrt{3}x $, combine with the definition of the ellipse to set up an equation, simplify and find the eccentricity. As shown in the figure, let the left focus be $ F_{1} $, connect $ AF_{1} $, $ BF_{1} $. By the symmetry of the ellipse and $ AF \\perp BF $, it follows that $ AF_{1}BF $ is a rectangle, $ \\therefore |OA| = |OF| = |OF| = c $. From the line $ y = \\sqrt{3}x $, we get $ \\angle AOF = 60^{\\circ} $, $ \\therefore |AF| = c $, and $ \\angle AF_{1}F = 30^{\\circ} $, $ |AF_{1}| = \\sqrt{3}c $. By the definition of the ellipse, $ |AF| + |AF_{1}| = c + \\sqrt{3}c = 2a $, $ \\therefore e = \\frac{c}{a} = \\frac{2}{\\sqrt{3}+1} = \\sqrt{3}-1 $" }, { "text": "In $\\triangle ABC$, $A(-2 , 0)$, $B(2 , 0)$, then the trajectory equation of point $C$ satisfying that the perimeter of $\\triangle ABC$ is $8$ is?", "fact_expressions": "A: Point;B: Point;C: Point;Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);Perimeter(TriangleOf(A,B,C))=8", "query_expressions": "LocusEquation(C)", "answer_expressions": " {(x^2/16+y^2/12=1)&(Negation(x=pm*4))}", "fact_spans": "[[[17, 28]], [[31, 41]], [[68, 72]], [[17, 28]], [[31, 41]], [[45, 67]]]", "query_spans": "[[[68, 79]]]", "process": "" }, { "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(4,0)$. A line passing through point $F$ intersects the ellipse $E$ at points $A$ and $B$. If the midpoint of $AB$ has coordinates $M(1,-1)$, then the equation of the ellipse $E$ is?", "fact_expressions": "E: Ellipse;b: Number;a: Number;G: Line;A: Point;B: Point;F: Point;M: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (4, 0);Coordinate(M) = (1, -1);RightFocus(E) = F;PointOnCurve(F, G);Intersection(G, E) = {A, B};MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Expression(E)", "answer_expressions": "x^2/24+y^2/8=1", "fact_spans": "[[[2, 59], [82, 87], [121, 126]], [[9, 59]], [[9, 59]], [[79, 81]], [[88, 91]], [[92, 95]], [[64, 72], [74, 78]], [[110, 119]], [[9, 59]], [[9, 59]], [[2, 59]], [[64, 72]], [[110, 119]], [[2, 72]], [[73, 81]], [[79, 97]], [[99, 119]]]", "query_spans": "[[[121, 131]]]", "process": "According to the problem, let A(x_{1},y_{1}), B(x_{2},y_{2}), substituting into the ellipse equation \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 gives \\begin{cases}\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1,\\\\\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{\\frac{2}{32}}=1,\\end{cases} subtracting the two equations yields \\frac{x^{2}-x^{2}}{a^{2}}+\\frac{y^{2}-y^{2}}{b^{2}}=0, rearranging gives k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{-(x_{1}+x_{2})b^{2}}{(y_{1}+y_{2})^{2}}. Also, the midpoint M of AB is (1), so x_{1}+x_{2}=2, y_{1}+y_{2}=-2, substituting into the above expression gives k_{AB}=\\frac{\\frac{b^{2}k_{AB}=k_{MF}, F(4,0), k_{MF}=\\frac{1}{3}, thus \\frac{b^{2}}{a^{2}}=\\frac{1}{3}, 3b^{2}=a^{2}. Since a^{2}=b^{2}+c^{2}, c^{2}=16, solving gives a^{2}=b^{2}+c^{2}, c^{2}=16, therefore the equation of ellipse E is \\frac{x^{2}}{24}+\\frac{y^{2}}{8}=1" }, { "text": "If $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $P$ is a moving point on the ellipse, then the maximum value of $|P F_{1}| \\cdot|P F_{2}|$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G)", "query_expressions": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))", "answer_expressions": "4", "fact_spans": "[[[17, 44], [55, 57]], [[17, 44]], [[1, 8]], [[9, 16]], [[1, 49]], [[50, 53]], [[50, 63]]]", "query_spans": "[[[65, 97]]]", "process": "From the ellipse equation $\\frac{x^{2}}{4}+y^{2}=1$, we know $a=2$. Since $P$ is a moving point on this ellipse, $|PF_{1}|+|PF_{2}|=2a=4$. Therefore, by the basic inequality, we have: $|PF_{1}|\\cdot|PF_{2}|\\leqslant\\left(\\frac{|PF_{1}|+|PF_{2}|}{2}\\right)^{2}=4$ (equality holds if and only if $|PF_{1}|=|PF_{2}|=2$)." }, { "text": "It is known that the vertex of the parabola $C$ is at the origin, the focus is on the $x$-axis, and there is a point $P(4 , m)$ on the parabola whose distance to the focus is $6$. Then the equation of the parabola $C$ is?", "fact_expressions": "C: Parabola;P: Point;O: Origin;m:Number;Coordinate(P) = (4, m);Vertex(C) = O;PointOnCurve(Focus(C), xAxis);PointOnCurve(P, C);Distance(P, Focus(C)) = 6", "query_expressions": "Expression(C)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[2, 8], [25, 28], [55, 61]], [[32, 42]], [[12, 14]], [[32, 42]], [[32, 42]], [[2, 14]], [[2, 23]], [[25, 42]], [[32, 52]]]", "query_spans": "[[[55, 66]]]", "process": "According to the problem, the equation of the parabola can be written as y^{2}=2px. Its directrix is x=-\\frac{p}{2}. By definition, we have 4+\\frac{p}{2}=6, solving for p gives p=4. Therefore, the equation of parabola C is y^{2}=8x." }, { "text": "Given that a moving point $P(x, y)$ satisfies $10 \\sqrt{(x-1)^{2}+(y-2)^{2}}=|3 x+4 y|$, what is the trajectory of point $P$?", "fact_expressions": "P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;10*sqrt((x1 - 1)^2 + (y1 - 2)^2) = Abs(3*x1 + 4*y1)", "query_expressions": "Locus(P)", "answer_expressions": "Ellipse", "fact_spans": "[[[4, 13], [58, 62]], [[4, 13]], [[4, 13]], [[4, 13]], [[15, 56]]]", "query_spans": "[[[58, 67]]]", "process": "" }, { "text": "Given that the focus of a parabola is $F(0,1)$, a line $l$ passing through the point $P(0,2)$ intersects the parabola at points $A$ and $B$. If $|A B|=4 \\sqrt{6}$, then $|A F|+|B F|$=?", "fact_expressions": "G: Parabola;Focus(G) = F;Coordinate(F) = (0, 1);F: Point;P: Point;Coordinate(P) = (0, 2);l: Line;PointOnCurve(P,l) = True;Intersection(l, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 4*sqrt(6)", "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))", "answer_expressions": "10", "fact_spans": "[[[2, 5], [35, 38]], [[2, 17]], [[9, 17]], [[9, 17]], [[19, 28]], [[19, 28]], [[29, 34]], [[18, 34]], [[29, 49]], [[40, 43]], [[44, 47]], [[51, 69]]]", "query_spans": "[[[71, 86]]]", "process": "Find the equation of the parabola, set up the equation of line $ l $, solve the system of equations of line $ l $ and the parabola, simplify and write down the relationship between roots and coefficients. Find the slope of line $ l $ given $ |AB| = 4\\sqrt{6} $, and use the definition of the parabola to find $ |AF| + |BF| $. [Detailed Solution] Since the focus of the parabola is $ F(0,1) $, the equation of the parabola is $ x^{2} = 4y $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $; clearly the slope of line $ l $ exists, let its equation be $ y = kx + 2 $. From \n\\[\n\\begin{cases}\ny = kx + 2 \\\\\nx^{2} = 4y\n\\end{cases}\n\\]\nwe obtain $ x^{2} - 4kx - 8 = 0 $, so $ x_{1} + x_{2} = 4k $, $ x_{1}x_{2} = -8 $. Thus,\n\\[\n|AB| = \\sqrt{1 + k^{2}} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{1 + k^{2}} \\cdot \\sqrt{16k^{2} + 32} = 4\\sqrt{6},\n\\]\nsolving gives $ k = \\pm 1 $. Therefore, $ |AF| + |BF| = y_1 + y_2 + 2 = k(x_{1} + x_{2}) + 6 = 4k^{2} + 6 = 10 $." }, { "text": "The line $y = kx + 1$ always has common points with the ellipse $\\frac{x^{2}}{9} + \\frac{y^{2}}{m} = 1$ whose foci lie on the $x$-axis. Find the range of real values for $m$.", "fact_expressions": "G: Ellipse;m: Real;H: Line;k: Number;Expression(G) = (x^2/9 + y^2/m = 1);Expression(H) = (y = k*x + 1);PointOnCurve(Focus(G),xAxis);IsIntersect(H,G)", "query_expressions": "Range(m)", "answer_expressions": "[1,9)", "fact_spans": "[[[21, 58]], [[65, 70]], [[0, 11]], [[2, 11]], [[21, 58]], [[0, 11]], [[12, 58]], [[0, 63]]]", "query_spans": "[[[65, 77]]]", "process": "The line y = kx + 1 always passes through the fixed point P(0,1). For the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{m}=1$ with foci on the x-axis, we have $0 < m < 9$, $\\textcircled{1}$. Since the line $y = kx + 1$ always intersects the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{m}=1$ with foci on the x-axis, point P lies on or inside the ellipse, that is, $\\frac{0}{9}+\\frac{1}{m}\\leqslant1$, solving gives $m\\geqslant1$, $\\textcircled{2}$. From $\\textcircled{1}$ and $\\textcircled{2}$, we obtain $1\\leqslant m < 9$." }, { "text": "If a point $P$ on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$ has a distance of $6$ to one of its foci $F_{1}$, then what is the distance from $P$ to the other focus $F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/100 + y^2/36 = 1);P: Point;PointOnCurve(P, G);F1: Point;F2: Point;OneOf(Focus(G)) = F1;Distance(P, F1) = 6;OneOf(Focus(G)) = F2;Negation(F1= F2)", "query_expressions": "Distance(P, F2)", "answer_expressions": "14", "fact_spans": "[[[1, 41], [48, 49]], [[1, 41]], [[44, 47], [67, 70]], [[1, 47]], [[51, 58]], [[75, 82]], [[48, 58]], [[44, 65]], [[48, 82]], [[48, 82]]]", "query_spans": "[[[67, 87]]]", "process": "" }, { "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)$, point $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;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);Focus(G)={F1,F2};P: Point;PointOnCurve(P, G);Area(TriangleOf(P, F1, F2)) =(sqrt(2)/2)*b^2", "query_expressions": "Cos(AngleOf(F1, P, F2))", "answer_expressions": "1/3", "fact_spans": "[[[32, 84], [95, 97]], [[32, 84]], [[34, 84]], [[34, 84]], [[34, 84]], [[34, 84]], [[2, 16]], [[18, 31]], [[2, 16]], [[2, 16]], [[18, 31]], [[2, 89]], [[90, 94]], [[90, 98]], [[100, 152]]]", "query_spans": "[[[154, 183]]]", "process": "Let |PF₁| = m, |PF₂| = n. In triangle PF₁F₂, by the definition of the ellipse, m + n = 2a. Applying the law of cosines, we have cos∠F₁PF₂ = (m² + n² - 4c²)/(2mn), which implies mn = (2b²)/(1 + cos∠F₁PF₂). Since the area of triangle PF₁F₂ is (√2/2)b², it follows that S = (1/2)mn sin∠F₁PF₂ = (1/2) × (2b²)/(1 + cos∠F₁PF₂) × sin∠F₁PF₂ = (b²)/(1 + cos∠F₁PF₂) × sin∠F₁PF₂ = (√2/2)b². Therefore, 1 + cos∠F₁PF₂ = √2 sin∠F₁PF₂. Using the identity cos²∠F₁PF₂ + sin²∠F₁PF₂ = 1, we obtain cos∠F₁PF₂ = 1/3. Hence, the answer is 1/3." }, { "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)$, $B_{1} B_{2}$ is the imaginary axis of the hyperbola, $M$ is the midpoint of $O B_{1}$, and the line passing through $F$ and $M$ intersects the hyperbola $C$ at point $A$, with $F M=2 MA$. 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;LeftFocus(C) = F;B1: Point;B2: Point;ImageinaryAxis(C) = LineSegmentOf(B1, B2);M: Point;O: Origin;MidPoint(LineSegmentOf(O, B1)) = M;G: Line;PointOnCurve(F, G);PointOnCurve(M, G);A: Point;Intersection(G, C) = A;LineSegmentOf(F, M) = 2*LineSegmentOf(M, A)", "query_expressions": "Eccentricity(C)", "answer_expressions": "5/2", "fact_spans": "[[[6, 68], [87, 90], [123, 129], [148, 154]], [[6, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[14, 68]], [[2, 5], [112, 115]], [[2, 72]], [[73, 86]], [[73, 86]], [[73, 93]], [[94, 97], [116, 119]], [[98, 107]], [[94, 110]], [[120, 122]], [[111, 122]], [[111, 122]], [[130, 134]], [[120, 134]], [[136, 146]]]", "query_spans": "[[[148, 160]]]", "process": "" }, { "text": "It is known that the slope of one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\sqrt{2}$, and the right focus coincides with the focus of the parabola $y^{2}=4 \\sqrt{3} x$. 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;Slope(OneOf(Asymptote(G))) = sqrt(2);H: Parabola;Expression(H) = (y^2 = 4*(sqrt(3)*x));RightFocus(G) = Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/2=1", "fact_spans": "[[[2, 58], [115, 118]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 78]], [[84, 107]], [[84, 107]], [[2, 112]]]", "query_spans": "[[[115, 123]]]", "process": "The focus of the parabola y^{2}=4\\sqrt{3}x is (\\sqrt{3},0), so c=\\sqrt{3}. According to the problem, the slope of one asymptote of the hyperbola is \\frac{b}{a}=\\sqrt{2}. Combining this with c^{2}=a^{2}+b^{2}, we can solve to get a=1, b=\\sqrt{2}. Therefore, the equation of the hyperbola is x^{2}-\\frac{y^{2}}{2}=1." }, { "text": "The number of intersection points of the curves $x^{2}+y^{2}+2 x=0$ and $y+|x|=0$ is?", "fact_expressions": "Z: Curve;Expression(Z) = (2*x + x^2 + y^2 = 0);G: Curve;Expression(G) = (y + Abs(x) = 0)", "query_expressions": "NumIntersection(Z, G)", "answer_expressions": "2", "fact_spans": "[[[0, 21]], [[0, 21]], [[22, 33]], [[22, 33]]]", "query_spans": "[[[0, 40]]]", "process": "From \\begin{cases}x^{2}+y^{2}+2x=0\\\\y+|x|=0\\end{cases} we get $x^{2}+x=0$, so \\begin{cases}x=0\\\\y=0\\end{cases} or \\begin{cases}x=-1\\\\y=-1\\end{cases}, therefore the number of intersection points is 2." }, { "text": "Given that the length of the major axis of an ellipse is $8$ and the eccentricity is $\\frac{3}{4}$, then the standard equation of this ellipse is?", "fact_expressions": "G: Ellipse;Length(MajorAxis(G)) = 8;Eccentricity(G) = 3/4", "query_expressions": "Expression(G)", "answer_expressions": "{(x^2/16+y^2/7=1),(x^2/7+y^2/16=1)}", "fact_spans": "[[[2, 4], [33, 35]], [[2, 12]], [[2, 30]]]", "query_spans": "[[[33, 42]]]", "process": "Test analysis: From the given conditions combined with the definition of an ellipse, we can obtain the standard equation of the ellipse. According to the problem, 2a=8, ∴a=4, and e=\\frac{c}{a}=\\frac{3}{4}, ∴c=3, then b^{2}=a^{2}-c^{2}=7. When the foci of the ellipse are on the x-axis, the equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1. When the foci of the ellipse are on the y-axis, the equation of the ellipse is \\frac{x^{2}}{7}+\\frac{y^{2}}{16}=1" }, { "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, 16]], [[0, 16]]]", "query_spans": "[[[0, 23]]]", "process": "Analysis: Obtain the value of p directly from the standard equation of the parabola, and get the focus coordinates (-\\frac{p}{2},0). For the standard equation of the parabola y^{2}=-2px (p>0), the focus is (-\\frac{p}{2},0); therefore, the focus coordinates of y^{2}=-12x are (-3,0)." }, { "text": "Given that the line $y = -x + 1$ intersects the ellipse $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ $(a > 0, b > 0)$ at points $A$ and $B$, and $OA \\perp OB$ ($O$ being the origin), if the eccentricity of the ellipse $e \\in \\left[\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right]$, then the maximum value of $a$ is?", "fact_expressions": "H: Line;Expression(H) = (y = 1 - x);G: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Intersection(H, G) = {A, B};A: Point;B: Point;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B)) = True;O: Origin;e: Number;Eccentricity(G) = e;In(e, [1/2, sqrt(3)/2])", "query_expressions": "Max(a)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 68], [111, 113]], [[13, 68]], [[15, 68]], [[159, 162]], [[15, 68]], [[15, 68]], [[2, 80]], [[71, 74]], [[75, 78]], [[82, 97]], [[100, 103]], [[117, 157]], [[111, 157]], [[117, 157]]]", "query_spans": "[[[159, 168]]]", "process": "Let A(x₁, y₁), B(x₂, y₂). From \n\\begin{cases} y = -x + 1 \\\\ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\end{cases}, \neliminating y, we obtain (a² + b²)x² - 2a²x + a²(1 - b²) = 0. \n∴ x₁ + x₂ = \\frac{2a^{2}}{a^{2}+b^{2}}, x₁x₂ = \\frac{a^{2}(1-b^{2})}{a^{2}+b^{2}}. \nFrom Δ = (-2a²)² - 4a²(a² + b²)(1 - b²) > 0, simplifying gives a² + b² > 1. \n∴ y₁y₂ = (-x₁ + 1)(-x₂ + 1) = x₁x₂ - (x₁ + x₂) + 1. \nOA ⊥ OB (where O is the origin), we get \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0. \n∴ x₁x₂ + y₁y₂ = 0, i.e., x₁x₂ + (-x₁ + 1)(-x₂ + 1) = 0, \nsimplifying yields 2x₁x₂ - (x₁ + x₂) + 1 = 0. \n∴ 2 \\cdot \\frac{a^{2}(1-b^{2})}{a^{2}+b^{2}} - \\frac{2a^{2}}{a^{2}+b^{2}} + 1 = 0, \nsimplifying gives a² + b² - 2a²b² = 0. \n∵ b² = a² - c² = a² - a²e², substituting into the above equation and simplifying gives 2a² = 1 + \\frac{1}{1-e^{2}}, \n∴ a² = \\frac{1}{2}(1 + \\frac{1}{1-e^{2}}). \n∵ e ∈ [\\frac{1}{2}, \\frac{\\sqrt{3}}{2}], squaring gives \\frac{1}{4} ≤ e² ≤ \\frac{3}{4}. \n∴ \\frac{1}{4} ≤ 1 - e² ≤ \\frac{3}{4}, we get \\frac{4}{3} ≤ \\frac{1}{1-e^{2}} ≤ 4, \ntherefore \\frac{7}{3} ≤ 2a² = 1 + \\frac{1}{1-e^{2}} ≤ 5, \\frac{7}{6} ≤ a² ≤ \\frac{5}{2}, \nwe obtain the maximum value of a² is \\frac{5}{2}, satisfying the condition a² + b² > 1, \n∴ when the ellipse's eccentricity e = \\frac{\\sqrt{3}}{2}, the maximum value of a is \\frac{\\sqrt{10}}{2}." }, { "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{16}=1$ $(a>0)$ passes through the point $(2,0)$, then what is the equation of the asymptotes of this hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;H: Point;a>0;Expression(G) = (-y^2/16 + x^2/a^2 = 1);Coordinate(H) = (2, 0);PointOnCurve(H, G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*2*x", "fact_spans": "[[[1, 49], [62, 65]], [[4, 49]], [[51, 59]], [[4, 49]], [[1, 49]], [[51, 59]], [[1, 59]]]", "query_spans": "[[[62, 73]]]", "process": "Substituting the coordinates of the point (2,0) into the hyperbola equation gives \\frac{4}{a2}=1,\\becausea>0, we obtain a=2. Therefore, the equation of the hyperbola is \\frac{x^{2}}{4}-\\frac{y^{2}}{\\frac{2}{16}}=1, thus the asymptotes of this hyperbola are y=\\pm2x" }, { "text": "Given the parabola equation $y^{2}=4 x$, the line $x=a$ intersects the parabola at points $A$ and $B$. The focus $F$ of the parabola is the orthocenter of $\\triangle O A B$ ($O$ being the origin). Then the real value of $a$ is?", "fact_expressions": "G: Parabola;H: Line;a: Real;O: Origin;A: Point;B: Point;F:Point;Expression(H) = (x = a);Expression(G) = (y^2 = 4*x);Intersection(H, G) = {A, B};Focus(G)=F;Orthocenter(TriangleOf(O,A,B))=F", "query_expressions": "a", "answer_expressions": "5", "fact_spans": "[[[2, 5], [28, 31], [43, 46]], [[20, 27]], [[86, 91]], [[71, 74]], [[33, 36]], [[37, 40]], [[49, 52]], [[20, 27]], [[2, 19]], [[20, 42]], [[43, 52]], [[49, 84]]]", "query_spans": "[[[86, 95]]]", "process": "From the given conditions, F(1,0), let A(a,b), B(a,-b), then b^{2}=4a, a\\neq0. Hence \\overrightarrow{FA}=(a-1,b), \\overrightarrow{OB}=(a,-b), then \\overrightarrow{FA}\\cdot\\overrightarrow{OB}=(a-1)a-b^{2}=a(a-5)=0, \\therefore a=5," }, { "text": "Given the parabola $C$: $y^{2}=16 x$ with focus $F$, point $A(-4 , 0)$, and point $P$ a moving point on the parabola $C$, find the minimum value of $\\frac{|P F|}{|P A|}$.", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 16*x);F: Point;Focus(C) = F;A: Point;Coordinate(A) = (-4, 0);P: Point;PointOnCurve(P, C)", "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[2, 22], [48, 54]], [[2, 22]], [[26, 29]], [[2, 29]], [[30, 42]], [[30, 42]], [[43, 47]], [[43, 58]]]", "query_spans": "[[[60, 87]]]", "process": "As shown in the figure, let point Q be the projection of point P on the directrix. The angle ∠PAQ is minimized when the line AP is tangent to the parabola C, and thus sin∠PAQ is also minimized. Then solve according to A=0. Point A(4,0) lies on the directrix x=4 of parabola C. Let point Q be the projection of point P on the directrix, then \\frac{|PF|}{|PA|}=\\frac{|PQ|}{|PA|}=\\sin\\anglePAQ. When the line AP is tangent to the parabola C, ∠PAQ is minimized and sin∠PAQ is also minimized. Let the equation of PA be y=k(x+4). Substituting into y^{2}=16x gives k^{2}x^{2}+(8k^{2}-16)x+16k^{2}=0. From \\Delta=(8k^{2}-16)^{2}-64k^{4}=0, we obtain k=\\pm1. When k=\\pm1, \\sin\\anglePAQ=\\frac{\\sqrt{2}}{2}." }, { "text": "If a point $M$ on the parabola $y=\\frac{1}{4} x^{2}$ is at a distance of $4$ from the focus $F$, then what is the value of the vertical coordinate of point $M$?", "fact_expressions": "G: Parabola;M: Point;F: Point;Expression(G) = (y = x^2/4);PointOnCurve(M, G);Focus(G) = F;Distance(M, F) = 4", "query_expressions": "YCoordinate(M)", "answer_expressions": "3", "fact_spans": "[[[1, 25]], [[28, 31], [46, 50]], [[34, 37]], [[1, 25]], [[1, 31]], [[1, 37]], [[28, 44]]]", "query_spans": "[[[46, 58]]]", "process": "Find the focus and the equation of the directrix according to the parabolic equation, and solve using the definition of a parabola. [Detailed solution] From y = \\frac{1}{4}x^{2}, we get x^{2} = 4y. Therefore, the focus of the parabola is F(0,1), and the equation of the directrix is y = -1. Let M(x_{M}, y_{M}). By the definition of the parabola, |MF| = y_{M} + 1 = 4, so y_{M} = 3." }, { "text": "The distance from point $M$ to point $F(2,0)$ is 1 less than its distance to the line $l$: $x+3=0$. What is the equation of the trajectory of point $M$?", "fact_expressions": "M: Point;F: Point;Coordinate(F) = (2, 0);l: Line;Expression(l) = (x + 3 = 0);Distance(M, F) = Distance(M, l) - 1", "query_expressions": "LocusEquation(M)", "answer_expressions": "y^2 = 8*x", "fact_spans": "[[[0, 4], [18, 19], [43, 47]], [[5, 14]], [[5, 14]], [[20, 34]], [[20, 34]], [[0, 41]]]", "query_spans": "[[[43, 54]]]", "process": "\\because the distance from point M to point F(2,0) is 1 less than its distance to the line l: x+3=0, \\therefore the distance from point M to the line x=-2 is equal to its distance to the point (2,0). Therefore, the trajectory of point M satisfies the definition of a parabola. According to the definition of a parabola, the trajectory of point M is a parabola with focus at point (2,0) and directrix the line x=-2. \\therefore p=4, the standard equation of the parabola is y^2=8x" }, { "text": "Let $A$ and $B$ be two points on the curve $C$: $y = \\frac{x^{2}}{4}$, with the sum of the x-coordinates of $A$ and $B$ equal to $4$. What is the slope of the line $AB$?", "fact_expressions": "A: Point;B: Point;C: Curve;Expression(C) = (y = x^2/4);PointOnCurve(A, C);PointOnCurve(B, C);XCoordinate(A) + XCoordinate(B) = 4", "query_expressions": "Slope(LineOf(A, B))", "answer_expressions": "1", "fact_spans": "[[[1, 4], [39, 42]], [[5, 8], [43, 46]], [[9, 35]], [[9, 35]], [[1, 38]], [[1, 38]], [[39, 56]]]", "query_spans": "[[[58, 69]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Substitute the coordinates of points A and B into the parabola equation, subtract the two equations and simplify to obtain the slope of line AB. Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since A and B are two points on the curve C: y = \\frac{x^{2}}{4}, we have \\begin{cases} y_{1} = \\frac{x_{1}^{2}}{4} \\\\ y_{2} = \\frac{x_{2}^{2}}{4} \\end{cases}, then y_{1} - y_{2} = \\frac{x_{1}^{2}}{4} - \\frac{x_{2}^{2}}{4} = \\frac{(x_{1} - x_{2})(x_{1} + x_{2})}{4}. Moreover, the sum of the x-coordinates of A and B is 4, that is, x_{1} + x_{2} = 4. Therefore, the slope of line AB is k_{AB} = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = \\frac{x_{1} + x_{2}}{4} = 1." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$, respectively. There is a point $P$ on the hyperbola $C$ such that $|P F_{1}|=7$. Find $|P F_{2}|$.", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/9 - y^2/7 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C) = True;P: Point;Abs(LineSegmentOf(P, F1)) = 7", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "{1,13}", "fact_spans": "[[[2, 45], [71, 77]], [[2, 45]], [[54, 61]], [[62, 69]], [[2, 69]], [[2, 69]], [[71, 84]], [[81, 84]], [[86, 99]]]", "query_spans": "[[[101, 114]]]", "process": "Use the definition of a hyperbola to solve. [Detailed solution] Since the hyperbola C: \\frac{x^{2}}{9}-\\frac{y^{2}}{7}=1, we have a=3, so ||PF_{2}|-|PF_{1}||=2a=6. Given that |PF_{1}|=7, it follows that |PF_{2}|=13 or |PF_{2}|=1." }, { "text": "Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, a line $l$ passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the parabola $C$ in the first and fourth quadrants at points $A$ and $B$ respectively. Then $\\frac{|A F|}{|B F|}$=?", "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C)=(y^2 = 2*(p*x));p:Number;p>0;Focus(C) = F;PointOnCurve(F, l);Inclination(l) = ApplyUnit(60, degree);Intersection(l, C) = {A, B};Quadrant(A)=1;Quadrant(B)=4", "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))", "answer_expressions": "3", "fact_spans": "[[[54, 59]], [[2, 23], [60, 66]], [[77, 80]], [[27, 30], [32, 36]], [[81, 84]], [[2, 23]], [[5, 23]], [[5, 23]], [[2, 30]], [[31, 59]], [[37, 59]], [[54, 86]], [[67, 86]], [[67, 86]]]", "query_spans": "[[[88, 111]]]", "process": "Problem Analysis: As shown in the figure, draw AM\\botl, BN\\botl, where M and N are the feet of the perpendiculars, then |AF|=|AM|, |BF|=|BN|. From the given information, in the right trapezoid AMNB, \\angleMAB=60^{\\circ}, so \\frac{|AM|-|BN|}{|AB|}=\\cos60^{\\circ}=\\frac{1}{2}, thus \\frac{|AF|-|BF|}{|AF|+|BF|}=\\frac{1}{2}, therefore \\frac{|AF|}{|BF|}=3. Key point: Definition of a parabola." }, { "text": "Given that the focus of the parabola $y^{2}=2 x$ is $F$, point $P$ is a moving point on the parabola, and there is a point $A(3 , 2)$. Then the minimum value of $|PA|+|PF|$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*x);Coordinate(A) = (3, 2);Focus(G) = F;PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "7/2", "fact_spans": "[[[2, 16], [29, 32]], [[39, 50]], [[24, 28]], [[20, 23]], [[2, 16]], [[39, 50]], [[2, 23]], [[24, 36]]]", "query_spans": "[[[53, 70]]]", "process": "" }, { "text": "The equation of the hyperbola whose vertices are the foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{8}=1$, and whose foci are the vertices of this ellipse, is?", "fact_expressions": "E: Hyperbola;C: Ellipse;Expression(C) = (x**2/5 + y**2/8 = 1);Vertex(E) = Focus(C);Focus(E) = Vertex(C)", "query_expressions": "Expression(E)", "answer_expressions": "y^2/3 - x^2/5 = 1", "fact_spans": "[[[57, 60]], [[1, 38], [48, 50]], [[1, 38]], [[0, 60]], [[46, 60]]]", "query_spans": "[[[57, 65]]]", "process": "" }, { "text": "Given that the left and right foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are $F_{1}$ and $F_{2}$ respectively, if a line $l$ passing through $F_{1}$ intersects the ellipse at points $A$ and $B$, then the perimeter of $\\triangle A B F_{2}$ is equal to?", "fact_expressions": "G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;PointOnCurve(F1,l) = True;Intersection(l,G) = {A,B};Expression(G) = (x^2/4 + y^2/3 = 1)", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "8", "fact_spans": "[[[2, 39], [84, 86]], [[89, 92]], [[93, 96]], [[58, 65]], [[50, 57], [70, 77]], [[2, 65]], [[2, 65]], [[78, 83]], [[68, 83]], [[78, 98]], [[2, 39]]]", "query_spans": "[[[99, 126]]]", "process": "According to the problem and the definition of an ellipse, we have: |AF₁| + |AF₂| = 2a = 4, |BF₁| + |BF₂| = 2a = 4. Also, the perimeter of triangle ABF₂ is: |AB| + |AF₂| + |BF₂| = |AF₁| + |AF₂| + |BF₁| + |BF₂| = 4a = 8," }, { "text": "If the foci of the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{16}=1$ are $F_{1}$ and $F_{2}$, and point $P$ is any point on $C$, then $|P F_{1}|+|P F_{2}|$=?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/36 + y^2/16 = 1);F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C)", "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))", "answer_expressions": "12", "fact_spans": "[[[1, 40], [69, 72]], [[1, 40]], [[46, 53]], [[55, 62]], [[1, 62]], [[64, 68]], [[64, 77]]]", "query_spans": "[[[79, 102]]]", "process": "The foci of the ellipse \\frac{x^{2}}{36}+\\frac{y^{2}}{16}=1 are F_{1} and F_{2}, and point P is any point on C. \\therefore a=6, then |PF|+|PF_{1}|=2a=12." }, { "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}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the two asymptotes of $C$ at points $A$ and $B$ respectively. If $\\overrightarrow{F_{1} A}=\\overrightarrow{A B}$ and $\\overrightarrow {F_{1} B} \\cdot \\overrightarrow {F_{2} B}=0$, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;F1: Point;A: Point;B: 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, G);L1:Line;L2:Line;Asymptote(C)={L1,L2};Intersection(G, L1) = A;Intersection(G,L2) =B;VectorOf(F1, A) = VectorOf(A, B);DotProduct(VectorOf(F1,B),VectorOf(F2,B))=0", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[2, 63], [100, 103], [237, 240]], [[10, 63]], [[10, 63]], [[97, 99]], [[89, 96], [72, 79]], [[113, 116]], [[117, 120]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 99]], [], [], [[100, 109]], [[97, 122]], [[97, 122]], [[125, 172]], [[174, 235]]]", "query_spans": "[[[237, 246]]]", "process": "As shown in the figure, from $\\overrightarrow{F_{1}A}=\\overrightarrow{AB}$, we get $F_{1}A=AB$. Also, since $OF_{1}=OF_{2}$, it follows that $OA$ is the midline of triangle $F_{1}F_{2}B$, i.e., $BF_{2}//OA$, $BF_{2}=2OA$. From $\\overrightarrow{F_{1}B}\\cdot\\overrightarrow{F_{2}B}=0$, we get $F_{1}B\\bot F_{2}B$, $OA\\bot F_{1}A$, then $OB=OF_{1}$, and $\\angle AOB=\\angle AOF_{1}$. Since both $OA$ and $OB$ are asymptotes, $\\angle BOF_{2}=\\angle AOF_{1}$. Also, $\\angle BOF_{2}+\\angle AOB+\\angle AOF_{1}=\\pi$, so $\\angle BOF_{2}=\\angle AOF_{1}=\\angle BOA=60^{\\circ}$. The slope of asymptote $OB$ is $\\frac{b}{a}=\\tan 60^{\\circ}=\\sqrt{3}$, thus the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{1+(\\frac{b}{a})^{2}}=\\sqrt{1+(\\sqrt{3})^{2}}=2$. This problem examines planar vectors combined with the asymptotes and eccentricity of a hyperbola, involving logical reasoning, intuitive imagination, and mathematical computation. It uses a geometric method, applying the idea of combining numbers and figures to solve the problem." }, { "text": "The asymptotes of the hyperbola are given by $x \\pm 2 y=0$, and the focal distance is $10$. What is the equation of this hyperbola?", "fact_expressions": "E: Hyperbola;Expression(Asymptote(E)) = (x + pm*2*y = 0);FocalLength(E) = 10", "query_expressions": "Expression(E)", "answer_expressions": "{x**2/20-y**2/5=1, y**2/5-x**2/20=1}", "fact_spans": "[[[0, 3], [33, 36]], [[0, 23]], [[0, 31]]]", "query_spans": "[[[33, 41]]]", "process": "" }, { "text": "The focus of the parabola $y^{2}=8 x$ is $F$, point $A(6,3)$, and $P$ is a point on the parabola such that $P$ does not lie on the line $A F$. Then the minimum value of the perimeter of $\\triangle P A F$ is?", "fact_expressions": "G: Parabola;F: Point;A: Point;P: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;Coordinate(A)=(6,3);PointOnCurve(P,G);Negation(PointOnCurve(P,LineOf(A,F)))", "query_expressions": "Min(Perimeter(TriangleOf(P, A, F)))", "answer_expressions": "13", "fact_spans": "[[[0, 14], [37, 40]], [[18, 21]], [[22, 31]], [[33, 36], [45, 48]], [[0, 14]], [[0, 21]], [[22, 31]], [[33, 43]], [[45, 58]]]", "query_spans": "[[[60, 85]]]", "process": "By the definition of a parabola, the distance PF from a point on the parabola to the focus is equal to the distance d from this point to the directrix, that is, FP = d. Therefore, the perimeter l = PA + PF + AF = PA + AF + d = PA + d + 5 \\geqslant 13, fill in 13" }, { "text": "The line intersects the parabola $y^{2}=4x$ at points $A$ and $B$, with $|AB|=8$. Then, the minimum value of the distance from the midpoint of segment $AB$ to the $y$-axis is?", "fact_expressions": "H: Line;G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 8", "query_expressions": "Min(Distance(MidPoint(LineSegmentOf(A, B)), yAxis))", "answer_expressions": "3", "fact_spans": "[[[0, 2]], [[3, 17]], [[3, 17]], [[19, 22]], [[23, 26]], [[0, 28]], [[29, 38]]]", "query_spans": "[[[40, 62]]]", "process": "" }, { "text": "The foci of the ellipse $\\frac{y^{2}}{16}+\\frac{x^{2}}{9}=1$ are $F_{1}$, $F_{2}$, and $P$ is a point on the ellipse different from the endpoints of the major axis. Then, what is the perimeter of $\\Delta P F_{1} F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/16 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Negation(P=Endpoint(MajorAxis(G)))", "query_expressions": "Perimeter(TriangleOf(P, F1, F2))", "answer_expressions": "8+2*sqrt(7)", "fact_spans": "[[[0, 38], [62, 64]], [[0, 38]], [[42, 49]], [[50, 57]], [[0, 57]], [[58, 61]], [[58, 75]], [[58, 75]]]", "query_spans": "[[[77, 104]]]", "process": "From $\\frac{y^{2}}{16}+\\frac{x^{2}}{9}=1$ we get $a^{2}=16$, $b^{2}=9$, so $c^{2}=a^{2}-b^{2}=16-9=7$, thus $c=\\sqrt{7}$, so $|F_{1}F_{2}|=2c=2\\sqrt{7}$. According to the definition of the ellipse, $|PF_{1}|+|PF_{2}|=2a=8$, so the perimeter of $\\triangle PF_{1}F_{2}$ is: $|PF_{1}|+|PF_{2}|+|F_{1}F_{2}|=8+2\\sqrt{7}$." }, { "text": "The directrix equation of the parabola $y=-\\frac{1}{4} x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/4)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=1", "fact_spans": "[[[0, 25]], [[0, 25]]]", "query_spans": "[[[0, 32]]]", "process": "The equation of the parabola can be written as: $x^{2}=-4y$. Therefore, the equation of the directrix of the parabola is: $y=1$." }, { "text": "The equation of a hyperbola that shares the same foci with the ellipse $x^{2}+4 y^{2}=16$ and has an asymptote given by $x+ \\sqrt {3} y=0$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H)=(x^2 + 4*y^2 = 16);Focus(H) = Focus(G);Expression(OneOf(Asymptote(G))) = (x + sqrt(3)*y = 0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9-y^2/3=1", "fact_spans": "[[[56, 59]], [[2, 22]], [[2, 22]], [[0, 59]], [[29, 59]]]", "query_spans": "[[[56, 64]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has its right focus at $F$. From point $F$, a perpendicular is drawn to one asymptote of the hyperbola, with foot of the perpendicular at $M$, intersecting the other asymptote at $N$. If $2 \\overrightarrow{M F}= \\overrightarrow{F N}$, then what is the eccentricity of the hyperbola?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Point;F: Point;N: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;L:Line;L1:Line;L2:Line;PointOnCurve(F, L);IsPerpendicular(L,L1);FootPoint(L,L1)=M;Intersection(L,L2)=N;2*VectorOf(M, F) = VectorOf(F, N);OneOf(Asymptote(C))=L1;OneOf(Asymptote(C))=L2;Negation(L1=L2)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 53], [68, 71], [149, 152]], [[10, 53]], [[10, 53]], [[84, 87]], [[58, 61], [63, 67]], [[96, 99]], [[2, 53]], [[2, 61]], [], [], [], [[62, 80]], [[62, 80]], [[62, 87]], [[62, 99]], [[101, 147]], [66, 74], [66, 108], [66, 108]]", "query_spans": "[[[149, 157]]]", "process": "As shown in the figure, the equation of asymptote OM is bx+ay=0, and the right focus is F(c,0), thus |FM|=\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=b. Draw a perpendicular line from point F to ON, with foot of perpendicular at P, then |FP|=|FM|=b. Since 2\\overrightarrow{MF}=\\overrightarrow{FN}, it follows that |FN|=2b. In the right triangle FPN, \\sin\\angleFNP=\\frac{|PF|}{|FN|}=\\frac{b}{2b}=\\frac{1}{2}, so \\angleFNP=\\frac{\\pi}{6}. Therefore, in triangle OMN, \\angleMON=\\frac{\\pi}{3}, hence \\angleFON=\\frac{\\pi}{6}, so \\frac{b}{a}=\\frac{\\sqrt{3}}{3}, i.e., a=\\sqrt{3}b, c=\\sqrt{a^{2}+b^{2}}=2b. Thus, the eccentricity of the hyperbola is e=\\frac{c}{a}=\\frac{2b}{\\sqrt{3}b}=\\frac{2\\sqrt{3}}{3}" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $A$, $B$ are two distinct points on the ellipse $C$. Let $P(1, \\frac{1}{2})$. If $\\overrightarrow{A P}=\\overrightarrow{P B}$, then the equation of the line $AB$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);A: Point;B: Point;PointOnCurve(A,C) = True;PointOnCurve(B,C) = True;Negation(A=B);P: Point;Coordinate(P) = (1, 1/2);VectorOf(A, P) = VectorOf(P, B)", "query_expressions": "Expression(LineOf(A,B))", "answer_expressions": "3*x+2*y-4=0", "fact_spans": "[[[2, 44], [57, 62]], [[2, 44]], [[47, 50]], [[53, 56]], [[47, 70]], [[47, 70]], [[47, 70]], [[72, 91]], [[72, 91]], [[93, 136]]]", "query_spans": "[[[138, 150]]]", "process": "From $\\overrightarrow{AP}=\\overrightarrow{PB}$, it follows that $P$ is the midpoint of $AB$, and $P(1,\\frac{1}{2})$ lies inside the ellipse $C: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\frac{x_{1}^{2}}{4}+\\frac{y_{1}^{2}}{3}=1$, $\\therefore\\underline{(x_{1}-x_{2})(x_{1}+x}$ then $\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{3(x_{1}}{4(y_{1}}\\frac{y_{1}-y_{2})(y_{1}+y_{2})}{4\\times1}=-\\frac{3}{2}$, that is, the slope of the line $AB$ is $-\\frac{3}{2}$. $\\therefore$ the equation of line $AB$ is $y-\\frac{1}{2}=-\\frac{3}{2}(x-])$, namely $3x+2y-4=0$." }, { "text": "Let the left and right foci of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{24}=1$ be $F_{1}$ and $F_{2}$, respectively, and let point $A$ be a point on the right branch of $C$. Then the minimum value of $|A F_{1}|+\\frac{8}{|A F_{2}|}$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/24 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;A: Point;PointOnCurve(A, RightPart(C))", "query_expressions": "Min(Abs(LineSegmentOf(A, F1)) + 8/Abs(LineSegmentOf(A, F2)))", "answer_expressions": "8", "fact_spans": "[[[1, 35], [67, 70]], [[1, 35]], [[46, 53]], [[54, 61]], [[1, 61]], [[1, 61]], [[62, 66]], [[62, 76]]]", "query_spans": "[[[78, 116]]]", "process": "From the hyperbola $ C: x^{2} - \\frac{y^{2}}{24} = 1 $, we obtain $ a^{2} = 1 $, $ b^{2} = 24 $, so $ c^{2} = a^{2} + b^{2} = 25 $, hence $ a = 1 $, $ c = 5 $. By the definition of the hyperbola, $ |AF_{1}| - |AF_{2}| = 2a = 2 $, so $ |AF_{1}| = |AF_{2}| + 2 $. Therefore, $ |AF_{1}| + \\frac{8}{|AF_{2}|} = |AF_{2}| + \\frac{8}{|AF_{2}|} + 2 $. From the properties of the hyperbola, we know: $ |AF_{2}| \\geqslant c - a = 4 $. Let $ |AF_{2}| = t $, then $ t \\geqslant 4 $, so $ |AF_{1}| + \\frac{8}{|AF_{2}|} = |AF_{2}| + \\frac{8}{|AF_{2}|} + 2 = t + \\frac{8}{t} + 2 $. Let $ y = t + \\frac{8}{t} + 2 $. Suppose $ 4 \\leqslant t_{1} < t_{2} $, then $ y_{1} - y_{2} = t_{1} + \\frac{8}{t_{1}} + 2 - (t_{2} + \\frac{8}{t_{2}} + 2) = \\frac{(t_{1} - t_{2})(t_{1}t_{2} - 8)}{t_{1}t_{2}} < 0 $, so $ y_{1} < y_{2} $, i.e., $ y = t + \\frac{8}{t} + 2 $ is monotonically increasing on $ [4, +\\infty) $. Therefore, when $ t = 4 $, it attains the minimum value $ 4 + \\frac{8}{4} + 2 = 8 $, at which point $ A $ is the right vertex of the hyperbola $ (1, 0) $." }, { "text": "The eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 34]]]", "process": "Analysis: From the given conditions, we have: a=1, c^{2}=1+2=3, c=\\sqrt{3}, e=\\frac{c}{a}=\\sqrt{3}." }, { "text": "Find the equation of a line passing through the point $P(1,1)$ that intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ at points $M$ and $N$, such that $P$ is exactly the midpoint of segment $MN$.", "fact_expressions": "P: Point;Coordinate(P) = (1, 1);L: Line;PointOnCurve(P, L) = True;G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);Intersection(L, G) = {M, N};M: Point;N: Point;MidPoint(LineSegmentOf(M, N)) = P", "query_expressions": "Expression(L)", "answer_expressions": "x+2*y-3=0", "fact_spans": "[[[1, 10], [66, 69]], [[1, 10]], [[13, 15], [84, 86]], [[0, 15]], [[16, 53]], [[16, 53]], [[13, 64]], [[55, 58]], [[59, 62]], [[66, 81]]]", "query_spans": "[[[84, 91]]]", "process": "Let M(x₁,y₁), N(x₂,y₂), then  \\frac{x_{1}^{2}}{4}+\\frac{y_{1}^{2}}{2}=1, \\frac{x_{2}^{2}}{4}+\\frac{y_{2}^{2}}{2}=1. Subtracting these two equations gives \\frac{x_{1}^{2}-x_{2}^{2}}{4}+\\frac{y_{1}^{2}-y_{2}^{2}}{2}=0, which simplifies to \\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{4}+\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{2}=0. Since point P(1,1) is the midpoint of segment MN, we have x₁+x₂=2, y₁+y₂=2. Therefore, \\frac{(x_{1}-x_{2}) \\cdot 2}{4}+\\frac{(y_{1}-y_{2}) \\cdot 2}{2}=0, which leads to \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}. Hence, the equation of the line is: y-1=-\\frac{1}{2}(x-1), or equivalently, x+2y-3=0." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, let the right vertex and the right focus be points $A$ and $F$, respectively. A line $l$ passing through point $A$ intersects one asymptote of $C$ at point $Q$. The line $QF$ intersects $C$ at a point $B$, such that $\\overrightarrow{A Q} \\cdot \\overrightarrow{A B}=\\overrightarrow{A Q} \\cdot \\overrightarrow{F B}$ and $\\overrightarrow{B Q}=4 \\overrightarrow{F Q}$. Find the eccentricity $e$ of the hyperbola.", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;F: Point;Q: Point;A: Point;B: Point;e: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightVertex(C) = A;RightFocus(C) = F;PointOnCurve(A,l);Intersection(l,OneOf(Asymptote(C)))=Q;OneOf(Intersection(LineOf(Q,F), C)) = B;DotProduct(VectorOf(A, Q), VectorOf(A, B)) = DotProduct(VectorOf(A, Q), VectorOf(F, B));VectorOf(B, Q) = 4*VectorOf(F, Q);Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "(3+sqrt(10))/4", "fact_spans": "[[[86, 91]], [[2, 63], [92, 95], [116, 119], [275, 278]], [[10, 63]], [[10, 63]], [[76, 79]], [[103, 107]], [[81, 85], [72, 75]], [[125, 128]], [[282, 285]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 79]], [[2, 79]], [[80, 91]], [[86, 107]], [[108, 128]], [[129, 226]], [[228, 273]], [[275, 285]]]", "query_spans": "[[[282, 287]]]", "process": "Hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $), $ A(a,0) $, asymptotes $ y = \\pm \\frac{b}{a}x $, let the right focus be $ F(c,0) $. From $ \\overrightarrow{AQ} \\cdot \\overrightarrow{AB} = \\overrightarrow{AQ} \\cdot \\overrightarrow{FB} \\Leftrightarrow \\overrightarrow{AQ} \\cdot (\\overrightarrow{AB} + \\overrightarrow{BF}) = 0 \\Leftrightarrow \\overrightarrow{AQ} \\cdot \\overrightarrow{AF} = 0 $, i.e., $ \\overrightarrow{AQ} \\perp \\overrightarrow{AF} $, line $ l: x = a $. By symmetry of the hyperbola, without loss of generality, let $ Q(a,b) $, let $ B(x_{0},y_{0}) $, then $ \\overrightarrow{BQ} = (a - x_{0}, b - y_{0}) $, $ \\overrightarrow{FQ} = (a - c, b) $. Since $ \\overrightarrow{BQ} = 4\\overrightarrow{FQ} $, then $ (a - x_{0}, b - y_{0}) = 4(a - c, b) $, solving gives $ x_{0} = 4c - 3a $, $ y_{0} = -3b $, i.e., point $ B(4c - 3a, -3b) $. Also, point $ B $ lies on hyperbola $ C $, so $ \\frac{(4c - 3a)^{2}}{a^{2}} - \\frac{(-3b)^{2}}{b^{2}} = 1 \\Leftrightarrow (4e - 3)^{2} = 10 $, solving gives $ e = \\frac{3 \\pm \\sqrt{10}}{4} $, since $ e > 1 $, then $ e = \\frac{3 + \\sqrt{10}}{4} $." }, { "text": "The parabola $y^{2}=2 px$ passes through the point $M(2 , 2)$. Then, the distance from point $M$ to the focus of the parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;M: Point;Coordinate(M) = (2, 2);PointOnCurve(M, G)", "query_expressions": "Distance(M, Focus(G))", "answer_expressions": "5/2", "fact_spans": "[[[0, 15], [34, 37]], [[0, 15]], [[3, 15]], [[16, 27], [29, 33]], [[16, 27]], [[0, 27]]]", "query_spans": "[[[29, 44]]]", "process": "" }, { "text": "Given that $O$ is the coordinate origin, a line passing through the point $M(a, 0)$ $(a \\neq 0)$ intersects the parabola $y^{2}=2 p x$ $(p>0)$ at points $A$ and $B$. Let the slopes of lines $OA$ and $OB$ be $k_{1}$ and $k_{2}$, respectively. If $k_{1} k_{2}=-2 p$, then the value of $a$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;H: Line;O: Origin;M: Point;a: Number;Negation(a=0);Coordinate(M) = (a, 0);PointOnCurve(M, H);A: Point;B: Point;Intersection(H, G) = {A, B};k1: Number;k2: Number;Slope(LineSegmentOf(O, A)) = k1;Slope(LineSegmentOf(O, B)) = k2;k1*k2 = -2*p", "query_expressions": "a", "answer_expressions": "1", "fact_spans": "[[[36, 57]], [[36, 57]], [[39, 57]], [[39, 57]], [[33, 35]], [[2, 5]], [[12, 32]], [[128, 131]], [[13, 32]], [[12, 32]], [[11, 35]], [[59, 62]], [[63, 66]], [[33, 68]], [[90, 97]], [[99, 106]], [[70, 106]], [[70, 106]], [[108, 126]]]", "query_spans": "[[[128, 135]]]", "process": "Since the line passes through the point M(a,0), let the equation of the line be x = my + a, A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}x = my + a\\\\y^2 = 2px\\end{cases}, we get y^2 - 2pmy - 2pa = 0, so y_{1}y_{2} = -2pa. Also, since points A(x_{1},y_{1}) and B(x_{2},y_{2}) lie on the parabola, we have y_{1}^2 = 2px_{1}, y_{2}^2 = 2px_{2}. Thus y_{1}^2 y_{2}^2 = 4p^2 x_{1}x_{2}, that is, \\frac{y_{1}y_{2}}{x_{1}x_{2}} = \\frac{4p^2}{y_{1}y_{2}} = \\frac{4p^2}{-2pa}, so \\frac{y_{1}y_{2}}{x_{1}x_{2}} = \\frac{2p}{-a}. Since k_{1}k_{2} = -2p, it follows that \\frac{y_{1}y_{2}}{x_{1}x_{2}} = \\frac{2p}{-a} = -2p^2, thus a = 1. Answer: 1" }, { "text": "A point $M$ on the parabola $y=\\frac{1}{4} x^{2}$ is at a distance of $3$ from the focus. What is the vertical coordinate of point $M$?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4);M: Point;PointOnCurve(M, G);Distance(M, Focus(G)) = 3", "query_expressions": "YCoordinate(M)", "answer_expressions": "2", "fact_spans": "[[[0, 24]], [[0, 24]], [[27, 30], [42, 46]], [[0, 30]], [[0, 40]]]", "query_spans": "[[[42, 52]]]", "process": "Find the standard equation of the parabola and the directrix of the parabola y = -1. According to the definition of a parabola, the distance from point M to the directrix y = -1 is 3, thus solve it. Rewrite the parabola equation as x^{2} = 4y. According to the definition of the parabola, the distance from point M to the directrix y = -1 is also 3, so the y-coordinate of M is 2." }, { "text": "Given that the ellipse $C$ with center at the origin passes through the point $(1, \\frac{2 \\sqrt{5}}{5})$, and one focus of $C$ is $(2,0)$, then the standard equation of $C$ is?", "fact_expressions": "C: Ellipse;G: Point;O:Origin;Center(C)=O;Coordinate(G) = (1, 2*sqrt(5)/5);Coordinate(OneOf(Focus(C)))=(2,0);PointOnCurve(G,C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/5+y^2=1", "fact_spans": "[[[10, 15], [46, 49], [64, 67]], [[16, 44]], [[5, 9]], [[2, 15]], [[16, 44]], [[46, 62]], [[10, 44]]]", "query_spans": "[[[64, 74]]]", "process": "Analysis: First, from the coordinate of one focus of the ellipse, obtain the other focus as (-2,0). Combining a given point through which the graph passes and using the definition of the ellipse, calculate the value of 2a by the distance between two points, thus obtaining a=\\sqrt{5}. Using the focus coordinates, get c=2. Then use the relationship among a, b, c in the ellipse to find b=\\sqrt{5-4}=1, thereby deriving the equation of the ellipse. According to the problem, the other focus of the ellipse is (-2,0), then \\frac{\\sqrt{5}+3\\sqrt{5}}{5}=2\\sqrt{5}, so a=\\sqrt{5}. Since c=2, it follows that b=\\sqrt{5-4}=1, thus obtaining the equation of the ellipse as \\frac{x^{2}}{5}+y^{2}=1." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ with left and right foci $F_{1}$, $F_{2}$. A line passing through point $F_{1}$ intersects ellipse $C$ at points $A$, $B$. Then the perimeter of $\\triangle A B F_{2}$ is?", "fact_expressions": "C: Ellipse;G: Line;A: Point;B: Point;F2: Point;F1: Point;Expression(C) = (x^2/2 + y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, G);Intersection(G, C) = {A, B}", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[2, 34], [69, 74]], [[66, 68]], [[75, 78]], [[79, 82]], [[48, 55]], [[40, 47], [57, 65]], [[2, 34]], [[2, 55]], [[2, 55]], [[56, 68]], [[66, 84]]]", "query_spans": "[[[86, 112]]]", "process": "According to the definition of an ellipse, |AF₁| + |AF₂| = 2a, |BF₁| + |BF₂| = 2a. ∴ The perimeter of ΔABF₂ is |AF₁| + |AF₂| + |BF₁| + |BF₂| = 4a. ∵ a = √2, ∴ the perimeter of △ABF₂ is 4√2." }, { "text": "$F$ is the focus of the parabola $y^{2}=2x$, and $A$, $B$ are two points on the parabola such that $|AF| + |BF| = 8$. What is the distance from the midpoint of segment $AB$ to the $y$-axis?", "fact_expressions": "F: Point;Focus(G) = F;G: Parabola;Expression(G) = (y^2 = 2*x);A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 8", "query_expressions": "Distance(MidPoint(LineSegmentOf(A, B)), yAxis)", "answer_expressions": "7/2", "fact_spans": "[[[0, 3]], [[0, 21]], [[4, 18], [30, 33]], [[4, 18]], [[22, 25]], [[26, 29]], [[22, 37]], [[22, 37]], [[38, 53]]]", "query_spans": "[[[55, 75]]]", "process": "Since F is the focus of the parabola \\( y^{2} = 2x \\), \\( \\therefore F\\left(\\frac{1}{2}, 0\\right) \\), the equation of the directrix is \\( x = -\\frac{1}{2} \\). Let \\( A(x_{1}, y_{1}) \\), \\( B(x_{2}, y_{2}) \\), \\( \\therefore AF + |BF| = x_{1} + \\frac{1}{2} + x_{2} + \\frac{1}{2} = 8 \\), \\( \\therefore x_{1} + x_{2} = 7 \\), \\( \\therefore \\) the horizontal coordinate of the midpoint of segment AB is \\( \\frac{7}{2} \\)." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $C$: $x^{2}-y^{2}=1$, and point $P$ lies on $C$ with $\\angle F_{1} P F_{2}=60^{\\circ}$, find $|P F_{1}| \\cdot |P F_{2}|$=?", "fact_expressions": "C: Hyperbola;F1: Point;P: Point;F2: Point;Expression(C) = (x^2 - y^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))", "answer_expressions": "4", "fact_spans": "[[[18, 41], [53, 56]], [[2, 9]], [[48, 52]], [[10, 17]], [[18, 41]], [[2, 47]], [[2, 47]], [[48, 57]], [[58, 91]]]", "query_spans": "[[[93, 121]]]", "process": "Test analysis: From the equation of the hyperbola, we have a=1, b=1, c=\\sqrt{2}. By the cosine theorem, \\therefore\\frac{1}{2}=\\frac{F_{2}^{2}+2|PF_{1}^{2}+|PF_{1}^{2}|FF_{P}|_{PFPF_{1}PF_{2}}^{2}}{2|PF_{1}PF_{1}-2\\sqrt{2}}|,\\therefore|PF_{1}|PF_{1}=4" }, { "text": "$F_{1}(-4,0)$, $F_{2}(4,0)$ are the two foci of the hyperbola $C$: $\\frac{x^{2}}{m}-\\frac{y^{2}}{4}=1$ $(m>0)$, and point $M$ is a point on the hyperbola $C$ such that $\\angle F_{1} M F_{2}=60^{\\circ}$. Then the area of $\\triangle F_{1} M F_{2}$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/4 + x^2/m = 1);m: Number;m>0;F1: Point;F2: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(C) = {F1, F2};M: Point;PointOnCurve(M, C);AngleOf(F1, M, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, M, F2))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[27, 75], [86, 92]], [[27, 75]], [[35, 75]], [[35, 75]], [[0, 13]], [[14, 26]], [[0, 13]], [[14, 26]], [[0, 80]], [[81, 85]], [[81, 95]], [[97, 130]]]", "query_spans": "[[[132, 162]]]", "process": "Since F_{1}(-4,0) and F_{2}(4,0) are the two foci of the hyperbola C: \\frac{x^{2}}{m} - \\frac{y^{2}}{4} = 1 (m > 0), \\therefore m + 4 = 16, \\therefore m = 12. Let |MF_{1}| = m', |MF_{2}| = n. Since point M is a point on the hyperbola and \\angle F_{1}MF_{2} = 60^{\\circ}, \\therefore |m' - n| = 4\\sqrt{3} \\textcircled{1}, m'^{2} + n^{2} - 2m'n\\cos60^{\\circ} = 64 \\textcircled{2}. From \\textcircled{2} - \\textcircled{1}^{2} we get m'n = 16, \\therefore the area S of \\triangle F_{1}MF_{2} = \\frac{1}{2}m'n\\sin60^{\\circ} = 4\\sqrt{3}." }, { "text": "Through the focus $F$ of the parabola $C$: $y^{2}=4x$, draw lines $AB$ and $DE$ intersecting the parabola $C$ at points $A$, $B$ and $D$, $E$ respectively. If the slopes of lines $AB$ and $DE$ are $k_{1}$, $k_{2}$ respectively, and satisfy $k_{1}^{2}+k_{2}^{2}=4$, then the minimum value of $|AB|+|DE|$ is?", "fact_expressions": "C: Parabola;B: Point;A: Point;D: Point;E: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(F,LineOf(A,B));PointOnCurve(F,LineOf(D,E));Intersection(LineOf(A,B),C)={A,B};Intersection(LineOf(D,E),C)={D,E};Slope(LineOf(A,B))=k1;Slope(LineOf(D,E))=k2;k1:Number;k2:Number;k1^2+k2^2=4", "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(D, E)))", "answer_expressions": "12", "fact_spans": "[[[1, 20], [43, 49]], [[55, 58]], [[51, 54]], [[59, 62]], [[63, 66]], [[23, 26]], [[1, 20]], [[1, 26]], [[0, 34]], [[0, 40]], [[27, 66]], [[27, 66]], [[68, 102]], [[68, 102]], [[87, 94]], [[95, 102]], [[106, 129]]]", "query_spans": "[[[131, 150]]]", "process": "The focus of the parabola $ y^2 = 4x $ is at $ (1,0) $. Let the equation of line $ AB $ be $ y = k_1(x - 1) $. Solving simultaneously with the parabola equation gives:\n\\[\n\\begin{cases}\ny^2 = 4x \\\\\ny = k_1(x - 1)\n\\end{cases}\n\\Rightarrow\nk_1^2 x^2 - 2(k_1^2 + 2)x + k_1^2 = 0.\n\\]\nLet $ A(x_1, y_1) $, $ B(x_2, y_2) $. Then $ x_1 + x_2 = \\frac{2(k_1^2 + 2)}{k_1^2} $, $ x_1 x_2 = 1 $,\n\\[\n|AB| = \\sqrt{1 + k_1^2} |x_1 - x_2| = \\sqrt{1 + k_1^2} \\sqrt{(x_1 + x_2)^2 - 4x_1 x_2} = \\sqrt{1 + k_1^2} \\sqrt{\\left( \\frac{2k_1^2 + 4}{k_1^2} \\right)^2 - 4} = 4\\left(1 + \\frac{1}{k_1^2}\\right).\n\\]\nSimilarly, we get: $ |DE| = 4\\left(1 + \\frac{1}{k_2^2}\\right) $, $ k_1^2 + k_2^2 = 4 $. Therefore,\n\\[\n\\left( \\frac{1}{k_1^2} + \\frac{1}{k_2^2} \\right)(k_1^2 + k_2^2) = 2 + \\frac{k_2^2}{k_1^2} + \\frac{k_1^2}{k_2^2} \\geqslant 2 + 2\\sqrt{ \\frac{k_2^2}{k_1^2} \\cdot \\frac{k_1^2}{k_2^2} } = 4,\n\\]\nwith equality if and only if $ k_1^2 = k_2^2 $. Thus $ |AB| + |DE| \\geqslant 8 + 4 = 12 $." }, { "text": "The equation of the directrix of the parabola $x^{2}=160 y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 160*y)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-40", "fact_spans": "[[[0, 16]], [[0, 16]]]", "query_spans": "[[[0, 23]]]", "process": "Since the equation of the parabola is \\( x^{2} = 160y \\), its directrix equation is \\( y = -\\frac{160}{4} = -40 \\)." }, { "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{3-m}+\\frac{y^{2}}{m-2}=1$ is $2 \\sqrt{3}$, what is its eccentricity?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(3 - m) + y^2/(m - 2) = 1);m: Number;FocalLength(G) = 2*sqrt(3)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[2, 44], [63, 64]], [[2, 44]], [[5, 44]], [[2, 60]]]", "query_spans": "[[[63, 69]]]", "process": "Analysis: Given that the hyperbola $\\frac{x^{2}}{3-m}+\\frac{y^{2}}{m-2}=1$ has a focal distance of $2\\sqrt{3}$, so $c=\\sqrt{3}$. Then, depending on the position of the foci, use $a^{2}+b^{2}=c^{2}$ to establish an equation to solve for $m$, and then find the eccentricity. \nDetailed solution: From the problem, when $m<2$, the foci lie on the $x$-axis, $3-m+[-(m-2)]=3 \\Rightarrow m=1$, at this time $e=\\frac{\\sqrt{3}}{\\sqrt{2}}=\\frac{\\sqrt{6}}{2}$; or when $m>3$, the foci lie on the $y$-axis, $-(3-m)+(m-2)=3 \\Rightarrow m=4$, at this time $e=\\frac{\\sqrt{3}}{\\sqrt{2}}=\\frac{\\sqrt{6}}{2}$. Therefore, combining both cases, the eccentricity is $\\frac{\\sqrt{6}}{2}$." }, { "text": "The left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-4}=1$ $(a>2)$ are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse $C$, $|O P|=2$ ($O$ is the origin), and $S_{\\Delta F_{1} P F_{2}}=2$. Find the length of the major axis of the ellipse $C$.", "fact_expressions": "C: Ellipse;a: Number;O: Origin;P: Point;F1: Point;F2: Point;a>2;Expression(C) = (y^2/(a^2 - 4) + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Abs(LineSegmentOf(O, P)) = 2;Area(TriangleOf(F1,P,F2)) = 2", "query_expressions": "Length(MajorAxis(C))", "answer_expressions": "2*sqrt(6)", "fact_spans": "[[[0, 57], [87, 92], [145, 150]], [[6, 57]], [[105, 108]], [[82, 86]], [[66, 73]], [[74, 81]], [[6, 57]], [[0, 57]], [[0, 81]], [[0, 81]], [[82, 93]], [[94, 103]], [[115, 143]]]", "query_spans": "[[[145, 156]]]", "process": "From the ellipse $ C: \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{a^{2}-4} = 1 $ $ (a > 2) $, we get $ c = \\sqrt{a^{2} - (a^{2} - 4)} = 2 $, then $ |OP| = |OF_{1}| = |OF_{2}| $, so $ \\angle F_{1}PF_{2} = 90^{\\circ} $. Let $ |PF_{1}| = m $, $ |PF_{2}| = n $, then $ m + n = 2a $, and $ m^{2} + n^{2} = 16 $, so $ mn = \\frac{(2a)^{2} - 16}{2} = 2a^{2} - 8 $. Therefore, $ S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2}mn = a^{2} - 4 = 2 $, solving gives $ a = \\sqrt{6} $. So the major axis length of ellipse $ C $ is $ 2\\sqrt{6} $." }, { "text": "Let the ellipse $ C $: $ \\frac{x^{2}}{4} + y^{2} = 1 $ have its left focus at $ F $, and let the line $ l $: $ y = kx $ ($ k \\neq 0 $) intersect the ellipse $ C $ at points $ A $ and $ B $. Then the range of the perimeter of $ \\triangle AFB $ is?", "fact_expressions": "l: Line;C: Ellipse;A: Point;F: Point;B: Point;k:Number;Negation(k=0);Expression(C) = (x^2/4 + y^2 = 1);LeftFocus(C) = F;Expression(l)=(y = k*x) ;Intersection(l, C) = {A, B}", "query_expressions": "Range(Perimeter(TriangleOf(A, F, B)))", "answer_expressions": "(6,8)", "fact_spans": "[[[42, 66]], [[1, 33], [67, 72]], [[74, 77]], [[38, 41]], [[78, 81]], [[49, 66]], [[49, 66]], [[1, 33]], [[1, 41]], [[42, 66]], [[42, 83]]]", "query_spans": "[[[85, 111]]]", "process": "Let the right focus of ellipse C be F. Connect AF and BF. Since the midpoints of AB and FF are both at the origin, quadrilateral AFBF is a parallelogram. Therefore, |BF| = |AF|. By the definition of an ellipse, |AF| + |AF| = 2a = 4. Since the line y = kx (k ≠ 0) is symmetric to ellipse C about the origin, points A and B are also symmetric about the origin. Let point A be (x₀, y₀), then 0 < x₀² < 4. Thus, |AB| = 2√(x₀² + y₀²) = 2√(x₀² + 1 - x₀²/4) = 2√(3x₀²/4 + 1) ∈ (2, 4). Therefore, the perimeter of quadrilateral AFBF is |AF| + |AF| + |AB| = 4 + |AB| ∈ (6, 8)." }, { "text": "Given that the parabola $y^{2}=4 x$ intersects the line $2 x+y-4=0$ at points $A$ and $B$, and the focus of the parabola is $F$, then $|\\overrightarrow{F A}|+|\\overrightarrow{F B}|$=?", "fact_expressions": "G: Parabola;H: Line;F: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (2*x + y - 4 = 0);Intersection(G, H) = {A, B};Focus(G) = F", "query_expressions": "Abs(VectorOf(F, A)) + Abs(VectorOf(F, B))", "answer_expressions": "7", "fact_spans": "[[[2, 16], [43, 46]], [[17, 30]], [[50, 53]], [[33, 36]], [[37, 40]], [[2, 16]], [[17, 30]], [[2, 42]], [[43, 53]]]", "query_spans": "[[[56, 105]]]", "process": "According to the definition of a parabola, express |\\overrightarrow{FA}|+|\\overrightarrow{FB}| in terms of the horizontal coordinates of points A and B. Then, solve by combining the equations of the line and the parabola and applying Vieta's formulas. [Detailed solution] Let A(x_{1},y_{1}), B(x_{2},y_{2}), then |\\overrightarrow{FA}|+|\\overrightarrow{FB}|=x_{1}+1+x_{2}+1=x_{1}+x_{2}+2. From y^{2}=4x and 2x+y-4=0, we obtain x^{2}-5x+4=0. \\therefore x_{1}+x_{2}=5, so |\\overrightarrow{FA}|+|\\overrightarrow{FB}|=7. This problem examines the definition of a parabola and the application of Vieta's formulas, as well as basic analytical and problem-solving abilities, belonging to fundamental problems." }, { "text": "The distance from point $M(3,2)$ to the directrix of the parabola $C$: $y=a x^{2}$ $(a>0)$ is $4$, and $F$ is the focus of the parabola. Point $N(1,1)$. When point $P$ moves on the line $l$: $x-y=2$, the minimum value of $\\frac{|P N|-1}{|P F|}$ is?", "fact_expressions": "l: Line;C: Parabola;a: Number;M: Point;N: Point;P: Point;F: Point;a>0;Expression(C) = (y = a*x^2);Expression(l)=(x-y=2);Coordinate(M) = (3, 2);Coordinate(N) = (1, 1);Distance(M,Directrix(C))=4;Focus(C)=F;PointOnCurve(P,l)", "query_expressions": "Min((Abs(LineSegmentOf(P, N)) - 1)/Abs(LineSegmentOf(P, F)))", "answer_expressions": "(2-sqrt(2))/4", "fact_spans": "[[[71, 85]], [[10, 34], [48, 51]], [[18, 34]], [[0, 9]], [[55, 64]], [[66, 70]], [[44, 47]], [[18, 34]], [[10, 34]], [[71, 85]], [[0, 9]], [[55, 64]], [[0, 43]], [[44, 54]], [[66, 86]]]", "query_spans": "[[[90, 119]]]", "process": "Analysis: First, find the equation of the parabola. Let AP = t, then AN = \\sqrt{2}, AF = 2\\sqrt{2}, PN = \\sqrt{t^{2}+2}, PF = \\sqrt{t^{2}+8}. Then express \\frac{|PN|-1}{|PF|}, and use substitution to obtain the conclusion. Since the point M(3,2) is at a distance of 4 from the directrix of the parabola C: y = ax^{2} (a > 0), we have 2 + \\frac{1}{4a} = 4, so a = \\frac{1}{8}. Therefore, the parabola C: x^{2} = 8y. The line l: x - y = 2 intersects the x-axis at A(2,0), then FA \\perp l. Let AP = t, then AN = \\sqrt{2}, AF = 2\\sqrt{2}, PN = \\sqrt{t^{2}+2}, PF = \\sqrt{t^{2}+8}. Let \\sqrt{t^{2}+2} = m (m \\geqslant \\sqrt{2}), then \\frac{|PN|-1}{|PF|} = \\frac{\\sqrt{t^{2}+2}-1}{\\sqrt{t^{2}+8}}. When m = \\sqrt{2}, i.e., t = 0, \\frac{|PN|-1}{|PF|} attains its minimum value." }, { "text": "Let $A_{1}$ and $A_{2}$ be the left and right vertices of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, respectively. If there exists a point $M$ on the hyperbola such that the product of the slopes of the two lines satisfies $k_{M A_{1}} k_{M A_{2}}<2$, then the range of the eccentricity of hyperbola $C$ is?", "fact_expressions": "A1: Point;A2: Point;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > 0;b > 0;LeftVertex(C) = A1;RightVertex(C) = A2;M: Point;PointOnCurve(M, C);k1: Number;k2: Number;Slope(LineOf(M, A1)) = k1;Slope(LineOf(M, A2)) = k2;k1 * k2 < 2", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(1,sqrt(3))", "fact_spans": "[[[1, 8]], [[9, 16]], [[19, 77], [84, 87], [130, 136]], [[19, 77]], [[22, 77]], [[22, 77]], [[22, 77]], [[22, 77]], [[1, 82]], [[1, 82]], [[90, 94]], [[84, 94]], [[101, 128]], [[101, 128]], [[96, 128]], [[96, 128]], [[101, 128]]]", "query_spans": "[[[130, 147]]]", "process": "From the given conditions, we have A_{1}(-a,0), A_{2}(a,0). Let M(m,n), then \\frac{m^{2}}{a^{2}} - \\frac{n^{2}}{b^{2}} = 1, which implies \\frac{n^{2}}{m^{2}-a^{2}} = \\frac{b^{2}}{a^{2}}, k_{MA_{1}}k_{MA_{2}} = \\frac{n}{m+a} \\cdot \\frac{n}{m-a} = \\frac{n^{2}}{m^{2}-a^{2}} = \\frac{b^{2}}{a^{2}} < 2. \\therefore e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} < \\sqrt{3}, and since e > 1, \\therefore 1 < e < \\sqrt{3}." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $e$, and its asymptotes are given by $y=\\pm \\frac{\\sqrt{6}}{e} x$, then $e=$?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;e: Number;Eccentricity(G) = e;Expression(Asymptote(G)) = (y = pm*x*sqrt(6))", "query_expressions": "e", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 58], [67, 68]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66], [104, 107]], [[2, 66]], [[67, 102]]]", "query_spans": "[[[104, 109]]]", "process": "From the given condition, $\\frac{b}{a} = \\frac{\\sqrt{6}}{e}$, $\\frac{b}{a} \\therefore \\frac{\\sqrt{6}}{a^{2}}$, combining with $c^{2} = a^{2} + b^{2}$ and transforming into an equation in $e$, the conclusion can be solved." }, { "text": "The equation of the hyperbola that shares foci with the curve $\\frac{x^{2}}{24}+\\frac{y^{2}}{49}=1$ and shares asymptotes with the curve $\\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1$ is?", "fact_expressions": "E: Hyperbola;C1: Curve;C2: Curve;Expression(C1) = (x**2/24 + y**2/49 = 1);Expression(C2) = (x**2/36 - y**2/64 = 1);Focus(E) = Focus(C1);Asymptote(E) = Asymptote(C2)", "query_expressions": "Expression(E)", "answer_expressions": "y^2/16-x^2/9=1", "fact_spans": "[[[90, 93]], [[1, 40]], [[47, 85]], [[1, 40]], [[46, 85]], [[0, 93]], [[45, 93]]]", "query_spans": "[[[90, 97]]]", "process": "" }, { "text": "The hyperbola $x^{2}-16 y^{2}=16$ has left and right foci denoted by $F_{1}$ and $F_{2}$, respectively. A line $l$ passes through the left focus $F_{1}$ of the hyperbola and intersects the left branch of the hyperbola at points $A$ and $B$, with $|A B|=12$. What is the perimeter of $\\triangle A B F_{2}$?", "fact_expressions": "l: Line;G: Hyperbola;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2 - 16*y^2 = 16);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l, LeftPart(G)) = {A, B};Abs(LineSegmentOf(A, B)) = 12", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "40", "fact_spans": "[[[47, 52]], [[0, 22], [53, 56], [68, 71]], [[75, 78]], [[79, 82]], [[39, 46]], [[29, 36], [60, 67]], [[0, 22]], [[0, 46]], [[0, 46]], [[47, 67]], [[47, 82]], [[84, 94]]]", "query_spans": "[[[96, 122]]]", "process": "" }, { "text": "The focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ coincides with the focus of the parabola $y^{2}=8x$, and the distance from the focus to the asymptote is $1$. Find the length of the real axis 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;H: Parabola;Expression(H) = (y^2 = 8*x);OneOf(Focus(G))=Focus(H);Distance(Focus(G),Asymptote(G))=1", "query_expressions": "Length(RealAxis(G))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[0, 56], [99, 102]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[62, 76]], [[62, 76]], [[0, 81]], [[0, 96]]]", "query_spans": "[[[99, 108]]]", "process": "From the hyperbola equation, its asymptotes are: $ y = \\pm\\frac{b}{a}x $, that is, $ \\pm bx - ay = 0 $. Since the focus of $ y^2 = 8x $ is $ (2,0) $, a focus of the hyperbola is $ (2,0) $, so $ c = 2 $. Therefore, the distance from $ (2,0) $ to the asymptote is $ d = \\frac{|\\pm 2b|}{\\sqrt{a^2 + b^2}}} = \\frac{2b}{c} = 1 $, so $ b = \\frac{c}{2} = 1 $, thus $ a = \\sqrt{c^2 - b^2} = \\sqrt{3} $. Therefore, the length of the real axis of the hyperbola is $ 2a = 2\\sqrt{3} $." }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{a}-y^{2}=1$ $(a>0)$ is $\\sqrt{3}$, then the value of $a$ is?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2 + x^2/a = 1);Eccentricity(G) = sqrt(3)", "query_expressions": "a", "answer_expressions": "1/2", "fact_spans": "[[[0, 33]], [[50, 53]], [[3, 33]], [[0, 33]], [[0, 48]]]", "query_spans": "[[[50, 57]]]", "process": "" }, { "text": "The equation $\\frac{x^{2}}{m+3}-\\frac{y^{2}}{2 m-1}=1$ represents a hyperbola with focus on the $y$-axis. Find the range of real values for $m$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 3) - y^2/(2*m - 1) = 1);m: Real;PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(m)", "answer_expressions": "(-oo, -3)", "fact_spans": "[[[54, 57]], [[0, 57]], [[59, 64]], [[45, 57]]]", "query_spans": "[[[59, 70]]]", "process": "Since the equation $\\frac{x^{2}}{m+3}-\\frac{y^{2}}{2m-1}=1$ represents a hyperbola with foci on the $y$-axis, we have $\\begin{cases}m+3<0\\\\2m-1<0\\end{cases}$, solving gives $m<-3$," }, { "text": "The abscissa of a point $M$ on the parabola $y^{2}=16x$ is $6$. Then, what is the distance from $M$ to the focus $F$?", "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 16*x);PointOnCurve(M, G);XCoordinate(M) = 6;F: Point;Focus(G) = F", "query_expressions": "Distance(M, F)", "answer_expressions": "10", "fact_spans": "[[[0, 15]], [[18, 21], [31, 34]], [[0, 15]], [[0, 21]], [[18, 29]], [[37, 40]], [[0, 40]]]", "query_spans": "[[[31, 45]]]", "process": "" }, { "text": "Let $P$ be a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{15}=1$. From point $P$, draw tangents to the circles $(x+4)^{2}+y^{2}=4$ and $(x-4)^{2}+y^{2}=1$, with points of tangency $M$ and $N$ respectively. Then the minimum value of $|P M|^{2}-|P N|^{2}$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/15 = 1);P: Point;PointOnCurve(P, RightPart(G));H: Circle;Expression(H) = (y^2 + (x + 4)^2 = 4);K: Circle;Expression(K) = (y^2 + (x - 4)^2 = 1);Z1: Line;Z2: Line;TangentOfPoint(P, H) = Z1;TangentOfPoint(P, K) = Z2;M: Point;TangentPoint(Z1, H) = M;N: Point;TangentPoint(Z2, K) = N", "query_expressions": "Min(Abs(LineSegmentOf(P, M))^2 - Abs(LineSegmentOf(P, N))^2)", "answer_expressions": "13", "fact_spans": "[[[5, 34]], [[5, 34]], [[1, 4], [42, 46]], [[1, 40]], [[49, 69]], [[49, 69]], [[70, 89]], [[70, 89]], [], [], [[41, 92]], [[41, 92]], [[98, 101]], [[41, 105]], [[102, 105]], [[41, 105]]]", "query_spans": "[[[107, 134]]]", "process": "Find the centers and radii of the two circles. Let the left and right foci of the hyperbola $ x^{2} - \\frac{y^{2}}{15} = 1 $ be $ F_{1}(-4,0) $, $ F_{2}(4,0) $. Connect $ PF_{1} $, $ PF_{2} $, $ F_{1}M $, $ F_{2}N $. Using the Pythagorean theorem and the definition of the hyperbola, combined with the fact that the sum of distances is minimized when three points are collinear, calculate to obtain the desired value. Solution: The circle $ C_{1}: (x+4)^{2} + y^{2} = 4 $ has center $ (-4,0) $ and radius $ r_{1} = 2 $; the circle $ C_{2}: (x-4)^{2} + y^{2} = 1 $ has center $ (4,0) $ and radius $ r_{2} = 1 $. Let the left and right foci of the hyperbola $ x^{2} - \\frac{y^{2}}{15} = 1 $ be $ F_{1}(-4,0) $, $ F_{2}(4,0) $. Connect $ PF_{1} $, $ PF_{2} $, $ F_{1}M $, $ F_{2}N $. Then we have $ |PM|^{2} \\cdot |PN|^{2} = (|PF_{1}|^{2} - r_{1}^{2}) \\cdot (|PF_{2}|^{2} - r_{2}^{2}) = (|PF_{1}|^{2} - 4) \\cdot (|PF_{2}|^{2} - 1) = |PF_{1}|^{2} \\cdot |PF_{2}|^{2} - 3 = (|PF_{1}| \\cdot |PF_{2}|)(|PF_{1}| + |PF_{2}|) \\cdot 3 = 2a(|PF_{1}| + |PF_{2}|) \\cdot 3 = 2(|PF_{1}| + |PF_{2}|) \\cdot 3 \\geqslant 2 \\cdot 2c \\cdot 3 = 2 \\cdot 8 \\cdot 3 = 13 $. The equality holds if and only if $ P $ is the right vertex, thus achieving the minimum value 13." }, { "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>0 , b>0)$, respectively, the line $l$ passing through $F_{1}$ intersects $C$ at points $A$ and $B$, $O$ is the origin, and if $O A \\perp B F_{1}$, $|A F_{2}|=|B F_{2}|$, then the eccentricity of $C$ is?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F2: Point;B: Point;O: Origin;F1: 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);Intersection(l, C) = {A, B};IsPerpendicular(LineSegmentOf(B, F1),LineSegmentOf(O, A));Abs(LineSegmentOf(A, F2)) = Abs(LineSegmentOf(B, F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(7)", "fact_spans": "[[[98, 103]], [[20, 82], [104, 107], [172, 175]], [[28, 82]], [[28, 82]], [[108, 111]], [[10, 17]], [[112, 115]], [[118, 121]], [[2, 9], [90, 97]], [[28, 82]], [[28, 82]], [[20, 82]], [[2, 88]], [[2, 88]], [[89, 103]], [[98, 117]], [[128, 148]], [[149, 170]]]", "query_spans": "[[[172, 181]]]", "process": "Draw the graph, take the midpoint E of AB, connect EF₂. Let F₁A = x. According to the hyperbola definition, we get x = 2a. Then by the Pythagorean theorem, we obtain c² = 7a², and thus find the value of e. Take the midpoint E of AB, connect EF₂. Then from the given condition, BF₁ ⊥ EF₂, F₁A = AE = EB. Let F₁A = x, then by the hyperbola definition, AF₂ = 2a + x, F₁B - BF₂ = 3x - 2a - x = 2a, so x = 2a. In right triangle △F₁EF₂, by the Pythagorean theorem, (4a)² + (2√3a)² = (2c)², then e = √7." }, { "text": "Given the parabola $C$: $y^{2}=4x$, a line passing through the focus $F$ with slope $1$ intersects $C$ at points $P$ and $Q$. The projections of points $P$ and $Q$ onto the directrix are $M$ and $N$, respectively. Then, the area of $\\Delta MFN$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);Slope(G) = 1;Intersection(G, C) = {P, Q};P: Point;Q: Point;Projection(P, Directrix(C)) = M;Projection(Q, Directrix(C)) = N;M: Point;N: Point", "query_expressions": "Area(TriangleOf(M, F, N))", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[2, 21], [40, 43]], [[2, 21]], [[26, 29]], [[2, 29]], [[37, 39]], [[23, 39]], [[30, 39]], [[37, 55]], [[46, 49], [57, 60]], [[50, 53], [61, 64]], [[40, 85]], [[40, 85]], [[76, 79]], [[80, 83]]]", "query_spans": "[[[87, 106]]]", "process": "The parabola $ C: y^2 = 4x $ has focus coordinates $ F(1,0) $, and the equation of the directrix is $ x = -1 $. The equation of the line passing through the focus $ F $ with slope 1 is $ y = x - 1 $, which simplifies to $ x = y + 1 $. The parabola $ C $ intersects the line at two points $ P $ and $ Q $. Let $ P(x_1, y_1) $, $ Q(x_2, y_2) $, and let the projections of points $ P $ and $ Q $ onto the directrix be $ M $ and $ N $, respectively. Then \n$$\n\\begin{cases}\nx = y + 1 \\\\\ny^2 = 4x\n\\end{cases}\n$$ \nSimplifying yields $ y^2 - 4y - 4 = 0 $. Thus, $ y_1 + y_2 = 4 $, $ y_1 \\cdot y_2 = -4 $. Then $ |MN| = |y_1 - y_2| = \\sqrt{(y_1 + y_2)^2 - 4y_1 \\cdot y_2} $. Therefore, $ S_{\\triangle MFN} = \\frac{1}{2} \\times 4\\sqrt{2} $" }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{m}=1$ ($m \\in \\mathbb{R}$, $m \\neq 0$) is $2$, find the value of $m$.", "fact_expressions": "G: Hyperbola;m: Real;Negation(m=0);Expression(G) = (x^2/9 - y^2/m = 1);Eccentricity(G) = 2", "query_expressions": "m", "answer_expressions": "27", "fact_spans": "[[[2, 60]], [[70, 73]], [[5, 60]], [[2, 60]], [[2, 68]]]", "query_spans": "[[[70, 77]]]", "process": "According to the standard equation of the hyperbola, we have: a^{2}=9, b^{2}=m>0. Since the eccentricity of the hyperbola is 2, \\therefore \\frac{c}{a}=2, and c^{2}=a^{2}+b^{2}. \\therefore m=27" }, { "text": "A point $M(x_{1}, y_{1})$ on the parabola $y^{2}=6x$ is at a distance of $\\frac{9}{2}$ from its focus. Then, what is the distance from point $M$ to the origin?", "fact_expressions": "G: Parabola;M: Point;x1: Number;y1: Number;Expression(G) = (y^2 = 6*x);Coordinate(M) = (x1, y1);PointOnCurve(M, G);Distance(M, Focus(G)) = 9/2;O: Origin", "query_expressions": "Distance(M, O)", "answer_expressions": "3*sqrt(3)", "fact_spans": "[[[0, 14], [35, 36]], [[17, 34], [57, 61]], [[17, 34]], [[17, 34]], [[0, 14]], [[17, 34]], [[0, 34]], [[17, 55]], [[62, 66]]]", "query_spans": "[[[57, 71]]]", "process": "From the given conditions, the focus is at $(\\frac{3}{2},0)$, and the directrix is $x=-\\frac{3}{2}$. Since the distance from $M(x_{1},y_{1})$ to the focus equals the distance to the directrix, we have $x_{1}+\\frac{3}{2}=\\frac{9}{2}$, then $x_{1}=3$. $\\therefore y^{2}=18$, hence $|OM|=\\sqrt{x^{2}+y^{2}}=3\\sqrt{3}$." }, { "text": "Through a focus $F$ of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$, draw a perpendicular line $FM$ to one of its asymptotes, with the foot of the perpendicular at $M$, intersecting the $y$-axis at $E$. If $M$ is the midpoint of $EF$, then $m=$?", "fact_expressions": "G: Hyperbola;m: Number;F: Point;M: Point;E: Point;Expression(G) = (x^2 - y^2/m = 1);OneOf(Focus(G))=F;PointOnCurve(F,LineOf(F,M));IsPerpendicular(LineOf(F,M),OneOf(Asymptote(G)));FootPoint(LineOf(F,M),OneOf(Asymptote(G)))=M;Intersection(LineOf(F,M),yAxis)=E;MidPoint(LineSegmentOf(E, F)) = M", "query_expressions": "m", "answer_expressions": "1", "fact_spans": "[[[1, 29], [38, 39]], [[86, 89]], [[34, 37]], [[58, 61], [74, 77]], [[69, 72]], [[1, 29]], [[1, 37]], [[0, 53]], [[38, 53]], [[38, 61]], [[48, 72]], [[74, 84]]]", "query_spans": "[[[86, 91]]]", "process": "" }, { "text": "The equation of the directrix of the parabola $x=\\frac{1}{4} y^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x = y^2/4)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x = -1", "fact_spans": "[[[0, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "" }, { "text": "The distance from a focus of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{4}=1$ to one of its asymptotes is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/4 = 1)", "query_expressions": "Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))", "answer_expressions": "3", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 54]]]", "process": "" }, { "text": "Given that point $A$ is the left vertex of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $O$ is the origin, and a line $l$ perpendicular to the $x$-axis is drawn through the right focus $F$ of the ellipse. If there exists a point $P$ on the line $l$ such that $\\angle A P O=30^{\\circ}$, then what is the maximum value of the eccentricity of the ellipse?", "fact_expressions": "l: Line;G: Ellipse;a: Number;b: Number;A: Point;F: Point;P: Point;O: Origin;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(G) = A;RightFocus(G) = F;PointOnCurve(F, l);IsPerpendicular(xAxis, l);PointOnCurve(P, l);AngleOf(A, P, O) = ApplyUnit(30, degree)", "query_expressions": "Max(Eccentricity(G))", "answer_expressions": "1/2", "fact_spans": "[[[92, 97], [99, 104]], [[7, 59], [74, 76], [140, 142]], [[9, 59]], [[9, 59]], [[2, 6]], [[80, 83]], [[107, 111]], [[64, 67]], [[9, 59]], [[9, 59]], [[7, 59]], [[2, 63]], [[74, 83]], [[73, 97]], [[84, 97]], [[99, 111]], [[113, 138]]]", "query_spans": "[[[140, 150]]]", "process": "By symmetry, we may assume P is above the x-axis. Let P(c, m), m > 0, ∠POF = α, ∠PAF = β. \n∴ tan∠APO = tan(α − β) = (m / ((m / (2)) − (m / (a + c)))) / (c(a + c) + m²) = (c(a + c) / c) ⋅ ((a + c) / m) + m^(a / (2√(c(a + c)))) \nEquality holds if and only if m = √(c(a + c)). \n∴ (tan∠APO)_max ≥ √3 / 3 \n∴ 4e² + 4e − 3 ≤ 0, i.e., (2e − 1)(2e + 3) ≤ 0, so 0 < e ≤ 1/2. \nHence, the maximum eccentricity of the ellipse is 1/2." }, { "text": "The line $ l $ has the equation $ y = x + 3 $. Take any point $ P $ on $ l $. If an ellipse is drawn passing through point $ P $ and having the foci of the hyperbola $ 12x^2 - 4y^2 = 3 $ as its foci, then what is the equation of the ellipse with the shortest major axis?", "fact_expressions": "l: Line;G: Hyperbola;H: Ellipse;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));P:Point", "query_expressions": "Expression(H)", "answer_expressions": "", "fact_spans": "[[[0, 5], [18, 21]], [[38, 61]], [[65, 67], [71, 73], [83, 85]], [[38, 61]], [[0, 16]], [[17, 29]], [[31, 73]], [[37, 73]], [[76, 85]], [[26, 29], [32, 36]]]", "query_spans": "[[[83, 89]]]", "process": "" }, { "text": "The equations of the asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$ are? (Express in general form)", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/9 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "3*x+pm*2*y=0", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 46]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be the foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and let $P$ be a point on the ellipse such that $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$. Then the area of the incircle of $\\Delta P F_{1} F_{2}$ is?", "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G) = True;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "Area(InscribedCircle(TriangleOf(P,F1,F2)))", "answer_expressions": "pi", "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 58]], [[17, 55], [63, 65]], [[17, 55]], [[59, 62]], [[59, 69]], [[73, 132]]]", "query_spans": "[[[134, 164]]]", "process": "From $\\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=0$, it follows that the angle between $PF_{1}$ and $PF_{2}$ is $\\frac{\\pi}{2}$. By the area formula of an elliptic focal triangle, the area $S_{\\triangle PF_{1}F_{2}}$ can be obtained. Let $r$ be the inradius of $\\triangle PF_{1}F_{2}$, then $S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\times r$, from which the value of $r$ can be found, leading to the answer. [Solution] From the ellipse equation $\\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1$, we obtain $a = 5$, $b = 3$, $c = 4$, $|F_{1}F_{2}| = 8$. Let $\\theta$ be the angle between $PF_{1}$ and $PF_{2}$. From $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$, we get $\\theta = \\frac{\\pi}{2}$, $\\frac{\\theta}{2} = \\frac{\\pi}{4}$. By the area formula of the elliptic focal triangle: $S_{\\triangle PF_{1}F_{2}} = b^{2} \\times \\tan\\frac{\\theta}{2} = 9$. Let $r$ be the inradius of $\\triangle PF_{1}F_{2}$. Then $S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2}(|PF_{1}| + |PF_{2}| + |F_{1}F_{2}|) \\times r = \\frac{1}{2} \\times (10 + 8) \\times r = 9$, so $r = 1$. Hence, the area of the incircle of $\\triangle PF_{1}F_{2}$ is $\\pi$." }, { "text": "Given that point $P$ is a moving point on the parabola $y=\\frac{1}{4} x^{2}$, find the minimum value of the sum of the distances from point $P$ to the lines $l_{1}: 4 x-3 y-12=0$ and $l_{2}: y+1=0$.", "fact_expressions": "G: Parabola;l1: Line;l2: Line;P: Point;Expression(G) = (y = x^2/4);Expression(l1)=(4*x - 3*y - 12 = 0);Expression(l2)=(y+1=0);PointOnCurve(P, G)", "query_expressions": "Min(Distance(P,l1)+Distance(P,l2))", "answer_expressions": "3", "fact_spans": "[[[7, 31]], [[44, 69]], [[70, 86]], [[2, 6], [39, 43]], [[7, 31]], [[44, 69]], [[70, 86]], [[2, 37]]]", "query_spans": "[[[39, 97]]]", "process": "Find the coordinates of the focus and the equation of the directrix of the parabola, and use the definition of the parabola to convert the problem into finding the distance from the focus to a line. [Detailed solution] The parabola $ y = \\frac{1}{4}x^2 $, or equivalently $ x^2 = 4y $, has focus coordinates $ (0,1) $ and directrix equation $ l_2: y + 1 = 0 $. By the definition of a parabola, the distance from any point on the parabola to the directrix is equal to the distance from that point to the focus. Therefore, the minimum value of the sum of the distances from a point $ P $ on the parabola to the lines $ l_1: 4x - 3y - 12 = 0 $ and $ l_2: y + 1 = 0 $ can be transformed into the minimum distance from the focus $ (0,1) $ to the line $ l_1: 4x - 3y - 12 = 0 $, which is $ d = \\frac{|0 - 3 - 12|}{\\sqrt{16 + 9}} = 3 $." }, { "text": "Given that the vertex of a parabola is at the origin of the coordinate system and its axis of symmetry is the $x$-axis, the point $A(2, k)$ lies on the parabola, and the distance from point $A$ to the focus is $5$. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;A: Point;O: Origin;k:Number;Coordinate(A) = (2, k);Vertex(G) = O;SymmetryAxis(G)=xAxis;PointOnCurve(A, G);Distance(A,Focus(G))=5", "query_expressions": "Expression(G)", "answer_expressions": "y^2=12*x", "fact_spans": "[[[2, 5], [33, 36], [55, 58]], [[22, 32], [39, 43]], [[8, 12]], [[23, 32]], [[22, 32]], [[2, 12]], [[2, 21]], [[22, 37]], [[33, 53]]]", "query_spans": "[[[55, 63]]]", "process": "The solution process is omitted" }, { "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}$ and $F_{2}$, respectively. If there exists a point $P$ on the ellipse such that $\\angle F_{1} P F_{2}=90^{\\circ}$, what is the range of values for the eccentricity $e$?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;e: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);AngleOf(F1, P, F2) = ApplyUnit(90, degree);Eccentricity(G)=e", "query_expressions": "Range(e)", "answer_expressions": "[sqrt(2)/2,1)", "fact_spans": "[[[1, 53], [79, 81]], [[3, 53]], [[3, 53]], [[61, 68]], [[84, 88]], [[69, 76]], [[128, 131]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 76]], [[1, 76]], [[79, 88]], [[90, 123]], [[79, 131]]]", "query_spans": "[[[128, 137]]]", "process": "Test analysis: Since the circle with diameter F_{1}F_{2} and the ellipse have a common point, it follows that b^{2}\\leqslant c^{2}, i.e., a^{2}-c^{2}\\leqslant c^{2}, \\frac{1}{2}\\leqslante^{2}, so \\frac{\\sqrt{2}}{3}\\leqslante<1" }, { "text": "Given that the parabola passes through the point $(1,1)$, then the standard equation of this parabola is?", "fact_expressions": "G: Parabola;H: Point;Coordinate(H) = (1, 1);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=x,x^2=y}", "fact_spans": "[[[2, 5], [18, 21]], [[6, 14]], [[6, 14]], [[2, 14]]]", "query_spans": "[[[18, 28]]]", "process": "If the parabola intersects the x-axis, let the equation of the parabola be y^{2}=mx, m\\neq0, \\because the parabola passes through the point (1,1), \\therefore1=m, that is, m=1, \\therefore at this time the equation of the parabola is y^{2}=x. If the parabola intersects the y-axis, let the equation of the parabola be x^{2}=ny, n\\neq0, \\because the parabola passes through the point (1,1), \\therefore1=n, that is, n=1, at this time the equation of the parabola is x^{2}=y. In summary: y^{2}=x or x^{2}=y." }, { "text": "The equation of the directrix of the parabola $x^{2}=a y(a \\neq 0)$ is?", "fact_expressions": "G: Parabola;a: Number;Expression(G) = (x^2 = a*y);Negation(a=0)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-a/4", "fact_spans": "[[[0, 24]], [[3, 24]], [[0, 24]], [[3, 24]]]", "query_spans": "[[[0, 31]]]", "process": "" }, { "text": "The coordinates of the point on the parabola $y=x^{2}$ that is closest to the line $2 x-y=4$ are?", "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y = x^2);Expression(H) = (2*x - y = 4);Z: Point;PointOnCurve(Z, G);WhenMin(Distance(Z, H))", "query_expressions": "Coordinate(Z)", "answer_expressions": "(1, 1)", "fact_spans": "[[[0, 12]], [[14, 25]], [[0, 12]], [[14, 25]], [[30, 31]], [[0, 31]], [[0, 31]]]", "query_spans": "[[[30, 36]]]", "process": "Let P(x, y) be any point on the parabola y = x^2. Then the distance from P to the line is d = |2x - y - 4| / \\sqrt{5} = |x^2 - 2x + 4| / \\sqrt{5} = (x - 1)^2 + 3 / \\sqrt{5}. Therefore, when x = 1, d takes the minimum value 3\\sqrt{5}/5. At this time, P(1, 1). Hence, choose B." }, { "text": "The coordinates of the focus of the parabola $x=y^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (x = y^2)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(1/4,0)", "fact_spans": "[[[0, 12]], [[0, 12]]]", "query_spans": "[[[0, 19]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$ has its right focus at point $F$. If a line passing through point $F$ intersects the right branch of the hyperbola at exactly one point, what is the range of possible values for the slope of this line?", "fact_expressions": "G: Hyperbola;H: Line;F: Point;Expression(G) = (x^2/12 - y^2/4 = 1);RightFocus(G) = F;PointOnCurve(F,H);NumIntersection(H,RightPart(G))=1", "query_expressions": "Range(Slope(H))", "answer_expressions": "[-sqrt(3)/3,sqrt(3)/3]", "fact_spans": "[[[2, 41], [60, 63]], [[57, 59], [77, 79]], [[46, 49], [52, 56]], [[2, 41]], [[2, 49]], [[51, 59]], [[57, 74]]]", "query_spans": "[[[77, 87]]]", "process": "" }, { "text": "The eccentricity of the ellipse $\\frac{y^{2}}{m}+\\frac{x^{2}}{4}=1$ with foci on the $y$-axis is $\\frac{\\sqrt{3}}{2}$, then the value of the real number $m$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/m = 1);m: Real;PointOnCurve(Focus(G), yAxis);Eccentricity(G) = sqrt(3)/2", "query_expressions": "m", "answer_expressions": "16", "fact_spans": "[[[9, 46]], [[9, 46]], [[73, 78]], [[0, 46]], [[9, 71]]]", "query_spans": "[[[73, 82]]]", "process": "Since the foci of the ellipse lie on the y-axis, we have a² = m, b² = 4, so c² = a² - b² = m - 4. From e² = c²/a² = (m - 4)/m = 3/4, we get m = 16. [Note: This question mainly examines the equation of an ellipse and its simple geometric properties, and is a medium-difficulty problem.]" }, { "text": "If a point $P$ on the parabola $y^{2}=4x$ is at a distance of $3$ from its focus, then the horizontal coordinate of point $P$ is equal to?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G);Distance(P, Focus(G)) = 3", "query_expressions": "XCoordinate(P)", "answer_expressions": "2", "fact_spans": "[[[1, 15], [22, 23]], [[1, 15]], [[18, 21], [34, 38]], [[1, 21]], [[18, 32]]]", "query_spans": "[[[34, 45]]]", "process": "" }, { "text": "Given the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$. A line $l$ passing through the origin $O$ intersects the ellipse $E$ at points $P$ and $Q$. If $|P F|=3|Q F|$ and $\\angle P F Q=90^{\\circ}$, then the eccentricity of the ellipse $E$ is?", "fact_expressions": "E: Ellipse;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;LeftFocus(E) = F;l: Line;O: Origin;PointOnCurve(O, l);P: Point;Q: Point;Intersection(l, E) = {P, Q};Abs(LineSegmentOf(P, F)) = 3*Abs(LineSegmentOf(Q, F));AngleOf(P, F, Q) = ApplyUnit(90, degree)", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(10)/4", "fact_spans": "[[[2, 59], [82, 87], [143, 148]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 67]], [[2, 67]], [[76, 81]], [[70, 75]], [[69, 81]], [[89, 92]], [[93, 96]], [[76, 98]], [[100, 114]], [[116, 141]]]", "query_spans": "[[[143, 154]]]", "process": "Take the right focus F of the ellipse, connect QF', PF'. By the symmetry of the ellipse, the quadrilateral PFQF' is a parallelogram, then |PF| = |QF'|, \\angle FPF' = \\pi - \\angle PF'F = 180^{\\circ} - 90^{\\circ} = 90^{\\circ}, |PF'| = 3|QF'|, and |PF| + |PF'| = 2a, so |PF| = \\frac{a}{2}, thus |PF'| = \\frac{3a}{2}. In the right triangle \\triangle PFF', (\\frac{a}{2})^{2} + (\\frac{3a}{2})^{2} = (2c)^{2}, solving gives: e = \\frac{\\sqrt{10}}{4}" }, { "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 $a$, then 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);Distance(Focus(G),Asymptote(G)) = a", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 59], [75, 78]], [[5, 59]], [[70, 73]], [[5, 59]], [[5, 59]], [[2, 59]], [[2, 73]]]", "query_spans": "[[[75, 86]]]", "process": "" }, { "text": "The distance from the focus of the hyperbola $\\frac{y^{2}}{5}-\\frac{x^{2}}{4}=1$ to its asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/4 + y^2/5 = 1)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "2", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 50]]]", "process": "" }, { "text": "Given that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $P$ is a point on $C$ such that $|P F|=1$, then the coordinates of point $P$ are?", "fact_expressions": "C: Ellipse;P: Point;F: Point;Expression(C) = (x^2/4 + y^2/3 = 1);RightFocus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 1", "query_expressions": "Coordinate(P)", "answer_expressions": "(2,0)", "fact_spans": "[[[6, 48], [57, 60]], [[53, 56], [77, 81]], [[2, 5]], [[6, 48]], [[2, 52]], [[53, 64]], [[66, 75]]]", "query_spans": "[[[77, 86]]]", "process": "From the given conditions, a=2, c=1, then a−c=1, a+c=3, so the distance from any point on the ellipse to F ranges from [1,3]. Since |PF|=1, P is the right vertex of the ellipse, that is, the coordinates of P are (2,0)." }, { "text": "Given that $F$ is the left focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $P$ is a point on the ellipse, then the range of values for $|P F|$ is?", "fact_expressions": "G: Ellipse;P: Point;F: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F;PointOnCurve(P, G)", "query_expressions": "Range(Abs(LineSegmentOf(P, F)))", "answer_expressions": "[1,3]", "fact_spans": "[[[6, 43], [52, 54]], [[48, 51]], [[2, 5]], [[6, 43]], [[2, 47]], [[48, 57]]]", "query_spans": "[[[59, 73]]]", "process": "By the given condition, F(-1,0), let P(x,y), then \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1\\Rightarrow y^{2}=3-\\frac{3}{4}x^{2}, so +4, because -2\\leqslant x\\leqslant 2, so the range of |PF| is [1,3]." }, { "text": "Let $O$ be the coordinate origin, $P$ be an arbitrary point on the parabola $y^{2}=2 p x(p>0)$ with focus $F$, and $M$ be a point on the segment $P F$ such that $|P M|=2|M F|$. Then the maximum value of the slope of the line $O M$ is?", "fact_expressions": "G: Parabola;p: Number;O: Origin;M: Point;F: Point;P: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G)=F;PointOnCurve(P, G);PointOnCurve(M, LineSegmentOf(P,F));Abs(LineSegmentOf(P, M)) = 2*Abs(LineSegmentOf(M, F))", "query_expressions": "Max(Slope(LineOf(O,M)))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[22, 43]], [[25, 43]], [[1, 4]], [[49, 52]], [[15, 18]], [[10, 13]], [[25, 43]], [[22, 43]], [[14, 43]], [[10, 48]], [[49, 63]], [[65, 79]]]", "query_spans": "[[[81, 96]]]", "process": "According to the problem, the parabola with focus $ F $ is given by $ y^{2} = 2px $ ($ p > 0 $), so $ F\\left(\\frac{p}{2}, 0\\right) $. Let $ P $ be any point on the parabola $ y^{2} = 2px $ ($ p > 0 $), and let $ M $ be a point on the segment $ PF $ such that $ |PM| = 2|MF| $. Let $ P\\left(\\frac{y_{0}^{2}}{2p}, y_{0}\\right) $ ($ y_{0} > 0 $). Then $ M\\left(\\frac{y_{0}^{2}}{6p} + \\frac{p}{3}, \\frac{y_{0}}{3}\\right) $, so $ k_{OM} = \\frac{\\frac{y_{0}}{3}}{\\frac{y_{0}^{2}}{6p} + \\frac{p}{3}} = \\frac{2}{\\frac{y_{0}}{p} + \\frac{2p}{y_{0}}} \\leqslant \\frac{2}{2\\sqrt{2}} = \\frac{\\sqrt{2}}{2} $, with equality if and only when $ y_{0}^{2} = 2p^{2} $. Therefore, the maximum value of the slope of line $ OM $ is $ \\frac{\\sqrt{2}}{2} $." }, { "text": "The length of the imaginary axis of the hyperbola $m x^{2}+y^{2}=1$ is $2$ times the length of the real axis. Find the value of the real number $m$.", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (m*x^2 + y^2 = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))", "query_expressions": "m", "answer_expressions": "-1/4", "fact_spans": "[[[0, 20]], [[35, 40]], [[0, 20]], [[0, 33]]]", "query_spans": "[[[35, 44]]]", "process": "" }, { "text": "Through the focus of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, draw a line perpendicular to the $x$-axis intersecting the ellipse at points $A$ and $B$. If $AB = \\frac{a}{2}$, then \nwhat is the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Ellipse;l: Line;A: Point;B: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a > b;b > 0;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Focus(H), l);IsPerpendicular(l,xAxis);Intersection(l, H) = {A, B};LineSegmentOf(A, B) = a/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[102, 148]], [[3, 53]], [[3, 53]], [[1, 53]], [[65, 67]], [[71, 74]], [[75, 78]], [[102, 148]], [[3, 53]], [[3, 53]], [[1, 53]], [[0, 67]], [[0, 67]], [[65, 80]], [[82, 99]]]", "query_spans": "[[[102, 154]]]", "process": "" }, { "text": "Given the standard equation of an ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{m^{2}}=1$ $(00;m<5", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[2, 4]], [[68, 73]], [[2, 57]], [[2, 66]], [[10, 57]], [[10, 57]]]", "query_spans": "[[[68, 77]]]", "process": "From the given condition, c = \\sqrt{25 - m^{2}}, so 2\\sqrt{25 - m^{2}} = 6. Since 0 < m < 5, it follows that m = 4." }, { "text": "If the ellipse $\\frac{x^{2}}{m}+y^{2}=1$ $(m>1)$ and the hyperbola $\\frac{x^{2}}{n}-y^{2}=1$ $(n>0)$ have the same foci $F_{1}$, $F_{2}$, and $P$ is an intersection point of the two curves, then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Hyperbola;n: Number;H: Ellipse;m: Number;F1: Point;P: Point;F2: Point;n>0;Expression(G) = (-y^2 + x^2/n = 1);m>1;Expression(H) = (y^2 + x^2/m = 1);Focus(G) = {F1,F2};Focus(H) = {F1,F2};OneOf(Intersection(G,H))=P", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "1", "fact_spans": "[[[34, 67]], [[37, 67]], [[1, 33]], [[3, 33]], [[73, 80]], [[89, 92]], [[81, 88]], [[37, 67]], [[34, 67]], [[3, 33]], [[1, 33]], [[1, 88]], [[1, 88]], [[89, 101]]]", "query_spans": "[[[103, 130]]]", "process": "Test analysis: Since the two curves have the same foci, $ c^{2} = m - 1 = n + 1 $, that is, $ m - n = 2 $. Let $ P $ be the intersection point of the two curves in the first quadrant, then by the definitions of the ellipse and hyperbola, we have \n\\[\n\\begin{cases}\n|PF_1| + |PF_2| = 2\\sqrt{m} \\\\\n|PF_1| - |PF_2| = 2\\sqrt{n}\n\\end{cases}\n\\],\nsolving which yields\n\\[\n\\begin{matrix}\n|PF_1| = \\sqrt{m} + \\sqrt{n} \\\\\n|PF_2| = \\sqrt{m} - \\sqrt{n}\n\\end{matrix}\n\\],\nso $ |PF_1||PF_2| = 2 $. In $ \\triangle F_1PF_2 $, by the cosine law, we get \n$ \\cos\\angle F_{1}PF_{2} = \\frac{|PF_1|^{2} + |PF_2|^{2} - |F_1F_2|^{2}}{2|PF_1|\\cdot|PF_2|} = \\frac{(\\sqrt{m}+\\sqrt{n})^{2}+(\\sqrt{m}-\\sqrt{n})^{2}-4(m-1)}{2\\times2} = \\frac{2(n-m)+4}{4} = 0 $,\nso $ \\angle F_{1}PF_{2} = \\frac{\\pi}{2} $, therefore $ S_{\\triangle F_1PF_2} = $" }, { "text": "Given the ellipse $\\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{9}=1$ $(m>0)$ and the hyperbola $\\frac{x^{2}}{n^{2}}-\\frac{y^{2}}{4}=1$ $(n>0)$ have the same foci $F_{1}$, $F_{2}$, and point $P$ is an intersection point of the ellipse and the hyperbola, then the value of $|P F_{1}| \\cdot |P F_{2}|$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;m: Number;n: Number;P: Point;F1: Point;F2: Point;m>0;n>0;Expression(H) = (y^2/9 + x^2/m^2 = 1);Expression(G) = (-y^2/4 + x^2/n^2 = 1);Focus(H) = {F1, F2};Focus(G) = {F1, F2};OneOf(Intersection(H,G))=P", "query_expressions": "Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2))", "answer_expressions": "13", "fact_spans": "[[[49, 96], [126, 129]], [[2, 48], [123, 125]], [[4, 48]], [[52, 96]], [[118, 122]], [[102, 109]], [[110, 117]], [[4, 48]], [[52, 96]], [[2, 48]], [[49, 96]], [[2, 117]], [[2, 117]], [[118, 134]]]", "query_spans": "[[[136, 166]]]", "process": "\\because given the ellipse \\frac{x^{2}}{m^{2}}+\\frac{y^{2}}{9}=1 (m>0) and the hyperbola \\frac{x^{2}}{n^{2}}-\\frac{y^{2}}{4}=1 (n>0) have the same foci F_{1}, F_{2}, \\therefore m^{2}-9=n^{2}+4, i.e., m^{2}-n^{2}=13. Assume P is in the first quadrant. According to the definitions of the ellipse and hyperbola, we have: |PF_{1}|+|PF_{2}|=2m, |PF_{1}|-|PF_{2}|=2n. Taking the difference of squares of the two equations yields 4|PF_{1}||PF_{2}|=4m^{2}-4n^{2}=4\\times13, \\therefore |PF_{1}||PF_{2}|=13" }, { "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. Point $A$ on the left branch of the hyperbola satisfies $A F_{1} \\perp A F_{2}$. Find the area of $\\triangle A F_{1} F_{2}$.", "fact_expressions": "G: Hyperbola;A: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(A,LeftPart(G));IsPerpendicular(LineSegmentOf(A,F1),LineSegmentOf(A,F2))", "query_expressions": "Area(TriangleOf(A, F1, F2))", "answer_expressions": "3", "fact_spans": "[[[2, 30], [55, 58]], [[61, 65]], [[39, 46]], [[47, 54]], [[2, 30]], [[2, 54]], [[2, 54]], [[55, 65]], [[67, 90]]]", "query_spans": "[[[92, 122]]]", "process": "The left and right foci of the hyperbola \\( x^{2} - \\frac{y^{2}}{3} = 1 \\) are denoted by \\( F_{1} \\) and \\( F_{2} \\), respectively. For a point \\( A \\) on the left branch of the hyperbola satisfying \\( AF_{1} \\perp AF_{2} \\), we have: \n\\( a = 1 \\), \\( b = \\sqrt{3} \\), \\( c = \\sqrt{a^{2} + b^{2}} = 2 \\). Then \n\\( |AF_{2}| - |AF_{1}| = 2a = 2 \\), and \n\\( |AF_{2}|^{2} + |AF_{1}|^{2} = (2c)^{2} = 16 \\). \nHence, \n\\( (|AF_{2}| - |AF_{1}|)^{2} = |AF_{2}|^{2} + |AF_{1}|^{2} - 2|AF_{2}| \\cdot |AF_{1}| = 4 \\), \nso \\( |AF_{2}| \\cdot |AF_{1}| = 6 \\), and therefore \n\\( S_{\\Delta AF_{1}F_{2}} = \\frac{1}{2} |AF_{2}| \\cdot |AF_{1}| = 3 \\)." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a focal distance of $6$, and the minor axis is $\\frac{\\sqrt{7}}{4}$ of the major axis. The line $l$ intersects the ellipse at points $A$ and $B$, and the midpoint of chord $AB$ is $M(2,1)$. Then the equation of line $l$ is?", "fact_expressions": "l: Line;G: Ellipse;a: Number;b: Number;A: Point;B: Point;M: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (2, 1);FocalLength(G)=6;MinorAxis(G)=(sqrt(7)/4)*MajorAxis(G);Intersection(l,G)={A,B};IsChordOf(LineSegmentOf(A,B),G);MidPoint(LineSegmentOf(A,B))=M", "query_expressions": "Expression(l)", "answer_expressions": "7*x+8*y-22=0", "fact_spans": "[[[89, 94], [89, 94]], [[2, 54], [95, 97]], [[4, 54]], [[4, 54]], [[99, 102]], [[103, 106]], [[119, 127]], [[4, 54]], [[4, 54]], [[2, 54]], [[119, 127]], [[2, 61]], [[2, 88]], [[89, 108]], [[95, 115]], [[110, 127]]]", "query_spans": "[[[129, 139]]]", "process": "From the given, the semi-focal length of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ is $c=3$, and the minor axis is $\\frac{\\sqrt{7}}{4}$ of the major axis, hence $2b=\\frac{\\sqrt{7}}{4}\\times2a$, so $a=4$, $b=\\sqrt{7}$, thus the equation of the ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$. Let the endpoints of the chord be $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Subtracting gives $\\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{16}+\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{7}=0$. Therefore, the slope of the line containing the chord is $-\\frac{7}{8}$, and its equation is $y-1=-\\frac{7}{8}(x-2)$, which simplifies to $7x+8y-22=0$." }, { "text": "From a point $P$ on the parabola $y^{2}=4x$, draw a perpendicular to the directrix of the parabola, with foot of 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;Expression(G) = (y^2 = 4*x);P: Point;PointOnCurve(P, G) = True;L: Line;PointOnCurve(P, L) = True;IsPerpendicular(L, Directrix(G)) = True;FootPoint(L, Directrix(G)) = M;M: Point;Abs(LineSegmentOf(P, M)) = 5;F: Point;Focus(G) = F", "query_expressions": "Area(TriangleOf(M, P, F))", "answer_expressions": "10", "fact_spans": "[[[1, 15], [22, 25], [50, 53]], [[1, 15]], [[18, 21]], [[1, 21]], [], [[0, 30]], [[0, 30]], [[0, 37]], [[34, 37]], [[39, 48]], [[57, 60]], [[50, 60]]]", "query_spans": "[[[62, 82]]]", "process": "" }, { "text": "The ellipse $\\frac{x^{2}}{81}+\\frac{y^{2}}{27}=1$ has two foci $F_{1}$, $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $P$, $Q$. What is the perimeter of $\\triangle P Q F_{2}$?", "fact_expressions": "G: Ellipse;H:Line;P: Point;Q: Point;F2: Point;F1: Point;Expression(G) = (x^2/81 + y^2/27 = 1);Focus(G) = {F1, F2};PointOnCurve(F1,H);Intersection(H, G) = {P, Q}", "query_expressions": "Perimeter(TriangleOf(P, Q, F2))", "answer_expressions": "36", "fact_spans": "[[[0, 39], [72, 74]], [[61, 63]], [[75, 78]], [[79, 82]], [[52, 59]], [[44, 51], [64, 71]], [[0, 39]], [[0, 59]], [[61, 71]], [[61, 82]]]", "query_spans": "[[[84, 110]]]", "process": "From the equation of the ellipse, we know that a=9. By the definition of the ellipse, the perimeter of \\trianglePQF_{2} is 4a. \\therefore the perimeter of \\trianglePQF_{2} is 36." }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $\\frac{\\sqrt{3}}{2}$, and the length of the minor axis is $4$, then what is the standard equation of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = sqrt(3)/2;Length(MinorAxis(G))=4", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16+y^2/4=1", "fact_spans": "[[[1, 53], [88, 90]], [[3, 53]], [[3, 53]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 78]], [[1, 86]]]", "query_spans": "[[[88, 97]]]", "process": "The length of the minor axis gives b=2, the eccentricity e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, and using a^{2}=b^{2}+c^{2}, a can be found, thus obtaining the equation of the ellipse. From the given conditions, e=\\frac{c}{a}=\\frac{\\sqrt{3}}{2}, 2b=4, so b=2, hence \\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{3}{4}, solving gives a=4, so the standard equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{4}=" }, { "text": "Given that a point on the parabola $y^{2}=4x$ is at a distance of $6$ from the focus, then the coordinates of this point are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);P:Point;Distance(P,Focus(G))=6;PointOnCurve(P,G)", "query_expressions": "Coordinate(P)", "answer_expressions": "(5,pm*2*sqrt(5))", "fact_spans": "[[[2, 16]], [[2, 16]], [[32, 33]], [[2, 29]], [[2, 19]]]", "query_spans": "[[[32, 38]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left focus at $F$, and point $A$ has coordinates $(0,2 b)$. If the inclination angle of line $AF$ is $45^{\\circ}$, 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;F: Point;LeftFocus(C) = F;A: Point;Coordinate(A) = (0, 2*b);Inclination(LineOf(A, F)) = ApplyUnit(45, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "(2*sqrt(3))/3", "fact_spans": "[[[2, 63], [117, 120]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71]], [[2, 71]], [[72, 76]], [[72, 89]], [[91, 115]]]", "query_spans": "[[[117, 126]]]", "process": "According to the problem, we have $ k_{AF} = \\frac{2b}{c} = 1 $, so $ c = 2b $, $ c^{2} = 4b^{2} = 4(c^{2} - a^{2}) $, that is, $ 3c^{2} = 4a^{2} $, therefore $ e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{3} $." }, { "text": "Given that point $P$ is a moving point on the parabola $C$: $y^{2}=4x$, find the minimum value of the sum of the distance from point $P$ to point $A(0,2)$ and the distance from point $P$ to the directrix of the parabola $C$.", "fact_expressions": "C: Parabola;A: Point;P: Point;Expression(C) = (y^2 = 4*x);Coordinate(A) = (0, 2);PointOnCurve(P, C)", "query_expressions": "Min(Distance(P, A) + Distance(P, Directrix(C)))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[7, 26], [56, 62]], [[38, 47]], [[2, 6], [33, 37], [51, 55]], [[7, 26]], [[38, 47]], [[2, 31]]]", "query_spans": "[[[33, 75]]]", "process": "From the problem: It is easy to see from the figure: AA + AA' = \\sqrt{5}." }, { "text": "Given that the two asymptotes of a hyperbola are perpendicular to each other, then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;l1: Line;l2: Line;Asymptote(G) = {l1, l2};IsPerpendicular(l1, l2)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 5], [18, 21]], [], [], [[2, 11]], [[2, 15]]]", "query_spans": "[[[18, 27]]]", "process": "" }, { "text": "Given that the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has a focal distance of $10$, and the point $P(2, 1)$ lies on an asymptote of $C$, then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (2, 1);FocalLength(C) = 10;PointOnCurve(P, Asymptote(C))", "query_expressions": "Expression(C)", "answer_expressions": "x^2/20-y^2/5=1", "fact_spans": "[[[2, 52], [74, 77], [84, 87]], [[9, 52]], [[9, 52]], [[63, 73]], [[2, 52]], [[63, 73]], [[2, 60]], [[62, 82]]]", "query_spans": "[[[84, 92]]]", "process": "Knowing the focal length is 10, we have c = 5, that is, a^{2} + b^{2} = 25. According to the hyperbola equation, the asymptotes are y = \\pm\\frac{b}{a}x. Substituting the coordinates of point P gives a = 2b. Solving the system of equations yields a^{2} = 20, b^{2} = 5. Therefore, the hyperbola equation is \\frac{x^{2}}{20} - \\frac{y^{2}}{5} = 1." }, { "text": "Given $a>0$, if the circle $x^{2}+(y-a)^{2}=4$ passes through the foci of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$, then $a=$?", "fact_expressions": "G: Hyperbola;a: Number;H: Circle;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (x^2 + (-a + y)^2 = 4);a > 0;PointOnCurve(Focus(G),H)", "query_expressions": "a", "answer_expressions": "sqrt(6)/2", "fact_spans": "[[[31, 63]], [[68, 71]], [[9, 29]], [[31, 63]], [[9, 29]], [[2, 7]], [[9, 66]]]", "query_spans": "[[[68, 73]]]", "process": "The coordinates of the foci of the hyperbola are $(\\pm\\sqrt{a^{2}+1},0)$; substituting into the equation of the circle gives $(\\pm\\sqrt{a^{2}+1})^{2}+(-a)^{2}=4$, $2a^{2}+1=4$, $a>0$, solving yields: $a=\\frac{\\sqrt{6}}{2}$" }, { "text": "The focus of the parabola $y^{2}=16 x$ coincides with the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m^{2}}=1(m>0)$. Then the length of the imaginary axis of this hyperbola is equal to?", "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;m>0;Expression(G) = (x^2/4 - y^2/m^2 = 1);Expression(H) = (y^2 = 16*x);Focus(H) = RightFocus(G)", "query_expressions": "Length(ImageinaryAxis(G))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[19, 66], [75, 78]], [[22, 66]], [[0, 15]], [[22, 66]], [[19, 66]], [[0, 15]], [[0, 72]]]", "query_spans": "[[[75, 85]]]", "process": "From the problem, we know: the focus of the parabola is at (4,0). Since the focus of the parabola coincides with the right focus of the hyperbola, c=4. Also, c^{2}=a^{2}+b^{2}, so 16=4+m^{2}, then m=2\\sqrt{3}. Therefore, the length of the imaginary axis of the hyperbola is 4\\sqrt{3}." }, { "text": "If the point $(a, 3)$ lies on the curve of the equation $(x-y+1)(x+y)=0$, then $a$=?", "fact_expressions": "G: Curve;H: Point;a:Number;Coordinate(H) = (a, 3);Expression(G)=((x-y+1)*(x+y)=0);PointOnCurve(H,G)", "query_expressions": "a", "answer_expressions": "2,-3", "fact_spans": "[[[30, 32]], [[1, 10]], [[35, 38]], [[1, 10]], [[11, 32]], [0, 33]]", "query_spans": "[[[35, 40]]]", "process": "Substituting the point (a,3) into the equation (x-y+1)(x+y)=0 gives (a-3+1)(a+3)=0. Solving yields: a=2 or a=-3. The answers are: 2 or -3" }, { "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;A: Point;Coordinate(A) = (3, sqrt(2));G:Hyperbola;Expression(G)=(x^2/12-y^2/8=1);Asymptote(G)=Asymptote(C);PointOnCurve(A,C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/6-y^2/4=1", "fact_spans": "[[[1, 7], [74, 80]], [[55, 72]], [[55, 72]], [[8, 48]], [[8, 48]], [[1, 52]], [[1, 72]]]", "query_spans": "[[[74, 85]]]", "process": "" }, { "text": "A line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$, and intersects its directrix $l$ at point $C$, with $|AF| > |BF|$. If $|BC| = 2|BF|$, then $|AF| =$ ?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H) = True;Intersection(H, G) = {A, B};A: Point;B: Point;l: Line;Directrix(G) = l;Intersection(H, l) = C;C: Point;Abs(LineSegmentOf(A, F)) > Abs(LineSegmentOf(B, F));Abs(LineSegmentOf(B, C)) = 2*Abs(LineSegmentOf(B, F))", "query_expressions": "Abs(LineSegmentOf(A, F))", "answer_expressions": "4", "fact_spans": "[[[1, 15], [25, 28], [39, 40]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[0, 24]], [[22, 37]], [[29, 33]], [[34, 37]], [[42, 45]], [[39, 45]], [[22, 50]], [[46, 50]], [[52, 65]], [[67, 81]]]", "query_spans": "[[[83, 92]]]", "process": "Draw AM, BN perpendicular to the directrix at points M, N. Then |BN| = |BF|. Also |BC| = 2|BF|, so |BC| = 2|BN| and |BC| = \\frac{8}{3}, \\frac{|AM|}{|AF|} = \\frac{4}{4+|AF|}, solving gives |AF| = 4." }, { "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 $(\\overrightarrow{O P}+\\overrightarrow{O F_{2}}) \\cdot \\overrightarrow{F_{2} P}=0$, where $O$ is the origin, and $|\\overrightarrow{PF_{1}}|=\\sqrt{3}|\\overrightarrow{P F_{2}}|$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Origin;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,RightPart(G));DotProduct((VectorOf(O, F2) + VectorOf(O, P)),VectorOf(F2, P)) = 0;Abs(VectorOf(P, F1)) = sqrt(3)*Abs(VectorOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)+1", "fact_spans": "[[[19, 75], [84, 87], [259, 262]], [[22, 75]], [[22, 75]], [[184, 187]], [[94, 97]], [[9, 16]], [[1, 8]], [[22, 75]], [[22, 75]], [[19, 75]], [[1, 82]], [[1, 82]], [[84, 97]], [[99, 182]], [[194, 256]]]", "query_spans": "[[[259, 268]]]", "process": "Take the midpoint A of PF₂, then (∀OP + ∀OF₂) ⋅ ∀F₂P = 0, ⋅2∀OA ⋅ ∀F₂P = 0, ∴ ∀OA ⊥ ∀F₂P. Since OA is the midline of triangle PF₁F₂, ∴ PF₁ ⊥ PF₂, OA = ½PF₁. By the definition of hyperbola, |PF₁| − |PF₂| = 2a. Since |PF₁| = √3|PF₂|, ∴ |PF₂| = 2a/(√3−1), |PF₁| = 2√3a/(√3−1). In triangle PF₁F₂, by Pythagorean theorem, |PF₁|² + |PF₂|² = 4c², (2a/(√3−1))² + (2√3a/(√3−1))² = 4c², ∴ e = √3 + 1" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$, the right focus is $F$, $O$ is the origin. There is exactly one point $P$ on $C$ such that $|O F|=|F P|$. Then the equation of $C$ is?", "fact_expressions": "C: Ellipse;a: Number;O: Origin;F: Point;P: Point;Expression(C) = (y^2/3 + x^2/a^2 = 1);RightFocus(C)=F;PointOnCurve(P,C);Abs(LineSegmentOf(O,F))=Abs(LineSegmentOf(F,P))", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4+y^2/3=1", "fact_spans": "[[[2, 48], [97, 100], [97, 100]], [[9, 48]], [[57, 60]], [[53, 56]], [[76, 80]], [[2, 48]], [[2, 56]], [[66, 80]], [[82, 95]]]", "query_spans": "[[[97, 105]]]", "process": "Let the coordinates of point P be (x, y), we have y^{2}= Using the distance formula between two points, we obtain |FP|=a-\\frac{c}{a}x=c. Combining with the given conditions, we get x=a, from which it follows that a=2c. The value of a can be found, and then the equation of the ellipse C can be determined. Let the coordinates of point P be (x, y), and point F(c, 0). From the given conditions, we have y^{2}=3-\\frac{3x^{2}}{a^{2}}. Since |FP|=\\sqrt{(x-c)^{2}+y^{2}}=\\sqrt{x^{2}-2cx+c^{2}+3-\\frac{3y^{2}}{a^{2}}}=\\sqrt{\\frac{c^{2}}{a^{2}}x^{2}-2cx+a^{2}}=|\\frac{c}{a}x-a|. Because -a\\leqslant x \\leqslant a, then \\frac{c}{a}x-a<0, so |FP|=a-\\frac{c}{a}x=c. Since there is exactly one point P on the ellipse C satisfying |OF|=|FP|, then x=\\pm a. If x=-a, then |FP|=a+c>c, which does not meet the condition. If x=a, then |FP|=a-c=c, giving a=2c, i.e., a=2\\sqrt{a^{2}-3}, solving yields a=2. Therefore, the equation of the ellipse C is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1" }, { "text": "Given the parabola $C$: $y^{2}=4x$, the distance from the focus $F$ of the parabola to the line $l$: $y=\\sqrt{3}x+b$ is $\\sqrt{3}$, and the line $l$ intersects the parabola $C$ at points $A$ and $B$. Then $|AF|+|BF|=$?", "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C) = (y^2 = 4*x);Expression(l) = (y = sqrt(3)*x + b);Focus(C) = F;Distance(F, l) = sqrt(3);Intersection(l, C) = {A, B};b:Number", "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F))", "answer_expressions": "28/3", "fact_spans": "[[[28, 51], [67, 72]], [[2, 21], [73, 79]], [[81, 84]], [[24, 27]], [[85, 88]], [[2, 21]], [[28, 51]], [[2, 27]], [[24, 65]], [[67, 90]], [[35, 51]]]", "query_spans": "[[[92, 107]]]", "process": "Given that the focus of parabola C is F(1,0), the distance from focus F to the line $ l: y = \\sqrt{3}x + b $ is $ d = \\frac{|\\sqrt{3} + b|}{2} = \\sqrt{3} $, solving gives $ \\begin{cases} b = -3\\sqrt{3} \\quad \\text{or} \\quad b = \\sqrt{3}. \\end{cases} $ Also, line $ l $ intersects parabola C at points A and B, let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, solving the system \n$ y^{2} = 4x $, $ y = \\sqrt{3}x + b $, eliminating $ x $ and simplifying yields $ \\sqrt{3}y^{2} - 4y + 4b = 0 $, then $ \\Delta = 16 - 16\\sqrt{3}b > 0 $, solving gives $ b < \\frac{\\sqrt{3}}{3} $, so $ b = -3\\sqrt{3} $, thus line $ l: y = \\sqrt{3}x - 3\\sqrt{3} $. Substituting into parabola $ C: y^{2} = 4x $, eliminating $ y $ and simplifying yields $ 3x^{2} - 22x + 27 = 0 $, then $ x_{1} + x_{2} = \\frac{22}{3} $, so $ |AF| + |BF| = x_{1} + x_{2} + 2 = \\frac{28}{3} $." }, { "text": "Given the parabola $C$: $x^{2}=8 y$ with focus $F$, and $O$ as the origin. Point $P$ lies on the parabola $C$ such that $P F \\perp O F$. Then $|\\overrightarrow{O F}-\\overrightarrow{P F}|$=?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 8*y);F: Point;Focus(C) = F;O: Origin;P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F), LineSegmentOf(O, F))", "query_expressions": "Abs(VectorOf(O, F) - VectorOf(P, F))", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 21], [43, 49]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32]], [[38, 42]], [[38, 50]], [[52, 67]]]", "query_spans": "[[[69, 116]]]", "process": "According to the problem, |OF| = 2, |PF| = 4, then |\\overrightarrow{OF} - \\overrightarrow{PF}| = |\\overrightarrow{OP}| = \\sqrt{|OF|^{2} + |FP|^{2}} = 2\\sqrt{5}" }, { "text": "The focal distance of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)", "query_expressions": "FocalLength(G)", "answer_expressions": "10", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 44]]]", "process": "Since $ c^{2} = a^{2} + b^{2} = 9 + 16 = 25^{\\circ} $, it follows that $ c = 5 $, so the focal distance is $ 2c = 10 $." }, { "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{8}=1$. If $P$ is a point on the left branch of $C$ and $A(0,6 \\sqrt{6})$ is a point on the $y$-axis, then the minimum perimeter of $\\triangle A P F$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2 - y^2/8 = 1);F: Point;RightFocus(C) = F;P: Point;PointOnCurve(P, LeftPart(C));A: Point;Coordinate(A) = (0, 6*sqrt(6));PointOnCurve(A, yAxis)", "query_expressions": "Min(Perimeter(TriangleOf(A, P, F)))", "answer_expressions": "32", "fact_spans": "[[[6, 39], [50, 53]], [[6, 39]], [[2, 5]], [[2, 43]], [[46, 49]], [[46, 59]], [[60, 77]], [[60, 77]], [[60, 85]]]", "query_spans": "[[[87, 112]]]", "process": "Problem Analysis: Since a=1, b^{2}=8 \\Rightarrow c=3, let the left focus of the hyperbola be F', and the perimeter of triangle APF be L=AF+AP+PF. Note that AF=\\sqrt{9+216}=15, and AP+PF=AP+PF'+2. Thus, when points A, P, F' are collinear, AP+PF=AP+PF'+2 is minimized. Let P(x,y), then y=2\\sqrt{6}(x+3). Substituting into the hyperbola equation yields x=-2, y=2\\sqrt{6}. Therefore, PF=\\sqrt{25+24}=7, PA=\\sqrt{4+96}=10. Hence, the minimum perimeter of the triangle is L=15+7+10=32. Fill in 32." }, { "text": "$P$ is a point on the right branch of the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, $F_{1}$ and $F_{2}$ are the left and right foci respectively, $I$ is the incenter of triangle $P F_{1} F_{2}$. If $S_{\\Delta P F_{1} F_{2}}=2 S_{\\Delta I P F_{2}}+\\left(1+\\frac{1}{\\lambda}\\right) S_{\\Delta I F_{1} F_{2}}$, then the value of the real number $\\lambda$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);P: Point;PointOnCurve(P, RightPart(G));F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;I: Point;Incenter(TriangleOf(P,F1,F2)) = I;lambda: Real;Area(TriangleOf(P,F1,F2)) = 2*Area(TriangleOf(I,P,F2)) + (1+1/lambda)*Area(TriangleOf(I,F1,F2))", "query_expressions": "lambda", "answer_expressions": "sqrt(3)", "fact_spans": "[[[4, 32]], [[4, 32]], [[0, 3]], [[0, 37]], [[38, 45]], [[46, 53]], [[4, 61]], [[4, 61]], [[62, 65]], [[62, 87]], [[187, 198]], [[89, 185]]]", "query_spans": "[[[187, 202]]]", "process": "According to the problem, let the focal distance of the hyperbola be 2c and the length of the real axis be 2a, then c=\\sqrt{3}, a=1. Since I is the incenter of triangle PF_{1}F_{2}, let r be the inradius of triangle PF_{1}F_{2}, then: S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}(|F_{1}F_{2}|+|PF_{1}|+|PF_{2}|)r. S_{\\Delta IPF_{2}}=\\frac{1}{2}|PF_{2}|r, S_{\\Delta IF_{1}F_{2}}=\\frac{1}{2}|F_{1}F_{2}|r. Since S_{\\Delta PF_{1}F_{2}}=2S_{\\Delta IPF_{2}}+(1+\\frac{1}{\\lambda})S_{\\Delta IF_{1}F_{2}} and S_{\\Delta PF_{1}F_{2}}=S_{\\Delta IPF_{2}}+S_{\\Delta IF_{1}F_{2}}+S_{\\Delta IPF_{1}}, therefore S_{\\Delta IPF_{2}}+\\frac{1}{2}S_{\\Delta IF_{1}F_{2}}=S_{\\Delta IPF_{1}}, that is, \\frac{1}{2}|PF_{2}|r+\\frac{1}{\\lambda}\\times\\frac{1}{2}|F_{1}F_{2}|r=\\frac{1}{2}|PF_{1}|r. Therefore |PF_{2}|+\\frac{1}{\\lambda}|F_{1}F_{2}|=|PF_{1}|. Since P is a point on the right branch of the hyperbola, \\frac{1}{\\lambda}=\\frac{|PF_{1}|-|PF_{2}|}{|F_{1}F_{2}|}=\\frac{2a}{2c}=\\frac{a}{c}=\\frac{1}{\\sqrt{3}}. Therefore \\lambda=\\sqrt{3}." }, { "text": "If one asymptote of a hyperbola is given by the equation $y=\\sqrt{2} x$, then what is its eccentricity?", "fact_expressions": "G: Hyperbola;Expression(OneOf(Asymptote(G))) = (y=sqrt(2)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{sqrt(6)/2,sqrt(3)}", "fact_spans": "[[[1, 4], [29, 30]], [[1, 27]]]", "query_spans": "[[[29, 35]]]", "process": "Problem Analysis: According to the given conditions, when the foci of the hyperbola lie on the x-axis, we have $\\frac{b}{a}=\\sqrt{2}\\Rightarrow b=\\sqrt{2}a$, and the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\frac{\\sqrt{3}a}{a}=\\sqrt{3}$. When the foci of the hyperbola lie on the y-axis, we have $\\frac{a}{b}=\\sqrt{2}\\Rightarrow a=\\sqrt{2}b$, and the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{\\sqrt{a^{2}+b^{2}}}{a}=\\frac{\\sqrt{6}}{2}$." }, { "text": "Given that line $l$ passes through the focus of the ellipse $\\frac{y^{2}}{2}+x^{2}=1$ and intersects the ellipse at points $P$ and $Q$, and the perpendicular bisector of segment $PQ$ intersects the $x$-axis at point $M$, then the maximum area of $\\triangle M P Q$ is?", "fact_expressions": "l: Line;PointOnCurve(Focus(G), l) = True;Intersection(l,G) = {P,Q};G: Ellipse;Expression(G) = (x^2 + y^2/2 = 1);P: Point;Q: Point;M: Point;Intersection(PerpendicularBisector(LineSegmentOf(P,Q)),xAxis) = M", "query_expressions": "Max(Area(TriangleOf(M, P, Q)))", "answer_expressions": "3*sqrt(6)/8", "fact_spans": "[[[2, 7]], [[2, 39]], [[2, 56]], [[9, 36], [42, 44]], [[9, 36]], [[47, 50]], [[51, 54]], [[78, 82]], [[57, 82]]]", "query_spans": "[[[84, 109]]]", "process": "" }, { "text": "The asymptotes of the hyperbola $C$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1$ are given by? Let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola $C$, and let $P$ be a point on $C$ such that $|PF_{1}|=4$. Then $|PF_{2}|$=?", "fact_expressions": "C: Hyperbola;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/16 - y^2/4 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(P, C);Abs(LineSegmentOf(P, F1)) = 4", "query_expressions": "Expression(Asymptote(C));Abs(LineSegmentOf(P, F2))", "answer_expressions": "y=pm*(1/2)*x\n12", "fact_spans": "[[[0, 43], [70, 76], [87, 90]], [[83, 86]], [[52, 59]], [[62, 69]], [[0, 43]], [[52, 82]], [[52, 82]], [[83, 93]], [[95, 107]]]", "query_spans": "[[[0, 51]], [[109, 121]]]", "process": "Problem Analysis: For the hyperbola C: \\frac{x^{2}}{16}-\\frac{y^{2}}{4}=1, a=4, b=2, then the asymptotes are y=\\pm\\frac{1}{2}x. Given that point P lies on the left branch of the hyperbola, |PF_{2}|\\cdot|PF_{1}|=2a=8, \\therefore|PF_{2}|=12" }, { "text": "The line $x=m$ intersects the ellipse $C$: $3 x^{2}+4 y^{2}=12$ at points $A$ and $B$. The left focus of the ellipse is $F$. When the perimeter of $\\triangle F A B$ is maximized, what is the area of $\\triangle F A B$?", "fact_expressions": "C: Ellipse;G: Line;m: Number;F: Point;A: Point;B: Point;Expression(C) = (3*x^2 + 4*y^2 = 12);Expression(G) = (x = m);Intersection(G, C) = {A, B};LeftFocus(C) = F;WhenMax(Perimeter(TriangleOf(F, A, B)))", "query_expressions": "Area(TriangleOf(F, A, B))", "answer_expressions": "3", "fact_spans": "[[[8, 35], [45, 47]], [[0, 7]], [[2, 7]], [[52, 55]], [[37, 40]], [[41, 44]], [[8, 35]], [[0, 7]], [[0, 44]], [[45, 55]], [[56, 80]]]", "query_spans": "[[[81, 103]]]", "process": "Let the right focus of the ellipse be $ F_{1} $. Clearly, $ F_{1}A + F_{1}B \\geqslant AB $ (equality holds when the line $ x = m $ passes through point $ F_{1} $). Therefore, $ F_{1}A + F_{1}B + FA + FB \\geqslant AB + FA + FB $, that is, $ AB + FA + FB \\leqslant 2a + 2a = 4a $. Thus, the maximum perimeter of $ \\triangle FAB $ is $ 4a $, and this occurs when the line $ x = m $ passes through $ F_{1} $. From $ 3x^{2} + 4y^{2} = 12 \\Rightarrow \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 \\Rightarrow a = 2, b = \\sqrt{3} $, so $ c = \\sqrt{a^{2} - b^{2}} = \\sqrt{4 - 3} = 1 $. Hence, the equation of the line $ x = m $ at this time is $ x = 1 $. When $ x = 1 $, $ 3 \\times 1^{2} + 4y^{2} = 12 \\Rightarrow y = \\pm \\frac{3}{2} $, so $ AB = 2 \\times \\frac{3}{2} = 3 $, $ FF_{1} = 2 $, therefore the area of $ \\triangle FAB $ is $ \\frac{1}{2} \\times 2 \\times 3 = 3 $." }, { "text": "Given that the focus of the parabola $x^{2}=4 y$ is $F$, the intersection point of the directrix and the $y$-axis is $M$, and $N$ is a point on the parabola satisfying $|N F|=\\lambda|M N|$, then the range of values for $\\lambda$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;Intersection(Directrix(G),yAxis) = M;M: Point;PointOnCurve(N,G) = True;N: Point;lambda: Number;Abs(LineSegmentOf(N, F)) = lambda*Abs(LineSegmentOf(M, N))", "query_expressions": "Range(lambda)", "answer_expressions": "[sqrt(2)/2,1]", "fact_spans": "[[[2, 16], [43, 46]], [[2, 16]], [[20, 23]], [[2, 23]], [[2, 38]], [[35, 38]], [[39, 50]], [[39, 42]], [[76, 85]], [[54, 74]]]", "query_spans": "[[[76, 92]]]", "process": "" }, { "text": "Given that points $A$, $B$, $F_{1}$, $F_{2}$ are the right vertex, lower vertex, left focus, and right focus of the ellipse $\\frac{x^{2}}{a^{2}}+y^{2}=1$ $(a>1)$, respectively, and points $M$, $N$ are arbitrary points on the ellipse. If the maximum area of $\\Delta M A B$ is $\\sqrt{2}+1$, then what is the maximum value of $\\frac{|N F_{1}| \\cdot|N F_{2}|}{|N F_{1}|+9|N F_{2}|}$?", "fact_expressions": "G: Ellipse;a: Number;M: Point;A: Point;B: Point;N: Point;F1: Point;F2: Point;a>1;Expression(G) = (y^2 + x^2/a^2 = 1);RightVertex(G) = A;LowerVertex(G) = B;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(M, G);PointOnCurve(N, G);Max(Area(TriangleOf(M, A, B))) = 1 + sqrt(2)", "query_expressions": "Max((Abs(LineSegmentOf(N, F1))*Abs(LineSegmentOf(N, F2)))/(Abs(LineSegmentOf(N, F1)) + 9*Abs(LineSegmentOf(N, F2))))", "answer_expressions": "1/4", "fact_spans": "[[[29, 65], [91, 93]], [[31, 65]], [[82, 86]], [[2, 6]], [[7, 10]], [[87, 90]], [[11, 18]], [[19, 26]], [[31, 65]], [[29, 65]], [[2, 81]], [[2, 81]], [[2, 81]], [[2, 81]], [[82, 98]], [[82, 98]], [[100, 133]]]", "query_spans": "[[[135, 196]]]", "process": "Let M(a\\cos\\theta,\\sin\\theta). Since A(a,0), B(0,-1), the equation of line AB is x-ay-a=0. The distance from point M to line AB is d=\\frac{|a\\cos\\theta-a\\sin\\theta-a|}{\\sqrt{1+a^{2}}}=\\frac{a}{\\sqrt{1+a^{2}}}|\\sqrt{2}\\cos(\\theta+\\frac{\\pi}{4})-1|\\leqslant\\frac{(\\sqrt{2}+1)a}{\\sqrt{1+a^{2}}}. Since the maximum area of AMAB is \\sqrt{2}+1, we have \\frac{1}{2}\\sqrt{1+a^{2}}\\frac{(\\sqrt{2}+1)a}{\\sqrt{1+a^{2}}}=\\sqrt{2}+1, so a=2. By the definition of the ellipse, |NF_{1}|+|NF_{2}|=2a=4. Let t=|NF_{2}|, then |NF_{1}|=4-t and 00 when 00)$ and an asymptote of the hyperbola $C_{2}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$. If the distance from point $A$ to the directrix of the parabola $C_{1}$ is $p$, then the eccentricity of the hyperbola $C_{2}$ is?", "fact_expressions": "C2: Hyperbola;C1: Parabola;A: Point;p: Number;a: Number;b: Number;a > 0;b > 0;p > 0;Expression(C1) = (y^2 = 2*p*x);Expression(C2) = (x^2/a^2 - y^2/b^2 = 1);Intersection(C1, OneOf(Asymptote(C2))) = A;Distance(A, Directrix(C1)) = p", "query_expressions": "Eccentricity(C2)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[36, 101], [139, 149]], [[5, 35], [117, 127]], [[0, 4], [112, 116]], [[134, 137]], [[48, 101]], [[48, 101]], [[48, 101]], [[48, 101]], [[17, 35]], [[5, 35]], [[36, 101]], [[0, 110]], [[112, 137]]]", "query_spans": "[[[139, 155]]]", "process": "Take one asymptote of the hyperbola: $ y = \\frac{b}{a}x $, solve the system \n$$\n\\begin{cases}\ny^{2} = 2px \\\\\ny = \\frac{b}{a}x\n\\end{cases}\n$$\nobtaining \n$$\n\\begin{cases}\nx = \\frac{2pa^{2}}{b^{2}} \\\\\ny = \\frac{2pa}{b}\n\\end{cases}\n$$\nthus $ A\\left( \\frac{2pa^{2}}{b^{2}}, \\frac{2pa}{b} \\right) $. Since the distance from point $ A $ to the directrix of the parabola is $ p $, \n$ \\frac{p}{2} + \\frac{2pa^{2}}{b^{2}} = p $, simplifying to $ \\frac{a^{2}}{b^{2}} = \\frac{1}{4} $. \nTherefore, the eccentricity of hyperbola $ C_{2} $ is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{5} $." }, { "text": "If one asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1(a>0)$ is given by $3 x-2 y=0$, then $a$=?", "fact_expressions": "G: Hyperbola;a: Number;a > 0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(OneOf(Asymptote(G))) = (3*x - 2*y = 0)", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[1, 48]], [[70, 73]], [[4, 48]], [[1, 48]], [[1, 68]]]", "query_spans": "[[[70, 75]]]", "process": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ are given by $\\frac{x}{a}\\pm\\frac{y}{3}=0$, that is, $3x\\pm ay=0$ $(a>0)$. It is also given that one of its asymptotes is $3x-2y=0$. Therefore, $a=2$." }, { "text": "Given the parabola $C$: $y^{2}=4x$ and the point $M(-1,2)$, 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;Expression(C) = (y^2 = 4*x);M: Point;Coordinate(M) = (-1, 2);PointOnCurve(Focus(C), G);Slope(G) = k;k: Number;G: Line;Intersection(G, C) = {A, B};A: Point;B: Point;DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0", "query_expressions": "k", "answer_expressions": "1", "fact_spans": "[[[2, 21], [34, 37], [52, 55]], [[2, 21]], [[22, 32]], [[22, 32]], [[33, 51]], [[42, 51]], [[45, 48], [121, 124]], [[49, 51]], [[49, 66]], [[57, 60]], [[61, 64]], [[68, 119]]]", "query_spans": "[[[121, 126]]]", "process": "By the given condition, the focus of the parabola is F(1,0). Let the equation of line AB be y = k(x−1). Solving the system \n\\begin{cases}y^{2}=4x\\\\y=k(x-1)\\end{cases} \nby eliminating y, we obtain k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=\\frac{2k^{2}+4}{k^{2}}, x_{1}x_{2}=1, so y_{1}+y_{2}=k(x_{1}+x_{2}-2)=\\frac{4}{k}, y_{1}y_{2}=k^{2}(x_{1}-1)(x_{2}-1)=k^{2}[x_{1}x_{2}-(x_{1}+x_{2})+1]=-4. Since M(-1,2), we have \\overrightarrow{MA}=(x_{1}+1,y_{1}-2), \\overrightarrow{MB}=(x_{2}+1,y_{2}-2). Because \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=0, it follows that (x_{1}+1)(x_{2}+1)+(y_{1}-2)(y_{2}-2)=0. Rearranging gives x_{1}x_{2}+(x_{1}+x_{2})+y_{1}y_{2}-2(y_{1}+y_{2})+5=0. Therefore, 1+\\frac{2k^{2}+4}{k^{2}}-4-2\\times\\frac{4}{k}+5=0, which simplifies to k^{2}-2k+1=0, so k=1" }, { "text": "If the constant $ a>0 $, and the major axis length of the ellipse $ x^{2}+a^{2} y^{2}=2 a^{2} $ is 3 times the minor axis length, then the value of the real number $ a $ is?", "fact_expressions": "a: Real;a>0;G: Ellipse;Expression(G) = (a^2*y^2 + x^2 = 2*a^2);Length(MajorAxis(G)) = 3*Length(MinorAxis(G))", "query_expressions": "a", "answer_expressions": "{3, 1/3}", "fact_spans": "[[[54, 59]], [[3, 8]], [[10, 39]], [[10, 39]], [[10, 52]]]", "query_spans": "[[[54, 63]]]", "process": "From the ellipse $x^{2}+a^{2}y^{2}=2a^{2}$, we obtain the ellipse $\\frac{x^{2}}{2a^{2}}+\\frac{y^{2}}{2}=1$. When $a>1$, $\\frac{x^{2}}{2a^{2}}+\\frac{y^{2}}{2}=1$ represents an ellipse with foci on the x-axis, $\\therefore 2\\sqrt{2a^{2}}=3\\times2\\sqrt{2}$, so $a=3$. When $00;b>0", "query_expressions": "Range(Abs(LineSegmentOf(M, N)))", "answer_expressions": "[2*a, (4*sqrt(3)*a)/3]", "fact_spans": "[[[20, 71], [103, 109], [78, 81]], [[20, 71]], [[28, 71]], [[28, 71]], [[2, 9]], [[10, 17], [92, 99]], [[2, 77]], [[2, 77]], [[85, 90]], [[78, 90]], [[85, 90]], [[100, 102]], [[91, 102]], [[114, 117], [126, 130]], [[118, 121]], [[100, 123]], [[126, 135]], [[138, 142]], [[143, 146]], [[138, 198]], [[138, 198]], [[20, 71]], [[20, 71]]]", "query_spans": "[[[200, 212]]]", "process": "" }, { "text": "Given that the point $(1,1)$ lies on the parabola $C$: $y^{2}=2 p x$ $(p>0)$, then the distance from the focus of $C$ to its directrix is?", "fact_expressions": "C: Parabola;p: Number;G: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(G) = (1, 1);PointOnCurve(G, C)", "query_expressions": "Distance(Focus(C), Directrix(C))", "answer_expressions": "1/2", "fact_spans": "[[[11, 37], [40, 43], [47, 48]], [[19, 37]], [[2, 10]], [[19, 37]], [[11, 37]], [[2, 10]], [[2, 38]]]", "query_spans": "[[[40, 55]]]", "process": "Substitute the point (1,1) into the parabola equation 1=2p\\times1, solving gives p=\\frac{1}{2}, the parabola equation is: y^{2}=x, the focus coordinates are (\\frac{1}{4},0), the directrix equation is: x=-\\frac{1}{4}. The distance from the focus to the directrix: \\frac{1}{4}-(-\\frac{1}{4})=\\frac{1}{2} The final answer is: \\underline{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}$ and $F_{2}$ respectively. A line passing through $F_{1}$ intersects the left branch of hyperbola $C$ at points $P$ and $Q$. If $\\overrightarrow{P F_{2}} \\perp \\overrightarrow{P Q}$ and the perimeter of $\\triangle P Q F_{2}$ is $12 a$, then the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;P: Point;Q: 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(F1, G);Intersection(G, LeftPart(C)) = {P, Q};IsPerpendicular(VectorOf(P,F2),VectorOf(P,Q));Perimeter(TriangleOf(P,Q,F2))=12*a", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[2, 63], [101, 107], [209, 215]], [[10, 63]], [[10, 63]], [[98, 100]], [[111, 114]], [[115, 118]], [[81, 88]], [[73, 80], [90, 97]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 88]], [[2, 88]], [[89, 100]], [[98, 120]], [[122, 175]], [[176, 207]]]", "query_spans": "[[[209, 221]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are 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$, and the distance from point $P$ to the $x$-axis is equal to $\\frac{\\sqrt{5}}{5}$. Then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0;Distance(P, xAxis) = sqrt(5)/5", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[18, 50], [62, 65], [167, 170]], [[21, 50]], [[56, 60], [129, 133]], [[2, 9]], [[10, 17]], [[18, 50]], [[2, 55]], [[56, 66]], [[67, 126]], [[129, 163]]]", "query_spans": "[[[167, 177]]]", "process": "According to the problem, we have\n\\begin{cases}\n|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2} \\\\\n|PF_{1}|-|PF_{2}|=\\pm2a,\n\\end{cases}\n|PF_{1}|^{2}+|PF_{2}|^{2}-(|PF_{1}|-|PF_{2}|)^{2}=2|PF_{1}||PF_{2}|=4c^{2}-4a^{2}=4b^{2}, |PF_{1}||PF_{2}|=2b^{2}=2.\nAlso, S_{\\triangle PF_{1}F_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|=\\frac{1}{2}|F_{1}F_{2}|\\times\\frac{\\sqrt{5}}{6},\ntherefore |F_{1}F_{2}|=2\\sqrt{5}, a=\\sqrt{(\\sqrt{5})^{2}-1}=2,\nthe eccentricity of this hyperbola is \\frac{|F_{1}F_{2}|}{2a}=\\frac{\\sqrt{5}}{2}." }, { "text": "Given that the center of a hyperbola is at the origin, one focus is at $F(10 , 0)$, and the equations of the two asymptotes are $y = \\pm \\frac{4}{3}x$, then what is the standard equation of this hyperbola?", "fact_expressions": "G: Hyperbola;F: Point;O:Origin;Center(G)=O;Coordinate(F) = (10, 0);OneOf(Focus(G))=F;Expression(Asymptote(G)) = (y = pm*(4/3)*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/36-y^2/64=1", "fact_spans": "[[[2, 5], [64, 67]], [[19, 30]], [[9, 13]], [[2, 13]], [[19, 30]], [[2, 30]], [[2, 61]]]", "query_spans": "[[[64, 74]]]", "process": "" }, { "text": "Given point $Q(-2 \\sqrt{2}, 0)$ and a moving point $P(x, y)$ on the parabola $x^{2} = -4y$, what is the minimum value of $|y| + |PQ|$?", "fact_expressions": "Q: Point;Coordinate(Q) = (-2*sqrt(2), 0);G: Parabola;Expression(G) = (x^2 = -4*y);P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;PointOnCurve(P, G) = True", "query_expressions": "Min(Abs(LineSegmentOf(P, Q)) + Abs(y1))", "answer_expressions": "2", "fact_spans": "[[[2, 23]], [[2, 23]], [[24, 39]], [[24, 39]], [[43, 52]], [[43, 52]], [[43, 52]], [[43, 52]], [[24, 52]]]", "query_spans": "[[[54, 71]]]", "process": "" }, { "text": "If the distance from the focus of the parabola $x^{2}=4 y$ to the asymptote of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is equal to $\\frac{1}{3}$, 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;G: Parabola;Expression(G) = (x^2 = 4*y);Distance(Focus(G), Asymptote(C)) = 1/3", "query_expressions": "Eccentricity(C)", "answer_expressions": "3", "fact_spans": "[[[19, 80], [103, 109]], [[19, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[1, 15]], [[1, 15]], [[1, 101]]]", "query_spans": "[[[103, 115]]]", "process": "First find that the focus of the parabola $ x^{2} = 4y $ is at $ (0,1) $, and one asymptote of the hyperbola is $ y = \\frac{b}{a}x $. Then use the point-to-line distance formula and the eccentricity formula to find the result. Detailed solution: The focus of the parabola $ x^{2} = 4y $ is $ (0,1) $, and one asymptote of the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) is $ y = \\frac{b}{a}x $. Therefore, $ e = \\frac{c}{a} = 3 $." }, { "text": "Given that point $P(m, 2)$ lies on the right branch of the hyperbola $C$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{2}=1$, and $F_{1}$, $F_{2}$ are its left and right foci respectively, the line segment $P F_{1}$ intersects the left branch of $C$ at point $Q$. Then $|P Q|+|Q F_{2}|$=?", "fact_expressions": "P: Point;m: Number;Coordinate(P) = (m, 2);C: Hyperbola;Expression(C) = (x^2/2 - y^2/2 = 1);PointOnCurve(P, RightPart(C));F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;Q: Point;Intersection(LineSegmentOf(P, F1), LeftPart(C)) = Q", "query_expressions": "Abs(LineSegmentOf(P, Q)) + Abs(LineSegmentOf(Q, F2))", "answer_expressions": "2*sqrt(3)+3*sqrt(2)", "fact_spans": "[[[2, 12]], [[3, 12]], [[2, 12]], [[13, 56], [81, 82], [99, 102]], [[13, 56]], [[2, 62]], [[63, 70]], [[71, 78]], [[63, 86]], [[63, 86]], [[106, 110]], [[87, 110]]]", "query_spans": "[[[112, 131]]]", "process": "" }, { "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$, respectively. If there exists a point $A$ on the hyperbola such that $\\angle F_{1} A F_{2}=90^{\\circ}$ and $|A F_{1}|=3|A 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;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;PointOnCurve(A, G) = True;AngleOf(F1, A, F2) = ApplyUnit(90, degree);Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(A, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(10)/ 2", "fact_spans": "[[[19, 65], [73, 76], [144, 147]], [[19, 65]], [[22, 65]], [[22, 65]], [[1, 8]], [[9, 16]], [[1, 71]], [[1, 71]], [[79, 83]], [[73, 83]], [[85, 118]], [[120, 142]]]", "query_spans": "[[[144, 153]]]", "process": "\\because\\angle\\angleF_{1}AF_{2}=90^{\\circ}, and |AF_{1}|=3|AF_{2}|, let |AF_{2}|=m, |AF_{1}|=3m \\therefore (3m)^{2}+m^{2}=4c^{2}, get_{m}=\\frac{\\sqrt{10}}{5}c^{2}, |AF_{1}|-|AF_{2}|=2a=2m=\\frac{2\\sqrt{10}}{5}c \\therefore \\frac{c}{a}=\\frac{\\sqrt{10}}{2}" }, { "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}$ and $F_{2}$ respectively. The line passing through $F_{1}$ and perpendicular to the $x$-axis intersects the left branch of the hyperbola at points $A$ and $B$. Lines $A F_{2}$ and $B F_{2}$ intersect the $y$-axis at points $P$ and $Q$ respectively. If the perimeter of $\\triangle P Q F_{2}$ is $16$, then the maximum value of $\\frac{b^{2}}{a+1}$ is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Line;P: Point;Q: Point;F2: Point;A: Point;B: 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(F1,H);IsPerpendicular(xAxis, H);Intersection(H, LeftPart(G)) = {A, B};Intersection(LineSegmentOf(A, F2), yAxis) = P ;Intersection(LineSegmentOf(B, F2), yAxis) = Q;Perimeter(TriangleOf(P, Q, F2)) = 16", "query_expressions": "Max(b^2/(a + 1))", "answer_expressions": "4", "fact_spans": "[[[2, 58], [107, 110]], [[5, 58]], [[5, 58]], [[103, 105]], [[152, 155]], [[156, 159]], [[77, 84]], [[115, 118]], [[119, 122]], [[67, 74], [87, 94]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 84]], [[2, 84]], [[86, 105]], [[95, 105]], [[103, 124]], [[125, 161]], [[125, 161]], [[163, 192]]]", "query_spans": "[[[194, 219]]]", "process": "the perimeter is 32, AB is the latus rectum of the hyperbola, |AB|=\\frac{2b^{2}}{a} \\because |AF_{2}|+|BF_{2}|+|AB|=32, |AF_{2}|+|BF_{2}|-|AB|=4a, |AB|=\\frac{2b^{2}}{a} we obtain \\frac{4b^{2}}{a}=32-4a, \\therefore b^{2}=a(8-a), we get a\\in(0,8) then \\frac{b^{2}}{a+1}=\\frac{8a-a^{2}}{a+1}=-(a+1+\\frac{9}{a+1}-10)\\leqslant4, equality holds if and only if a+1=\\frac{9}{a+1}, i.e., when a=2, thus fill in 4. Qing] This question mainly examines the definition of hyperbola, application of basic inequality, and tests students' ability to analyze and solve problems, it is a difficult problem" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ has asymptotes with equations $3 x \\pm 2 y=0$, then the value of the positive number $a$ is?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/9 + x^2/a^2 = 1);Expression(Asymptote(G)) = ((2*y)+(3*(pm*x)) = 0)", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[1, 48]], [[72, 77]], [[4, 48]], [[1, 48]], [[1, 70]]]", "query_spans": "[[[72, 81]]]", "process": "Test analysis: Determine the asymptote equations of the hyperbola, compare with the given condition, and then obtain the conclusion. The asymptote equations of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{9}=1$ $(a>0)$ are $y=\\pm\\frac{3}{a}x$, i.e., $3x+ay=0$. $\\because$ the asymptote equations of the hyperbola $\\frac{x^{2}}{2}-\\frac{y^{2}}{9}=1$ $(a>0)$ are $3x\\pm2$. $\\therefore a=2$" }, { "text": "Let the ellipse $ C $: $\\frac{x^{2}}{100}+\\frac{y^{2}}{48}=1$ have left and right foci $ F_{1} $, $ F_{2} $ respectively. Point $ Q $ lies on ellipse $ C $ and satisfies $|Q F_{1}|=\\frac{2}{3}|Q F_{2}|$. Then the area of $ \\Delta Q F_{1} F_{2} $ is?", "fact_expressions": "C: Ellipse;Q: Point;F1: Point;F2: Point;Expression(C) = (x^2/100 + y^2/48 = 1);LeftFocus(C) =F1;RightFocus(C)=F2;PointOnCurve(Q, C);Abs(LineSegmentOf(Q, F1)) = (2/3)*Abs(LineSegmentOf(Q, F2))", "query_expressions": "Area(TriangleOf(Q, F1, F2))", "answer_expressions": "48", "fact_spans": "[[[1, 45], [76, 81]], [[71, 75]], [[54, 61]], [[63, 70]], [[1, 45]], [[1, 70]], [[1, 70]], [[71, 82]], [[86, 118]]]", "query_spans": "[[[120, 147]]]", "process": "Ellipse C: \\frac{x^{2}}{100}+\\frac{y^{2}}{48}=1, so a=10. {|QF_{1}|=\\frac{2}{3}|QF_{2}|, |QF_{1}|+|QF_{2}|=20, \\therefore |QF_{1}|=8, |QF_{2}|=12, also |F_{1}F_{2}|=4\\times(100-48)=208, \\therefore |QF_{1}|^{2}+|QF_{2}|^{2}=|F_{1}F_{2}|^{2}, \\therefore \\triangle QF_{1}F_{2} is a right triangle, \\therefore S_{\\Delta QF_{1}F_{2}}=\\frac{1}{2}\\times|QF_{1}|\\times|QF_{2}|=\\frac{1}{2}\\times8\\times12=48" }, { "text": "Given the parabola $C$: $x^{2}=2 p y(p>0)$, the distance from the focus $F$ to the directrix is $4$. The line $l$ passing through point $F$ and $R(m, 0)$ intersects the parabola $C$ at points $P$ and $Q$. If $\\overrightarrow{R P}=\\overrightarrow{P F}$, then $|P Q|$=?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*p*y);p: Number;p > 0;F: Point;Focus(C) = F;Distance(F, Directrix(C)) = 4;PointOnCurve(F, l) = True;PointOnCurve(R, l) = True;R: Point;Coordinate(R) = (m, 0);m: Number;l: Line;Intersection(C, l) = {P, Q};P: Point;Q: Point;VectorOf(R, P) = VectorOf(P, F)", "query_expressions": "Abs(LineSegmentOf(P, Q))", "answer_expressions": "9", "fact_spans": "[[[2, 28], [67, 73]], [[2, 28]], [[10, 28]], [[10, 28]], [[31, 34], [46, 50]], [[2, 34]], [[2, 44]], [[45, 66]], [[45, 66]], [[51, 60]], [[51, 60]], [[51, 60]], [[61, 66]], [[61, 84]], [[75, 78]], [[79, 82]], [[86, 129]]]", "query_spans": "[[[131, 140]]]", "process": "Given that the distance from the focus F to the directrix of the parabola C: x^{2}=2py (p>0) is 4, obtain the parabola equation x^{2}=8y. Then, using \\overrightarrow{RP}=\\overrightarrow{PF} and F(0,2), find the coordinates of point P, then derive the equation of line l, solve simultaneously with the parabola equation to find the coordinates of Q, and finally use the distance formula between two points to find the solution. (Detailed solution) Since the distance from the focus F to the directrix of the parabola C: x^{2}=2py (p>0) is 4, we have p=4, so the parabola equation is x^{2}=8y. Because \\overrightarrow{RP}=\\overrightarrow{PF} and F(0,2), the y-coordinate of point P is 1. Substituting into the parabola equation gives the x-coordinate of point P as \\pm2\\sqrt{2}. Without loss of generality, assume P(-2\\sqrt{2},1). Then k_{PF}=\\frac{2-1}{0-(-2\\sqrt{2})}=\\frac{\\sqrt{2}}{4}. Hence, the equation of the line is y=\\frac{\\sqrt{2}}{4}x+2^{x}. Substituting into x^{2}=8y yields x^{2}-2\\sqrt{2}x-16=0, giving Q(4\\sqrt{2},4). Therefore, |PQ|=\\sqrt{(-2\\sqrt{2}-4\\sqrt{2})^{2}+(1-4)^{2}}=9" }, { "text": "If the major axis of an ellipse is $2$ times the minor axis, and the ellipse passes through the point $(3,0)$, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (3, 0);MajorAxis(G) = 2*MinorAxis(G);PointOnCurve(H,G)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[1, 3], [28, 30]], [[18, 26]], [[18, 26]], [[1, 14]], [[1, 26]]]", "query_spans": "[[[28, 36]]]", "process": "Since the ellipse passes through the point (3,0), when the foci are on the x-axis, we know that a=3, b=\\frac{3}{2}, so c=\\sqrt{a^{2}-b^{2}}=\\frac{3\\sqrt{3}}{2}; therefore, when the foci are on the y-axis, similarly we get c=\\frac{\\sqrt{3}}{2}." }, { "text": "Given that $F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25} + \\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $| F_{2} A|+|F_{2} B|=12$, then $|AB|$=?", "fact_expressions": "G: Ellipse;H: Line;F2: Point;A: Point;B: Point;F1: Point;Expression(G) = (x^2/25 + y^2/9 = 1);Focus(G)={F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {A,B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[19, 59], [78, 80]], [[75, 77]], [[10, 18]], [[81, 84]], [[87, 90]], [[2, 9], [67, 74]], [[19, 59]], [[2, 64]], [[66, 77]], [[75, 92]], [[95, 120]]]", "query_spans": "[[[123, 131]]]", "process": "" }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $\\frac{1}{2}$. Then the minimum value of $\\frac{b^{2}+1}{3 a}$ is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 1/2", "query_expressions": "Min((b^2 + 1)/(3*a))", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[0, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 70]]]", "query_spans": "[[[73, 100]]]", "process": "" }, { "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ is $\\frac{1}{2}$, then the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ is?", "fact_expressions": "H: Ellipse;Expression(H) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(H) = 1/2;b: Number;a: Number;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "{sqrt(7)/2, sqrt(21)/3}", "fact_spans": "[[[2, 47]], [[2, 47]], [[2, 65]], [[4, 47]], [[4, 47]], [[67, 113]], [[67, 113]]]", "query_spans": "[[[67, 119]]]", "process": "" }, { "text": "The focus of the parabola $y^{2}=4x$ is $F$. A line passing through $F$ intersects the parabola at points $A$ and $B$. What is the minimum value of $|AF|+2|BF|$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);Focus(G) = F;F: Point;PointOnCurve(F,H) = True;H: Line;Intersection(H,G) = {A,B};A: Point;B: Point", "query_expressions": "Min(Abs(LineSegmentOf(A, F)) + 2*Abs(LineSegmentOf(B, F)))", "answer_expressions": "3+2*sqrt(2)", "fact_spans": "[[[0, 14], [31, 34]], [[0, 14]], [[0, 21]], [[18, 21], [23, 26]], [[22, 29]], [[27, 29]], [[27, 44]], [[35, 38]], [[39, 42]]]", "query_spans": "[[[46, 66]]]", "process": "The focus F of the parabola y^{2}=4x has coordinates (1,0), and the equation of the directrix is x=-1. Let the line passing through F be y=k(x-1), and let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system \\begin{cases}y=k(x-1)\\\\y^{2}=4x\\end{cases}, simplifying yields k^{2}x^{2}-(2k^{2}+4)x+k^{2}=0, then x_{1}+x_{2}=\\frac{2k^{2}+4}{k^{2}}, x_{1}x_{2}=1. According to the definition of the parabola, |AF|=x_{1}+1, |BF|=x_{2}+1. \\frac{1}{|AF|}+\\frac{+1}{+1}=\\frac{x_{1}+x_{2}+2}{x_{1}+x_{2}+x_{1}x_{2}+}\\frac{1+x_{2}+2}{x_{2}+x_{1}x_{2}+1}=1 if and only if \\frac{2|BF|}{|AF|}=\\frac{|AF|}{|BF|}, that is, |AF|=1+\\sqrt{2}, |l\\because|AF|+2|BF|=(|AF|+2|BF|(\\frac{1}{|AF|}+\\frac{1}{|BF|})=3+\\frac{2|BF|}{|AF|}+\\frac{|AF|}{|BF|}\\geqslant3+2\\sqrt{2}B|=\\frac{\\sqrt{2}(1+\\sqrt{2})}{2} when, \\therefore the minimum value of |AF|+2|BF| is 3+2\\sqrt{2}." }, { "text": "If the equation $\\frac{x^{2}}{m}-\\frac{y^{2}}{2 m-3}=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 - 3) + x^2/m = 1)", "query_expressions": "Range(m)", "answer_expressions": "(0,1)+(1,3/2)", "fact_spans": "[[[44, 46]], [[48, 53]], [[1, 46]]]", "query_spans": "[[[48, 60]]]", "process": "According to the problem, the equation represents an ellipse. Set up the system of inequalities and solve it. The equation $\\frac{x^{2}}{m}-\\frac{y^{2}}{2m-3}=1$ represents an ellipse, that is, the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{-(2m-3)}=1$ represents an ellipse. Then \n$$\n\\begin{cases}\nm>0\\\\\n-(2m-3)>0\\\\\nm\\neq-(2m-3)\n\\end{cases}\n$$\nSolving gives $00 \\\\ -m-1>0 \\end{cases} $ we get $ m<-1 $, hence $ m\\in(-\\infty,-1) $." }, { "text": "The moving chord $AB$ of the parabola $y^{2}=2 p x(p>0)$ has length $a(a \\geq 2 p)$. What is the minimum distance from the midpoint $M$ of chord $AB$ to the $y$-axis?", "fact_expressions": "G: Parabola;p: Number;a:Number;A: Point;B: Point;M: Point;p>0;a>=2*p;Expression(G) = (y^2 = 2*(p*x));IsChordOf(LineSegmentOf(A,B),G);Length(LineSegmentOf(A,B))=a;MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Min(Distance(M, yAxis))", "answer_expressions": "a/2-p/2", "fact_spans": "[[[0, 21]], [[3, 21]], [[31, 46]], [[24, 29]], [[24, 29]], [[57, 60]], [[3, 21]], [[31, 46]], [[0, 21]], [[0, 29]], [[24, 46]], [[49, 60]]]", "query_spans": "[[[57, 72]]]", "process": "" }, { "text": "Given that point $A(-2,0)$ lies on the directrix of the parabola $C$: $y^{2}=2 p x$, then $p$=?", "fact_expressions": "C: Parabola;p: Number;A: Point;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (-2, 0);PointOnCurve(A, Directrix(C))", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[13, 34]], [[40, 43]], [[2, 12]], [[13, 34]], [[2, 12]], [[2, 38]]]", "query_spans": "[[[40, 45]]]", "process": "According to the problem, the equation of the directrix of the parabola is $ x = -\\frac{p}{2} $, so $ -\\frac{p}{2} = -2 $, yielding $ p = 4 $." }, { "text": "The line $y = kx - 2$ intersects the parabola $y^2 = 8x$ at points $A$ and $B$. If the abscissa of the midpoint of $AB$ is $2$, then $|AB| = $?", "fact_expressions": "H: Line;Expression(H) = (y = k*x - 2);k: Number;G: Parabola;Expression(G) = (y^2 = 8*x);A: Point;B: Point;Intersection(H, G) = {A, B};XCoordinate(MidPoint(LineSegmentOf(A, B))) = 2", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "2*sqrt(15)", "fact_spans": "[[[0, 11]], [[0, 11]], [[2, 11]], [[12, 26]], [[12, 26]], [[27, 30]], [[31, 34]], [[0, 36]], [[38, 52]]]", "query_spans": "[[[55, 64]]]", "process": "" }, { "text": "It is known that the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ coincides with the focus of the parabola $y^{2}=-12x$. Then, the equation of the right directrix of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (y^2 = -12*x);LeftFocus(G) = Focus(H)", "query_expressions": "Expression(RightDirectrix(G))", "answer_expressions": "x=8/3", "fact_spans": "[[[2, 34], [61, 64]], [[5, 34]], [[38, 54]], [[2, 34]], [[38, 54]], [[2, 59]]]", "query_spans": "[[[61, 72]]]", "process": "\\because the focus of the parabola y^{2}=-12x is (-3,0), \\therefore the foci of the hyperbola are (-3,0) and (3,0), \\therefore for the hyperbola, c=3. \\because c^{2}=a^{2}+b^{2}, b=1, \\therefore a=2\\sqrt{2}, \\therefore the equation of the right directrix of the hyperbola is x=\\frac{a^{2}}{c}=\\frac{8}{3}" }, { "text": "Given a line $ l $ with slope $ 1 $ and positive $ y $-intercept $ b $ that intersects the circle $ C: x^{2} + y^{2} = 4 $ at points $ A $ and $ B $, and $ O $ is the origin. If the area of $ \\triangle AOB $ is $ \\sqrt{3} $, then $ b = $?", "fact_expressions": "l: Line;C: Circle;A: Point;O: Origin;B: Point;b: Number;Expression(C) = (x^2 + y^2 = 4);Slope(l)=1;b>0;Intercept(l,yAxis)=b;Intersection(l, C) = {A, B};Area(TriangleOf(A, O, B)) = sqrt(3)", "query_expressions": "b", "answer_expressions": "{sqrt(6),sqrt(2)}", "fact_spans": "[[[25, 30]], [[31, 52]], [[54, 57]], [[64, 67]], [[58, 61]], [[19, 22], [107, 110]], [[31, 52]], [[2, 30]], [[19, 24]], [[10, 30]], [[25, 63]], [[74, 105]]]", "query_spans": "[[[107, 112]]]", "process": "According to the problem, the equation of line $ l $ is $ y = x + b $, the center of circle $ C $ is $ C(0,0) $, and the radius is $ r = 2 $. Using the point-to-line distance formula, $ \\frac{1}{2} \\cdot 2\\sqrt{4 - \\frac{b^{2}}{2}} \\cdot \\frac{|b|}{\\sqrt{3}} = \\sqrt{3} $. Given $ b > 0 $, solving yields $ b = \\sqrt{6} $ or $ \\sqrt{2} $." }, { "text": "Let the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0)$ have left and right foci $F_{1}$ and $F_{2}$ respectively, with eccentricity $\\sqrt{5}$. $P$ is a point on $C$ such that $F_{1} P \\perp F_{2} P$. If the area of $\\triangle P F_{1} F_{2}$ is $8$, then $a=?$", "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;P: 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);IsPerpendicular(LineSegmentOf(F1, P), LineSegmentOf(F2, P));Area(TriangleOf(P, F1, F2)) = 8;Eccentricity(C)=sqrt(5)", "query_expressions": "a", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 64], [108, 111]], [[9, 64]], [[176, 179]], [[73, 80]], [[104, 107]], [[81, 88]], [[9, 64]], [[9, 64]], [[1, 64]], [[1, 88]], [[1, 88]], [[104, 114]], [[116, 139]], [[142, 174]], [[1, 103]]]", "query_spans": "[[[176, 181]]]", "process": "Using the definition of a hyperbola, the area of a triangle, and the eccentricity of the hyperbola, transform and solve for a. Without loss of generality, assume P is a point on the left branch of the hyperbola. According to the problem, let |PF_{2}| = m, |PF_{1}| = n, then we have m - n = 2a, \\frac{1}{2}mn = 8, m^{2} + n^{2} = 4c^{2}, e = \\frac{c}{a} = \\sqrt{5}. It follows that m^{2} + n^{2} - 2mn = 4a^{2}, i.e., 4c^{2} - 32 = 4a^{2}, \\therefore 20a^{2} - 32 = 4a^{2}, thus a^{2} = 2, then a = \\sqrt{2}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, if a line passing through the right focus $F$ with an inclination angle of $30^{\\circ}$ intersects the right branch of the hyperbola at two points, then what is the range of values for the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(30, degree);NumIntersection(H, RightPart(G)) = 2", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,(2/3)*sqrt(3))", "fact_spans": "[[[2, 58], [88, 91], [103, 106]], [[5, 58]], [[5, 58]], [[85, 87]], [[64, 67]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 67]], [[60, 87]], [[68, 87]], [[85, 100]]]", "query_spans": "[[[103, 116]]]", "process": "" }, { "text": "Given that $P$ is a moving point on the parabola $y^{2}=12x$, and $Q$ is a moving point on the circle $x^{2}+(y-4)^{2}=1$, what is the minimum value of the sum of the distance from point $P$ to point $Q$ and the distance from point $P$ to the line $x=-3$?", "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = 12*x);PointOnCurve(P, G) = True;Q: Point;H: Circle;Expression(H) = (x^2 + (y - 4)^2 = 1);PointOnCurve(Q, H) = True;I: Line;Expression(I) = (x = -3)", "query_expressions": "Min(Distance(P, Q) + Distance(P, I))", "answer_expressions": "4", "fact_spans": "[[[2, 5], [59, 63], [72, 76]], [[6, 21]], [[6, 21]], [[2, 26]], [[27, 30], [64, 68]], [[31, 51]], [[31, 51]], [[27, 56]], [[77, 85]], [[77, 85]]]", "query_spans": "[[[59, 96]]]", "process": "The focus of the parabola y^{2}=12x is F(3,0), the equation of the directrix is x=-3, the center of the circle x^{2}+(y-4)^{2}=1 is E(0,4), and the radius is 1. According to the definition of a parabola, the distance from point P to the directrix x=-3 is equal to the distance from point P to the focus F(3,0). Thus, when points P, Q, and F are collinear, the minimum value of the sum of the distance from point P to point Q and the distance from point P to the line x=-3 is |QF|=|EF|-r=\\sqrt{3^{2}+4^{2}}-1=4. Therefore, the answer is: 4." }, { "text": "The vertex of the parabola is at the origin, the axis of symmetry is the $x$-axis, and the point $(-5,2 \\sqrt{5})$ lies on the parabola. Find the equation of the parabola?", "fact_expressions": "O: Origin;G: Parabola;Vertex(G) = O;H: Point;Coordinate(H) = (-5, 2*sqrt(5));PointOnCurve(H, G);SymmetryAxis(G) = xAxis", "query_expressions": "Expression(G)", "answer_expressions": "y^2=-4*x", "fact_spans": "[[[7, 9]], [[0, 3], [38, 41], [44, 47]], [[0, 9]], [[19, 37]], [[19, 37]], [[19, 42]], [[0, 18]]]", "query_spans": "[[[44, 52]]]", "process": "Let the standard equation of the parabola be y^{2}=-2px. Substituting the point (-5,2\\sqrt{5}) gives (2\\sqrt{5})^{2}=-2\\times(-5)p, solving for p yields: p=2. The standard equation of the parabola is y^{2}=-4x." }, { "text": "Given that $AB$ is a chord of the parabola $y^{2}=2x$ passing through its focus, $|AB|=4$, then the horizontal coordinate of the midpoint of $AB$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);PointOnCurve(Focus(G), LineSegmentOf(A, B));Abs(LineSegmentOf(A, B)) = 4", "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "3/2", "fact_spans": "[[[9, 23]], [[9, 23]], [[2, 7]], [[2, 7]], [[2, 27]], [[2, 27]], [[28, 37]]]", "query_spans": "[[[39, 52]]]", "process": "Problem Analysis: Since the distance from any point on the parabola to the focus is equal to the distance to the directrix, let the distances from A and B to the directrix be d_{1}, d_{2}. Therefore, d_{1}+d_{2}=x_{A}+x_{B}+1=4, so x_{A}+x_{B}=3. Hence, the horizontal coordinate of the midpoint of AB is x_{0}=\\frac{x_{A}+x_{B}}{2}=\\frac{3}{2}" }, { "text": "If the line $\\frac{x}{a}+\\frac{y}{2-a}=1$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively, then what is the equation of the locus of the midpoint of $AB$?", "fact_expressions": "G: Line;a: Number;A: Point;B: Point;Expression(G) = (y/(2 - a) + x/a = 1);Intersection(G, xAxis) = A;Intersection(G, yAxis) = B", "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "(x + y = 1)&(Negation(x = 0))&(Negation(x = 2))", "fact_spans": "[[[1, 32]], [[3, 32]], [[48, 51]], [[52, 55]], [[1, 32]], [[1, 55]], [[1, 55]]]", "query_spans": "[[[57, 72]]]", "process": "A(a,0), B(0,2-a), where a≠0, 2-a≠0, that is, a≠0 and a≠2. Let the midpoint of AB be Q(x₀,y₀), then x₀=a/2, y₀=(2-a)/2, so x₀+y₀=a/2+(2-a)/2=1, that is, x+y=1 (x≠0 and x≠1)." }, { "text": "Given point $A(1,0)$, $M$ and $N$ are moving points on the $x$-axis and $y$-axis respectively, satisfying $\\overrightarrow{A N} \\cdot \\overrightarrow{M N}=0$. If point $P$ satisfies $\\overrightarrow{M P}=2 \\overrightarrow{N P}$, then what is the trajectory equation of point $P$?", "fact_expressions": "A: Point;Coordinate(A) = (1, 0);M: Point;PointOnCurve(M, xAxis) ;N: Point;PointOnCurve(N, yAxis) ;DotProduct(VectorOf(A, N), VectorOf(M, N)) = 0;P: Point;VectorOf(M, P) = 2*VectorOf(N, P)", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2 = 4*x", "fact_spans": "[[[2, 12]], [[2, 12]], [[14, 17]], [[14, 39]], [[20, 23]], [[14, 39]], [[43, 94]], [[97, 101], [150, 154]], [[103, 148]]]", "query_spans": "[[[150, 161]]]", "process": "Let point M have coordinates (a,0), point N have coordinates (0,b), and point P have coordinates (x,y). Then \\overrightarrow{AN}=(-1,b), \\overrightarrow{MN}=(-a,b), \\therefore \\overrightarrow{AN}\\cdot\\overrightarrow{MN}=a+b^{2}=0 \\Rightarrow a=-b^{2}. While \\overrightarrow{MP}=(x-a,y), \\overrightarrow{NP}=(x,y-b), \\overrightarrow{MP}=2\\overrightarrow{NP} \\Rightarrow \\begin{cases} 2x=x-a \\\\ 2(y-b)=y \\end{cases} \\Rightarrow \\begin{cases} x=-a \\\\ y=2b \\end{cases}, substituting a=-b^{2} yields y^{2}=4x" }, { "text": "The distance from a point $A(2 \\sqrt{2}, 2)$ on the parabola $x^{2}=4 y$ to the focus is?", "fact_expressions": "G: Parabola;A: Point;Expression(G) = (x^2 = 4*y);Coordinate(A) = (2*sqrt(2), 2);PointOnCurve(A,G)", "query_expressions": "Distance(A, Focus(G))", "answer_expressions": "3", "fact_spans": "[[[0, 14]], [[17, 35]], [[0, 14]], [[17, 35]], [[0, 35]]]", "query_spans": "[[[0, 43]]]", "process": "Since $(2\\sqrt{2})^{2} = 4 \\times 2$, the point $A(2\\sqrt{2}, 2)$ lies on the parabola. The equation of the directrix of the parabola is $y = -1$, so the distance from point $A(2\\sqrt{2}, 2)$ to the focus is $2 - (-1) = 3$. The answer is $3$." }, { "text": "The distance from the focus $F$ of the parabola $x^{2}=2 y$ to the directrix $l$ is?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 2*y);F: Point;Focus(C) = F;l: Line;Directrix(C) = l", "query_expressions": "Distance(F, l)", "answer_expressions": "", "fact_spans": "[[[0, 3]], [[0, 15]], [[17, 23]], [[0, 23]], [[25, 31]], [[0, 31]]]", "query_spans": "[[[0, 37]]]", "process": "" }, { "text": "The standard equation of an ellipse with eccentricity $e=\\frac{\\sqrt{5}}{3}$ and a directrix $x=3$ is?", "fact_expressions": "G: Ellipse;e:Number;Eccentricity(G)=e;e=sqrt(5)/3;Expression(OneOf(Directrix(G)))=(x=3)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5+9*y^2/20=1", "fact_spans": "[[[38, 40]], [[3, 26]], [[3, 26]], [[3, 25]], [[27, 40]]]", "query_spans": "[[[38, 47]]]", "process": "" }, { "text": "A moving circle is tangent to the line $x=5$ and externally tangent to the circle $x^{2}+y^{2}+2 x-15=0$. What is the equation of the locus of the center of the moving circle?", "fact_expressions": "G: Circle;H: Line;Z: Circle;Expression(G) = (2*x + x^2 + y^2 - 15 = 0);Expression(H) = (x = 5);IsTangent(Z, H);IsOutTangent(Z, G)", "query_expressions": "LocusEquation(Center(Z))", "answer_expressions": "(y^2 = 80 - 20*x)&(x <= 4)", "fact_spans": "[[[17, 40]], [[5, 12]], [[2, 4], [44, 46]], [[17, 40]], [[5, 12]], [[2, 14]], [[2, 42]]]", "query_spans": "[[[44, 55]]]", "process": "Let the coordinates of the center of the moving circle be P(x, y), and its radius be r. \nx^{2}+y^{2}+2x-15=0 \\Rightarrow (x+1)^{2}+y^{2}=16, the radius of this circle is 4, and the center has coordinates M(-1, 0). \nSince the moving circle is tangent to the line x=5, |x-5|=r. \nAlso, since the moving circle is externally tangent to x^{2}+y^{2}+2x-15=0, then PM=4+r. \n\\sqrt{(x+1)^{2}+y^{2}}=4+|x-5|. \nWhen x<5, \\sqrt{(x+1)^{2}+y^{2}}=4-(x-5) \\Rightarrow (x+1)^{2}+y^{2}=4+5-x \\Rightarrow (x+1)^{2}+y^{2}=9-x, which clearly does not hold. \nWhen x\\geqslant5, \\sqrt{(x+1)^{2}+y^{2}}=4+x-5 \\Rightarrow y^{2}=-4x', which clearly does not hold." }, { "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ and the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ have the same foci, and the eccentricity of the hyperbola is twice the eccentricity of the ellipse, 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;H: Ellipse;Expression(H) = (x^2/16 + y^2/9 = 1);Focus(G) = Focus(H);Eccentricity(G) = Eccentricity(H)*2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2/3=1", "fact_spans": "[[[2, 58], [105, 108], [123, 126]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[59, 97], [113, 115]], [[59, 97]], [[2, 103]], [[105, 121]]]", "query_spans": "[[[123, 131]]]", "process": "From the given condition, the coordinates of the foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are $(\\sqrt{7},0)$, $(-\\sqrt{7},0)$, so $c=\\sqrt{7}$; and the eccentricity of the hyperbola is $2\\times\\frac{\\sqrt{7}}{4}=\\frac{\\sqrt{7}}{2}=\\frac{c}{a} \\Rightarrow a=2$. Then $b^{2}=c^{2}-a^{2}=3$. The equation of the hyperbola is $\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$." }, { "text": "The coordinates of the foci of the ellipse $9 x^{2}+4 y^{2}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (9*x^2 + 4*y^2 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,pm*sqrt(5)/6)", "fact_spans": "[[[0, 21]], [[0, 21]]]", "query_spans": "[[[0, 27]]]", "process": "" }, { "text": "The point $P(8, 1)$ bisects a chord of the ellipse $x^{2} + 4y^{2} = 4$. Then the equation of the line containing this chord is?", "fact_expressions": "G: Ellipse;H: LineSegment;P: Point;Expression(G) = (x^2 + 4*y^2 = 4);Coordinate(P) = (8, 1);MidPoint(H)=P;IsChordOf(H,G)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "2*x+y-17=0", "fact_spans": "[[[13, 32]], [], [[0, 11]], [[13, 32]], [[0, 11]], [[0, 36]], [[13, 36]]]", "query_spans": "[[[40, 50]]]", "process": "The equation of the ellipse can be written as \\frac{x^{2}}{4}+y^{2}=1. Let A(x_{1},y_{1}), B(x_{2},y_{2}) be points on the ellipse, and P be the midpoint of chord AB. Then \\begin{cases}\\frac{x_{1}^{2}}{4}+y_{1}^{2}=1\\\\\\frac{x_{2}^{2}}{4}+y_{2}^{2}=1\\end{cases}. Subtracting these two equations and simplifying yields -\\frac{1}{4}=\\frac{y_{1}+y_{2}}{x_{1}+x_{2}}\\cdot\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, that is, -\\frac{1}{4}=\\frac{1}{8}\\cdot\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-2. Therefore, the equation of the line containing chord AB is y-1=-2(x-8), i.e., 2x+y-17=0." }, { "text": "$F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, point $P$ moves on the ellipse, then the maximum value of $|P F_{1}| \\cdot |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": "Max(Abs(LineSegmentOf(P, F1))*Abs(LineSegmentOf(P, F2)))", "answer_expressions": "4", "fact_spans": "[[[17, 44], [56, 58]], [[51, 55]], [[0, 7]], [[9, 16]], [[17, 44]], [[0, 50]], [[0, 50]], [[51, 59]]]", "query_spans": "[[[63, 95]]]", "process": "" }, { "text": "If point $P(x, y)$ is any point on the curve $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, then the minimum value of $2x+\\sqrt{3}y$ is?", "fact_expressions": "P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;G: Curve;PointOnCurve(P, G);Expression(G) = (x^2/4 + y^2/3 = 1)", "query_expressions": "Min(2*x1 + sqrt(3)*y1)", "answer_expressions": "-5", "fact_spans": "[[[1, 12]], [[1, 12]], [[2, 12]], [[2, 12]], [[13, 50]], [[1, 55]], [[13, 50]]]", "query_spans": "[[[57, 80]]]", "process": "" }, { "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{m-2}=1$ represents a hyperbola, then what is the range of values for $m$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/m +y^2/(m-2) = 1);m: Number", "query_expressions": "Range(m)", "answer_expressions": "(0,2)", "fact_spans": "[[[43, 46]], [[2, 46]], [[49, 52]]]", "query_spans": "[[[49, 59]]]", "process": "From the given condition, we have m(m-2)<0, solving yields 00, b>0)$, and let $F_{1}$, $F_{2}$ be the left and right foci of the hyperbola, respectively. Given that $\\angle F_{1} P F_{2}=90^{\\circ}$ and $|P F_{1}|=2|P F_{2}|$, find the eccentricity of the hyperbola.", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(90, degree);Abs(LineSegmentOf(P, F1)) = 2*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[6, 62], [85, 88], [158, 161]], [[9, 62]], [[9, 62]], [[67, 74]], [[1, 5]], [[75, 82]], [[9, 62]], [[9, 62]], [[6, 62]], [[1, 66]], [[67, 94]], [[67, 94]], [[97, 131]], [[134, 156]]]", "query_spans": "[[[158, 167]]]", "process": "By the definition of a hyperbola, we have 2a = |PF_{1}| - |PF_{2}| = |PF_{2}|, hence |PF_{1}| = 4a. By the Pythagorean theorem, we have |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2}, that is, (4a)^{2} + (2a)^{2} = (2c)^{2}, which gives c = \\sqrt{5}a. Therefore, the eccentricity of this hyperbola is e = \\frac{c}{a} = \\sqrt{5}." }, { "text": "Given that $P$ is a moving point on the parabola $y^{2}=4x$, $A(2, \\sqrt{15})$, if the distance from point $P$ to the $y$-axis is $d_{1}$, and the distance from point $P$ to point $A$ is $d_{2}$, then the minimum value of $d_{1}+d_{2}$ is?", "fact_expressions": "G: Parabola;A: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (2, sqrt(15));PointOnCurve(P, G);d1:Number;d2:Number;Distance(P, yAxis) = d1;Distance(P, A) = d2", "query_expressions": "Min(d1 + d2)", "answer_expressions": "3", "fact_spans": "[[[6, 20]], [[70, 74], [25, 42]], [[2, 5], [44, 48], [65, 69]], [[6, 20]], [[25, 42]], [[2, 24]], [[57, 64]], [[78, 85]], [[44, 64]], [[65, 85]]]", "query_spans": "[[[87, 106]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$ $(m>0)$ has eccentricity $e=\\frac{\\sqrt{10}}{5}$, then the value of $m$ is?", "fact_expressions": "G: Ellipse;m: Number;m>0;e: Number;Expression(G) = (x^2/5 + y^2/m = 1);Eccentricity(G) = e;e = sqrt(10)/5", "query_expressions": "m", "answer_expressions": "25/3, 3", "fact_spans": "[[[2, 44]], [[73, 76]], [[4, 44]], [[48, 71]], [[2, 44]], [[2, 71]], [[48, 71]]]", "query_spans": "[[[73, 80]]]", "process": "The equation of the ellipse is given by \\frac{x^2}{5} + \\frac{y^2}{m} = 1 (m > 0 and m \\neq 5). When the foci are on the x-axis, i.e., 0 < m < 5, we have a = \\sqrt{5}, b = \\sqrt{m}, then c = \\sqrt{5 - m}. According to the problem, \\frac{\\sqrt{5 - m}}{\\sqrt{5}} = \\frac{\\sqrt{10}}{5}, solving gives m = 3. When the foci are on the y-axis, i.e., m > 5, we have a = \\sqrt{m}, b = \\sqrt{5}, then c = \\sqrt{m - 5}. According to the problem, \\frac{\\sqrt{m - 5}}{\\sqrt{m}} = \\frac{\\sqrt{10}}{5}, solving gives m = \\frac{25}{2}, so the value of m is 3 or \\frac{25}{2}." }, { "text": "Given $F_{1}(-5,0)$, $F_{2}(5,0)$, and a moving point $P$ on the hyperbola such that the difference of distances from $P$ to $F_{1}$ and $F_{2}$ is $6$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-5, 0);Coordinate(F2) = (5, 0);PointOnCurve(P,H);Distance(P,F1)-Distance(P,F2)=6;P:Point", "query_expressions": "Expression(G)", "answer_expressions": "(x^2/9-y^2/16=1)&(x>0)", "fact_spans": "[[[30, 33], [66, 69]], [[2, 15], [41, 48]], [[16, 29], [49, 56]], [[2, 15]], [[16, 29]], [[30, 40]], [[37, 64]], [37, 39]]", "query_spans": "[[[66, 74]]]", "process": "" }, { "text": "Given that the focus $F$ of the parabola $y^{2}=4x$ is one focus of an ellipse $C$ centered at the origin, $Q$ is the other focus of the ellipse $C$, the eccentricity of the ellipse $C$ is $\\frac{\\sqrt{2}}{2}$, $P$ is a point on the ellipse $C$, and $\\angle FPQ=60^{\\circ}$, then the area of $\\Delta FPQ$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;O: Origin;Center(C) = O;C: Ellipse;OneOf(Focus(C)) = F;Q: Point;OneOf(Focus(C)) = Q;Negation(F = Q);Eccentricity(C) = sqrt(2)/2;P: Point;PointOnCurve(P, C) = True;AngleOf(F, P, Q) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F, P, Q))", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 16]], [[2, 16]], [[19, 22]], [[2, 22]], [[26, 28]], [[23, 34]], [[29, 34], [44, 49], [56, 61], [92, 97]], [[19, 39]], [[40, 43]], [[40, 55]], [[19, 55]], [[56, 86]], [[88, 91]], [[88, 101]], [[103, 128]]]", "query_spans": "[[[130, 149]]]", "process": "\\because F(1,0), \\frac{c}{a} = \\frac{\\sqrt{2}}{2}, \\therefore a = \\sqrt{2}, b = 1, the equation of the ellipse is \\frac{x^{2}}{2} + y^{2} = 1. Let |PF| = m, |PQ| = n, then m + n = 2a = 2\\sqrt{2}. In \\triangle PQF, by the cosine law, we have m^{2} + n^{2} - 2mn\\cos60^{\\circ} = 4, \\therefore mn = \\frac{4}{3}, \\therefore s_{\\Delta FPQ} = \\frac{1}{2}mn\\sin60^{\\circ} = \\frac{\\sqrt{3}}{3}" }, { "text": "Let the foci of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ be $F_{1}$, $F_{2}$, and let $P$ be a point on the ellipse such that $\\angle F_{1} P F_{2}=\\frac{\\pi}{3}$. If the circumradius and inradius of $\\Delta F_{1} P F_{2}$ are $R$ and $r$ respectively, and $R=4r$, then what is the eccentricity of the ellipse?", "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) ;AngleOf(F1, P, F2) = pi/3;R:Number;r:Number;Radius(CircumCircle(TriangleOf(F1,P,F2)))=R;Radius(InscribedCircle(TriangleOf(F1,P,F2)))=r;R=4*r", "query_expressions": "Eccentricity(G)", "answer_expressions": "2/3", "fact_spans": "[[[1, 53], [77, 79], [177, 179]], [[3, 53]], [[3, 53]], [[57, 64]], [[73, 76]], [[65, 72]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 72]], [[73, 82]], [[84, 120]], [[158, 161]], [[163, 166]], [[122, 166]], [[122, 166]], [[168, 175]]]", "query_spans": "[[[177, 185]]]", "process": "In $\\triangle F_{1}PF_{2}$, using the Law of Sines: $2R = \\frac{|F_{1}F_{2}|}{\\sin\\angle F_{1}PF_{2}}$, we obtain $R = \\frac{\\sqrt[2]{3}}{3}c'$, $r = \\frac{1}{4}R = \\frac{\\sqrt{3}}{6}c'$. Let $|PF_{1}| = m$, $|PF_{2}| = n$, then use the Law of Cosines to find $mn$, and then solve using $S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2}mn\\sin\\frac{\\pi}{3} = \\frac{1}{2}(m+n+2c)r$. The foci of the ellipse are $F_{1}(-c,0)$, $F_{2}(c,0)$, $|F_{1}F_{2}| = 2c$. In $\\triangle F_{1}PF_{2}$, by the Law of Sines: $2R = \\frac{|F_{1}F_{2}|}{\\sin\\angle F_{1}PF_{2}} = \\frac{2c}{\\sin\\frac{\\pi}{3}} = \\frac{4\\sqrt{3}}{3}c$, solving gives $R = \\frac{2\\sqrt{3}}{3}c'$, $r = \\frac{1}{4}R = \\frac{\\sqrt{3}}{6}c'$. Let $|PF_{1}| = m$, $|PF_{2}| = n$. In $\\triangle F_{1}PF_{2}$, by the Law of Cosines: $4c^{2} = m^{2} + n^{2} - 2mn\\cos\\frac{\\pi}{3} = (m+n)^{2} - 3mn$, solving gives $mn = \\frac{4(a^{2}-c^{2})}{3}$. Therefore, $S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2}mn\\sin\\frac{\\pi}{3} = \\frac{\\sqrt{3}(a^{2}-c^{2})}{3}$, and $S_{\\Delta F_{1}PF_{2}} = \\frac{1}{2}(m+n+2c)r = \\frac{\\sqrt{3}c(a+c)}{6}$. Hence, $\\frac{\\sqrt{3}(a^{2}-c^{2})}{3} = \\frac{\\sqrt{3}c(a+c)}{6}$, simplifying yields $2a^{2} - 3c^{2} - ac = 0$, i.e., $3e^{2} + e - 2 = 0$, solving gives $e = \\frac{2}{3}$ or $e = -1$ (discarded)." }, { "text": "Through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$), a line with inclination angle $30^{\\circ}$ intersects the parabola at points $A$ and $B$. If the length of segment $AB$ is $8$, then $p=$?", "fact_expressions": "G: Parabola;p: Number;l: Line;B: Point;A: Point;F: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;PointOnCurve(F,l);Inclination(l)=ApplyUnit(30,degree);Intersection(l, G) = {A, B};Length(LineSegmentOf(A,B))=8", "query_expressions": "p", "answer_expressions": "1", "fact_spans": "[[[1, 22]], [[79, 82]], [[46, 48]], [[57, 60]], [[53, 56]], [[25, 28]], [[4, 22]], [[1, 22]], [[1, 28]], [[0, 48]], [[29, 48]], [[46, 62]], [[64, 77]]]", "query_spans": "[[[79, 84]]]", "process": "According to the problem, the equation of the line passing through the focus with an inclination angle of 30^{\\circ} is y=\\frac{\\sqrt{3}}{3}(x-\\frac{p}{2}). Solving the system \\begin{cases}y^{2}=2px\\\\y=\\frac{\\sqrt{3}}{3}(x-\\frac{p}{2})\\end{cases} gives: \\Rightarrow x^{2}-7px+\\frac{p^{2}}{4}=0. \\therefore x_{1}+x_{2}=7p, x_{1}x_{2}=\\frac{p^{2}}{4}, \\therefore |x_{1}-x_{2}|=\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{(7p)^{2}-4\\times\\frac{p^{2}}{4}}=4\\sqrt{3}p. \\therefore |AB|=\\sqrt{1+\\frac{\\sqrt{3}}{3}}|x_{1}-x_{2}|=\\frac{2\\sqrt{3}}{3}\\times4\\sqrt{3}p=8. Solving gives: p=1. Therefore, the correct answer is: 1" }, { "text": "If the line $ l $ passes through the focus of the parabola $ y^{2} = 4x $, intersects the parabola at points $ A $ and $ B $, and the x-coordinate of the midpoint of segment $ AB $ is $ 2 $, then what is the length of segment $ AB $?", "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))) = 2", "query_expressions": "Length(LineSegmentOf(A,B))", "answer_expressions": "6", "fact_spans": "[[[1, 6]], [[8, 22], [27, 30]], [[32, 35]], [[36, 39]], [[8, 22]], [[1, 25]], [[1, 41]], [[43, 60]]]", "query_spans": "[[[62, 73]]]", "process": "According to the focal radius formula, the length of the focal chord is obtained; calculate as follows. [Detailed solution] Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=2\\times2=4. In the parabola, 2p=4, p=2, so |AB|=x_{1}+\\frac{p}{2}+x_{2}+\\frac{p}{2}=x_{1}+x_{2}+p=4+2=6." }, { "text": "Given that the line $x - y = 2$ intersects the parabola $y^2 = 4x$ at points $A$ and $B$, what are the coordinates of the midpoint of segment $AB$?", "fact_expressions": "A: Point;B: Point;G: Parabola;H: Line;Expression(G) = (y^2 = 4*x);Expression(H) = (x - y = 2);Intersection(H, G) = {A,B}", "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "(4,2)", "fact_spans": "[[[29, 32]], [[35, 38]], [[12, 26]], [[2, 11]], [[12, 26]], [[2, 11]], [[2, 40]]]", "query_spans": "[[[43, 56]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $P$ is a point on the line $l$: $y=x+m$ $(m \\in \\mathbb{R})$, if the minimum value of $|P F_{1}|+|P F_{2}|$ is $4$, then the real number $m$=?", "fact_expressions": "C: Ellipse;l: Line;m: Real;P: Point;F1: Point;F2: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Expression(l) = (y = x + m);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, l);Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))) = 4", "query_expressions": "m", "answer_expressions": "pm*sqrt(7)", "fact_spans": "[[[18, 60]], [[70, 93]], [[129, 134]], [[66, 69]], [[2, 9]], [[10, 17]], [[18, 60]], [[70, 93]], [[2, 65]], [[2, 65]], [[66, 96]], [[98, 127]]]", "query_spans": "[[[129, 136]]]", "process": "According to the problem, the ellipse $ C: \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 $, then $ 2a = 4 $, $ a = 2 $. Also, since point $ P $ is on the line $ l: y = x + m $ ($ m \\in \\mathbb{R} $), if the minimum value of $ |PF_{1}| + |PF_{2}| $ is 4, then this line is tangent to the ellipse. From $ \\begin{cases} \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 \\\\ y = x + m \\end{cases} $, eliminating $ y $ and simplifying yields $ 7x^{2} + 8mx + 4m^{2} - 12 = 0 $. The discriminant $ \\Delta = 48(7 - m^{2}) = 0 $, solving gives $ m = \\pm\\sqrt{7} $." }, { "text": "The asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ is parallel to the line $2 x-y+1=0$, then its eccentricity 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: Line;Expression(H) = (2*x - y + 1 = 0);IsParallel(OneOf(Asymptote(G)), H) = True", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[0, 56], [80, 81]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[63, 76]], [[63, 76]], [[0, 78]]]", "query_spans": "[[[80, 87]]]", "process": "Since the asymptote is parallel to the line $2x - y + 1 = 0$, we have $\\frac{b}{a} = 2$. According to the formula for the eccentricity of a hyperbola, we get $e = \\sqrt{1 + \\left(\\frac{b}{a}\\right)^2} = \\sqrt{5}$." }, { "text": "The left focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ is $F$, and the line $x=t$ intersects the ellipse at points $M$ and $N$. When the perimeter of $\\Delta F M N$ is maximized, what is the area of $\\Delta F M N$?", "fact_expressions": "G: Ellipse;H: Line;t: Number;F: Point;M: Point;N: Point;Expression(G) = (x^2/5 + y^2/4 = 1);Expression(H) = (x = t);LeftFocus(G)=F;Intersection(H, G) = {M, N};WhenMax(Perimeter(TriangleOf(F,M,N)))", "query_expressions": "Area(TriangleOf(F, M, N))", "answer_expressions": "8*sqrt(5)/5", "fact_spans": "[[[0, 37], [54, 56]], [[46, 53]], [[48, 53]], [[42, 45]], [[59, 63]], [[64, 67]], [[0, 37]], [[46, 53]], [[0, 45]], [[46, 67]], [[68, 89]]]", "query_spans": "[[[90, 109]]]", "process": "Let the right focus of the ellipse be $ F_{2} $, then $ |MF_{2}| + |NF_{2}| \\geqslant |MN| $, with equality if and only if points $ M $, $ N $, $ F_{2} $ are collinear. Therefore, the perimeter of $ \\triangle FMN $, $ |MF| + |NF| + |MN| \\leqslant |MF| + |NF| + |MF_{2}| + |NF_{2}| = 4a = 4\\sqrt{5} $. At this time, $ |MN| = \\frac{2b^{2}}{a^{2}} = \\frac{8\\sqrt{5}}{5} $, so the area of $ \\triangle FMN $ is $ s = \\frac{1}{2} \\times \\frac{8\\sqrt{5}}{5} \\times 2c = \\frac{1}{2} \\times \\frac{8\\sqrt{5}}{5} \\times 2 = \\frac{8\\sqrt{5}}{5} $." }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ with two foci $F_{1}$ and $F_{2}$, and a point $P$ on this ellipse. If $|P F_{1}|-|P F_{2}|=2$, then what is the area of $\\Delta P F_{1}F_{2}$?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) - Abs(LineSegmentOf(P, F2)) = 2", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 39], [67, 69]], [[61, 65]], [[45, 52]], [[53, 60]], [[2, 39]], [[2, 60]], [[61, 70]], [[72, 95]]]", "query_spans": "[[[97, 123]]]", "process": "" }, { "text": "Given that the foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ are $F_{1}$ and $F_{2}$, and $A$ is a point on the ellipse, then $|A F_{1}|+|A F_{2}|$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);Focus(G) = {F1, F2};F1: Point;F2: Point;A: Point;PointOnCurve(A, G)", "query_expressions": "Abs(LineSegmentOf(A, F1)) + Abs(LineSegmentOf(A, F2))", "answer_expressions": "4", "fact_spans": "[[[2, 39], [65, 67]], [[2, 39]], [[2, 60]], [[45, 52]], [[53, 60]], [[61, 64]], [[61, 70]]]", "query_spans": "[[[72, 95]]]", "process": "Since the foci of the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1 are F_{1} and F_{2}, and A is a point on the ellipse, it follows that |AF_{1}|+|AF_{2}|=2a=4." }, { "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0 , b>0)$ intersects the line $y=2 x$, then the range of the eccentricity is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y = 2*x);IsIntersect(G, H)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(5),+oo)", "fact_spans": "[[[1, 58]], [[4, 58]], [[4, 58]], [[59, 68]], [[4, 58]], [[4, 58]], [[1, 58]], [[59, 68]], [[1, 71]]]", "query_spans": "[[[1, 83]]]", "process": "As shown in the figure, since the asymptotes of the hyperbola are given by $ y = \\pm\\frac{b}{a}x $, if the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0 $, $ b>0 $) intersects the line $ y = 2x $, then it must hold that $ \\frac{b}{a} > 2 $, hence $ \\frac{b^{2}}{a^{2}} > 4 $, $ \\frac{c^{2}-a^{2}}{a^{2}} > 4 $, solving yields $ e^{2} = \\frac{c^{2}}{a^{2}} > 5 $, $ e > \\sqrt{5} $." }, { "text": "If one asymptote of the hyperbola $C$: $\\frac{y^{2}}{m}-x^{2}=1$ $(m>0)$ is $m x+\\sqrt{3} y=0$, then what is the focal length of $C$?", "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2 + y^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(C))) = (m*x + sqrt(3)*y = 0)", "query_expressions": "FocalLength(C)", "answer_expressions": "4", "fact_spans": "[[[1, 39], [66, 69]], [[1, 39]], [[9, 39]], [[9, 39]], [[1, 64]]]", "query_spans": "[[[66, 74]]]", "process": "The asymptotes of hyperbola C are given by the equations $ y = \\pm\\sqrt{m}x $. From the problem, we have $ \\sqrt{m} = \\frac{m}{\\sqrt{3}} $. Since $ m > 0 $, solving gives $ m = 3 $. Thus, $ a = \\sqrt{3} $, $ b = 1 $, then $ c = \\sqrt{a^{2} + b^{2}} = 2 $. Therefore, the focal length of hyperbola C is $ 2c = 4 $." }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the ellipse. When $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, the area of $\\triangle F_{1} P F_{2}$ is?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 + y^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "1", "fact_spans": "[[[2, 29], [59, 61]], [[38, 45]], [[54, 58]], [[46, 53]], [[2, 29]], [[2, 53]], [[2, 53]], [[54, 62]], [[64, 123]]]", "query_spans": "[[[125, 155]]]", "process": "" }, { "text": "Given the hyperbola equation $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, let points $F_{1}$ and $F_{2}$ be its left and right foci, respectively, let point $A$ be on the circle $x^{2}+(y-5)^{2}=1$, and let point $M$ lie on the right branch of the hyperbola. Then the minimum value of $|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 - 5)^2 = 1);Expression(G) = (x^2/9 - y^2/16 = 1);PointOnCurve(A, H);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(M, RightPart(G))", "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F1)))", "answer_expressions": "5*sqrt(2)+5", "fact_spans": "[[[2, 5], [63, 64], [103, 106]], [[73, 93]], [[98, 102]], [[46, 54]], [[55, 62]], [[69, 72]], [[73, 93]], [[2, 45]], [[69, 97]], [[46, 68]], [[46, 68]], [[98, 110]]]", "query_spans": "[[[112, 135]]]", "process": "As shown in the figure, the coordinates of the right focus are F_{2}(5,0). Connect AF_{2}, MF_{2}. By the definition of the hyperbola, |MF_{1}|-|MF_{2}|=2a=6. Therefore, |MF_{1}|+|MA|=6+|MF_{2}|+|MA|\\geqslant6+|AF_{2}|. Since point A lies on the circle x^{2}+(y-5)^{2}=1, whose center is (0,5) and radius is 1, it follows that |AF_{2}|\\geqslant|CF_{2}|-1=5\\sqrt{2}-1. Hence, |MF_{1}|+|MA|\\geqslant6+|AF_{2}|\\geqslant5\\sqrt{2}+5. The equality holds when points M and A lie on the segment CF_{2}, so the minimum value of |MF_{1}|+|MA| is 5\\sqrt{2}+5." }, { "text": "On the parabola $y^{2}=12x$ with focus $F$, a point $M$ lies on the parabola. A perpendicular is drawn from $M$ to its directrix, with foot of perpendicular at $N$. If $|NF|=10$, then $|MF|=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 12*x);F: Point;Focus(G)=F;M: Point;PointOnCurve(M,G);L:Line;PointOnCurve(M,L);IsPerpendicular(Directrix(G),L);N: Point;FootPoint(Directrix(G),L)=N;Abs(LineSegmentOf(N, F)) = 10", "query_expressions": "Abs(LineSegmentOf(M, F))", "answer_expressions": "25/3", "fact_spans": "[[[8, 23], [30, 31]], [[8, 23]], [[4, 7]], [[1, 23]], [[26, 29]], [[8, 29]], [], [[0, 36]], [[0, 36]], [[40, 43]], [[0, 43]], [[45, 55]]]", "query_spans": "[[[57, 66]]]", "process": "" }, { "text": "Given that a line passing through the focus $F$ of the parabola $y^{2}=2 p x$ ($p>0$) intersects the parabola at points $A$ and $B$, and the projections of points $A$ and $B$ onto the directrix of the parabola are $A_{1}$ and $B_{1}$ respectively, then what is $\\angle A_{1} F B_{1}$?", "fact_expressions": "G: Parabola;p: Number;H: Line;A1: Point;F: Point;B1: Point;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*p*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Projection(A,Directrix(G))=A1;Projection(B,Directrix(G))=B1", "query_expressions": "AngleOf(A1,F,B1)", "answer_expressions": "pi/2", "fact_spans": "[[[3, 24], [33, 36], [57, 60]], [[6, 24]], [[30, 32]], [[67, 74]], [[26, 29]], [[76, 84]], [[37, 40], [47, 50]], [[41, 44], [51, 54]], [[6, 24]], [[3, 24]], [[3, 29]], [[2, 32]], [[30, 46]], [[47, 84]], [[47, 84]]]", "query_spans": "[[[85, 109]]]", "process": "From the definition of a parabola, we have $ AA_{1} = AF $, $ BB_{1} = BF $. Therefore, $ \\angle AFA_{1} = \\angle AA_{1}F $, $ \\angle BFB_{1} = \\angle BB_{1}F $. Since $ \\angle FAA_{1} + \\angle FBB_{1} = \\pi $, it follows that $ \\angle AFA_{1} + \\angle AA_{1}F + \\angle BFB_{1} + \\angle BB_{1}F = 2\\pi - (\\angle FAA_{1} + \\angle FBB_{1}) = \\pi $. Therefore, $ \\angle AFA_{1} + \\angle BFB_{1} = \\frac{\\pi}{2} \\Rightarrow \\angle A_{1}FB_{1} = \\frac{\\pi}{2} $." }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ is equal to $\\frac{\\sqrt{3}}{2}$, then $m=?$", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/4 + y^2/m = 1);Eccentricity(G) = sqrt(3)/2", "query_expressions": "m", "answer_expressions": "{1,16}", "fact_spans": "[[[1, 38]], [[66, 69]], [[1, 38]], [[1, 64]]]", "query_spans": "[[[66, 71]]]", "process": "" }, { "text": "Given that the moving line $ l $: $ y = \\frac{1}{m} x + 2m $ ($ m \\neq 0 $) intersects the ellipse $ C $: $ \\frac{x^{2}}{8} + \\frac{y^{2}}{4} = 1 $ at two points $ M $, $ N $. If there exists a point $ Q $ on the ellipse $ C $ such that $ \\overrightarrow{O M} + \\overrightarrow{O N} = \\lambda \\overrightarrow{O Q} $ ($ \\lambda \\neq 0 $), then what is the range of real values of $ \\lambda^{2} $?", "fact_expressions": "l: Line;C: Ellipse;O: Origin;M: Point;N: Point;Q: Point;lambda:Real;Expression(C) = (x^2/8 + y^2/4 = 1);Expression(l) = (y = 2*m + x/m);m:Number;Negation(m=0);Intersection(l, C) = {M, N};PointOnCurve(Q, C);VectorOf(O, M) + VectorOf(O, N) = lambda*VectorOf(O, Q);Negation(lambda=0)", "query_expressions": "Range(lambda^2)", "answer_expressions": "(0,4)", "fact_spans": "[[[3, 41]], [[42, 84], [98, 103]], [[113, 201]], [[87, 90]], [[91, 94]], [[106, 110]], [[203, 218]], [[42, 84]], [[3, 41]], [[10, 41]], [[10, 41]], [[3, 96]], [[98, 110]], [[113, 201]], [[113, 201]]]", "query_spans": "[[[203, 224]]]", "process": "Let the midpoint of MN be P(x_{0},y_{0}), Q(x_{Q},y_{Q}), then \\overrightarrow{OM}+\\overrightarrow{ON}=\\lambda\\overrightarrow{OQ}, \\overrightarrow{OQ}=\\frac{2}{\\lambda}\\overrightarrow{OP}, so x_{Q}=\\frac{2}{\\lambda}x_{0}, y_{Q}=\\frac{2}{\\lambda}y_{0}. Substitute into the ellipse equation: \\lambda^{2}=\\frac{x_{0}^{2}}{2}+y_{0}^{2}=(\\frac{1}{2}+\\frac{1}{m^{2}})x_{0}^{2}+4x_{0}+4m^{2}. Substitute the line equation y=\\frac{1}{m}x+2m into the ellipse equation \\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1, yielding: (1+\\frac{2}{m^{2}})x^{2}+8x+8m^{2}-8=0. Therefore, \\Delta=64-4(1+\\frac{2}{m^{2}})(8m^{2}-8)>0, solving gives 0\\frac{1}{2}, so 2(\\frac{1}{m^{2}}+\\frac{1}{4})^{2}-\\frac{1}{8}>2(\\frac{1}{2}+\\frac{1}{4})^{2}-\\frac{1}{8}=1, therefore 0<\\lambda^{2}=\\frac{4}{2(\\frac{1}{m^{2}}+\\frac{1}{4})^{2}-\\frac{1}{8}}, \\lambda^{2}\\in(0,4)." }, { "text": "Through the focus $F$ of the parabola $C$: $y^{2}=4x$, draw a line $l$ intersecting the parabola $C$ at points $A$ and $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, 20], [33, 39]], [[1, 20]], [[23, 26]], [[1, 26]], [[27, 32], [66, 71]], [[0, 32]], [[40, 43]], [[44, 47]], [[27, 47]], [[50, 64]]]", "query_spans": "[[[66, 76]]]", "process": "\\because the equation of parabola C is y^{2}=4x, its focus is F(1,0), \\therefore let the equation of line l be y=k(x-1), from \\begin{cases}y=k(x-1)\\\\y^{2}=4x\\end{cases} eliminating x gives \\frac{k}{4}y^{2}-y-k=0, let A(x_{1},y_{1}), B(x_{2},y_{2}), we get y_{1}+y_{2}=\\frac{4}{k}, y_{1}y_{2}=-4\\textcircled{1} \\because |AF|=4|BF|, \\therefore y_{1}+4y_{2}=0, we get y_{1}=-4y_{2}, substituting into \\textcircled{1} yields -3y_{1}=\\frac{4}{k} and -4y_{2}^{2}=-4, solving gives y_{2}=\\pm1, then k=\\pm\\frac{4}{3}," }, { "text": "If a focus of the hyperbola $\\frac{x^{2}}{m}+\\frac{y^{2}}{12}=1$ has coordinates $F_{1}(0,4)$, then what is the value of the real number $m$?", "fact_expressions": "G: Hyperbola;m: Real;F1: Point;Expression(G) = (y^2/12 + x^2/m = 1);Coordinate(F1) = (0, 4);OneOf(Focus(G)) = F1", "query_expressions": "m", "answer_expressions": "-4", "fact_spans": "[[[1, 40]], [[61, 66]], [[47, 59]], [[1, 40]], [[47, 59]], [[1, 59]]]", "query_spans": "[[[61, 70]]]", "process": "" }, { "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)$, and point $M$ lies on the left branch of the hyperbola such that $|M F_{1}|=2|F_{1} F_{2}|$ and $\\cos \\angle M F_{1} F_{2}=-\\frac{5}{16}$, find the eccentricity $e$ of the hyperbola.", "fact_expressions": "G: Hyperbola;b: Number;a: Number;M: 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(M, LeftPart(G));Abs(LineSegmentOf(M, F1)) = 2*Abs(LineSegmentOf(F1, F2));Cos(AngleOf(M, F1, F2)) = -5/16;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "2", "fact_spans": "[[[18, 74], [86, 89], [171, 174]], [[21, 74]], [[21, 74]], [[81, 85]], [[2, 9]], [[10, 17]], [[178, 181]], [[21, 74]], [[21, 74]], [[18, 74]], [[2, 80]], [[2, 80]], [[81, 95]], [[98, 124]], [[127, 168]], [[171, 181]]]", "query_spans": "[[[178, 183]]]", "process": "By the given conditions, |F₁F₂| = 2c, |MF₁| = 4c, |MF₂| = 4c + 2a. Then in △MF₁F₂, by the cosine theorem, we get (4c + 2a)² = 4c² + 16c² - 2 × 2c × 4c × (-5/16), that is, 9c² - 16ac - 4a² = 0. Therefore, 9e² - 16e - 4 = 0, solving gives e = 2 or e = -2/9 (discarded)." }, { "text": "Given that the midpoint of the chord intercepted by the line $l$ on the ellipse $\\frac{x^{2}}{36}+\\frac{y^{2}}{9}=1$ is $M(4 , 2)$, find the length of the chord intercepted by the ellipse on $l$.", "fact_expressions": "G: Ellipse;l: Line;M: Point;Expression(G) = (x^2/36 + y^2/9 = 1);Coordinate(M) = (4, 2);MidPoint(InterceptChord(l, G)) = M", "query_expressions": "Length(InterceptChord(l, G))", "answer_expressions": "sqrt(10)", "fact_spans": "[[[8, 46], [71, 73]], [[2, 7], [67, 70]], [[55, 65]], [[8, 46]], [[55, 65]], [[2, 65]]]", "query_spans": "[[[67, 80]]]", "process": "" }, { "text": "The line $x - 2y + 3 = 0$ intersects the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ at points $A$ and $B$, and $P(-1, 1)$ is exactly the midpoint of $AB$. Then the eccentricity of the ellipse is?", "fact_expressions": "H: Line;Expression(H) = (x - 2*y + 3 = 0);G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Intersection(H, G) = {A, B};A: Point;B: Point;P: Point;Coordinate(P) = (-1, 1);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[0, 13]], [[0, 13]], [[14, 66], [100, 102]], [[14, 66]], [[16, 66]], [[16, 66]], [[16, 66]], [[16, 66]], [[0, 78]], [[69, 72]], [[73, 76]], [[80, 89]], [[80, 89]], [[80, 98]]]", "query_spans": "[[[100, 108]]]", "process": "From \\begin{cases}x-2y+3=0\\\\b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}\\end{cases}, eliminating $x$, we get $(4b^{2}+a^{2})y^{2}-12b^{2}y+9b^{2}-a^{2}b^{2}=0$, $M=144b^{2}-4(a^{2}+4b^{2})(9b^{2}-a^{2}b^{2})>0\\Rightarrow a^{2}+4b^{2}>9$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $y_{1}+y_{2}=$. Since the midpoint of segment $AB$ is $(-1, 2)$, we obtain $a^{2}=2b^{2}$. Also $a^{2}=b^{2}+c^{2}$, $\\therefore a^{2}=2c^{2}$, $\\frac{c}{a}=\\frac{\\sqrt{2}}{2}$." }, { "text": "Given the parabola $y^{2}=2 p x (p>0)$ has focus $F$, directrix $l$, and a point $P$ on the parabola. Draw $P M \\perp l$ from point $P$, with foot of perpendicular at $M$, such that $|M P|=| M F |$. If the area of $\\triangle P M F$ is $\\sqrt{3}$, then what is the value of $p$?", "fact_expressions": "G: Parabola;p: Number;P: Point;M: Point;F: Point;l: Line;p>0;Expression(G) = (y^2 = 2*(p*x));Focus(G) = F;Directrix(G) = l;PointOnCurve(P, G);PointOnCurve(P,LineSegmentOf(P,M));IsPerpendicular(LineSegmentOf(P,M),l);FootPoint(LineSegmentOf(P,M),l)=M;Abs(LineSegmentOf(M, P)) = Abs(LineSegmentOf(M, F));Area(TriangleOf(P, M, F)) = sqrt(3)", "query_expressions": "p", "answer_expressions": "1", "fact_spans": "[[[2, 25], [40, 43]], [[129, 132]], [[47, 50], [52, 56]], [[74, 77]], [[29, 32]], [[36, 39]], [[5, 25]], [[2, 25]], [[2, 32]], [[2, 39]], [[40, 50]], [[51, 70]], [[57, 70]], [[57, 77]], [[79, 94]], [[96, 127]]]", "query_spans": "[[[129, 135]]]", "process": "As shown in the figure, |MP| = |MF|, according to the definition of the parabola, |PM| = |PF|, so triangle PMF is equilateral, and the area of triangle PMF is √3. S_{\\triangle PMF} = \\frac{1}{2}|PF|^{2}\\sin60^{\\circ} = \\sqrt{3}, solving gives: |PF| = 2. Therefore, |MF| = 2, \\angle MFO = 60^{\\circ}, p = |OF| = |MF|\\cos60^{\\circ} = 2 \\times \\frac{1}{2} = 1" }, { "text": "Given the parabola $C$: $x^{2}=4 y$ with focus $F$, draw a perpendicular from a point $A$ on $C$ to the directrix $l$ of $C$, with foot of perpendicular at $B$. Connect $F B$ and let it intersect the $x$-axis at point $D$. If $|A F|=5$, then $|A D|$=?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = 4*y);F: Point;Focus(C) = F;A: Point;PointOnCurve(A, C) = True;PointOnCurve(A, Z) = True;l: Line;Directrix(C) = l;IsPerpendicular(Z, l) = True;Z: Line;FootPoint(Z, l) = B;B: Point;Intersection(LineSegmentOf(F, B), xAxis) = D;D: Point;Abs(LineSegmentOf(A, F)) = 5", "query_expressions": "Abs(LineSegmentOf(A, D))", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[2, 21], [30, 33], [40, 43]], [[2, 21]], [[25, 28]], [[2, 28]], [[36, 39]], [[30, 39]], [[29, 52]], [[46, 49]], [[40, 49]], [[29, 52]], [], [[29, 59]], [[56, 59]], [[62, 77]], [[73, 77]], [[79, 88]]]", "query_spans": "[[[90, 99]]]", "process": "Let A(x_{0},y_{0}). Since |AF|=5, it follows that |AB|=y_{0}+1=5, so y_{0}=4. Substituting into x^{2}=4y gives x_{0}=\\pm4. When A(4,4), B(4,-1), then the line BF is y=; let y=0, we get D(2,0), so AD=2\\sqrt{5}. When A(-4,4), similarly |AD|=2\\sqrt{5}." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, the inclination angle of one of its asymptotes is $120^{\\circ}$. What is its eccentricity?", "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))) = ApplyUnit(120, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[2, 63], [88, 89]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]]]", "query_spans": "[[[88, 95]]]", "process": "From the given condition, $-\\frac{b}{a}=\\tan120^{\\circ}=-\\sqrt{3}$, the eccentricity $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=2$" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, $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 $|ME|=\\sqrt{11}$, then $|\\overrightarrow{AB}|=$?", "fact_expressions": "C: Parabola;G: Line;A: Point;B: Point;M: Point;E: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = E;PointOnCurve(F, G);Intersection(G, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = M;Abs(LineSegmentOf(M, E)) = sqrt(11)", "query_expressions": "Abs(VectorOf(A, B))", "answer_expressions": "6", "fact_spans": "[[[6, 25], [53, 59], [33, 34]], [[50, 52]], [[60, 63]], [[64, 67]], [[70, 73]], [[29, 32]], [[46, 49], [2, 5]], [[6, 25]], [[2, 28]], [[29, 44]], [[45, 52]], [[50, 69]], [[70, 84]], [[86, 103]]]", "query_spans": "[[[105, 131]]]", "process": "Analysis: Find the focus and directrix equations of the parabola to obtain the coordinates of E. Let the line passing through F be y = k(x - 1). Substitute into the parabola equation y^{2} = 4x, apply Vieta's formulas and the midpoint coordinate formula to find the coordinates of M. Use the distance formula between two points to solve for k, then use the focal chord formula of the parabola to compute the desired value. Detailed solution: F(1,0) is the focus of the parabola C: y^{2} = 4x, and E(-1,0) is the intersection point of its directrix with the x-axis. Let the line passing through F be y = k(x - 1). Substituting into the parabola equation y^{2} = 4x yields k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}). Then x_{1} + x_{2} = 2 + \\frac{4}{k^{2}}, leading to \\sqrt{(2 + \\frac{2}{k^{2}})^{2} + \\frac{4}{k^{2}}} = \\sqrt{11}. Solving gives k^{2} = 2. Then x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} = 4. By the definition of the parabola, |\\overrightarrow{AB}| = x_{1} +" }, { "text": "The hyperbola $C$: $x^{2}-\\frac{y^{2}}{3}=1$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Point $P$ lies on $C$ such that $\\tan \\angle F_{1} P F_{2}=4 \\sqrt{3}$, and $O$ is the origin. Then $|O P|$=?", "fact_expressions": "C: Hyperbola;F1: Point;P: Point;F2: Point;O: Origin;Expression(C) = (x^2 - y^2/3 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);Tan(AngleOf(F1, P, F2)) = 4*sqrt(3)", "query_expressions": "Abs(LineSegmentOf(O, P))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[0, 33], [63, 66]], [[42, 49]], [[58, 62]], [[50, 57]], [[108, 112]], [[0, 33]], [[0, 57]], [[0, 57]], [[58, 67]], [[68, 107]]]", "query_spans": "[[[119, 128]]]", "process": "Let point $ P(x_{P}, y_{P}) $, $ \\angle F_{1}PF_{2} = \\theta $, then $ \\tan\\theta = 4\\sqrt{3} $. Since $ \\tan\\theta = \\tan 2 \\cdot \\frac{\\theta}{2} = \\frac{2\\tan\\frac{\\theta}{2}}{1 - \\tan^{2}\\frac{\\theta}{2}} = 4\\sqrt{3} $, $ \\therefore \\tan\\frac{\\theta}{2} = \\frac{\\sqrt{3}}{2} $. Also, $ \\because a = 1, b = \\sqrt{3}, c = 2^{n} $, $ \\therefore S_{\\triangle F_{1}PF_{2}} = \\frac{b^{2}}{\\tan\\frac{\\theta}{2}} = \\frac{3}{\\frac{\\sqrt{3}}{2}} = 2\\sqrt{3} $. As shown in the figure below, draw a perpendicular from point $ P $ to the $ x $-axis, with foot $ M $, then $ S_{\\triangle F_{1}PF_{2}} = \\frac{1}{2}|F_{1}F_{2}||PM| = \\frac{1}{2} \\times 4|PM| = 2\\sqrt{3} $, so $ |PM| = \\sqrt{3} $, i.e., $ |y_{P}| = \\sqrt{3} $, $ \\therefore y_{P}^{2} = 3 $. Substitute into $ x^{2} - \\frac{y^{2}}{3} = 1 $, $ (|OP| = \\sqrt{|PM|^{2} + |OM|^{2}} = \\sqrt{3 + 2} = \\sqrt{5}) $. $ x_{P}^{2} = 2 $," }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$, and point $M$ lies on $C$, then the maximum value of $|M F_{1}| \\cdot|M F_{2}|$ is?", "fact_expressions": "C: Ellipse;M: Point;F1: Point;F2: Point;Expression(C) = (x^2/9 + y^2/4 = 1);Focus(C) = {F1, F2};PointOnCurve(M, C)", "query_expressions": "Max(Abs(LineSegmentOf(M, F1))*Abs(LineSegmentOf(M, F2)))", "answer_expressions": "9", "fact_spans": "[[[18, 60], [71, 74]], [[66, 70]], [[2, 9]], [[10, 17]], [[18, 60]], [[2, 65]], [[66, 75]]]", "query_spans": "[[[77, 109]]]", "process": "Since M lies on the ellipse C, we have |MF₁| + |MF₂| = 2 × 3 = 6. Therefore, by the AM-GM inequality, |MF₁| + |MF₂| = 6 ≥ 2√(|MF₁|·|MF₂|), which implies |MF₁|·|MF₂| ≤ 9, with equality if and only if |MF₁| = |MF₂| = 3." }, { "text": "Through a point $P$ on the ellipse $x^{2}+\\frac{y^{2}}{2}=1$, draw a perpendicular to the $x$-axis, with foot $Q$. Then the equation of the locus of the midpoint $M$ of the segment $P Q$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2/2 = 1);P: Point;L: Line;PointOnCurve(P,G);PointOnCurve(P, L);IsPerpendicular(L, xAxis);Q: Point;FootPoint(L, xAxis) = Q;M: Point;MidPoint(LineSegmentOf(P, Q)) = M", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2+2*y^2=1", "fact_spans": "[[[1, 28]], [[1, 28]], [[31, 34]], [], [[1, 34]], [[0, 42]], [[0, 42]], [[46, 49]], [[0, 49]], [[60, 63]], [[51, 63]]]", "query_spans": "[[[60, 70]]]", "process": "Let P(x_{0},y_{0}), M(x,y), then Q(x_{0},0), then \\begin{cases}x=x_{0}\\\\y=\\frac{1}{2}y_{0}\\end{cases}, that is, \\begin{cases}x_{0}=x\\\\y_{0}=2y\\end{cases}, since x_{0}^{2}+\\frac{y_{0}^{2}}{2}=1, substituting gives x^{2}+2y^{2}=1, so the trajectory equation of M is x^{2}+2y^{2}=1" }, { "text": "Let $ m $ be a constant. If the point $ F(0, 5) $ is a focus of the hyperbola $ \\frac{y^{2}}{m} - \\frac{x^{2}}{9} = 1 $, then $ m = $?", "fact_expressions": "G: Hyperbola;m: Number;F: Point;Expression(G) = (-x^2/9 + y^2/m = 1);Coordinate(F) = (0, 5);OneOf(Focus(G)) = F", "query_expressions": "m", "answer_expressions": "16", "fact_spans": "[[[21, 59]], [[66, 69], [1, 4]], [[9, 20]], [[21, 59]], [[9, 20]], [[9, 64]]]", "query_spans": "[[[66, 71]]]", "process": "" }, { "text": "The maximum length of the chord intercepted by the line $y=k x+1$ on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ is?", "fact_expressions": "G: Ellipse;H: Line;k: Number;Expression(G) = (x^2/4 + y^2 = 1);Expression(H) = (y = k*x + 1)", "query_expressions": "Max(Length(InterceptChord(H,G)))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[12, 39]], [[0, 11]], [[2, 11]], [[12, 39]], [[0, 11]]]", "query_spans": "[[[0, 50]]]", "process": "Solving the system of the line and ellipse equations, combining the chord length formula and Vieta's Theorem, we can find |AB| = \\sqrt{\\frac{16}{3} - 12\\left(\\frac{1}{1+4k^{2}} - \\frac{1}{3}\\right)^{2}}, and then determine the maximum chord length. (Detailed solution) Solve the system \\begin{cases} \\frac{x^{2}}{4} + y^{2} = 1 \\\\ y = kx + 1 \\end{cases}, eliminating y gives (1+4k^{2})x^{2} + 8kx = 0. Let the chord be AB, where A(x_{1}, y_{1}), B(x_{2}, y_{2}), then x_{1} + x_{2} = -\\frac{8k}{1+4k^{2}}\\frac{2}{2}, x_{1}x_{2} = 0, thus |AB| = \\sqrt{1+k^{2}} = \\sqrt{1+k^{2}}, \\times \\sqrt{(x_{1}} + x_{2})^{2} - 5 - 12(- Therefore when \\frac{1}{1+4k^{2}} - \\frac{1}{3} = 0, i.e., k = \\pm\\frac{\\sqrt{2}}{1}, the chord length reaches its maximum value \\underline{4\\sqrt{3}}" }, { "text": "Given that the distance from the focus of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ to its asymptote is $1$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2 - y^2/m = 1);Distance(Focus(G), Asymptote(G)) = 1", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 30], [46, 49]], [[5, 30]], [[2, 30]], [[2, 44]]]", "query_spans": "[[[46, 55]]]", "process": "From the given conditions, we have a^{2}=1, b^{2}=m, then c=\\sqrt{1+m}, the asymptotes are y=\\pm\\frac{b}{a}x, so the distance from the focus to the asymptote is d=\\frac{bc}{\\sqrt{a^{2}+b^{2}}}=\\frac{bc}{c}=b=1, hence c=\\sqrt{2}, therefore the eccentricity e=\\frac{\\sqrt{2}}{1}=\\sqrt{2}." }, { "text": "Given that $A$ and $B$ are two points on the parabola $y^{2}=2 p x(p>0)$, and $O$ is the origin. If $|O A|=|O B|$ and the orthocenter of $\\triangle A O B$ is exactly the focus of this parabola, then what is the equation of line $A B$?", "fact_expressions": "G: Parabola;p: Number;B: Point;A: Point;O: Origin;p>0;Expression(G) = (y^2 = 2*p*x);PointOnCurve(A,G);PointOnCurve(B,G);Abs(LineSegmentOf(O, A)) = Abs(LineSegmentOf(O, B));Orthocenter(TriangleOf(A,O,B))=Focus(G)", "query_expressions": "Expression(LineOf(A, B))", "answer_expressions": "x=5*p/2", "fact_spans": "[[[10, 31], [83, 86]], [[13, 31]], [[6, 9]], [[2, 5]], [[34, 37]], [[13, 31]], [[10, 31]], [[2, 33]], [[2, 33]], [[45, 58]], [[60, 89]]]", "query_spans": "[[[91, 103]]]", "process": "By the property of the parabola, A and B are symmetric with respect to the x-axis. Let A(x, y), then B(x, -y), and the focus is F(\\frac{p}{2}). From the given condition, AF \\perp OB, \\therefore k_{AF} \\cdot k_{OB} = -1. So y^{2} = x(x - \\frac{p}{2}), that is, 2px = x(x - \\frac{p}{2}). Since x \\neq 0, we have 2p = x - \\frac{p}{2}, thus x = \\frac{5p}{2}. Therefore, the equation of line AB is x = \\frac{5p}{2}." }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, and $P$ is a point on $C$, if $|PF|=3$, then what are the coordinates of point $P$?", "fact_expressions": "C: Parabola;P: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 3", "query_expressions": "Coordinate(P)", "answer_expressions": "(2,pm*2*sqrt(2))", "fact_spans": "[[[6, 25], [33, 36]], [[29, 32], [53, 57]], [[2, 5]], [[6, 25]], [[2, 28]], [[29, 40]], [[42, 51]]]", "query_spans": "[[[53, 62]]]", "process": "According to the problem, the focus of the parabola C is F(1,0), and the equation of the directrix is x = -1. If P(x, y), by the definition of a parabola, we have: |PF| = x + 1 = 3, so x = 2. Hence, y^{2} = 8. Therefore, the coordinates of P are (2, \\pm2\\sqrt{2})." }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ passes through $F$ and intersects $C$ at points $A$ and $B$. If $|AF|=|BF|$, then what is the length of the chord cut from the $y$-axis by the circle with diameter $AB$?", "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)) = Abs(LineSegmentOf(B, F));G: Circle;IsDiameter(LineSegmentOf(A, B), G)", "query_expressions": "Length(InterceptChord(yAxis, G))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 21], [39, 42]], [[2, 21]], [[25, 28], [35, 38]], [[2, 28]], [[29, 34]], [[29, 38]], [[44, 47]], [[48, 51]], [[29, 53]], [[55, 68]], [[87, 88]], [[75, 88]]]", "query_spans": "[[[70, 95]]]", "process": "Since |AF| = |BF|, it follows that l ⊥ x-axis, so the center of the circle is at F(1,0), the radius is r = 2, and the chord length is 2\\sqrt{2^{2}-1}^{2} = 2\\sqrt{3}." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{9}+y^{2}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(2)/3", "fact_spans": "[[[0, 27]], [[0, 27]]]", "query_spans": "[[[0, 33]]]", "process": "Using the standard equation given in the problem, find a and b, then solve for c, and thus find the eccentricity. For the ellipse \\frac{x^2}{9}+y^{2}=1, the semi-major axis is a=3, the semi-minor axis is b=1, then the semi-focal distance is c=\\sqrt{9-1}=2\\sqrt{2}. Therefore, the eccentricity of the ellipse is: e=\\frac{c}{a}=\\frac{2\\sqrt{2}}{3}." }, { "text": "The left and right foci of the ellipse $\\frac{x^{2}}{a}+\\frac{y^{2}}{b}=1$ are $F_{1}$ and $F_{2}$ respectively, $P$ is a point on the ellipse, and $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$, $\\angle P F_{2} F_{1}=\\frac{\\pi}{3}$. Then the eccentricity $e$ of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b + x^2/a = 1);b: Number;a: Number;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);AngleOf(P, F1, F2) = pi/6;AngleOf(P, F2, F1) = pi/3;e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[0, 37], [65, 67], [149, 151]], [[0, 37]], [[2, 37]], [[2, 37]], [[45, 52]], [[53, 60]], [[0, 60]], [[0, 60]], [[61, 64]], [[61, 70]], [[72, 108]], [[110, 147]], [[155, 158]], [[149, 158]]]", "query_spans": "[[[155, 160]]]", "process": "" }, { "text": "Given that $A$ and $F$ are the right vertex and the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with eccentricity $2$, and that the distances from $A$ and $F$ to the line $b x - a y = 0$ are $d_{1}$ and $d_{2}$ respectively, then $\\frac{d_{1}}{d_{2}}$=?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;A: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-a*y + b*x = 0);Eccentricity(G) = 2;RightVertex(G) = A;RightFocus(G) = F;d1: Number;d2: Number;Distance(A, H) = d1;Distance(F, H) = d2", "query_expressions": "d1/d2", "answer_expressions": "1/2", "fact_spans": "[[[18, 74]], [[21, 74]], [[21, 74]], [[92, 105]], [[2, 5], [84, 87]], [[6, 9], [88, 91]], [[21, 74]], [[21, 74]], [[18, 74]], [[92, 105]], [[10, 74]], [[2, 82]], [[2, 82]], [[111, 118]], [[119, 126]], [[84, 126]], [[84, 126]]]", "query_spans": "[[[128, 151]]]", "process": "From the given conditions, we can derive $ e = \\frac{c}{a} = 2 $, then $ c = 2a $, so $ \\frac{d_{1}}{d_{2}} = \\frac{|OA|}{|OF|} = \\frac{a}{c} = \\frac{1}{2} $." }, { "text": "Given a line $ l $ with slope $ k = \\frac{3}{4} $ passing through the focus of the parabola $ x^{2} = 4y $, intersecting the parabola at points $ A $ and $ B $. If two tangents to the parabola are drawn at points $ A $ and $ B $, intersecting at point $ M $, then the area of $ \\triangle MAB $ is?", "fact_expressions": "k: Number;Slope(l) = k;k = 3/4;l: Line;G: Parabola;Expression(G) = (x^2 = 4*y);PointOnCurve(Focus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};M: Point;Intersection(TangentOfPoint(A, G), TangentOfPoint(B, G)) = M", "query_expressions": "Area(TriangleOf(M, A, B))", "answer_expressions": "125/16", "fact_spans": "[[[4, 19]], [[2, 25]], [[4, 19]], [[20, 25]], [[26, 40], [46, 49], [75, 78]], [[26, 40]], [[20, 43]], [[52, 55], [65, 69]], [[56, 59], [70, 73]], [[20, 61]], [[86, 90]], [[62, 90]]]", "query_spans": "[[[92, 114]]]", "process": "l: y - 1 = \\frac{3}{4}x, let A(x_{1},\\frac{x_{1}^{2}}{4}), B(x_{2},\\frac{x^{2}}{4}) \\because y = \\frac{x^{2}}{4} \\therefore y' = \\frac{x}{2} therefore the tangent at A is y - \\frac{x_{1}^{2}}{4} = \\frac{x_{1}}{2}(x - x_{1}) \\Rightarrow y = \\frac{x_{1}}{2}x - \\frac{x_{1}^{2}}{4}, similarly the tangent at B is y = \\frac{x_{2}}{2}x - \\frac{x_{2}^{2}}{4} from \\begin{cases} y - 1 = \\frac{3}{4}x \\\\ x^{2} = 4y \\end{cases} solving gives x_{1} = -1, x_{2} = 4, so from \\begin{cases} y = \\\\ y \\end{cases} -\\frac{1}{2}x - \\frac{1}{4} obtain M(\\frac{3}{2}, -1) so d_{M-AB} = \\frac{\\frac{9}{8} + 2}{\\frac{5}{4}} = \\frac{5}{2} D. = \\frac{2}{5} \\therefore S = \\frac{1}{2} \\times \\frac{5}{2} \\times \\frac{25}{4} = \\frac{125}{16}" }, { "text": "If on the parabola $y^{2}=2 p x(p>0)$ the distance from the point with abscissa $6$ to the focus is equal to $8$, then what is the distance from the focus to the directrix?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Point;XCoordinate(H) = 6;PointOnCurve(H, G);Distance(H, Focus(G)) = 8", "query_expressions": "Distance(Focus(G), Directrix(G))", "answer_expressions": "4", "fact_spans": "[[[1, 22]], [[1, 22]], [[4, 22]], [[4, 22]], [[31, 32]], [[23, 32]], [[1, 32]], [[1, 43]]]", "query_spans": "[[[1, 55]]]", "process": "From the definition of the parabola, we get 6+\\frac{p}{2}=8 \\therefore p=4, therefore the distance from the focus to the directrix is p=4." }, { "text": "Given the ellipse $E$: $\\frac{x^{2}}{11}+\\frac{y^{2}}{2}=1$ and the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1$ $(a>0)$ have the same foci, then the equations of the asymptotes of hyperbola $C$ are?", "fact_expressions": "E: Ellipse;Expression(E) = (x^2/11 + y^2/2 = 1);C: Hyperbola;Expression(C) = (-y^2/5 + x^2/a^2 = 1);a: Number;a>0;Focus(E) = Focus(C)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(sqrt(5)/2)*x", "fact_spans": "[[[2, 45]], [[2, 45]], [[46, 98], [106, 112]], [[46, 98]], [[54, 98]], [[54, 98]], [[2, 104]]]", "query_spans": "[[[106, 120]]]", "process": "The foci of ellipse E are (\\pm3,0), thus a^{2}=3^{2}-5=4, the asymptotes of hyperbola C are given by y=\\pm\\frac{\\sqrt{5}}{2}x" }, { "text": "What are the coordinates of the foci of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{m}=1$?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/5 + y^2/m = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "", "fact_spans": "[[[0, 37]], [[2, 37]], [[0, 37]]]", "query_spans": "[[[0, 44]]]", "process": "When 0 < m < 5, the foci lie on the x-axis, then a^{2} = 5, b^{2} = m, so c^{2} = 5 - m, hence the coordinates of the foci are (\\pm\\sqrt{5 - m}, 0). When m > 5, the foci lie on the y-axis, then a^{2} = m, b^{2} = 5, so c^{2} = m - 5, hence the coordinates of the foci are (0, \\pm\\sqrt{m - 5})." }, { "text": "The coordinates of the focus of the parabola $x^{2}=-2 y$ are?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -2*y)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, -1/2)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "Test analysis: From the equation it follows that 2p=2, ∴ p/2=1/2, ∴ the focus is (0,-1/2)" }, { "text": "One of the foci of the hyperbola $\\frac{x^{2}}{k}-\\frac{y^{2}}{4}=1$ has coordinates $(\\sqrt{6}, 0)$. Find the real number $k=?$", "fact_expressions": "G: Hyperbola;k: Real;Expression(G) = (-y^2/4 + x^2/k = 1);Coordinate(OneOf(Focus(G)))=(sqrt(6),0)", "query_expressions": "k", "answer_expressions": "2", "fact_spans": "[[[0, 38]], [[65, 70]], [[0, 38]], [[0, 63]]]", "query_spans": "[[[65, 72]]]", "process": "From the hyperbola equation, we obtain a^{2}=k, b^{2}=4. According to c^{2}=a^{2}+b^{2}, the solution can be found. [Detailed Solution] From the hyperbola \\frac{x^{2}}{k}-\\frac{y^{2}}{4}=1, we get a^{2}=k, b^{2}=4. Then from c^{2}=a^{2}+b^{2}, we have k+4=6, solving gives k=2." }, { "text": "If $a>0$, then the coordinates of the focus of the parabola $x^{2}=4 a y$ are?", "fact_expressions": "G: Parabola;a: Number;Expression(G) = (x^2 = 4*a*y);a > 0", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,a)", "fact_spans": "[[[8, 24]], [[11, 24]], [[8, 24]], [[1, 6]]]", "query_spans": "[[[8, 31]]]", "process": "The coordinates of the focus of the parabola x^{2}=4ay are (0,a)." }, { "text": "Given that one focus of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ coincides with the focus of the parabola $y^{2}=a x$, then $a$=?", "fact_expressions": "G: Parabola;a: Number;H: Ellipse;Expression(G) = (y^2 = a*x);Expression(H) = (x^2/4 + y^2/3 = 1);OneOf(Focus(H)) = Focus(G)", "query_expressions": "a", "answer_expressions": "pm*4", "fact_spans": "[[[44, 58]], [[64, 67]], [[2, 39]], [[44, 58]], [[2, 39]], [[2, 62]]]", "query_spans": "[[[64, 69]]]", "process": "" }, { "text": "Given that a line $ l $ with slope $ 1 $ intersects the hyperbola $ C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a>0, b>0 $) at points $ B $ and $ D $, and the midpoint of $ BD $ is $ M(1,3) $, then the eccentricity of $ C $ is?", "fact_expressions": "l: Line;Slope(l) = 1;C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;B: Point;D: Point;Intersection(l, C) = {B, D};M: Point;Coordinate(M)=(1,3);MidPoint(LineSegmentOf(B, D)) = M", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[9, 14]], [[2, 14]], [[15, 76], [109, 112]], [[15, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[23, 76]], [[79, 82]], [[83, 86]], [[9, 88]], [[99, 107]], [[99, 107]], [[90, 107]]]", "query_spans": "[[[109, 118]]]", "process": "" }, { "text": "Given the line $l$: $\\sqrt{3} x - y - \\sqrt{3} = 0$ intersects the parabola $\\Gamma$: $y^{2} = 4x$ at points $A$, $B$ and intersects the $x$-axis at $F$. If $\\overrightarrow{O F} = \\lambda \\overrightarrow{O A} + \\mu \\overrightarrow{O B}$ $(\\lambda \\leq \\mu)$, then $\\frac{\\lambda}{\\mu} = $?", "fact_expressions": "Gamma: Parabola;l: Line;O: Origin;F: Point;A: Point;B: Point;Expression(Gamma) = (y^2 = 4*x);Expression(l) = (sqrt(3)*x - y - sqrt(3) = 0);Intersection(l,Gamma)={A,B};Intersection(l,xAxis)=F;lambda:Number;mu:Number;lambda<=mu;VectorOf(O, F) = lambda*VectorOf(O, A) + mu*VectorOf(O, B)", "query_expressions": "lambda/mu", "answer_expressions": "1/3", "fact_spans": "[[[35, 59]], [[2, 34]], [[83, 177]], [[77, 80]], [[61, 64]], [[65, 68]], [[35, 59]], [[2, 34]], [[2, 70]], [[2, 80]], [[83, 177]], [[83, 177]], [[83, 177]], [[83, 177]]]", "query_spans": "[[[179, 202]]]", "process": "Solve the system of equations \\begin{cases}y=\\sqrt{3}x-\\sqrt{3}\\\\y^{2}=4x\\end{cases} to obtain \\begin{cases}x_{1}=3\\\\y_{1}=2\\sqrt{3}\\end{cases} or \\begin{cases}x_{2}=\\frac{1}{3}\\\\y_{2}=-\\frac{2\\sqrt{3}}{3}\\end{cases}. From (1,0)=\\lambda(3,2\\sqrt{3})+\\mu(\\frac{1}{3},-\\frac{2\\sqrt{3}}{3}), we get: \\begin{cases}1=3\\lambda+\\frac{1}{3}\\mu\\\\0=2\\sqrt{3}\\lambda-\\frac{2\\sqrt{3}}{3}\\mu\\end{cases}\\Rightarrow a=\\frac{1}{3}." }, { "text": "Given that the distance from one endpoint of the minor axis of an ellipse to a focus is $5$, and the distance from the focus to the center of the ellipse is $3$, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;Distance(OneOf(Endpoint(MinorAxis(G))), OneOf(Focus(G))) = 5;Distance(Focus(G), Center(G)) = 3", "query_expressions": "Expression(G)", "answer_expressions": "{x^2/25 + y^2/16 = 1, y^2/25 + x^2/16 = 1}", "fact_spans": "[[[2, 4], [27, 29], [40, 42]], [[2, 23]], [[2, 38]]]", "query_spans": "[[[40, 49]]]", "process": "Problem Analysis: According to the given conditions, the foci of the ellipse may lie on the x-axis or the y-axis, and a=5, c=3, ∴b²=16. ∴ the required equation of the ellipse is $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ or $\\frac{y^{2}}{25}+\\frac{x^{2}}{16}=$" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with left and right foci $F_{1}$, $F_{2}$. A line passing through $F_{2}$ is tangent to the circle $x^{2}+y^{2}=b^{2}$ at point $A$, and intersects the ellipse $C$ at two points $P$, $Q$. If $P F_{1} \\perp P F_{2}$, then the eccentricity of the ellipse is?", "fact_expressions": "C: Ellipse;G: Circle;a:Number;b: Number;H: Line;P: Point;F1: Point;F2: Point;A: Point;Q: Point;Expression(C)=(y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = b^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, H);TangentPoint(H,G)=A;Intersection(H,C) = {P,Q};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));a>b;b>0", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/3", "fact_spans": "[[[2, 54], [121, 126], [164, 166]], [[91, 111]], [[4, 54]], [[4, 54]], [[88, 90]], [[130, 133]], [[60, 68]], [[69, 78], [80, 87]], [[114, 118]], [[134, 137]], [[2, 54]], [[91, 111]], [[2, 77]], [[2, 77]], [[79, 90]], [[88, 118]], [[88, 137]], [[139, 162]], [[4, 54]], [[4, 54]]]", "query_spans": "[[[164, 172]]]", "process": "Since OA\\botPQ, PF_{1}\\botPF_{2}, |OF_{1}|=|OF_{2}|, \\therefore|PF_{1}|=2b, |PF_{2}|=2a-2b, so |AF_{2}|=a-b, \\therefore b^{2}+(a-b)^{2}=c^{2}=a^{2}-b^{2}, so _{3b}=2a, \\therefore9b^{2}=4a^{2}, \\therefore9a^{2}-9c^{2}=4a^{2}, \\therefore e=\\frac{\\sqrt{5}}{3}" }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, let $A$ be a point on $C$ such that $|AF|=5$, and let $O$ be the origin. Then the area of $\\triangle OAF$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;A: Point;PointOnCurve(A, C);Abs(LineSegmentOf(A, F)) = 5;O: Origin", "query_expressions": "Area(TriangleOf(O, A, F))", "answer_expressions": "2", "fact_spans": "[[[2, 21], [33, 36]], [[2, 21]], [[25, 28]], [[2, 28]], [[29, 32]], [[29, 39]], [[41, 50]], [[53, 56]]]", "query_spans": "[[[63, 85]]]", "process": "Find the intersection points according to the standard equation of the parabola, then use the focal radius formula to find the vertical coordinate of point A, and use the triangle area formula to solve. According to the problem, the parabola C: y^{2}=4x has focus F(1,0). Let A(m,n), then |AF|=m+1=5, \\therefore m=4, \\therefore n=\\pm4. \\therefore S_{\\DeltaAOF}=\\frac{1}{2}\\times1\\times4=2." }, { "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ with left focus $F$, and the lines $x-y-2=0$, $x-y+2=0$ intersecting the ellipse at points $A$, $B$, $C$, $D$ respectively, then $|A F|+|B F|+|C F|+|D F|$=?", "fact_expressions": "G: Ellipse;l1: Line;l2:Line;A: Point;F: Point;B: Point;C: Point;D: Point;Expression(G) = (x^2/9 + y^2/5 = 1);Expression(l1) = (x - y - 2 = 0);Expression(l2)=(x-y+2=0);LeftFocus(G) = F;Intersection(l1,G)={A,B};Intersection(l2,G)={C,D}", "query_expressions": "Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) + Abs(LineSegmentOf(C, F)) + Abs(LineSegmentOf(D, F))", "answer_expressions": "12", "fact_spans": "[[[2, 39], [72, 74]], [[48, 59]], [[61, 71]], [[79, 82]], [[44, 47]], [[83, 86]], [[87, 90]], [[91, 94]], [[2, 39]], [[48, 59]], [[61, 71]], [[2, 47]], [[48, 94]], [[48, 94]]]", "query_spans": "[[[96, 123]]]", "process": "Let the right focus of the ellipse be F'. By the definition of an ellipse, |AF| + |AF'| = |BF| + |BF'| = 2a = 6. Since the lines x - y - 2 = 0 and x - y + 2 = 0 are symmetric about the origin, and the ellipse is centrally symmetric with symmetry center at the origin, DF = |AF|, |CF| = |BF|. Therefore, |AF| + |BF| + |CF| + |DF| = |AF| + |AF| + |BF| + |BF'| = 4a = 12." }, { "text": "It is known that the center of the ellipse is at the origin of the coordinate system, the foci lie on the $x$-axis, the focal distance is $6 \\sqrt{3}$, and the sum of the distances from a point on the ellipse to the two foci is $12$. Then what is the equation of the ellipse?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;PointOnCurve(Focus(G), xAxis);FocalLength(G) = 6*sqrt(3);P: Point;F1: Point;F2: Point;Focus(G) = {F1, F2};PointOnCurve(P, G);Distance(P, F1) + Distance(P, F2) = 12", "query_expressions": "Expression(G)", "answer_expressions": "x^2/36 + y^2/9 = 1", "fact_spans": "[[[2, 4], [39, 41], [61, 63]], [[8, 12]], [[2, 12]], [[2, 21]], [[2, 37]], [], [], [], [[39, 49]], [[39, 44]], [[39, 59]]]", "query_spans": "[[[61, 68]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $y=8 x^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y = 8*x^2)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, 1/32)", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "The parabola y=8x^{2} can be rewritten as x^{2}=\\frac{1}{8}y, with the focus on the y-axis. Since 2p=\\frac{1}{8}, \\frac{1}{2}p=\\frac{1}{32}, the coordinates of the focus of the parabola y=8x^{2} are (0,\\frac{1}{32})." }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, and $M$ is a point on the directrix of the parabola with its $y$-coordinate equal to $3 \\sqrt{3}$. $N$ is an intersection point of line $M F$ and the parabola, and $\\overrightarrow{M N}=2 \\overrightarrow{N F}$. Find $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, Directrix(C));YCoordinate(M) =3*sqrt(3);N: Point;OneOf(Intersection(LineOf(M, F), C)) = N;VectorOf(M, N) = 2*VectorOf(N, F)", "query_expressions": "p", "answer_expressions": "3", "fact_spans": "[[[2, 28], [40, 43], [86, 89]], [[2, 28]], [[143, 146]], [[10, 28]], [[32, 35]], [[2, 35]], [[36, 39], [51, 54]], [[36, 49]], [[51, 71]], [[74, 77]], [[74, 94]], [[96, 141]]]", "query_spans": "[[[143, 148]]]", "process": "】According to the properties of the parabola and the relation \\overrightarrow{MN}=2\\overrightarrow{NF}, express the coordinates of point N, then substitute them into the parabola equation to find the value of p. The parabola C: y^{2}=2px (p>0) has focus F, M is a point on the directrix of the parabola with its y-coordinate equal to 3\\sqrt{3}. N is an intersection point of line MF and the parabola. Given \\overrightarrow{MN}=2\\overrightarrow{NF}, \\therefore \\overrightarrow{MN}=(x+\\frac{p}{2}-\\frac{P}{2},3\\sqrt{3}), F(,y-3\\sqrt{3}), \\overrightarrow{NF}\\frac{p}{2},0), |x+\\frac{p}{2}==2(\\frac{p}{2}-.\\begin{cases}\\\\y-3\\sqrt{3}=2(-,\\end{cases} we obtain N(\\frac{p}{6},\\sqrt{3}). Substituting into the parabola equation yields: 3=2\\timesp\\times\\frac{p}{6}, solving gives p=3." }, { "text": "Given that the line $k x - y + 1 = 0$ intersects the hyperbola $\\frac{x^{2}}{2} - y^{2} = 1$ at two distinct points $A$ and $B$, if the point $M(3,0)$ on the $x$-axis is equidistant from points $A$ and $B$, then the value of $k$ is?", "fact_expressions": "H: Line;Expression(H) = (k*x - y + 1 = 0);k: Number;G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1);Intersection(H, G) = {A, B};A: Point;B: Point;Negation(A = B);M: Point;Coordinate(M) = (3, 0);PointOnCurve(M, xAxis);Distance(M, A) = Distance(M, B)", "query_expressions": "k", "answer_expressions": "1/2", "fact_spans": "[[[2, 15]], [[2, 15]], [[94, 97]], [[16, 44]], [[16, 44]], [[2, 60]], [[52, 56], [78, 81]], [[57, 60], [82, 85]], [[47, 60]], [[68, 77]], [[68, 77]], [[62, 77]], [[68, 92]]]", "query_spans": "[[[94, 101]]]", "process": "Solve the system \\begin{cases} y = kx + 1 \\\\ x^2 - 2y^2 = 1 \\end{cases}, we get (1 - 2k^2)x^2 - 4kx - 4 = 0. Since the line intersects the hyperbola at two distinct points, (1 - 2k^2) \\neq 0 and \\begin{cases} 16k^2 + 16 > 0 \\\\ (1 - 2k^2) \\end{cases}, solving gives -1 < k < 1 and k \\neq \\pm \\frac{\\sqrt{2}}{2}. Let A(x_1, y_1), B(x_2, y_2), then x_1 + x_2 = \\frac{4k}{1 - 2k^2}. Let P be the midpoint of AB, then P\\left( \\frac{2k}{1 - 2k^2}, \\frac{1}{1 - 2k^2} \\right). Since M(3, 0) is equidistant from A and B, \\therefore MP \\perp AB, \\therefore k_{MP} \\cdot k_{AB} = k \\cdot \\frac{\\frac{1}{1 - 2k^2}}{\\frac{2k}{1 - 2k^2} - 3} = -1, solving gives k = \\frac{1}{2} or k = -1, \\therefore k = \\frac{1}{2}." }, { "text": "Given that point $P(x, y)$ lies on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, what is the maximum value of $2x+y$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);x1: Number;y1: Number;P: Point;Coordinate(P) = (x1, y1);PointOnCurve(P, G)", "query_expressions": "Max(2*x1 + y1)", "answer_expressions": "sqrt(17)", "fact_spans": "[[[13, 40]], [[13, 40]], [[3, 12]], [[3, 12]], [[2, 12]], [[2, 12]], [[2, 41]]]", "query_spans": "[[[43, 56]]]", "process": "" }, { "text": "If the ellipse $C$ has coordinate axes as its axes of symmetry, foci on the $y$-axis, eccentricity $\\frac{{\\sqrt{5}}}{3}$, and area $6\\pi$, then what is the equation of the ellipse $C$?", "fact_expressions": "C: Ellipse;SymmetryAxis(C) = axis;PointOnCurve(Focus(C), yAxis);Eccentricity(C) = sqrt(5)/3;Area(C) = 6*pi", "query_expressions": "Expression(C)", "answer_expressions": "y^2/9 + x^2/4 = 1", "fact_spans": "[[[1, 6], [25, 30], [69, 74]], [[1, 14]], [[1, 23]], [[25, 57]], [[25, 67]]]", "query_spans": "[[[69, 79]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and points $M(x_{1}, y_{1})$, $N(x_{2}, y_{2})$ moving on the parabola $C$. If $x_{1}+x_{2}+2=2|M N|$, then the maximum value of $\\angle M F N$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;M: Point;Coordinate(M) = (x1, y1);x1: Number;y1: Number;N: Point;Coordinate(N) = (x2, y2);x2: Number;y2: Number;PointOnCurve(M, C);PointOnCurve(N, C);x1 + x2 + 2 = 2*Abs(LineSegmentOf(M, N))", "query_expressions": "Max(AngleOf(M, F, N))", "answer_expressions": "pi/3", "fact_spans": "[[[2, 21], [69, 75]], [[2, 21]], [[25, 28]], [[2, 28]], [[31, 48]], [[31, 48]], [[31, 48]], [[31, 48]], [[51, 68]], [[51, 68]], [[51, 68]], [[51, 68]], [[31, 81]], [[31, 81]], [[83, 105]]]", "query_spans": "[[[107, 127]]]", "process": "The directrix of the parabola C: y^{2}=4x is x=-1, so from the given x_{1}+x_{2}+2=2|MN|, we obtain |MF|+|NF|=2|MN|. Then \\cos\\angle MFN=\\frac{|MF|^{2}+|NF|}{2|MF|}\\frac{2-|MN|^{2}}{}\\frac{3}{4}\\times2|MF||NF|-\\frac{1}{2}|MF||NF|=\\frac{|MF||NF|}{2|MF||NF|}=\\frac{1}{2}. Since \\angle MFN\\in(0,\\pi), it follows that \\angle MFN\\in(0,\\frac{\\pi}{3}], therefore the maximum value of \\angle MFN is \\frac{\\pi}{3}." }, { "text": "Let the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ have common foci $F_{1}$, $F_{2}$, and let $P$ be a common point of the two curves. Then $S_\\Delta F_{1} F_{2} P$=?", "fact_expressions": "G: Hyperbola;a: Number;H: Ellipse;F1: Point;F2: Point;P:Point;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(H,G))=P", "query_expressions": "Area(TriangleOf(F1,F2,P))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[39, 71]], [[42, 71]], [[1, 38]], [[77, 84]], [[85, 92]], [[93, 96]], [[39, 71]], [[1, 38]], [[1, 92]], [[1, 92]], [[93, 106]]]", "query_spans": "[[[108, 134]]]", "process": "The ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ has foci at $(\\pm2,0)$, and the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$ has foci at $(\\pm2,0)$, so $a=\\sqrt{3}$. Let $|AF_{1}|=m$, $|AF_{2}|=n$. Without loss of generality, let point $P$ be in the first quadrant. By the definition of the ellipse, we have $m+n=2\\sqrt{6}$, $\\textcircled{1}$; by the definition of the hyperbola, we have $m-n=2\\sqrt{3}$, $\\textcircled{2}$. From $\\textcircled{1}$ and $\\textcircled{2}$, we obtain $m=\\sqrt{6}+\\sqrt{3}$, $n=\\sqrt{6}-\\sqrt{3}$. $\\cos\\angle F_{1}PF_{2}=\\frac{6+3+6\\sqrt{2}+6+3-6\\sqrt{2}-16}{2(\\sqrt{6}+\\sqrt{3})(\\sqrt{6}-\\sqrt{3})}=\\frac{1}{3}$, so $\\sin\\angle F_{1}PF_{2}=\\frac{2\\sqrt{2}}{3}$. Therefore, the area of the triangle is: $\\frac{1}{2}mn\\sin\\angle F_{1}PF_{2}=\\frac{1}{2}\\times(\\sqrt{6}+\\sqrt{3})(\\sqrt{6}-\\sqrt{3})\\times\\frac{2\\sqrt{2}}{3}=\\sqrt{2}$." }, { "text": "Given the hyperbola $C$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\sqrt{5}$, then the equations of its asymptotes are?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(C) = (-x^2/b^2 + y^2/a^2 = 1);Eccentricity(C) = sqrt(5)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y = pm*x/2", "fact_spans": "[[[2, 63], [80, 81]], [[9, 63]], [[9, 63]], [[9, 63]], [[9, 63]], [[2, 63]], [[2, 78]]]", "query_spans": "[[[80, 88]]]", "process": "From $\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{5}$, we get $\\frac{b^{2}}{a^{2}}=4$, $\\therefore\\frac{b}{a}=2$. Also, the asymptotes are given by $y=\\pm\\frac{a}{b}$. $\\therefore$ the asymptotes of the hyperbola are $y=\\pm\\frac{1}{2}x$." }, { "text": "Given the parabola $y^{2}=8x$, a line passing through the point $D(2,0)$ intersects the parabola at points $A$ and $B$, and the length of $AB$ is $10$. Let $C$ be the midpoint of $AB$. What is the distance from $C$ to the $y$-axis?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);H: Line;D: Point;Coordinate(D) = (2, 0);PointOnCurve(D, H);A: Point;B: Point;Intersection(H, G) = {A, B};Length(LineSegmentOf(A, B)) = 10;C: Point;MidPoint(LineSegmentOf(A, B)) = C", "query_expressions": "Distance(C, yAxis)", "answer_expressions": "3", "fact_spans": "[[[2, 16], [31, 34]], [[2, 16]], [[28, 30]], [[18, 27]], [[18, 27]], [[17, 30]], [[36, 39]], [[40, 43]], [[28, 43]], [[45, 57]], [[68, 71], [73, 76]], [[59, 71]]]", "query_spans": "[[[73, 86]]]", "process": "From the parabola equation, we know p=4, |AB|=|AF|+|BF|=x_{1}+\\frac{p}{2}+x_{2}+\\frac{p}{2}=x_{1}+x_{2}+4=10, \\therefore x_{1}+x_{2}=6. The distance from the midpoint C of segment AB to the y-axis is \\frac{1}{2}(x_{1}+x_{2})=3." }, { "text": "The minimum distance from points on the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ to the line $x-y+6=0$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2 = 1);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (x - y + 6 = 0)", "query_expressions": "Min(Distance(P, H))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[0, 27]], [[0, 27]], [[29, 30]], [[0, 30]], [[31, 42]], [[31, 42]]]", "query_spans": "[[[29, 51]]]", "process": "" }, { "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, $P$ is a moving point on the right branch of the hyperbola, and $M(4,3)$ is a fixed point, then the minimum value of $|P M|+|P F|$ is?", "fact_expressions": "G: Hyperbola;M: Point;P: Point;F: Point;Expression(G) = (x^2/3 - y^2 = 1);Coordinate(M) = (4, 3);LeftFocus(G) = F;PointOnCurve(P, RightPart(G))", "query_expressions": "Min(Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "sqrt(13)+2*sqrt(3)", "fact_spans": "[[[6, 34], [43, 46]], [[55, 63]], [[39, 42]], [[2, 5]], [[6, 34]], [[55, 63]], [[2, 38]], [[39, 52]]]", "query_spans": "[[[65, 84]]]", "process": "Let the right focus of the hyperbola \\frac{x^{2}}{3}-y^{2}=1 be F, then F(-2,0), F(2,0). Since P is a moving point on the right branch of the hyperbola, by definition we have |PF| - |PF| = 2a = 2\\sqrt{3}, that is, |PF| = |PF| + 2\\sqrt{3}. Hence, |PM| + |PF| = |PM| + |PF| + 2\\sqrt{3} \\geqslant |MF| + 2\\sqrt{3} = \\sqrt{(4-2)^{2}+3^{2}} + 2\\sqrt{3} = \\sqrt{13} + 2\\sqrt{3}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right vertices are denoted as $A$ and $B$ respectively, point $C(0,2b)$. If the perpendicular bisector of segment $AC$ passes through point $B$, 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;A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;C: Point;Coordinate(C) = (0, 2*b);PointOnCurve(B, PerpendicularBisector(LineSegmentOf(A, C)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(7)/2", "fact_spans": "[[[2, 58], [109, 112]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 70]], [[71, 74], [102, 106]], [[2, 74]], [[2, 74]], [[75, 86]], [[75, 86]], [[88, 106]]]", "query_spans": "[[[109, 118]]]", "process": "From the given conditions, |BC| = |BA|, which gives 3a^{2} = 4b^{2}. Then using c^{2} = a^{2} + b^{2} for hyperbolas, the eccentricity can be found. Since the left and right vertices of the hyperbola \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0) are A and B respectively, |AB| = 2a, A(-a, 0), B(a, 0). Also C(0, 2b), and the perpendicular bisector of segment AC passes through point B. Therefore |BC| = |BA|, so \\sqrt{a^{2} + 4b^{2}} = 2a, then b^{2} = \\frac{3}{4}a^{2}. Thus c^{2} = a^{2} + b^{2} = a^{2} + \\frac{3}{4}a^{2} = \\frac{7}{4}a^{2}, therefore e = \\frac{c}{a} = \\sqrt{\\frac{7}{4}} = \\frac{\\sqrt{7}}{2}" }, { "text": "The hyperbola that shares the same asymptotes as $x^{2}-4 y^{2}=4$ and passes through the point $(2,3)$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - 4*y^2 = 4);Z: Hyperbola;Asymptote(G) = Asymptote(Z);H: Point;Coordinate(H) = (2, 3);PointOnCurve(H, Z)", "query_expressions": "Expression(Z)", "answer_expressions": "y^2/8-x^2/32=1", "fact_spans": "[[[1, 21]], [[1, 21]], [[42, 45]], [[0, 45]], [[33, 41]], [[33, 41]], [[31, 45]]]", "query_spans": "[[[42, 47]]]", "process": "" }, { "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are $F_{1}$ and $F_{2}$, respectively. The line passing through $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at points $A$ and $B$. If $|F_{1} F_{2}|=\\frac{\\sqrt{3}}{2}|A B|$, then what is the equation of the asymptotes 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;H: Line;PointOnCurve(F2, H) ;IsPerpendicular(H, xAxis) ;Intersection(H, G) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(F1, F2)) = (sqrt(3)/2)*Abs(LineSegmentOf(A, B))", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*sqrt(2)*x", "fact_spans": "[[[0, 56], [101, 104], [158, 161]], [[0, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[65, 72]], [[73, 80], [82, 89]], [[0, 80]], [[0, 80]], [[98, 100]], [[81, 100]], [[90, 100]], [[98, 115]], [[106, 109]], [[110, 113]], [[117, 156]]]", "query_spans": "[[[158, 169]]]", "process": "\\because AB \\bot x\text{-axis and line }AB\text{ passes through focus }F_{2},\\therefore AB\text{ is the latus rectum, then }|AB|=\\frac{2b^{2}}{a}\\because|F_{1}F_{2}|=\\frac{\\sqrt{3}}{2}|AB|,\\therefore 2c=\\frac{\\sqrt{3}b^{2}}{a}=\\frac{\\sqrt{3}(c^{2}-a^{2})}{a},\text{ i.e. }\\sqrt{3}c^{2}-2ac-\\sqrt{3}a^{2}=0\\therefore\\sqrt{3}e^{2}-2e-\\sqrt{3}=0,\text{ solving: }e=\\sqrt{3}\text{ Also }e^{2}-\\frac{b^{2}}{a^{2}}=1,\\therefore\\frac{b^{2}}{a^{2}}=2,\\therefore\\frac{b}{a}=\\sqrt{2},\\therefore\text{ the asymptote equations of the hyperbola are }y=\\pm\\frac{b}{a}x=\\pm\\sqrt{2}x" }, { "text": "The hyperbola passing through the point $P(2,1)$ shares foci with the ellipse $\\frac{x^{2}}{4}+y^{2}=1$. What is its asymptote equation?", "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;Expression(H) = (x^2/4 + y^2 = 1);Coordinate(P) = (2, 1);Focus(G) = Focus(H);PointOnCurve(P,G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(2)/2*x", "fact_spans": "[[[11, 14], [47, 48]], [[15, 42]], [[1, 10]], [[15, 42]], [[1, 10]], [[11, 45]], [[0, 14]]]", "query_spans": "[[[47, 55]]]", "process": "" }, { "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)$, and there is a point $M$ on the hyperbola such that $|M F_{1}|=\\lambda|M F_{2}|$ $\\left(\\frac{1}{3} \\leq \\lambda \\leq \\frac{1}{2}\\right)$, and $\\angle F_{1} M F_{2}=60^{\\circ}$, find the range of values for the eccentricity of this hyperbola.", "fact_expressions": "G: Hyperbola;b: Number;a: Number;M: Point;F1: Point;F2: Point;a>0;b>0;lambda:Number;lambda>=1/3;lambda<=1/2;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(M, G);Abs(LineSegmentOf(M, F1)) = lambda*Abs(LineSegmentOf(M, F2));AngleOf(F1, M, F2) = ApplyUnit(60, degree)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(7)/2,sqrt(3)]", "fact_spans": "[[[20, 77], [84, 87], [206, 209]], [[23, 77]], [[23, 77]], [[91, 94]], [[2, 9]], [[10, 17]], [[23, 77]], [[23, 77]], [[97, 168]], [[97, 168]], [[97, 168]], [[20, 77]], [[2, 83]], [[2, 83]], [[84, 94]], [[97, 168]], [[170, 203]]]", "query_spans": "[[[206, 219]]]", "process": "From the definition in the diagram, we have: $ t - \\lambda t = 2a \\Rightarrow t = \\frac{2a}{1 - \\lambda} $, and $ \\angle F_{1}MF_{2} = 60^{\\circ} $. By the law of cosines: $ \\cos 60^{\\circ} = \\frac{t^{2} + \\lambda^{2}t^{2} - 4c^{2}}{2\\lambda t^{2}} = \\frac{1}{2} \\Rightarrow (\\lambda^{2} - \\lambda + 1)t^{2} = 4c^{2} $. Substituting $ t = \\frac{2a}{1 - \\lambda} $, we get $ (\\lambda^{2} - \\lambda + 1)\\left( \\frac{2a}{1 - \\lambda} \\right)^{2} = 4c^{2} \\Rightarrow e^{2} = \\frac{c^{2}}{a^{2}} = \\frac{\\lambda^{2} - \\lambda + 1}{(1 - \\lambda)^{2}} $ for $ \\frac{1}{3} \\leqslant \\lambda \\leqslant \\frac{1}{2} $. Let $ f(\\lambda) = \\frac{\\lambda^{2} - \\lambda + 1}{(1 - \\lambda)^{2}} $ for $ \\frac{1}{3} \\leqslant \\lambda \\leqslant \\frac{1}{2} $. Then $ f(\\lambda) = \\frac{\\lambda^{2} - \\lambda + 1}{(1 - \\lambda)^{2}} = \\frac{(1 - \\lambda)^{2} + \\lambda}{(1 - \\lambda)^{2}} = 1 + \\frac{\\lambda}{(1 - \\lambda)^{2}} $. From the properties of the function, the function $ y = \\lambda + \\frac{1}{\\lambda} - 2 $ is monotonically decreasing on $ [\\frac{1}{3}, \\frac{1}{2}] $, hence the function $ f(\\lambda) = 1 + \\frac{\\lambda}{(1 - \\lambda)^{2}} $ is monotonically increasing on $ [\\frac{1}{3}, \\frac{1}{2}] $. Therefore, $ f(\\lambda)_{\\min} = f(\\frac{1}{3}) = \\frac{7}{4} $, $ f(\\lambda)_{\\max} = f(\\frac{1}{2}) = 3 $, so $ e^{2} \\in [\\frac{7}{4}, 3] \\Rightarrow e \\in [\\frac{\\sqrt{7}}{2}, \\sqrt{3}] $." }, { "text": "Given the parabola $y^{2}=8x$, a line passing through the point $M(1,0)$ intersects the parabola at points $A$ and $B$, $F$ is the focus of the parabola. If $|AF|=6$, and $O$ is the origin, then the area of $\\triangle OAB$ is?", "fact_expressions": "G: Parabola;H: Line;M: Point;O: Origin;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 8*x);Coordinate(M) = (1, 0);PointOnCurve(M, H);Intersection(H, G) = {A, B};Focus(G) = F;Abs(LineSegmentOf(A, F)) = 6", "query_expressions": "Area(TriangleOf(O, A, B))", "answer_expressions": "5*sqrt(2)/2", "fact_spans": "[[[2, 16], [31, 34], [49, 52]], [[28, 30]], [[18, 27]], [[69, 72]], [[35, 38]], [[39, 42]], [[45, 48]], [[2, 16]], [[18, 27]], [[17, 30]], [[28, 44]], [[45, 55]], [[57, 66]]]", "query_spans": "[[[79, 101]]]", "process": "By the property of the parabola, the distance to the focus equals the distance to the directrix, so find the coordinates of A. Then, since the line passes through point M, find the equation of line AB, solve simultaneously with the parabola to find the coordinates of B, and use the area formula to find the area of triangle AOB. The directrix of the parabola $ y^{2} = 8x $ is $ x = -2 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Draw a perpendicular from point A to the directrix, meeting at H. As shown in the figure, by the definition of the parabola, $ |AF| = |AH| = 6 $. Therefore, $ x_{1} + 2 = 6 $, so $ x_{1} = 4 $, $ y_{1} = 4\\sqrt{2} $. Let the equation of line AB be $ y = k(x - 1) $ ($ k \\neq 0 $). From \n$$\n\\begin{cases}\ny = k(x - 1) \\\\\ny^{2} = 8x\n\\end{cases}\n$$\nwe obtain $ k^{2}x^{2} - (2k^{2} + 8)x + k^{2} = 0 $, $ x_{1}x_{2} = 1 $, $ x_{2} = \\frac{1}{x_{1}} = \\frac{1}{4} $, so $ y_{2} = -\\sqrt{2} $, thus the area of quadrilateral AOAB is $ S_{AOAB} = S_{AAOM} + S_{ABOM} = \\frac{1}{2}|y_{1}| \\times 1 + \\frac{1}{2}|y_{2}| \\times 1 = \\frac{1}{2} \\times (4\\sqrt{2} + \\sqrt{2}) = \\frac{5\\sqrt{2}}{2} $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1$, and point $P$ lies on the hyperbola such that $P F_{1}=2 P F_{2}$, find the area of $\\triangle P F_{1} F_{2}$.", "fact_expressions": "F1: Point;F2: Point;G: Hyperbola;Expression(G) = (x^2/4 - y^2/8 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2)", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "8*sqrt(3)", "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 56], [68, 71]], [[18, 56]], [[2, 62]], [[2, 62]], [[63, 67]], [[63, 72]], [[75, 94]]]", "query_spans": "[[[96, 126]]]", "process": "" }, { "text": "Given that the line $y=k(x+2)$ intersects the parabola $C$: $y^{2}=8x$ at two points $A$ and $B$, and $F$ is the focus of the parabola $C$. If $|\\overrightarrow{F A}|=2|\\overrightarrow{F B}|$, then the real number $k$=?", "fact_expressions": "C: Parabola;G: Line;F: Point;A: Point;B: Point;k: Real;Expression(C) = (y^2 = 8*x);Expression(G) = (y = k*(x + 2));Intersection(G, C) = {A, B};Focus(C) = F;Abs(VectorOf(F, A)) = 2*Abs(VectorOf(F, B))", "query_expressions": "k", "answer_expressions": "pm*2*sqrt(2)/3", "fact_spans": "[[[15, 34], [51, 57]], [[2, 14]], [[47, 50]], [[37, 40]], [[41, 44]], [[112, 117]], [[15, 34]], [[2, 14]], [[2, 46]], [[47, 60]], [[62, 110]]]", "query_spans": "[[[112, 119]]]", "process": "Let P(-2,0), and x = -2 be the directrix of the parabola. Draw perpendiculars from points A and B to the directrix, with feet at M and N respectively. Then |BN| = |FB|, |AM| = |FA|, so |BN| : |AM| = 1 : 2, hence |BP| = |BA|. Let B(a,b), then A(2+2a,2b). Thus \n\\begin{cases}b^{2}=8a\\\\4b^{2}=8(2+2a)\\end{cases} \nSolving gives \n\\begin{cases}a=1\\\\b=\\pm2\\sqrt{2}\\end{cases} \nHence _{k=\\pm\\frac{2\\sqrt{2}}{3}}. Fill in \\underline{2\\sqrt{2}}. For problems involving foci or directrices in conic sections, consider using their geometric properties. For example, the distance from a point on a parabola to the focus can be converted to the distance to the directrix; the distance from a point on an ellipse to one focus can be converted to the distance to the other focus, or to the corresponding directrix." }, { "text": "The coordinates of the focus of the parabola $y=-\\frac{1}{6} x^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/6)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,-3/2)", "fact_spans": "[[[0, 25]], [[0, 25]]]", "query_spans": "[[[0, 32]]]", "process": "" }, { "text": "If the curve represented by the equation $\\frac{x^{2}}{k-1}+\\frac{y^{2}}{2}=1$ is an ellipse, then the range of real values for $k$ is?", "fact_expressions": "G: Curve;Expression(G) = (x^2/(k - 1) + y^2/2 = 1);H: Ellipse;G = H;k: Real", "query_expressions": "Range(k)", "answer_expressions": "(1,3)+(3,+oo)", "fact_spans": "[[[43, 45]], [[1, 45]], [[46, 48]], [[43, 48]], [[50, 55]]]", "query_spans": "[[[50, 62]]]", "process": "The equation $\\frac{x^2}{k-1} + \\frac{y^2}{2} = 1$ represents an ellipse, then: \n\\begin{cases} k-1>0 \\\\ k-1\\neq2 \\end{cases}, \nsolving gives: $k>1$ and $k\\neq3$." }, { "text": "The focal distance of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G)=(x^2/9+y^2/4=1)", "query_expressions": "FocalLength(G)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 42]]]", "process": "From the given condition, $ c = \\sqrt{a^{2} - b^{2}} = \\sqrt{9 - 4} = \\sqrt{5} $, hence the focal distance is $ 2c = 2\\sqrt{5} $." }, { "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 $A$ on the right branch such that the distance from point $F_{2}$ to the line $A F_{1}$ is $2a$, 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;F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;A: Point;PointOnCurve(A, RightPart(G));Distance(F2, LineOf(A, F1)) = 2*a", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(2),+oo)", "fact_spans": "[[[19, 76], [129, 132]], [[19, 76]], [[22, 76]], [[22, 76]], [[22, 76]], [[22, 76]], [[1, 8]], [[9, 16], [97, 105]], [[1, 82]], [[1, 82]], [[90, 94]], [[19, 94]], [[97, 126]]]", "query_spans": "[[[129, 143]]]", "process": "Test Analysis: As shown in the figure below, let $ AF_{2} = x $, then $ AF_{1} = x + 2a_{n} $. Therefore, $ F_{1}H = \\sqrt{4c^{2} - 4a^{2}} = 2b $, $ AH = x + 2a - 2b $. In right triangle $ \\triangle AHF_{2} $, $ (x + 2a - 2b)^{2} + 4a^{2} = x^{2} \\Rightarrow x = \\frac{a^{2}}{b - a} + b - a \\geqslant c - a \\Rightarrow \\frac{a^{2}}{b - a} + b - c \\geqslant 0 \\Rightarrow \\frac{a^{2} + b^{2} - bc - ab + ac}{b - a} \\geqslant 0 \\Rightarrow \\frac{c^{2} - bc - ab + ac}{b - a} \\geqslant 0 \\Rightarrow \\frac{(c + a)(c - b)}{b - a} \\geqslant 0 \\Rightarrow b > a \\Rightarrow c^{2} - a^{2} > a^{2} $" }, { "text": "Given that the eccentricity of ellipse $E$ is $e$, the left and right foci are $F_{1}$, $F_{2}$ respectively, parabola $C$ has $F_{1}$ as vertex and $F_{2}$ as focus, $P$ is an intersection point of the two curves, $\\frac{|P F_{1}|}{|P F_{2}|}=e$, then the value of $e$ is?", "fact_expressions": "E: Ellipse;e: Number;Eccentricity(E) = e;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;C: Parabola;Vertex(C) = F1;Focus(C) = F2;P: Point;OneOf(Intersection(E, C)) = P;Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2)) = e", "query_expressions": "e", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 7]], [[12, 15], [113, 116]], [[2, 15]], [[23, 30], [46, 53]], [[31, 38], [56, 63]], [[2, 38]], [[2, 38]], [[39, 45]], [[39, 55]], [[39, 66]], [[67, 70]], [[67, 79]], [[80, 111]]]", "query_spans": "[[[113, 120]]]", "process": "" }, { "text": "The distance from the focus of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ 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": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 40]]]", "process": "The foci of the hyperbola are $(\\pm2,0)$, the equations of the asymptotes are $y=\\pm\\frac{\\sqrt{3}}{3}x$, or equivalently $x\\pm\\sqrt{3}y=0$, and the distance from a focus to an asymptote is $\\frac{2}{\\sqrt{1+3}}=1$." }, { "text": "Given that the center of an ellipse is at the origin and its eccentricity is $\\frac{4}{5}$, and one focus is $(0,4)$, what is the equation of the directrix of this ellipse?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;Eccentricity(G) = 4/5;Coordinate(OneOf(Focus(G)))=(0,4)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y = pm*25/4", "fact_spans": "[[[28, 30], [46, 48]], [[5, 9]], [[2, 30]], [[10, 30]], [27, 41]]", "query_spans": "[[[46, 55]]]", "process": "" }, { "text": "The line passing through the center of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ intersects the ellipse at points $M$ and $N$ (point $M$ is in the first quadrant). A perpendicular is drawn from point $M$ to the $x$-axis, with foot at point $E$. The line $N E$ intersects the ellipse again at point $P$. Then the value of $\\cos \\angle N M P$ is?", "fact_expressions": "G: Ellipse;H: Line;E: Point;N: Point;M: Point;P: Point;l:Line;Expression(G) = (x^2/2 + y^2 = 1);PointOnCurve(Center(G),H);Intersection(H,G)={M,N};Quadrant(M)=1;PointOnCurve(M,l);IsPerpendicular(l,xAxis);FootPoint(l,xAxis)=E;OneOf(Intersection(LineOf(N,E),G))=P", "query_expressions": "Cos(AngleOf(N, M, P))", "answer_expressions": "0", "fact_spans": "[[[2, 29], [35, 37], [93, 95]], [[32, 34]], [[78, 82]], [[44, 47]], [[40, 43], [50, 54], [62, 66]], [[102, 105]], [], [[2, 29]], [[0, 34]], [[32, 49]], [[50, 59]], [[61, 74]], [[61, 74]], [[61, 82]], [[85, 105]]]", "query_spans": "[[[108, 131]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left and right vertices are points $A$ and $B$, respectively. Point $C(0, \\sqrt{2} b)$. If the perpendicular bisector of segment $AC$ passes through point $B$, 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;A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;C: Point;Coordinate(C) = (0, sqrt(2)*b);PointOnCurve(B, PerpendicularBisector(LineSegmentOf(A, C)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[2, 58], [118, 121]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[67, 70]], [[71, 74], [112, 116]], [[2, 76]], [[2, 76]], [[77, 96]], [[77, 96]], [[98, 116]]]", "query_spans": "[[[118, 127]]]", "process": "From the given conditions, △ABC is an equilateral triangle, then $\\sqrt{2}b=\\sqrt{3}c$, so the eccentricity of the hyperbola $\\sqrt{1+(\\frac{b}{a})^{2}}=\\frac{\\sqrt{10}}{2}$" }, { "text": "Given that point $F$ is the focus of the parabola $C$: $y^{2}=2 p x(p>0)$, a line $l$ passing through point $F$ intersects $C$ at points $A$ and $B$, and intersects the directrix of $C$ at point $M$. If $\\overrightarrow{A B}+\\overrightarrow{A M}=\\overrightarrow{0}$, then the value of $|A B|$ equals?", "fact_expressions": "l: Line;C: Parabola;p: Number;A: Point;B: Point;M: Point;F: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Intersection(l,Directrix(C)) = M;VectorOf(A, B) + VectorOf(A, M) = 0", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "(9/4)*p", "fact_spans": "[[[43, 48]], [[7, 33], [49, 52], [64, 67]], [[14, 33]], [[53, 56]], [[57, 60]], [[72, 76]], [[2, 6], [38, 42]], [[14, 33]], [[7, 33]], [[2, 36]], [[37, 48]], [[43, 62]], [[43, 76]], [[78, 140]]]", "query_spans": "[[[142, 154]]]", "process": "As shown in the figure, draw perpendiculars from points A and B to the directrix l of the parabola C, with feet at points D and E respectively. \n∵ $\\overrightarrow{AB} + \\overrightarrow{AM} = \\overrightarrow{0}$, then point A is the midpoint of segment BM, \n∴ $|BE| = 2|AD|$, by the definition of the parabola, we have $|AD| = |AF|$, $|BE| = |BF|$, \n∴ $|BF| = 2|AF|$, $|AM| = |AB| = 3|AF|$. \n∵ $AD \\parallel FN$, ∴ $\\frac{|AD|}{|FN|} = \\frac{|AM|}{|FM|} = \\frac{3}{4}$, \n∴ $|AD| = \\frac{3}{4}|FN| = \\frac{3p}{4}$. \nTherefore, $|AB| = 3|AF| = 3|AD| = \\frac{9}{4}p$" }, { "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=4x$. A line $l$ passing through the point $P(-1,0)$ intersects the parabola $C$ at points $A$ and $B$. Point $Q$ is the midpoint of segment $AB$. If $|FQ|=2$, then the slope of line $l$ equals?", "fact_expressions": "l: Line;C: Parabola;B: Point;A: Point;P: Point;F: Point;Q: Point;Expression(C) = (y^2 = 4*x);Coordinate(P) = (-1, 0);Focus(C) = F;PointOnCurve(P, l);Intersection(l, C) = {A, B};MidPoint(LineSegmentOf(A,B)) = Q;Abs(LineSegmentOf(F, Q)) = 2", "query_expressions": "Slope(l)", "answer_expressions": "pm*1", "fact_spans": "[[[40, 45], [92, 97]], [[5, 24], [46, 52]], [[57, 60]], [[53, 56]], [[29, 39]], [[1, 4]], [[63, 67]], [[5, 24]], [[29, 39]], [[1, 27]], [[28, 45]], [[40, 62]], [[63, 78]], [[80, 90]]]", "query_spans": "[[[92, 103]]]", "process": "" }, { "text": "Through the focus $F$ of the parabola $y^{2}=2x$, draw a line $l$ intersecting the parabola at points $A$ and $B$. If $\\frac{1}{|A F|}-\\frac{1}{|B F|}=1$, then the inclination angle $\\theta$ of the line $l$ is?", "fact_expressions": "l: Line;G: Parabola;A: Point;F: Point;B: Point;theta: Number;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l, G) = {A, B};-1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F)) = 1;Inclination(l) = theta", "query_expressions": "theta", "answer_expressions": "{pi/3,2*pi/3}", "fact_spans": "[[[22, 27], [80, 85]], [[1, 15], [28, 31]], [[32, 35]], [[18, 21]], [[36, 39]], [[89, 97]], [[1, 15]], [[1, 21]], [[0, 27]], [[22, 41]], [[43, 78]], [[80, 97]]]", "query_spans": "[[[89, 99]]]", "process": "" }, { "text": "Given that point $A(-1,0)$ is the intersection of the directrix of the parabola $y^{2}=2 p x$ and the $x$-axis, $F$ is the focus of the parabola, and $P$ is a moving point on the parabola, then the minimum value of $\\frac{|P F|}{|P A|}$ is?", "fact_expressions": "G: Parabola;p: Number;A: Point;P: Point;F: Point;Expression(G) = (y^2 = 2*p*x);Coordinate(A) = (-1, 0);Intersection(Directrix(G), xAxis) = A;Focus(G) = F;PointOnCurve(P, G)", "query_expressions": "Min(Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(P, A)))", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[13, 29], [45, 48], [56, 59]], [[16, 29]], [[2, 12]], [[52, 55]], [[41, 44]], [[13, 29]], [[2, 12]], [[2, 40]], [[41, 51]], [[52, 63]]]", "query_spans": "[[[65, 91]]]", "process": "From the given conditions, we have p=2. Let point P be (x, y), and let d be the distance from P to the line x=-1. Then d=x+1, so \\frac{|PF|}{|PA|}=\\frac{d}{|PA|}=\\frac{x+1}{\\sqrt{(x+1)^{2}+4x}}=\\frac{1}{\\sqrt{1+\\frac{4x}{(x+1)^{2}}}\\frac{\\sqrt{2}}{2}=\\frac{4}{2}\\sqrt{1+\\frac{4}{x+\\frac{1}{x}+2}}. The minimum value of \\frac{|PF|}{|PA|} is \\frac{\\sqrt{2}}{2}, achieved if and only if x=\\frac{1}{x}, at which point x=1." }, { "text": "Given $A(-1,0)$, $B(1,0)$, and $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$, then the equation of the locus of the moving point $M$ is?", "fact_expressions": "A: Point;Coordinate(A) = (-1, 0);B: Point;Coordinate(B) = (1, 0);M: Point;DotProduct(VectorOf(M, A), VectorOf(M, B)) = 0", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2+y^2=1", "fact_spans": "[[[2, 11]], [[2, 11]], [[13, 21]], [[13, 21]], [[78, 81]], [[23, 74]]]", "query_spans": "[[[78, 88]]]", "process": "From the coordinate operation of vector perpendicularity, we obtain the answer. Let M(x, y), then \\overrightarrow{MA}=(-1-x,-y), \\overrightarrow{MB}=(1-x,-y). Since \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=0, it follows that \\overrightarrow{MA}\\cdot\\overrightarrow{MB}=x^{2}-1+y^{2}=0, so the trajectory equation is x^{2}+y^{2}=1." }, { "text": "Given the ellipse $E$: $\\frac{x^{2}}{2 m}+\\frac{y^{2}}{m}=1$ $(m>0)$ has its right focus at point $F$. A line passing through point $F$ intersects the ellipse $E$ at points $A$ and $B$. If the coordinates of the midpoint of $AB$ are $(1,-1)$, then what is the equation of the ellipse $E$?", "fact_expressions": "E: Ellipse;m: Number;G: Line;A: Point;B: Point;F: Point;m>0;Expression(E) = (x^2/(2*m) + y^2/m = 1);Coordinate(MidPoint(LineSegmentOf(A, B))) = (1, -1);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, 51], [69, 74], [107, 112]], [[9, 51]], [[66, 68]], [[75, 78]], [[79, 82]], [[56, 59]], [[9, 51]], [[2, 51]], [[86, 105]], [[2, 59]], [[60, 68]], [[66, 84]]]", "query_spans": "[[[107, 117]]]", "process": "From the given conditions, point F is at $(\\sqrt{m}, 0)$, so the slope of line AB is $k = \\frac{1}{\\sqrt{m}-1}$. Let the coordinates of points A and B be $(x_{1}, y_{1})$, $(x_{2}, y_{2})$, respectively. Then \n$$\n\\begin{cases}\n\\frac{x^{2}}{2m} + \\frac{y^{2}}{m} = 1 \\\\\n\\frac{x^{2}}{2} + \\frac{y^{2}}{2} = 1\n\\end{cases}\n$$\nSubtracting the two equations and simplifying yields \n$\\frac{y_{1}-y_{2}}{x_{1}} = \\frac{x_{1}+x_{2}}{2(y_{1}+y_{2})} = $. Therefore, $k = \\frac{m}{\\sqrt{m}-1} = \\frac{1}{2}$, solving gives $m = 9$, $\\therefore$ the equation of ellipse E is $\\frac{x^{2}}{18} + \\frac{y^{2}}{9} = 1$." }, { "text": "If the foci of a hyperbola lie on the $y$-axis and the eccentricity is $e=2$, then what is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), yAxis);Eccentricity(G)=e;e=2;e:Number", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(3)/3)*x", "fact_spans": "[[[1, 4], [25, 26]], [[1, 13]], [[1, 22]], [[17, 22]], [[17, 22]]]", "query_spans": "[[[25, 33]]]", "process": "\\because the eccentricity is 2, i.e., \\frac{c}{a}=2, let c=2k then a=k; and \\because c^{2}=a^{2}+b^{2}, \\therefore b=\\sqrt{3}k; also \\because the foci of the hyperbola lie on the y-axis, \\therefore the asymptotes of the hyperbola are y=\\pm\\frac{a}{b}x=\\pm\\frac{\\sqrt{3}}{2}x" }, { "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, and $P$ is a point on the right branch of the hyperbola such that $|P F_{2}|=|F_{1} F_{2}|$, and the line $P F_{1}$ intersects the circle $x^{2}+y^{2}=a^{2}$, then what is the range of values for the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Circle;F1: Point;P: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));IsIntersect(LineOf(P,F1), H)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,5/3]", "fact_spans": "[[[20, 76], [87, 90], [162, 165]], [[23, 76]], [[23, 76]], [[136, 156]], [[2, 9]], [[83, 86]], [[10, 17]], [[23, 76]], [[23, 76]], [[20, 76]], [[136, 156]], [[2, 82]], [[2, 82]], [[83, 95]], [[98, 123]], [[124, 160]]]", "query_spans": "[[[162, 176]]]", "process": "Draw OM ⊥ PF₁ from point O to M, and draw F₂N ⊥ PF₁ from point F₂ to N. Using the definition of a hyperbola and the Pythagorean theorem, |OM|² can be found. Given that |OM|² ≤ a², a homogeneous inequality in terms of a and c can be obtained. Combining with e > 1, the range of e can be determined. [Detailed solution] Draw OM ⊥ PF₁ from point O to M, and draw F₂N ⊥ PF₁ from point F₂ to N. Since |PF₂| = |F₁F₂|, it follows that |PN| = |F₁N|. Also, since |OF₂| = |OF₁|, we have |MN| = |F₁M|, hence |F₁M| = (1/4)|F₁P|. Since |PF₁| − |PF₂| = 2a and |PF₂| = |F₁F₂| = 2c, it follows that |PF₁| = 2a + 2c. Therefore, |F₁M| = (a + c)/2, so |OM|² = c² − ((a + c)/2)². Since line PF₁ has a common point with the circle x² + y² = a², we have |OM|² ≤ a², thus c² − ((a + c)/2)² ≤ a². That is, 3c² − 2ac − 5a² ≤ 0, then 3e² − 2e − 5 ≤ 0, so −1 ≤ e ≤ 5/3. Since the eccentricity of the hyperbola satisfies e > 1, we obtain 1 < e ≤ 5/3." }, { "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 segment connecting points $A$ and $B$ passes through the origin. If the product of the slopes of lines $PA$ and $PB$ is $k_{PA} \\cdot k_{PB} = \\frac{2}{3}$, then the eccentricity $e$ of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;A: Point;P: Point;B: Point;O:Origin;e: Number;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));Slope(LineOf(P, B)) = k2;Slope(LineOf(P,A))=k1;k1:Number;k2:Number;k1*k2=2/3;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(15)/3", "fact_spans": "[[[14, 71], [155, 158]], [[17, 71]], [[17, 71]], [[2, 5], [79, 82]], [[10, 13]], [[6, 9], [83, 86]], [[92, 96]], [[162, 165]], [[17, 71]], [[17, 71]], [[14, 71]], [[2, 77]], [[2, 77]], [[2, 77]], [[2, 77]], [[2, 77]], [[2, 77]], [[79, 96]], [[98, 152]], [[98, 152]], [[117, 152]], [[117, 152]], [[117, 152]], [[155, 165]]]", "query_spans": "[[[162, 167]]]", "process": "" }, { "text": "The parabola takes the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ as its vertex and the left directrix of the ellipse as its directrix. This parabola intersects the right directrix of the ellipse at points $A$ and $B$. Then $|A B|=$?", "fact_expressions": "G: Parabola;H: Ellipse;A: Point;B: Point;Expression(H) = (x^2/25 + y^2/16 = 1);Vertex(G)=Center(H);LeftDirectrix(H)=Directrix(G);Intersection(RightDirectrix(H),G)={A,B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "100/3", "fact_spans": "[[[58, 61]], [[1, 40], [48, 50], [62, 64]], [[69, 73]], [[74, 78]], [[1, 40]], [[0, 61]], [[47, 61]], [[58, 80]]]", "query_spans": "[[[82, 91]]]", "process": "" }, { "text": "The standard equation of the ellipse passing through the point $(\\sqrt{3},\\sqrt{5})$ and having the same foci as the ellipse $\\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1$ is?", "fact_expressions": "G: Ellipse;H: Point;Expression(G) = (x^2/9 + y^2/25 = 1);Coordinate(H) = (sqrt(3), sqrt(5));C:Ellipse;Focus(C)=Focus(G);PointOnCurve(H,C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/20+x^2/4=1", "fact_spans": "[[[26, 64]], [[1, 23]], [[26, 64]], [[1, 23]], [[70, 72]], [[25, 72]], [[0, 72]]]", "query_spans": "[[[70, 79]]]", "process": "According to the given conditions, the foci of the desired ellipse lie on the y-axis, and c^{2}=25-9=16. Let its standard equation be \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 (a>b>0). Based on the given conditions, set up a system of equations and solve it to obtain the result. \nBecause the desired ellipse has the same foci as the ellipse \\frac{y^{2}}{25}+\\frac{x^{2}}{9}=1, its foci lie on the y-axis, and c^{2}=25-9=16. Let its standard equation be \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 (a>b>0). Since c^{2}=16 and c^{2}=a^{2}-b^{2}, it follows that a^{2}-b^{2}=16\\textcircled{1}. Also, the point (\\sqrt{3},\\sqrt{5}) lies on the desired ellipse, so \\frac{5}{a^{2}}+\\frac{3}{b^{2}}=1\\textcircled{2}. From \\textcircled{1}\\textcircled{2}, we get b^{2}=4, a^{2}=20. Therefore, the standard equation of the desired ellipse is \\frac{y^{2}}{20}+\\frac{x^{2}}{4}=1." }, { "text": "Given the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ with left and right foci $F_{1}$, $F_{2}$, $P$ is a point on the ellipse, $O$ is the origin, $M$ is the midpoint of $P F_{1}$, if $|P F_{1}|=4$, then $|O M|$=?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;O: Origin;M: Point;Expression(G) = (x^2/9 + y^2/5 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);MidPoint(LineSegmentOf(P, F1)) = M;Abs(LineSegmentOf(P, F1)) = 4", "query_expressions": "Abs(LineSegmentOf(O, M))", "answer_expressions": "1", "fact_spans": "[[[2, 39], [65, 67]], [[61, 64]], [[45, 52]], [[53, 60]], [[71, 74]], [[80, 83]], [[2, 39]], [[2, 60]], [[2, 60]], [[61, 70]], [[80, 96]], [[98, 111]]]", "query_spans": "[[[113, 122]]]", "process": "By the definition of an ellipse, |PF_{1}| + |PF_{2}| = 6, and |PF_{1}| = 4, so |PF_{2}| = 2. O and M are the midpoints of F_{1}F_{2} and PF_{1}, respectively, so OM is the midline of triangle PF_{2}, hence |OM| = \\frac{1}{2}|PF_{2}| = 1" }, { "text": "The hyperbola has foci at $F_{1}(-3 , 0)$, $F_{2}(3 , 0)$ and the length of the imaginary axis is $\\sqrt{2}$ times the length of the real axis. Then the standard equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;F1: Point;F2: Point;Coordinate(F1) = (-3, 0);Coordinate(F2) = (3, 0);Focus(G) = {F1, F2};Length(ImageinaryAxis(G)) = sqrt(2)*Length(RealAxis(G))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/3 - y^2/6 = 1", "fact_spans": "[[[0, 3], [62, 65]], [[4, 19]], [[22, 36]], [[4, 19]], [[22, 36]], [[0, 39]], [[0, 59]]]", "query_spans": "[[[62, 72]]]", "process": "" }, { "text": "The line $ l $ passes through the focus of the parabola $ y^{2} = 2x $ and intersects the parabola at points $ A $ and $ B $. The circle $ M $, with diameter $ |AB| $, is tangent to the directrix of the parabola. If the area of circle $ M $ is $ 16\\pi $, then what is the slope of line $ l $?", "fact_expressions": "l: Line;G: Parabola;Expression(G) = (y^2 = 2*x);PointOnCurve(Focus(G), l);A: Point;B: Point;Intersection(l, G) = {A, B};M: Circle;IsDiameter(Abs(LineSegmentOf(A, B)), M);IsTangent(Directrix(G), M);Area(M) = 16*pi", "query_expressions": "Slope(l)", "answer_expressions": "pm*(sqrt(3)/3)", "fact_spans": "[[[0, 5], [85, 90]], [[6, 20], [25, 26], [57, 60]], [[6, 20]], [[0, 23]], [[28, 32]], [[33, 36]], [[0, 36]], [[51, 55], [67, 71]], [[37, 55]], [[51, 65]], [[67, 83]]]", "query_spans": "[[[85, 95]]]", "process": "From the problem, it is clear that the slope of line $ l $ exists, so the equation of the line can be written as $ y = k\\left(x - \\frac{1}{2}\\right) $, with $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Given that the circle $ M $ with diameter $ |AB| $ is tangent to the directrix of the parabola and the area of circle $ M $ is $ 16\\pi $, the radius of the circle is $ r = 4 $, hence $ |AB| = 8 $. From \n\\[\n\\begin{cases}\ny = k\\left(x - \\frac{1}{2}\\right) \\\\\ny^{2} = 2x\n\\end{cases}\n\\]\nwe obtain $ k^{2}x^{2} - (k^{2} + 2)x + \\frac{k^{2}}{4} = 0 $, so $ x_{1} + x_{2} = \\frac{k^{2} + 2}{k^{2}} $. Thus, $ |AB| = x_{1} + x_{2} + 1 = \\frac{k^{2} + 2}{k^{2}} + 1 = 8 $, solving gives $ k = \\pm\\frac{\\sqrt{3}}{3} $." }, { "text": "If the parabola $C$ has its vertex at the origin, its focus on the $y$-axis, and passes through the point $(2,1)$, then what is the standard equation of $C$?", "fact_expressions": "C: Parabola;O: Origin;Vertex(C) = O;PointOnCurve(Focus(C), yAxis);G: Point;Coordinate(G) = (2, 1);PointOnCurve(G, C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2=4*y", "fact_spans": "[[[1, 7], [34, 37]], [[10, 12]], [[1, 12]], [[1, 21]], [[24, 32]], [[24, 32]], [[1, 32]]]", "query_spans": "[[[34, 44]]]", "process": "Since the vertex of parabola C is at the origin and the focus is on the y-axis, the equation of the parabola is assumed to be x^{2}=my. Since the parabola passes through the point (2,1), we have 2^{2}=m, so m=4. Therefore, the equation of the parabola is x^{2}=4y." }, { "text": "Given two fixed points in the plane $A(-5,0)$, $B(5,0)$, a moving point $M$ satisfies $||MA| - |MB|| = 6$, then the equation of the trajectory of point $M$ is?", "fact_expressions": "A: Point;B: Point;Coordinate(A) = (-5, 0);Coordinate(B) = (5, 0);M: Point;Abs(Abs(LineSegmentOf(M, A)) - Abs(LineSegmentOf(M, B))) = 6", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2/9-y^2/16=1", "fact_spans": "[[[8, 17]], [[20, 28]], [[8, 17]], [[20, 28]], [[31, 34], [57, 61]], [[37, 55]]]", "query_spans": "[[[57, 68]]]", "process": "From the given condition: |AB| = 10, ||MA| - |MB|| = 6 < |AB|, thus the locus of M is a hyperbola with foci at A and B and a major axis length of 2a = 6. Let the equation of the hyperbola be \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1. Given a = 3, c = 5, we obtain b^{2} = c^{2} - a^{2} = 16. Hence, the equation of the locus of point M is \\frac{x^{2}}{9} - \\frac{y^{2}}{16} = 1" }, { "text": "The coordinates of the foci of the ellipse $x^{2}+4 y^{2}=4$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 4*y^2 = 4)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*sqrt(3),0)", "fact_spans": "[[[0, 19]], [[0, 19]]]", "query_spans": "[[[0, 26]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left vertex is $A$, the upper vertex is $B$, and the right focus is $F$. Let $M$ be the midpoint of segment $AB$. If $2 \\overrightarrow{M A} \\cdot \\overrightarrow{M F}+\\overrightarrow{B F}^2 \\geq 0$, 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;A: Point;LeftVertex(G) = A;B: Point;UpperVertex(G) = B;F: Point;RightFocus(G) = F;MidPoint(LineSegmentOf(A,B)) = M;M: Point;DotProduct(VectorOf(M, F), 2*VectorOf(M, A)) + VectorOf(B, F)^2 >= 0", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(0,sqrt(3)-1]", "fact_spans": "[[[2, 54], [181, 183]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[59, 62]], [[2, 62]], [[67, 70]], [[2, 70]], [[75, 78]], [[2, 78]], [[81, 95]], [[92, 95]], [[97, 178]]]", "query_spans": "[[[181, 193]]]", "process": "" }, { "text": "Given that the line $x - y = 2$ intersects the parabola $y^2 = 4x$ at points $A$ and $B$, what are the coordinates of the midpoint of segment $AB$?", "fact_expressions": "G: Parabola;H: Line;B: Point;A: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x - y = 2);Intersection(H, G) = {A, B}", "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "(4, 2)", "fact_spans": "[[[12, 26]], [[2, 11]], [[32, 35]], [[28, 31]], [[12, 26]], [[2, 11]], [[2, 37]]]", "query_spans": "[[[40, 54]]]", "process": "" }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$ is $\\frac{1}{2}$, then $m$=?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/4 + y^2/m = 1);Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "{16/3,3}", "fact_spans": "[[[1, 38]], [[58, 61]], [[1, 38]], [[1, 56]]]", "query_spans": "[[[58, 63]]]", "process": "" }, { "text": "If the focus of the parabola ${y}^{2}=2 px$ has coordinates $(1,0)$, then what is the equation of the directrix?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;Coordinate(Focus(G)) = (1,0)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-1", "fact_spans": "[[[1, 18]], [[1, 18]], [[4, 18]], [[1, 31]]]", "query_spans": "[[[1, 39]]]", "process": "" }, { "text": "Let the focus of the parabola $y=4 x^{2}$ be $F$, then what are the coordinates of point $F$?", "fact_expressions": "G: Parabola;F: Point;Expression(G) = (y = 4*x^2);Focus(G) = F", "query_expressions": "Coordinate(F)", "answer_expressions": "(0,1/16)", "fact_spans": "[[[1, 15]], [[19, 22], [24, 28]], [[1, 15]], [[1, 22]]]", "query_spans": "[[[24, 33]]]", "process": "" }, { "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}}{12-m}-\\frac{y^{2}}{m+4}=1$, then the value of the real number $p$ is?", "fact_expressions": "G: Hyperbola;m: Number;H: Parabola;p: Real;Expression(G) = (x^2/(12 - m) - y^2/(m + 4) = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Focus(H) = RightFocus(G)", "query_expressions": "p", "answer_expressions": "8", "fact_spans": "[[[28, 71]], [[31, 71]], [[1, 22]], [[77, 82]], [[28, 71]], [[4, 22]], [[1, 22]], [[1, 75]]]", "query_spans": "[[[77, 86]]]", "process": "From the problem, the right focus of the hyperbola is (4,0), so the focus of the parabola is (4,0). Therefore, \\frac{p}{2}=4,\\therefore p=8." }, { "text": "The line passing through the point $P(2,1)$ intersects the parabola $y^{2}=16x$ at points $A$ and $B$, and $\\overrightarrow{P A}+\\overrightarrow{P B}=\\overrightarrow{0}$. Then the equation of this line is?", "fact_expressions": "G: Parabola;H: Line;P: Point;A: Point;B: Point;Expression(G) = (y^2 = 16*x);Coordinate(P) = (2, 1);PointOnCurve(P, H);Intersection(H, G) = {A, B};VectorOf(P, A) + VectorOf(P, B) = 0", "query_expressions": "Expression(H)", "answer_expressions": "8*x - y - 15 = 0", "fact_spans": "[[[14, 29]], [[11, 13], [106, 108]], [[1, 10]], [[31, 34]], [[35, 38]], [[14, 29]], [[1, 10]], [[0, 13]], [[11, 40]], [[42, 104]]]", "query_spans": "[[[106, 113]]]", "process": "\\overrightarrow{PA}+\\overrightarrow{PB}=\\overrightarrow{0}, hence P is the midpoint of AB. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then y_{1}=16x_{1}, y_{2}^{2}=16x_{2}. Subtracting gives (y_{1}+y_{2})(y_{1}-y_{2})=16(x_{1}-x_{2}). Thus 2k=16, k=8. Hence the equation of the line is: y=8(x-2)+1, or 8x-y-15=0." }, { "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": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;H: Circle;e: Number;a > b;b > 0;Expression(H) = (y^2 + (x - 2)^2 = 1);PointOnCurve(OneOf(Focus(G)),H);PointOnCurve(OneOf(Vertex(G)),H);Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "1/3", "fact_spans": "[[[24, 78], [91, 93]], [[24, 78]], [[26, 78]], [[26, 78]], [[2, 22]], [[97, 100]], [[26, 78]], [[26, 78]], [[2, 22]], [[2, 88]], [[2, 83]], [[91, 100]]]", "query_spans": "[[[97, 102]]]", "process": "" }, { "text": "Given a point $P$ on the parabola $y^{2}=4x$ such that the distance from $P$ to the focus is twice the distance from $P$ to the $y$-axis, then what is the distance from point $P$ to the focus?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(P,G);Distance(P, Focus(G)) = Distance(P, yAxis)*2", "query_expressions": "Distance(P,Focus(G))", "answer_expressions": "2", "fact_spans": "[[[2, 16]], [[29, 30], [19, 22], [45, 49]], [[2, 16]], [[2, 22]], [[2, 43]]]", "query_spans": "[[[2, 57]]]", "process": "Let the horizontal coordinate of point P be m (m > 0). Use the definition of the parabola and the given conditions to set up an equation to find m. Let the horizontal coordinate of point P be m (m > 0). Since the equation of the parabola is y^{2} = 4x, its directrix equation is x. Therefore, according to the definition of the parabola, the distance from point P to the focus is m + 1. Thus, m + 1 = 2m, solving gives m = 1. Therefore, the distance from point P to the focus is 2." }, { "text": "If the equation $\\frac{x^{2}}{2+m}-\\frac{y^{2}}{m+1}=1$ represents a hyperbola, then what is the range of values for $m$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 2) - y^2/(m + 1) = 1);m: Number", "query_expressions": "Range(m)", "answer_expressions": "(-oo,-2)+(-1,+oo)", "fact_spans": "[[[44, 47]], [[1, 47]], [[49, 52]]]", "query_spans": "[[[49, 59]]]", "process": "Since the equation \\frac{x2}{2+m}-\\frac{y^{2}}{m+1}=1 represents a hyperbola, then (m+2)(m+1)>0, solving gives m<-2 or m>-1" }, { "text": "Given the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1(b>0)$ has eccentricity $\\frac{\\sqrt{5}}{2}$, $F_{1}$, $F_{2}$ are the two foci of the hyperbola, $A$ is the left vertex, $B(0, b)$, and point $P$ lies on the segment $AB$. Then the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?", "fact_expressions": "A: Point;B: Point;G: Hyperbola;b: Number;P: Point;F1: Point;F2: Point;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Coordinate(B) = (0, b);Eccentricity(G)=sqrt(5)/2;Focus(G)={F1,F2};LeftVertex(G)=A;PointOnCurve(P, LineSegmentOf(A, B))", "query_expressions": "Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))", "answer_expressions": "-21/5", "fact_spans": "[[[104, 107]], [[112, 121]], [[2, 49], [95, 98]], [[5, 49]], [[122, 126]], [[77, 84]], [[87, 94]], [[5, 49]], [[2, 49]], [[112, 121]], [[2, 74]], [[77, 103]], [[95, 111]], [[122, 134]]]", "query_spans": "[[[136, 199]]]", "process": "It is easy to get $ c=\\sqrt{5} $, $ b=1 $. Let $ P(x,y) $, then $ \\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}}=(-\\sqrt{5}-x,-y)\\cdot(\\sqrt{5}-x,-y)=x^{2}+y^{2}-5 $. $ x^{2}+y^{2} $ represents the square of the distance from the origin to point $ P $. Obviously, when $ OP\\perp AB $, $ x^{2}+y^{2} $ reaches its minimum value. The equation of line $ AB $ is $ x+2y-2=0 $. By the point-to-line distance formula, it is easy to obtain $ (x^{2}+y^{2})_{\\min}=\\frac{4}{5} $. Hence, the minimum value of $ \\overrightarrow{PF}_{1}\\cdot\\overrightarrow{PF_{2}} $ is $ \\frac{4}{5}-5=-\\frac{21}{5} $." }, { "text": "It is known that a focus of the hyperbola $x^{2}-y^{2}=k$ is the focus of the parabola $y^{2}=16x$. What is the value of $k$?", "fact_expressions": "G: Hyperbola;k: Number;H: Parabola;Expression(G) = (x^2 - y^2 = k);Expression(H) = (y^2 = 16*x);OneOf(Focus(G)) = Focus(H)", "query_expressions": "k", "answer_expressions": "8", "fact_spans": "[[[2, 20]], [[46, 49]], [[26, 41]], [[2, 20]], [[26, 41]], [[2, 44]]]", "query_spans": "[[[46, 53]]]", "process": "The focus of the parabola \\( y^{2} = 16x \\) is at \\( (4,0) \\). Therefore, the foci of the hyperbola \\( x^{2} - y^{2} = k \\) lie on the x-axis, that is, \\( k > 0 \\) and \\( c = 4 \\). The standard form of the hyperbola \\( x^{2} - y^{2} = k \\) is \\( \\frac{x^{2}}{k} - \\frac{y^{2}}{k} = 1 \\). Hence, \\( 2k = c^{2} = 16 \\), solving gives \\( k = 8 \\)." }, { "text": "The minimum distance from points on the curve $x y=2$ to the line $x+y+\\sqrt{2}=0$ is?", "fact_expressions": "G: Line;H: Curve;P0: Point;Expression(G) = (x + y + sqrt(2) = 0);Expression(H) = (x*y = 2);PointOnCurve(P0, H)", "query_expressions": "Min(Distance(P0, G))", "answer_expressions": "1", "fact_spans": "[[[13, 31]], [[0, 9]], [[11, 12]], [[13, 31]], [[0, 9]], [[0, 12]]]", "query_spans": "[[[11, 40]]]", "process": "Let the equation of the line parallel to the line $ x + y + \\sqrt{2} = 0 $ be $ x + y + m = 0 $. Solving the system of equations \n\\[\n\\begin{cases}\nx + y + m = 0 \\\\\nxy = 2\n\\end{cases}\n\\]\nyields $ x^{2} + mx + 2 = 0 $. Let $ A = m^{2} - 4 \\times 2 = 0 $, solving gives $ m = \\pm 2\\sqrt{2} $. At this time, the lines $ x + y \\pm 2\\sqrt{2} = 0 $ are tangent to the curve $ xy = 2 $. Combining with the graph, it can be seen that when $ m = 2\\sqrt{2} $, the minimum distance from points on the curve $ xy = 2 $ to the line $ x + y + \\sqrt{2} = 0 $ is the distance between the two parallel lines $ x + y + 2\\sqrt{2} = 0 $ and $ x + y + \\sqrt{2} = 0 $. Therefore, the minimum value is $ d = \\frac{|2\\sqrt{2} - \\sqrt{2}|}{\\sqrt{1^{2}} + 1^{2}} = 1 $." }, { "text": "Given that the asymptotes of hyperbola $C$ are $y = \\pm 2x$, write a standard equation of hyperbola $C$?", "fact_expressions": "C: Hyperbola;Expression(Asymptote(C)) = (y = pm*2*x)", "query_expressions": "Expression(C)", "answer_expressions": "x^2-y^2/4=1", "fact_spans": "[[[2, 8], [29, 35]], [[2, 26]]]", "query_spans": "[[[29, 43]]]", "process": "According to the asymptote equations, find $\\frac{b}{a}$, thus the standard equation satisfying the condition can be written. By the given condition, the asymptote equations of hyperbola $C$ are $y = \\pm 2x$. Without loss of generality, assume the foci of the hyperbola lie on the $x$-axis, then $\\frac{b}{a} = 2$, $\\frac{b^{2}}{a^{2}} = 4$. Let $b^{2} = 4a^{2} = 4$, then one standard equation of hyperbola $C$ is $x^{2} - \\frac{y^{2}}{4} = 1$. Alternatively, let $b^{2} = 4a^{2} = 8$, and so on." }, { "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $2$, and the distance from a focus to an asymptote is $\\sqrt{3}$, then the focal distance of this hyperbola is equal to?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = 2;Distance(Focus(G),Asymptote(G))=sqrt(3)", "query_expressions": "FocalLength(G)", "answer_expressions": "4", "fact_spans": "[[[2, 58], [90, 93]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 66]], [[2, 87]]]", "query_spans": "[[[90, 99]]]", "process": "The distance from the focus to the asymptote is b, so b = \\sqrt{3}. Given \\frac{c^{2}}{a^{2}} = a^{2} + 3 and c = 2, the focal length is 4." }, { "text": "If the line passing through the two points $A(a, 0)$ and $B(0, a)$ does not intersect the parabola $y=x^{2}-2 x-3$, then what is the range of real values of $a$?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;a: Real;Expression(G) = (y = x^2 - 2*x - 3);Coordinate(A)=(a,0);Coordinate(B)=(0,a);PointOnCurve(A,H);PointOnCurve(B,H);NumIntersection(H,G)=0", "query_expressions": "Range(a)", "answer_expressions": "(-oo,-13/4)", "fact_spans": "[[[28, 46]], [[25, 27]], [[5, 14]], [[15, 24]], [[53, 58]], [[28, 46]], [[5, 14]], [[15, 24]], [[2, 27]], [[2, 27]], [[25, 50]]]", "query_spans": "[[[53, 65]]]", "process": "" }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{k+4}+\\frac{y^{2}}{9}=1$ with foci on the $x$-axis is $\\frac{1}{2}$, then the value of the real number $k$ is?", "fact_expressions": "PointOnCurve(Focus(G), xAxis);G: Ellipse;Expression(G) = (x^2/(k + 4) + y^2/9 = 1);k: Real;Eccentricity(G) = 1/2", "query_expressions": "k", "answer_expressions": "8", "fact_spans": "[[[1, 49]], [[10, 49]], [[10, 49]], [[69, 74]], [[10, 67]]]", "query_spans": "[[[69, 78]]]", "process": "" }, { "text": "The line $ l $: $ y = 2x + 10 $ passes through a focus of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $ and is parallel to one of its asymptotes. Then the equation of the hyperbola is?", "fact_expressions": "l: Line;Expression(l) = (y = 2*x + 10);G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;PointOnCurve(OneOf(Focus(G)), l);IsParallel(l, OneOf(Asymptote(G)))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5-y^2/20=1", "fact_spans": "[[[0, 17]], [[0, 17]], [[18, 74], [90, 93], [80, 81]], [[18, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[21, 74]], [[0, 78]], [[0, 88]]]", "query_spans": "[[[90, 97]]]", "process": "According to the equations of the asymptotes and the coordinates of the foci, use the relationships among a, b, c and the given conditions to set up equations, solve for a², b², and substitute into the equation of the hyperbola. From the given conditions, \n\\begin{cases}\\frac{b}{a}=2\\\\-2c+10=0\\\\c^{2}=a^{2}+b^{2}\\end{cases}, \nsolving gives a^{2}=5, b^{2}=20, \n\\therefore the equation of the hyperbola is \\frac{x^{2}}{5}-\\frac{y^{2}}{20}=1" }, { "text": "The equation $\\frac{x^{2}}{m+2}+\\frac{y^{2}}{4}=1$ represents a hyperbola with foci on the $y$-axis. What is the range of values for $m$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(m + 2) + y^2/4 = 1);m: Number;PointOnCurve(Focus(G), yAxis) = True", "query_expressions": "Range(m)", "answer_expressions": "(-oo,-2)", "fact_spans": "[[[50, 53]], [[0, 53]], [[55, 58]], [[41, 53]]]", "query_spans": "[[[55, 64]]]", "process": "Since the equation \\frac{x^{2}}{m+2}+\\frac{y^{2}}{4}=1 represents a hyperbola with foci on the y-axis, then m+2<0, solving gives m<-2" }, { "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ intersect the circle $x^{2}+y^{2}-4 x+2=0$, then the range of the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-4*x + x^2 + y^2 + 2 = 0);IsIntersect(Asymptote(G),H)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, sqrt(2)]", "fact_spans": "[[[2, 58], [91, 94]], [[5, 58]], [[5, 58]], [[63, 85]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 85]], [[2, 88]]]", "query_spans": "[[[91, 105]]]", "process": "" }, { "text": "Let point $F$ and point $B$ be the right focus and the upper vertex of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{3}=1$ $(a>\\sqrt{3})$, respectively, and let $O$ be the origin. Given that the perimeter of $\\triangle O F B$ is $3+\\sqrt{3}$, then the value of the real number $a$ is?", "fact_expressions": "C: Ellipse;a: Real;O: Origin;F: Point;B: Point;a > sqrt(3);Expression(C) = (y^2/3 + x^2/a^2 = 1);RightFocus(C) = F;UpperVertex(C) = B;Perimeter(TriangleOf(O, F, B)) = sqrt(3) + 3", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[12, 70]], [[124, 129]], [[79, 82]], [[1, 5]], [[6, 9]], [[19, 70]], [[12, 70]], [[1, 78]], [[1, 78]], [[89, 122]]]", "query_spans": "[[[124, 133]]]", "process": "Analysis: Given the perimeter of AOFB is $ l = a + b + c = 3 + \\sqrt{3} $, we obtain $ a + c = 3 $. According to the property of the ellipse, $ a^{2} - c^{2} = b^{2} = 3 $. Solving these equations simultaneously yields $ \\begin{cases} a = 2 \\\\ c = 1 \\end{cases} $, thus obtaining the answer. Specifically, according to the given conditions, the perimeter of AOFB is $ a + b + c = 3 + \\sqrt{3} $, and since $ b = \\sqrt{3} $, it follows that $ a + c = 3 $. Combining this with $ a^{2} - c^{2} = b^{2} = 3 $, we can solve to get $ \\begin{cases} a = 2 \\\\ c = 1 \\end{cases} $. Therefore, the value of the real number $ a $ is 2." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{5}}{2}$, then the asymptotes of $C$ are?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;a > b;b > 0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(C) = sqrt(5)/2", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(1/2)*x", "fact_spans": "[[[2, 60], [87, 90]], [[10, 60]], [[10, 60]], [[10, 60]], [[10, 60]], [[2, 60]], [[2, 85]]]", "query_spans": "[[[87, 98]]]", "process": "Since the asymptotes of the hyperbola are given by: $ y = \\pm\\frac{b}{a}x $, and $ e = \\frac{c}{a} = \\frac{\\sqrt{5}}{2} $, so $ \\frac{c^{2}}{a^{2}} = \\frac{a^{2}+b^{2}}{a^{2}} = \\frac{5}{4} \\Rightarrow a^{2} = 4b^{2} \\Rightarrow \\frac{b}{a} = \\frac{1}{2} $" }, { "text": "Given an ellipse $C$: $\\frac{x^{2}}{4}+y^{2}=1$ with two points $A$ and $B$ on it. If the midpoint of $A B$ is $D$, and the slope of the line $O D$ is equal to $1$, then what is the slope of the line $A B$?", "fact_expressions": "C: Ellipse;O: Origin;D: Point;B: Point;A: Point;Expression(C) = (x^2/4 + y^2 = 1);PointOnCurve(A,C);PointOnCurve(B,C);MidPoint(LineSegmentOf(A, B)) = D;Slope(LineOf(O,D))=1", "query_expressions": "Slope(LineOf(A,B))", "answer_expressions": "-1/4", "fact_spans": "[[[2, 34]], [[61, 66]], [[55, 58]], [[41, 44]], [[37, 40]], [[2, 34]], [[2, 44]], [[2, 44]], [[46, 58]], [[59, 74]]]", "query_spans": "[[[76, 89]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then D(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}) \\begin{cases} \\frac{x_{1}^{2}}{4}+y_{1}^{2}=1 \\\\ \\frac{x_{2}^{2}}{4}+y_{2}^{2}=1 \\end{cases}, subtracting the two equations gives \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{4}+(y_{1}+y_{2})(y_{1}-y_{2})=0, that is, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{4}\\times\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}=-\\frac{1}{4}\\times1, so k_{AB}=-\\frac{1}{4}\\times\\frac{1}{k_{OD}}=-\\frac{1}{4}\\times1=-\\frac{1}{4}" }, { "text": "If the distance from point $A(x_0, \\sqrt{2})$ on the parabola $y^2 = 2px$ $(p > 0)$ to its focus is 3 times the distance from $A$ to the $y$-axis, then $p =$?", "fact_expressions": "G: Parabola;p: Number;A: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(A) = (x0, sqrt(2));x0:Number;PointOnCurve(A,G);Distance(A, Focus(G)) = Distance(A, yAxis) * 3", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[1, 22], [46, 47]], [[70, 73]], [[24, 45], [53, 56]], [[4, 22]], [[1, 22]], [[24, 45]], [[25, 45]], [[1, 45]], [[24, 68]]]", "query_spans": "[[[70, 75]]]", "process": "Problem Analysis: From the given condition, 3x_{0} = x_{0} + \\frac{p}{2}, x_{0} = \\frac{p}{4}, then \\frac{p^{2}}{2} = 2, \\because p > 0, \\therefore p = 2." }, { "text": "Through the focus $F$ of the parabola $x^{2}=4 y$, draw a line with inclination angle $30^{\\circ}$, intersecting the parabola at points $A$ and $B$ respectively. Then $|A B|=$?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;H: Line;PointOnCurve(F, H);Inclination(H) = ApplyUnit(30, degree);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16/3", "fact_spans": "[[[1, 15], [43, 46]], [[1, 15]], [[18, 21]], [[1, 21]], [[39, 41]], [[0, 41]], [[22, 41]], [[50, 53]], [[54, 57]], [[39, 59]]]", "query_spans": "[[[61, 70]]]", "process": "The focus of the parabola is F(0,1). From the given conditions, the equation of the line is y=\\frac{\\sqrt{3}}{3}x+1. Combining this with the equation of the parabola yields 3y^{2}-10y+3=0, giving y_{1}+y_{2}=\\frac{10}{3}, |AB|=y_{1}+y_{2}+p=\\frac{10}{3}+2=\\frac{16}{3}" }, { "text": "Given point $Q(m , 0)$, and $P$ is a moving point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$. If the length of segment $PQ$ reaches its minimum when $P$ is exactly at the right vertex of the ellipse, then the range of real number $m$ is?", "fact_expressions": "Q: Point;m: Real;Coordinate(Q) = (m, 0);P: Point;PointOnCurve(P, G);G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);RightVertex(G) = P;WhenMin(Length(LineSegmentOf(P, Q)))", "query_expressions": "Range(m)", "answer_expressions": "[3/2, +oo)", "fact_spans": "[[[2, 13]], [[81, 86]], [[2, 13]], [[15, 18], [52, 56]], [[15, 49]], [[19, 46], [58, 60]], [[19, 46]], [[52, 65]], [[66, 79]]]", "query_spans": "[[[81, 93]]]", "process": "" }, { "text": "Let the center of the circle $x^{2}+y^{2}-4x+2y-11=0$ be $A$, and let point $P$ lie on the circle. Then the equation of the locus of the midpoint $M$ of segment $PA$ is?", "fact_expressions": "P: Point;A: Point;G: Circle;M: Point;Expression(G) = (2*y - 4*x + x^2 + y^2 - 11 = 0);PointOnCurve(P, G);Center(G) = A;MidPoint(LineSegmentOf(P, A)) = M", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2 + y^2 - 4*x + 2*y + 1 = 0", "fact_spans": "[[[36, 40]], [[32, 35]], [[1, 28], [41, 42]], [[54, 57]], [[1, 28]], [[36, 43]], [[1, 35]], [[45, 57]]]", "query_spans": "[[[54, 64]]]", "process": "Let the coordinates of the midpoint M of PA be (x, y), P(x_{1}, y_{1}), and the center A of the circle x^{2}+y^{2}-4x+2y-11=0 have coordinates (2, -1). Then \\begin{cases}x_{1}=2x-2\\\\y_{1}=2y+2\\end{cases}. Since point P lies on the circle, x_{1}^{2}+y_{1}^{2}-4x_{1}+2y_{1}-11=0. Therefore, from the given conditions, \\begin{cases}z=\\\\\\frac{y_{1}-2}{2}=y\\end{cases}(2x-2)^{2}+(2y+2)^{2}-4(2x-2)+2(2y+2)-11=0, which simplifies to x^{2}+y^{2}-4x+2y+1=0." }, { "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, what is the length of the imaginary axis of this curve?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1)", "query_expressions": "Length(ImageinaryAxis(G))", "answer_expressions": "2", "fact_spans": "[[[2, 30], [32, 34]], [[2, 30]]]", "query_spans": "[[[32, 40]]]", "process": "From the given condition: b^{2}=1, then the length of the imaginary axis is: 2b=2. The correct answer to this question: 2. This question examines the basic definitions in the standard equation of a hyperbola, and is a fundamental problem." }, { "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=\\sqrt{3} x$, and one of its foci coincides with the focus of the parabola $y^{2}=16 x$, then the equation of the 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 = 16*x);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);OneOf(Focus(G))=Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2/15=1", "fact_spans": "[[[2, 58], [83, 84], [112, 115]], [[5, 58]], [[5, 58]], [[90, 105]], [[5, 58]], [[5, 58]], [[2, 58]], [[90, 105]], [[2, 82]], [[83, 110]]]", "query_spans": "[[[112, 120]]]", "process": "" }, { "text": "Given that the distance from a point $M(1, m)$ on the parabola $y^{2}=2 p x(p>0)$ to its focus is $5$, and the left vertex of the hyperbola $x^{2}-\\frac{y^{2}}{a}=1$ is $A$. If one asymptote of the hyperbola is perpendicular to the line $AM$, then the real number $a$=?", "fact_expressions": "A: Point;M: Point;G: Hyperbola;a: Real;H: Parabola;p: Number;Expression(G) = (x^2 - y^2/a = 1);p>0;Expression(H) = (y^2 = 2*(p*x));Coordinate(M) = (1, m);PointOnCurve(M, H);Distance(M, Focus(H)) = 5;LeftVertex(G) = A;IsPerpendicular(OneOf(Asymptote(G)), LineOf(A, M));m:Number", "query_expressions": "a", "answer_expressions": "1/4", "fact_spans": "[[[80, 83]], [[26, 35]], [[47, 75], [85, 88]], [[104, 109]], [[36, 37], [2, 23]], [[5, 23]], [[47, 75]], [[5, 23]], [[2, 23]], [[26, 35]], [[2, 35]], [[26, 46]], [[47, 83]], [[85, 102]], [[26, 35]]]", "query_spans": "[[[104, 111]]]", "process": "" }, { "text": "Let the vertex of the parabola be at the origin, and the equation of the directrix be $x = -2$. Then the standard equation of the parabola is?", "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;Expression(Directrix(G)) = (x=-2)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[1, 4], [24, 27]], [[8, 10]], [[1, 10]], [[1, 22]]]", "query_spans": "[[[24, 34]]]", "process": "From the given condition: \\frac{p}{2}=2,\\therefore p=4, and the focus of the standard equation of the parabola lies on the positive x-axis, so the standard equation of the parabola can be written as: y^{2}=2px. Substituting p gives y^{2}=8x." }, { "text": "Given that the coordinates of points $A$ and $B$ are $(-2,0)$ and $(2,0)$, respectively, lines $AM$ and $BM$ intersect at point $M$, and the difference between the slope of line $AM$ and the slope of line $BM$ is $4$. Then the equation of the trajectory of point $M$ is?", "fact_expressions": "A: Point;M: Point;B: Point;Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);Intersection(LineOf(A,M),LineOf(B,M))=M;Slope(LineOf(A,M))-Slope(LineOf(B,M))=4", "query_expressions": "LocusEquation(M)", "answer_expressions": "{(y=4-x^2)&(Negation(x=pm*2))}", "fact_spans": "[[[2, 5]], [[52, 56], [87, 91]], [[6, 9]], [[2, 34]], [[2, 34]], [[35, 56]], [[58, 85]]]", "query_spans": "[[[87, 98]]]", "process": "Let M(x,y), express the slopes of lines AM and BM, and establish a relationship based on the difference of the slopes being 4. Let point M(x,y), where x\\neq\\pm2, then k_{AM}=\\frac{y}{x+2}, k_{BM}=\\frac{y}{x-2}. From the condition, we have k_{AM}-k_{BM}=\\frac{y}{x+2}-\\frac{y}{x-2}=4. Simplifying yields y=4-x^{2} (x\\neq\\pm2), so the trajectory equation of point M is y=4-x^{2} (x\\neq\\pm2)." }, { "text": "The ellipse $C$ is centered at the origin of the coordinate system, with left and right foci $F_{1}$, $F_{2}$ on the $x$-axis. Let $A$, $B$ be the upper vertex and the right vertex of the ellipse respectively, and let $P$ be 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": "C: Ellipse;O: Origin;Center(C) = O;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F1, xAxis);PointOnCurve(F2, xAxis);A: Point;B: Point;UpperVertex(C) = A;RightVertex(C) = B;P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), xAxis);IsParallel(LineSegmentOf(P, F2), LineSegmentOf(A, B))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/5", "fact_spans": "[[[0, 5], [53, 55], [68, 70], [114, 116]], [[9, 13]], [[0, 13]], [[19, 26]], [[27, 34]], [[0, 34]], [[0, 34]], [[19, 40]], [[27, 40]], [[43, 46]], [[47, 50]], [[43, 63]], [[43, 63]], [[64, 67]], [[64, 73]], [[75, 93]], [[94, 111]]]", "query_spans": "[[[114, 122]]]", "process": "Since PF₁ ⊥ x-axis, the coordinates of point P are (-c, \\frac{b^{2}}{a}). From the fact that parallel lines have equal slopes, we get \\underline{b^{2}} = -\\frac{b}{a}, simplifying yields b = 2c. Hence, the eccentricity is e = \\sqrt{\\frac{c^{2}}{a^{2}}} = \\sqrt{\\frac{c^{2}}{b^{2} + c}} = \\frac{\\sqrt{5}}{5}." }, { "text": "Given that the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ and the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$ have the same foci, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/m = 1);m: Number;H: Ellipse;Expression(H) = (x^2/12 + y^2/4 = 1);Focus(G) = Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/7-y^2=1", "fact_spans": "[[[2, 30], [76, 79]], [[2, 30]], [[5, 30]], [[31, 69]], [[31, 69]], [[2, 73]]]", "query_spans": "[[[76, 84]]]", "process": "Analysis: According to the given conditions, find the coordinates of the foci of the ellipse. By the geometric properties of the hyperbola, if the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ and the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$ have the same foci, then $m+1=8$. Solving for $m$, substitute the value of $m$ into the equation of the hyperbola to obtain the answer. According to the conditions, the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$ has its foci on the $x$-axis, and the coordinates of the foci are $(\\pm2\\sqrt{2},0)$. If the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ and the ellipse $\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$ have the same foci, then $m+1=8$, solving gives $m=7$. Thus, the equation of the hyperbola is $\\frac{x^{2}}{7}-y^{2}=1$." }, { "text": "The hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{m}=1$ has a focal distance of $18$. What is the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/36 - y^2/m = 1);FocalLength(G) = 18", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "", "fact_spans": "[[[0, 39], [49, 52]], [[3, 39]], [[0, 39]], [[0, 47]]]", "query_spans": "[[[49, 60]]]", "process": "" }, { "text": "A focus of the hyperbola $3my^2 - mx^2 = 3$ is $(0, 2)$. What is the value of $m$?", "fact_expressions": "G: Hyperbola;m: Number;H: Point;Expression(G) = (-m*x^2 + 3*(m*y^2) = 3);Coordinate(H) = (0, 2);OneOf(Focus(G)) = H", "query_expressions": "m", "answer_expressions": "1", "fact_spans": "[[[0, 22]], [[39, 42]], [[28, 37]], [[0, 22]], [[28, 37]], [[0, 37]]]", "query_spans": "[[[39, 46]]]", "process": "" }, { "text": "Let the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ be $F$, and the left and right vertices be $A_{1}$, $A_{2}$, respectively. Draw a line through $F$ perpendicular to the $x$-axis, intersecting the hyperbola at points $B$ and $C$. If $A_{1} B \\perp A_{2} C$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;A1: Point;B: Point;A2: Point;C: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;LeftVertex(G) = A1;RightVertex(G) = A2;L:Line;PointOnCurve(F, L);IsPerpendicular(L,xAxis);Intersection(L, G) = {B, C};IsPerpendicular(LineSegmentOf(A1, B), LineSegmentOf(A2, C))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 57], [143, 146], [103, 106]], [[4, 57]], [[4, 57]], [[74, 81]], [[107, 110]], [[82, 89]], [[111, 114]], [[62, 65], [91, 94]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[1, 89]], [[1, 89]], [], [[90, 102]], [[90, 102]], [[90, 116]], [[118, 141]]]", "query_spans": "[[[143, 152]]]", "process": "From the given conditions, A_{1}(-a,0), A_{2}(a,0), B(c,\\frac{b^{2}}{a}), C(c,-\\frac{b^{2}}{a})," }, { "text": "A point $M(1, m)$ $(m>0)$ on the parabola $y^{2}=2 p x$ $(p>0)$ has a distance of 5 to its focus. The left vertex of the hyperbola $\\frac{x^{2}}{a}-y^{2}=1$ is $A$. If one asymptote of the hyperbola is parallel to the line $AM$, then the real number $a$ equals?", "fact_expressions": "A: Point;M: Point;G: Hyperbola;a: Real;H: Parabola;p: Number;m: Number;Expression(G) = (-y^2 + x^2/a = 1);p>0;m > 0;Expression(H) = (y^2 = 2*(p*x));Coordinate(M) = (1, m);PointOnCurve(M, H);Distance(M, Focus(H)) = 5;LeftVertex(G) = A;IsParallel(OneOf(Asymptote(G)), LineOf(A, M))", "query_expressions": "a", "answer_expressions": "1/9", "fact_spans": "[[[84, 87]], [[24, 38]], [[51, 79], [90, 93]], [[111, 116]], [[0, 21], [39, 40]], [[3, 21]], [[24, 38]], [[50, 79]], [[3, 21]], [[24, 38]], [[0, 21]], [[24, 38]], [[0, 38]], [[24, 49]], [[51, 87]], [[90, 108]]]", "query_spans": "[[[111, 119]]]", "process": "" }, { "text": "Given that an ellipse is tangent to the $x$-axis and has foci at $F_{1}(1,1)$ and $F_{2}(5,2)$, what is the length of its major axis?", "fact_expressions": "G: Ellipse;IsTangent(G,xAxis) = True;F1: Point;F2: Point;Coordinate(F1) = (1, 1);Coordinate(F2) = (5, 2);Focus(G) = {F1,F2}", "query_expressions": "Length(MajorAxis(G))", "answer_expressions": "", "fact_spans": "[[[2, 4], [47, 48]], [[2, 11]], [[19, 31]], [[33, 45]], [[19, 31]], [[33, 45]], [[2, 45]]]", "query_spans": "[[[47, 53]]]", "process": "\\because perpendiculars are drawn from F_{1}(1,1) and F_{2}(5,2) to the x-axis, intersecting the x-axis at points A and B respectively. Let M be the point where the ellipse is tangent to the x-axis. Draw a perpendicular from M intersecting F_{1}F_{2} at Q. Connect F_{1}M and F_{2}M. Extend F_{2}F_{1} to intersect the x-axis at K, so the coordinates of K are K(-3,0), and \\tan\\angle F_{2}KM = \\frac{1}{4} (slope). Since \\angle F_{1}MQ = \\angle QMF_{2} (optical property of ellipses: angle of incidence equals angle of reflection), it follows that \\triangle F_{1}MA \\sim \\triangle F_{2}MB, with similarity ratio \\lambda = 2. \\therefore the coordinates of M are (\\frac{7}{3},0). Hence |F_{2}M| = 2|F_{1}M|. The major axis length 2a = |F_{1}M| + |F_{2}M| = \\sqrt{(\\frac{7}{3}-1)^{2}+1^{2}} + \\sqrt{(\\frac{7}{3}-5)^{2}+2^{2}} = \\frac{5}{3} + \\frac{10}{3} = 5. The answer is: 5" }, { "text": "If a focus of the ellipse $\\frac{x^{2}}{k+4}+\\frac{y^{2}}{9}=1$ is $(0,2)$, then the real number $k$=?", "fact_expressions": "G: Ellipse;k: Real;H: Point;Expression(G) = (x^2/(k + 4) + y^2/9 = 1);Coordinate(H) = (0, 2);OneOf(Focus(G))=H", "query_expressions": "k", "answer_expressions": "1", "fact_spans": "[[[1, 40]], [[55, 60]], [[46, 53]], [[1, 40]], [[46, 53]], [[1, 53]]]", "query_spans": "[[[55, 62]]]", "process": "From the problem, it can be seen that the foci of the ellipse lie on the y-axis, ∴ 9 - (k + 4) = 4, solving gives k = 1." }, { "text": "Given that the distance between the foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2c$, and the distance from a focus to an asymptote of the hyperbola is $\\frac{c}{2}$, then 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);c: Number;FocalLength(G) = 2*c;Distance(Focus(G),Asymptote(G))=c/2", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[2, 58], [71, 74], [97, 100]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[82, 95]], [[2, 67]], [[2, 95]]]", "query_spans": "[[[97, 106]]]", "process": "According to the problem, for the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the right focus is $F(c,0)$, and the asymptotes are given by $bx\\pm ay=0$. By the symmetry of the hyperbola, without loss of generality, take the right focus $F$. Then $\\frac{|bc|}{\\sqrt{b^{2}+(\\pm a)^{2}}}=b=\\frac{c}{2}$, so $c=2b$, $a=\\sqrt{c^{2}-b^{2}}=\\sqrt{3}b$. Therefore, the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{2\\sqrt{3}}{3}$." }, { "text": "The standard equation of the hyperbola passing through points $P(3,2 \\sqrt{7})$, $Q(-6 \\sqrt{2}, 7)$ is?", "fact_expressions": "G: Hyperbola;P: Point;Q: Point;Coordinate(P) = (3, 2*sqrt(7));Coordinate(Q) = (-6*sqrt(2), 7);PointOnCurve(P, G);PointOnCurve(Q, G)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/25-x^2/75=1", "fact_spans": "[[[42, 45]], [[2, 20]], [[22, 41]], [[2, 20]], [[22, 41]], [[0, 45]], [[0, 45]]]", "query_spans": "[[[42, 52]]]", "process": "Let the hyperbola equation be: mx^{2}+ny^{2}=1 (mn<0). Substituting the two points gives the following. Let the hyperbola equation be mx^{2}+ny^{2}=1 (mn<0). Since the hyperbola passes through points P(3,2\\sqrt{7}) and Q(-6\\sqrt{2},7), we have \\begin{cases}9m+28n=1,\\\\72m+49n=1,\\end{cases} solving yields \\begin{cases}m=-\\frac{1}{75},\\\\n=\\frac{1}{25}.\\end{cases} Therefore, the required hyperbola equation is \\frac{y^{2}}{25}-\\frac{x^{2}}{75}=1" }, { "text": "If the focus of the parabola $y^{2}=2 p x$ has coordinates $(1,0)$, then $p=$? What is the equation of the directrix?", "fact_expressions": "G: Parabola;p: Number;Expression(G) = (y^2 = 2*(p*x));Coordinate(Focus(G)) = (1, 0)", "query_expressions": "p;Expression(Directrix(G))", "answer_expressions": "2\nx=-1", "fact_spans": "[[[1, 17]], [[32, 35]], [[1, 17]], [[1, 30]]]", "query_spans": "[[[32, 38]], [[1, 44]]]", "process": "Since the equation of the parabola is y^{2}=2px, the focus coordinates are (\\frac{p}{2},0). Given that the focus coordinates are (1,0), it follows that p=2, and the equation of the directrix is x=-1." }, { "text": "Given that point $P$ lies on the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{3}=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 $\\triangle F_{1}PF_{2}$.", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/8 + y^2/3 = 1);PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[7, 44], [65, 67]], [[49, 56]], [[2, 6]], [[57, 64]], [[7, 44]], [[2, 48]], [[49, 72]], [[73, 105]]]", "query_spans": "[[[107, 135]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=2 p x(p>0)$, the focus $F$ of the parabola is the right vertex of the ellipse $\\frac{4 x^{2}}{9}+\\frac{y^{2}}{b^{2}}=1$. The line $l$ is the directrix of the parabola $C$, and point $A$ lies on the parabola $C$. From $A$, draw $A B \\perp l$, with foot of perpendicular at $B$. If the slope of line $B F$ is $k_{B F}=-\\sqrt{3}$, then the area of $\\triangle A F B$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;F: Point;Focus(C) = F;G: Ellipse;Expression(G) = (4*x^2/9 + y^2/b^2 = 1);b: Number;RightVertex(G) = F;l: Line;Directrix(C) = l;A: Point;PointOnCurve(A, C) = True;PointOnCurve(A, LineSegmentOf(A, B)) = True;IsPerpendicular(LineSegmentOf(A, B), l) = True;FootPoint(LineSegmentOf(A, B), l) = B;B: Point;Slope(LineOf(B, F)) = k_BF;k_BF: Number;k_BF = -sqrt(3)", "query_expressions": "Area(TriangleOf(A, F, B))", "answer_expressions": "9*sqrt(3)", "fact_spans": "[[[2, 28], [89, 95], [104, 110]], [[2, 28]], [[10, 28]], [[10, 28]], [[31, 34]], [[2, 34]], [[35, 78]], [[35, 78]], [[37, 78]], [[31, 82]], [[83, 88]], [[83, 98]], [[99, 103], [113, 116]], [[99, 111]], [[112, 130]], [[117, 130]], [[117, 137]], [[134, 137]], [[139, 168]], [[149, 168]], [[149, 168]]]", "query_spans": "[[[170, 192]]]", "process": "\\because the focus F of the parabola C: y^{2}=2px (p>0) is the right vertex of the ellipse \\frac{4x^{2}}{9}+\\frac{y^{2}}{b^{2}}=1, we get m=3\\sqrt{3}; hence A(x_{0},3\\sqrt{3}) lies on y^{2}=6x, and we obtain x_{0}=\\frac{9}{2}; \\therefore |AB|=x_{0}+\\frac{p}{2}=6; then the area of \\triangle AFB is S=\\frac{1}{3}\\times6\\times3\\sqrt{3}=9\\sqrt{3}" }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, respectively, $P$ a point on the ellipse, $M$ the midpoint of $F_{1} P$, and $|O M|=3$. Then the distance from point $P$ to the left focus of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G);M: Point;MidPoint(LineSegmentOf(F1, P)) = M;O: Origin;Abs(LineSegmentOf(O, M)) = 3", "query_expressions": "Distance(P,LeftFocus(G))", "answer_expressions": "4", "fact_spans": "[[[19, 58], [69, 71], [108, 110]], [[19, 58]], [[1, 8]], [[9, 16]], [[1, 64]], [[1, 64]], [[65, 68], [103, 107]], [[65, 74]], [[75, 78]], [[75, 91]], [[92, 101]], [[92, 101]]]", "query_spans": "[[[103, 117]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, from one of its foci, perpendicular segments are drawn to the two asymptotes. The sum of the lengths of these two perpendicular segments is $a$. Find the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: LineSegment;L:LineSegment;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(F,H);PointOnCurve(F,L);OneOf(Focus(G)) = F ;F: Point;L1:Line;L2:Line;Asymptote(G)={L1,L2};IsPerpendicular(H,L1);IsPerpendicular(L,L2);H + L = a", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[2, 58], [92, 95]], [[5, 58]], [[87, 90]], [], [], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 78]], [[2, 78]], [[2, 66]], [], [], [], [[2, 74]], [[2, 78]], [[2, 78]], [[2, 90]]]", "query_spans": "[[[92, 101]]]", "process": "Without loss of generality, let the hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) have focus at (c,0), asymptotes y=\\pm\\frac{b}{a}x, i.e., bx\\pm ay=0. The length of the perpendicular segment is the distance from the focus to the asymptote, which is \\frac{|bc\\pm a\\cdot0|}{\\sqrt{b^{2}+a^{2}}}=b. Thus, from the given condition we obtain a=2b. Therefore, the eccentricity of the hyperbola satisfies e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=\\frac{5}{4}, so e=\\frac{\\sqrt{5}}{2}." }, { "text": "The equation of a parabola with vertex at the origin and focus at the right focus of the hyperbola $\\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1$ is?", "fact_expressions": "Vertex(H) = O;O: Origin;G: Hyperbola;Expression(G) = (x^2/12 - y^2/4 = 1);RightFocus(G) = Focus(H);H: Parabola", "query_expressions": "Expression(H)", "answer_expressions": "y^2=16*x", "fact_spans": "[[[0, 57]], [[3, 5]], [[7, 46]], [[7, 46]], [[6, 57]], [[54, 57]]]", "query_spans": "[[[54, 61]]]", "process": "Find the right focus of the hyperbola. The equation of the parabola can be set as y^{2}=mx, m>0. Using the focus coordinates of the parabola, m can be obtained, thus giving the required equation. According to the problem, the right focus of the hyperbola \\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1 is (4,0). The equation of the parabola is set as y^{2}=mx, m>0. Since the vertex of the parabola is at the origin and its focus is the right focus of the hyperbola \\frac{x^{2}}{12}-\\frac{y^{2}}{4}=1, \\therefore \\frac{m}{4}=4, so m=16. \\therefore The equation of the parabola is y^{2}=16x" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has two foci $F_{1}(-\\frac{\\sqrt{3}}{2}, 0)$, $F_{2}(\\frac{\\sqrt{3}}{2}, 0)$. Point $P$ lies on the hyperbola in the first quadrant, and $\\tan \\angle P F_{1} F_{2}=\\frac{1}{2}$, $\\tan \\angle P F_{2} F_{1}=-2$. Find the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;F1: Point;F2: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F1) = (-sqrt(3)/2, 0);Coordinate(F2) = (sqrt(3)/2, 0);Focus(G) = {F1, F2};Quadrant(P) = 1;PointOnCurve(P, G);Tan(AngleOf(P, F1, F2)) = 1/2;Tan(AngleOf(P, F2, F1)) = -2", "query_expressions": "Eccentricity(G)", "answer_expressions": "3*sqrt(5)/5", "fact_spans": "[[[2, 58], [137, 140], [217, 220]], [[5, 58]], [[5, 58]], [[64, 95]], [[96, 126]], [[127, 131]], [[5, 58]], [[5, 58]], [[2, 58]], [[64, 95]], [[96, 126]], [[2, 126]], [[127, 143]], [[127, 143]], [[145, 184]], [[185, 215]]]", "query_spans": "[[[217, 226]]]", "process": "" }, { "text": "If the hyperbola $\\frac{x^{2}}{8}-\\frac{y^{2}}{b^{2}}=1$ and the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ share the same directrices, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;H: Ellipse;Expression(G) = (x^2/8 - y^2/b^2 = 1);Expression(H) = (x^2/2 + y^2 = 1);Directrix(G)=Directrix(H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 43], [76, 79]], [[4, 43]], [[44, 71]], [[1, 43]], [[44, 71]], [[1, 74]]]", "query_spans": "[[[76, 85]]]", "process": "" }, { "text": "The minor axis length of an ellipse is $\\sqrt{5}$, and the eccentricity is $e = \\frac{2}{3}$. The foci of the ellipse are $F_{1}$ and $F_{2}$. A line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. Find the perimeter of $\\triangle ABF_{2}$.", "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;F2: Point;F1: Point;e:Number;Length(MinorAxis(G))=sqrt(5);Eccentricity(G)=e;e=2/3;Focus(G)={F1,F2};PointOnCurve(F1, H);Intersection(H, G) = {A,B}", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "6", "fact_spans": "[[[34, 36], [68, 70]], [[65, 67]], [[71, 74]], [[75, 78]], [[48, 55]], [[40, 47], [57, 64]], [[18, 33]], [[0, 36]], [[15, 36]], [[18, 33]], [[34, 55]], [[56, 67]], [[65, 80]]]", "query_spans": "[[[82, 106]]]", "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}=3 \\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) = 3*VectorOf(F2, B)", "query_expressions": "Coordinate(A)", "answer_expressions": "{(-sqrt(2)/2,sqrt(30)/6),(-sqrt(2)/2,-sqrt(30)/6)}", "fact_spans": "[[[19, 46], [62, 64]], [[1, 8]], [[53, 57], [122, 126]], [[9, 16]], [[58, 61]], [[19, 46]], [[1, 52]], [[1, 52]], [[53, 65]], [[58, 65]], [[67, 120]]]", "query_spans": "[[[122, 131]]]", "process": "In the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, $a=\\sqrt{3}$, $b=1$, $c=\\sqrt{2}$, then the left focus is $F_{1}(-\\sqrt{2},0)$, the right focus is $F_{2}(\\sqrt{2},0)$. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $\\overrightarrow{F_{1}A}=(x_{1}+\\sqrt{2},y_{1})$, $\\overrightarrow{F_{2}B}=(x_{2}-\\sqrt{2},y_{2})$. Since $\\overrightarrow{F_{1}A}=3\\overrightarrow{F_{2}B}$, we have $\\begin{cases}x_{1}+\\sqrt{2}=3(x_{2}-\\sqrt{2})\\\\y_{1}=3y_{2}\\end{cases}$, solving gives $\\begin{cases}x_{1}=3x_{2}-4\\sqrt{2}\\\\y_{1}=3y_{2}\\end{cases}$. Since points $A$, $B$ lie on the ellipse, we have $\\begin{cases}\\frac{x_{2}^{2}}{3}+y_{2}^{2}=1\\\\\\frac{(3x_{2}-4\\sqrt{2})^{2}}{3}+(3y_{2})^{2}=1\\end{cases}$. Solving yields $\\begin{cases}x_{2}=\\frac{7\\sqrt{2}}{6}\\\\y_{2}=\\frac{\\sqrt{30}}{18}\\end{cases}$, $\\begin{cases}x_{2}=\\frac{7\\sqrt{2}}{6}\\\\y_{2}=-\\frac{\\sqrt{30}}{18}\\end{cases}$. Hence $\\begin{cases}x_{1}=-\\frac{\\sqrt{2}}{2}\\\\y_{1}=\\frac{\\sqrt{30}}{6}\\end{cases}$ or $\\begin{cases}x_{1}=-\\frac{\\sqrt{2}}{2}\\\\y_{1}=-\\frac{\\sqrt{30}}{6}\\end{cases}$, that is $A(-\\frac{\\sqrt{2}}{2},\\frac{\\sqrt{30}}{6})$ or $A(-\\frac{\\sqrt{2}}{2},-\\frac{\\sqrt{30}}{6})$." }, { "text": "Given the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-y^{2}=1$ ($a>0$), the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively, and point $M$ lies on the left branch of $E$. If $\\angle F_{1} M F_{2} \\in\\left[\\frac{\\pi}{6}, \\frac{\\pi}{3}\\right]$, then the range of values for $\\overrightarrow{M F_{1}} \\cdot \\overrightarrow{M F_{2}}$ is?", "fact_expressions": "E: Hyperbola;Expression(E) = (-y^2 + x^2/a^2 = 1);a: Number;a>0;F1: Point;F2: Point;LeftFocus(E) = F1;RightFocus(E) = F2;M: Point;PointOnCurve(M,LeftPart(E)) = True;In(AngleOf(F1, M, F2),[pi/6, pi/3]) = True", "query_expressions": "Range(DotProduct(VectorOf(M, F1), VectorOf(M, F2)))", "answer_expressions": "[2,4*sqrt(3)+6]", "fact_spans": "[[[2, 44], [73, 76]], [[2, 44]], [[10, 44]], [[10, 44]], [[53, 60]], [[61, 68]], [[2, 68]], [[2, 68]], [[69, 72]], [[69, 80]], [[81, 137]]]", "query_spans": "[[[139, 203]]]", "process": "Let |MF₁| = m, |MF₂| = n, ∠F₁MF₂ = θ, then 4(a² + 1) = m² + n² - 2mn cosθ. Also n - m = 2a, so m² + n² - 2mn = 4a², solving gives mn = 2 / (1 - cosθ). Therefore, $\\overrightarrow{MF_{1}} \\cdot \\overrightarrow{MF_{2}} = |\\overrightarrow{MF_{1}}| \\cdot |\\overrightarrow{MF_{2}}| \\cdot \\cos\\theta = mn\\cos\\theta = \\frac{2\\cos\\theta}{1 - \\cos\\theta} = \\frac{2}{\\cos\\theta} - 1$. Since θ ∈ [π/6, π/3], we have 1/2 ≤ cosθ ≤ √3/2, and (2√3)/3 - 1 ≤ 2 / cosθ - 1 ≤ 4√3 + 6. Thus the range of $\\overrightarrow{MF}_{1} \\cdot \\overrightarrow{MF_{2}}$ is [2, 4√3 + 6]." }, { "text": "Given line $l_{1}$: $2x - y + 2 = 0$ and line $l_{2}$: $x = -1$, a moving point $P$ on the parabola $y^{2} = 4x$; what is the minimum value of the sum of the distances from $P$ to line $l_{1}$ and line $l_{2}$?", "fact_expressions": "G: Parabola;l1:Line;l2:Line;P:Point;Expression(G) = (y^2 = 4*x);Expression(l1)=(2*x-y+2=0);Expression(l2)=(x=-1);PointOnCurve(P,G)", "query_expressions": "Min(Distance(P,l1)+Distance(P,l2))", "answer_expressions": "4*sqrt(5)/5", "fact_spans": "[[[43, 57]], [[2, 24], [65, 74]], [[25, 42], [75, 84]], [[61, 64]], [[43, 57]], [[2, 24]], [[25, 42]], [[43, 64]]]", "query_spans": "[[[61, 95]]]", "process": "By the definition of a parabola, the problem can be transformed into finding the minimum value of the sum of the distance from a point on the parabola to the focus and the distance to the line $ l_{1} $. Since the line $ x = -1 $ is the directrix of the parabola, by the definition of a parabola, the distance from any point on the parabola to the focus is equal to the distance to the directrix. Therefore, the problem is transformed into finding the minimum value of the sum of the distance from a point on the parabola to the focus and the distance to the line $ 2x - y + 2 = 0 $. It is clear that this minimum value is the distance $ d $ from the focus $ F(1,0) $ to the line $ 2x - y + 2 = 0 $, given by $ d = \\frac{|2 - 0 + 2|}{\\sqrt{5}} = \\frac{4\\sqrt{5}}{5} $." }, { "text": "The circle with radius equal to the eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ and centered at its right focus is tangent to one of the asymptotes of the hyperbola. Then $m$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/m = 1);m: Number;Radius(H) = Eccentricity(G);Center(H) = RightFocus(G);H: Circle;IsTangent(H, OneOf(Asymptote(G))) = True", "query_expressions": "m", "answer_expressions": "4/3", "fact_spans": "[[[1, 39], [56, 59]], [[1, 39]], [[69, 72]], [[0, 55]], [[0, 55]], [[54, 55]], [[54, 67]]]", "query_spans": "[[[69, 74]]]", "process": "Slightly" }, { "text": "The eccentricity of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{m}=1$ is $\\frac{5}{4}$, then what is the value of $m$?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/16 - y^2/m = 1);Eccentricity(G) = 5/4", "query_expressions": "m", "answer_expressions": "9", "fact_spans": "[[[0, 39]], [[60, 63]], [[0, 39]], [[0, 57]]]", "query_spans": "[[[60, 66]]]", "process": "The hyperbola $\\frac{x^2}{16}-\\frac{y^{2}}{m}=1$ has its foci on the $x$-axis, so its real semi-axis length is $a=4$. Let the semi-focal distance be $c$, then $c^{2}=16+m$. The eccentricity is $e$, and $e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{16+m}{16}=\\frac{25}{16}$. Solving gives $m=9$, so $m$ equals 9." }, { "text": "Given that point $Q$ moves on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1$, from point $Q$ two tangents are drawn to the circle $(x-1)^{2}+y^{2}=1$, with points of tangency $A$ and $B$ respectively. Then the minimum value of $|A B|$ is?", "fact_expressions": "G: Ellipse;H: Circle;A: Point;B: Point;Q: Point;l1: Line;l2: Line;Expression(G) = (x^2/16 + y^2/8 = 1);Expression(H) = (y^2 + (x - 1)^2 = 1);PointOnCurve(Q, G);TangentOfPoint(Q, H) = {l1, l2};TangentPoint(l1,H)=A;TangentPoint(l2,H)=B", "query_expressions": "Min(Abs(LineSegmentOf(A, B)))", "answer_expressions": "2*sqrt(42)/7", "fact_spans": "[[[7, 45]], [[55, 75]], [[86, 89]], [[90, 93]], [[2, 6], [50, 54]], [], [], [[7, 45]], [[55, 75]], [[2, 46]], [[49, 80]], [[49, 93]], [[49, 93]]]", "query_spans": "[[[95, 108]]]", "process": "As shown in the figure, connect PC, intersecting AB at H, yielding H as the midpoint of AB. The center of the circle (x-1)^{2}+y^{2}=1 is C(1,0). Connect AC and BC, we obtain AC\\botPA, BC\\botPB, and |PA|=|PB|=\\sqrt{|PC|}. Also |AB|=2|AH|=\\frac{2|PA|\\cdot|AC|}{|PC|}=\\frac{2\\sqrt{|PC|^{2}-1}}{|PC|}=2. Let P(4\\cos\\theta,2\\sqrt{2}\\sin\\theta), \\theta\\in[0,2\\pi], then |PC|^{2}=(4\\cos\\theta-1)^{2}+(2\\sqrt{2}\\sin\\theta)^{2}=16\\cos^{2}\\theta-8\\cos\\theta+1+8\\sin^{2}\\theta=8(\\cos\\theta-\\frac{1}{2})^{2}+7. When \\cos\\theta=\\frac{1}{2}, |PC|^{2} attains its minimum value 7, so the minimum value of |AB| is 2\\sqrt{1-\\frac{1}{7}}=\\frac{2\\sqrt{42}}{7}." }, { "text": "The line $y = x + b$ intersects the parabola $x^{2} = 2y$ at points $A$ and $B$, $O$ is the origin, and $OA \\perp OB$. Then $b = $?", "fact_expressions": "H: Line;Expression(H) = (y = b + x);b: Number;G: Parabola;Expression(G) = (x^2 = 2*y);Intersection(H, G) = {A, B};A: Point;B: Point;O: Origin;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B)) = True", "query_expressions": "b", "answer_expressions": "2", "fact_spans": "[[[0, 9]], [[0, 9]], [[63, 66]], [[10, 24]], [[10, 24]], [[0, 35]], [[26, 29]], [[30, 33]], [[36, 39]], [[46, 61]]]", "query_spans": "[[[63, 68]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, its two asymptotes intersect the directrix of the parabola $y^{2}=-8x$ at points $A$ and $B$, respectively, and $O$ is the origin. If the area of $\\triangle ABO$ is $4\\sqrt{3}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Parabola;A: Point;B: Point;O: Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = -8*x);l1:Line;l2:Line;Asymptote(G)={l1,l2};Intersection(l1, Directrix(G)) = A;Intersection(l2, Directrix(G)) = B;Area(TriangleOf(A, B, O)) = 4*sqrt(3)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[2, 58], [142, 145]], [[5, 58]], [[5, 58]], [[65, 80]], [[87, 90]], [[91, 94]], [[97, 100]], [[5, 58]], [[5, 58]], [[2, 58]], [[65, 80]], [], [], [[2, 64]], [[2, 96]], [[2, 96]], [[107, 140]]]", "query_spans": "[[[142, 151]]]", "process": "Since the two asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ are $y=\\pm\\frac{b}{a}x$, and the directrix of the parabola $y^{2}=-8x$ is $x=2$, it follows that $A(2,\\frac{2b}{a})$, $B(2,-\\frac{2b}{a})$. Since the area of $\\triangle AOB$ is $4\\sqrt{3}$, we have $S_{\\triangle AOB}=\\frac{1}{2}\\times2\\times\\frac{4b}{a}=4\\sqrt{3}$, so $b=\\sqrt{3}a$, $c=2a$, $e=\\frac{c}{a}=2$." }, { "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, and $AB$ is any diameter of the circle $x^{2}+(y+2)^{2}=1$, then the maximum value of $\\overrightarrow{PA} \\cdot \\overrightarrow{PB}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2 = 1);P: Point;PointOnCurve(P, G);H: Circle;Expression(H) = (x^2 + (y + 2)^2 = 1);A: Point;B: Point;IsDiameter(LineSegmentOf(A, B), H)", "query_expressions": "Max(DotProduct(VectorOf(P, A), VectorOf(P, B)))", "answer_expressions": "25/3", "fact_spans": "[[[7, 34]], [[7, 34]], [[2, 6]], [[2, 37]], [[44, 64]], [[44, 64]], [[38, 43]], [[38, 43]], [[38, 70]]]", "query_spans": "[[[72, 127]]]", "process": "As shown in the figure, let P(x,y) satisfy \\frac{x^{2}}{4}+y^{2}=1, and let the center of the circle x^{2}+(y+2)^{2}=1 be C(0,-2). \\overrightarrow{PA}=\\overrightarrow{PC}+\\overrightarrow{CA}, \\overrightarrow{PB}=\\overrightarrow{PC}+\\overrightarrow{CB}, \\overrightarrow{CA}+\\overrightarrow{CB}=\\overrightarrow{0}, \\overrightarrow{CA}\\cdot\\overrightarrow{CB}=-r^{2}=-1. \\therefore \\overrightarrow{PA}\\cdot\\overrightarrow{PB}=(\\overrightarrow{PC}+\\overrightarrow{CA})\\cdot(\\overrightarrow{PC}+\\overrightarrow{CB})=\\overrightarrow{PC}^{2}+\\overrightarrow{PC}\\cdot(\\overrightarrow{CA}+\\overrightarrow{CB})+\\overrightarrow{CA}\\cdot\\overrightarrow{CB}=\\overrightarrow{PC}^{2}-1=x^{2}+(y+2)^{2}-1=4-4y^{2}+(y+2)^{2}-1=-3y^{2}+4y+7=-3(y-\\frac{2}{3})^{2}+\\frac{25}{3}. Since -1\\leqslant y\\leqslant 1, \\therefore \\overrightarrow{PA}\\cdot\\overrightarrow{PB} reaches its maximum value \\frac{25}{3} if and only if y=\\frac{2}{3}. \\therefore The maximum value of \\overrightarrow{PA}\\cdot\\overrightarrow{PB} is \\frac{25}{3}." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, its two asymptotes intersect the directrix of the parabola $y^{2}=2 p x(p>0)$ at points $A$ and $B$, respectively, and $O$ is the origin. If the eccentricity of the hyperbola is $2$, and the area of $\\Delta A O B$ is $\\sqrt{3}$, then $p=$?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Parabola;p: Number;A: Point;B:Point;O: Origin;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);p>0;Expression(H) = (y^2 = 2*(p*x));L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(L1,Directrix(H))=A;Intersection(L2,Directrix(H))=B;Eccentricity(G)=2;Area(TriangleOf(A,O,B))=sqrt(3)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 58], [113, 116]], [[5, 58]], [[5, 58]], [[65, 86]], [[155, 158]], [[93, 96]], [[97, 100]], [[103, 106]], [[5, 58]], [[5, 58]], [[2, 58]], [[68, 86]], [[65, 86]], [], [], [[2, 64]], [[2, 102]], [[2, 102]], [[113, 124]], [[125, 153]]]", "query_spans": "[[[155, 160]]]", "process": "Test analysis: Given $ e = \\frac{c}{a} = 2 $, we have $ c = 2a $, $ b = \\sqrt{3}a $, so the asymptotes of the hyperbola are $ y = \\pm\\sqrt{3}x $. The directrix of the parabola is $ x = -\\frac{p}{2} $. Solving simultaneously the asymptotes of the hyperbola and the directrix of the parabola gives $ -\\frac{\\sqrt{3}p}{2} $. In $ \\triangle AOB $, $ |AB| = \\sqrt{3}p $, the distance from $ O $ to $ AB $ is $ \\frac{p}{2} $, $ p \\cdot \\frac{p}{2} = \\sqrt{3} $, $ p = 2 $." }, { "text": "If one of the asymptotes of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ is given by $x-2 y=0$, then its eccentricity equals?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Expression(OneOf(Asymptote(G))) = (x - 2*y = 0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[1, 57], [77, 78]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 75]]]", "query_spans": "[[[77, 85]]]", "process": "Test analysis: Since one asymptote of the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0,b>0)$ is $x-2y=0$, $\\therefore a=2b$, $\\therefore c=\\sqrt{5}b$, so the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\frac{\\sqrt{5}}{2}$." }, { "text": "The line with slope $k$ passing through the left vertex $A$ of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the ellipse $C$ at another point $B$, and the line connecting point $B$ and the right focus $F$ is perpendicular to the $x$-axis. If $\\frac{1}{3} b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C)=A;PointOnCurve(A,H);Slope(H)=k;Intersection(H,C)={A,B};RightFocus(C)=F;IsPerpendicular(LineSegmentOf(B,F),xAxis);1/3 0). Substituting the coordinates of point B into the ellipse equation yields c²/a² + y₀²/b² = 1. Solving gives y₀² = (1 − c²/a²)b² = b⁴/a² ⇒ y₀ = ±b²/a. Without loss of generality, take y₀ = b²/a. Therefore, k = tanθ = |BF| / |AF| = (b²/a) / (a + c) = b²/(a² + ac) = (a² − c²)/(a² + ac). Given 1/3 < k < 1/2, we obtain \n{ \na² − c² > (1/3)(a² + ac) > (a² − c²)/3, i.e., \n(a² − c²)/(a² + ac) < 1/2 \n} \n⇒ \n{ \n3c² + ac − 2a² < 0 \n2c² + ac − a² > 0 \n} \n⇒ \n{ \n3e² + e − 2 < 0 \n2e² + e − 1 > 0 \n} \n⇒ \n{ \n−1 < e < 2/3 \ne > 1/2 or e < −1 \n} \n⇒ 1/2 < e < 2/3" }, { "text": "The left and right foci of the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is any point on the ellipse $M$. The range of the maximum value of $|P F_{1}| \\cdot |P F_{2}|$ is $[2 c^{2}, 3 c^{2}]$, where $c=\\sqrt{a^{2}-b^{2}}$. What is the range of the eccentricity $e$ of the ellipse $M$?", "fact_expressions": "M: Ellipse;b: Number;a: Number;P: Point;F1: Point;F2: Point;e: Number;c:Number;a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(M) =F1;RightFocus(M)=F2;PointOnCurve(P,M);Range(Max(Abs(LineSegmentOf(P,F1))*Abs(LineSegmentOf(P,F2))))=[2*c^2,3*c^2];Eccentricity(M) = e;c=sqrt(a^2-b^2)", "query_expressions": "Range(e)", "answer_expressions": "[sqrt(3)/3,sqrt(2)/2]", "fact_spans": "[[[0, 57], [86, 91], [180, 185]], [[7, 57]], [[7, 57]], [[82, 85]], [[66, 73]], [[74, 81]], [[189, 192]], [[156, 178]], [[7, 57]], [[7, 57]], [[0, 57]], [[0, 81]], [[0, 81]], [[82, 95]], [[97, 153]], [[180, 192]], [[156, 178]]]", "query_spans": "[[[189, 199]]]", "process": "\\because|PF_{1}|\\cdot|PF_{2}|=(a+ex)(a-ex)=a^{2}-e^{2}x^2\\leqslanta^{2}\\therefore the maximum value of |PF_{1}|\\cdot|PF_{2}| is a^{2}\\therefore by the problem 2c^{2}\\leqslanta^{2}\\leqslant3c^{2}\\therefore\\sqrt{2}c\\leqslanta\\leqslant\\sqrt{3}c\\therefore\\frac{\\sqrt{3}}{3}\\leqslante\\leqslant\\frac{\\sqrt{2}}{2} Therefore the range of eccentricity e of ellipse M is [\\frac{\\sqrt{3}}{3},\\frac{\\sqrt{2}}{2}]. The correct result of this problem: [\\frac{\\sqrt{3}}{3},\\frac{\\sqrt{2}}{2}]" }, { "text": "Given that the focus of the parabola $x^{2}=2 p y(p>0)$ coincides with one focus of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{12}=1$, then $p$=?", "fact_expressions": "G: Parabola;p: Number;H: Ellipse;p>0;Expression(G) = (x^2 = 2*(p*y));Expression(H) = (x^2/3 + y^2/12 = 1);Focus(G) = OneOf(Focus(H))", "query_expressions": "p", "answer_expressions": "6", "fact_spans": "[[[2, 23]], [[74, 77]], [[27, 65]], [[5, 23]], [[2, 23]], [[27, 65]], [[2, 72]]]", "query_spans": "[[[74, 79]]]", "process": "" }, { "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, a line $l$ passing through point $F$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $A$, and intersects the right branch of the hyperbola at point $B$. If $\\overrightarrow{F A}=\\frac{1}{3} \\overrightarrow{F B}$, then the eccentricity of the 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;LeftFocus(G) = F;PointOnCurve(F, l) = True;l: Line;H: Circle;Expression(H) = (x^2 + y^2 = a^2);IsTangent(l, H) = True;TangentPoint(l, H) = A;A: Point;Intersection(l, RightPart(G)) = B;B: Point;VectorOf(F, A) = VectorOf(F, B)/3", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(13)/2", "fact_spans": "[[[2, 5], [68, 72]], [[6, 62], [109, 112], [180, 183]], [[6, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[2, 66]], [[67, 78]], [[73, 78]], [[79, 99]], [[79, 99]], [[73, 101]], [[73, 106]], [[102, 106]], [[73, 121]], [[117, 121]], [[123, 178]]]", "query_spans": "[[[180, 189]]]", "process": "" }, { "text": "Given that point $P$ lies on the right branch of the hyperbola $x^{2}-y^{2}=1$, and the distance from point $P$ to the line $y=x$ is $\\sqrt{2}$, then the coordinates of point $P$ are?", "fact_expressions": "G: Hyperbola;H: Line;P: Point;Expression(G) = (x^2 - y^2 = 1);Expression(H) = (y = x);PointOnCurve(P, RightPart(G));Distance(P, H) = sqrt(2)", "query_expressions": "Coordinate(P)", "answer_expressions": "(5/4,-3/4)", "fact_spans": "[[[7, 25]], [[36, 43]], [[2, 6], [31, 35], [60, 64]], [[7, 25]], [[36, 43]], [[2, 29]], [[31, 57]]]", "query_spans": "[[[60, 69]]]", "process": "" }, { "text": "Given that the focus of the parabola $y^{2}=4 x$ is $F$, the directrix is the line $l$, and from a point $P$ on the parabola, $PE \\perp l$ at $E$. If the inclination angle of the line $EF$ is $150^{\\circ}$, then $|P F|$=?", "fact_expressions": "l: Line;G: Parabola;E: Point;F: Point;P: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Directrix(G) = l;PointOnCurve(P, G);PointOnCurve(P, LineSegmentOf(P, E));IsPerpendicular(LineSegmentOf(P, E), l);FootPoint(LineSegmentOf(P, E), l) = E;Inclination(LineOf(E, F)) = ApplyUnit(150, degree)", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "4/3", "fact_spans": "[[[27, 32]], [[2, 16], [34, 37]], [[59, 62]], [[20, 23]], [[41, 44]], [[2, 16]], [[2, 23]], [[2, 32]], [[34, 44]], [[33, 58]], [[45, 59]], [[45, 62]], [[64, 89]]]", "query_spans": "[[[91, 100]]]", "process": "From the parabola equation $ y^{2} = 4x $, we obtain the focus $ F(1,0) $ and the directrix $ l $ with equation $ x = -1 $. Since the inclination angle of line $ EF $ is $ 150^{\\circ} $, $ k_{EF} = \\tan 150^{\\circ} = -\\frac{\\sqrt{3}}{3} $. Thus, the equation of line $ EF $ is $ y = -\\frac{\\sqrt{3}}{3}(x-1) $. Solving the system \n\\[\n\\begin{cases}\nx = -1 \\\\\ny = -\\frac{\\sqrt{3}}{3}(x-1)\n\\end{cases}\n\\]\nyields $ y = \\frac{2\\sqrt{3}}{3} $, so $ E\\left(-1, \\frac{2\\sqrt{3}}{3}\\right) $. Since $ PE \\perp l $ at $ E $, $ y_{P} = \\frac{2\\sqrt{3}}{3} $. Substituting into the parabola equation gives $ \\left(\\frac{2\\sqrt{3}}{3}\\right)^{2} = 4x_{P} $, solving which yields $ x_{P} = \\frac{1}{3} $. Therefore, $ |PF| = |PE| = x_{P} + 1 = \\frac{4}{3} $." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the left focus is $F$, the right vertex is $A$, the point $B(0, b)$, $P$ is the midpoint of segment $AB$, and $|P F|=\\sqrt{6}|O P|$ ($O$ is the origin), then the eccentricity 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;F: Point;LeftFocus(G) = F;A: Point;RightVertex(G) = A;B: Point;Coordinate(B) = (0, b);P: Point;MidPoint(LineSegmentOf(A, B)) = P;Abs(LineSegmentOf(P, F)) = sqrt(6)*Abs(LineSegmentOf(O, P));O: Origin", "query_expressions": "Eccentricity(G)", "answer_expressions": "4", "fact_spans": "[[[2, 58], [135, 138]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[63, 66]], [[2, 66]], [[71, 74]], [[2, 74]], [[75, 85]], [[75, 85]], [[97, 100]], [[86, 100]], [[102, 123]], [[124, 127]]]", "query_spans": "[[[135, 144]]]", "process": "From the problem, we know A(a,0), let F(-c,0), and point B(0,b). Let P be the midpoint of segment AB, so point P(\\frac{a}{2},\\frac{b}{2}). Therefore, by the distance formula, we have \\sqrt{\\frac{|FF|=\\sqrt{6}}{4}+ac}|OP|." }, { "text": "Given the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, what are its asymptotes? What is its eccentricity?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1)", "query_expressions": "Expression(Asymptote(G));Eccentricity(G)", "answer_expressions": "y=pm*(1/2)*x\nsqrt(5)/2", "fact_spans": "[[[2, 30], [32, 33]], [[2, 30]]]", "query_spans": "[[[32, 40]], [[32, 45]]]", "process": "" }, { "text": "Given a point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ such that the distance from $M$ to the left focus $F_{1}$ is $6$, and $N$ is the midpoint of $M F_{1}$, then $|O N|=$?", "fact_expressions": "G: Ellipse;M: Point;F1: Point;O: Origin;N: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(M,G);LeftFocus(G)=F1;Distance(M,F1)=6;MidPoint(LineSegmentOf(M,F1))=N", "query_expressions": "Abs(LineSegmentOf(O, N))", "answer_expressions": "2", "fact_spans": "[[[2, 40]], [[43, 46]], [[50, 57]], [[83, 90]], [[65, 68]], [[2, 40]], [[2, 46]], [[2, 57]], [[43, 64]], [[65, 81]]]", "query_spans": "[[[83, 92]]]", "process": "Let the foci of the ellipse be $ F_{2} $, connect $ F_{2}M $. Since $ M $ is the midpoint of $ F_{1}F_{2} $, then $ ON $ is the midline of triangle $ F_{1}F_{2}M $, so $ |ON| = \\frac{1}{2}|MF_{2}| $. By the definition of the ellipse: $ |MF_{1}| + |MF_{2}| = 2a = 10 $, $ |MF_{1}| = 6 $, then $ |MF_{2}| = 4 $, thus $ |ON| = 2 $." }, { "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=\\sqrt{3} x$, and one of its foci lies on the directrix of the parabola $y^{2}=24 x$, find the equation of the hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;H: Parabola;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 24*x);Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x);PointOnCurve(OneOf(Focus(G)), Directrix(H))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9 - y^2/27 = 1", "fact_spans": "[[[2, 58], [82, 83], [110, 113]], [[5, 58]], [[5, 58]], [[89, 104]], [[5, 58]], [[5, 58]], [[2, 58]], [[89, 104]], [[2, 81]], [[82, 108]]]", "query_spans": "[[[110, 117]]]", "process": "From the given conditions, $\\frac{b}{a}=\\sqrt{3}, c=6$, $\\therefore a=3, b=3\\sqrt{3}$, so the equation of the hyperbola is $\\frac{x^2}{9}-\\frac{y^{2}}{27}=1$." }, { "text": "The distance from the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1(a>0)$ to its asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2 + x^2/a^2 = 1);a: Number;a>0", "query_expressions": "Distance(RightFocus(G), Asymptote(G))", "answer_expressions": "1", "fact_spans": "[[[0, 37], [42, 43]], [[0, 37]], [[3, 37]], [[3, 37]]]", "query_spans": "[[[0, 52]]]", "process": "" }, { "text": "A line passing through the focus of the parabola $y^{2}=2 x$ intersects the parabola at points $P(x_{1} , y_{1})$ and $Q(x_{2}, y_{2})$. If $x_{1}+x_{2}=3$, then $PQ$=?", "fact_expressions": "G: Parabola;H: Line;P: Point;Q: Point;Expression(G) = (y^2 = 2*x);Coordinate(P)=(x1,y1);Coordinate(Q)=(x2,y2);PointOnCurve(Focus(G), H);Intersection(H, G)={P,Q};x1:Number;x2:Number;y1:Number;y2:Number;x1+x2=3", "query_expressions": "LineSegmentOf(P, Q)", "answer_expressions": "4", "fact_spans": "[[[1, 15], [22, 25]], [[19, 21]], [[26, 45]], [[47, 64]], [[1, 15]], [[26, 44]], [[47, 64]], [[0, 21]], [[19, 66]], [[26, 44]], [[47, 64]], [[26, 44]], [[47, 64]], [[68, 83]]]", "query_spans": "[[[85, 91]]]", "process": "According to the problem, the parabola has p=1. Since the length of a chord passing through the focus of a parabola is given by x_{1}+x_{2}+p, it follows that |PQ|=x_{1}+x_{2}+p=3+1=4." }, { "text": "Given the ellipse: $\\frac{x^{2}}{4}+\\frac{y^{2}}{b^{2}}=1$, with left and right foci $F_{1}$ and $F_{2}$ respectively. A line $l$ passing through $F_{1}$ intersects the ellipse at points $A_{1}$ and $B$. If the maximum value of $A_{1} F_{2}+B F_{2}$ is $5$, then what is the equation of the ellipse?", "fact_expressions": "l: Line;G: Ellipse;b: Number;F2: Point;B: Point;F1: Point;A1: Point;Expression(G) = (x^2/4 + y^2/b^2 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(F1, l);Intersection(l, G) = {A1, B};Max(LineSegmentOf(A1, F2) + LineSegmentOf(B, F2)) = 5", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4+y^2/3=1", "fact_spans": "[[[77, 82]], [[2, 44], [83, 85], [132, 134]], [[5, 44]], [[60, 67]], [[94, 97]], [[52, 59], [69, 76]], [[86, 93]], [[2, 44]], [[2, 67]], [[2, 67]], [[68, 82]], [[77, 99]], [[101, 130]]]", "query_spans": "[[[132, 138]]]", "process": "Problem Analysis: |AF₂| + |BF₂| = 4a·|AB| = 8 - |AB|. According to the maximum value of |AF₂| + |BF₂| being 5, the minimum value of |AB| is 3. As per the problem, assume the equation of line l is: my = x + c (the case where the slope of line l is 0 need not be considered), A(x₁, y₁), B(x₂, y₂). Combining with the ellipse equation yields: (b²m² + 4)y² · 2mcb²y + b²c² - 4b² = 0. Using relationships between roots and coefficients, and the chord length formula, we obtain |AF₂| + |BF₂| = 4a·|AB| = 8 - |AB|. Since the maximum value of |AF₂| + |BF₂| is 5, the minimum value of |AB| is 3. As per the problem, assume the equation of line l is: my = x + c (the case where the slope of line l is 0 need not be considered), A(x₁, y₁), B(x₂, y₂). Solving the system: \n\\begin{cases} my = x + c \\\\ \\frac{x^2}{4} + \\frac{y^2}{b^2} = 1 \\end{cases} \nleads to: (b²m² + 4)y² · 2mcb²y + b²c² · 4b² = 0, c² = 4·b²\\frac{2)[(y₁ + y₂)² - 4y₁y₂]}{(b²c²b⁴ - \\frac{4(b²c² - 4b²)}{b²m² + 4}] = \\frac{4b²(1 + m²)}{b²m² + 4}. When m ≠ 0, |AB| = 4 + \\frac{4b² - 16}{b²m² + 4} > b² ∴ b² = 3." }, { "text": "Given point $P(0,1)$ and the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{1}=1$, $Q$ is a moving point on the ellipse. Then the maximum value of $|P Q|$ is?", "fact_expressions": "G: Ellipse;P: Point;Q: Point;Expression(G) = (x^2/4 + y^2/1 = 1);Coordinate(P) = (0, 1);PointOnCurve(Q, G)", "query_expressions": "Max(Abs(LineSegmentOf(P, Q)))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[12, 49], [56, 58]], [[2, 11]], [[52, 55]], [[12, 49]], [[2, 11]], [[52, 62]]]", "query_spans": "[[[64, 77]]]", "process": "" }, { "text": "Through a focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, draw a line $l$ parallel to one of its asymptotes. The line $l$ intersects the $y$-axis at point $P$. If the midpoint of segment $O P$ is an endpoint of the imaginary axis of the hyperbola ($O$ being the origin), 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;F:Point;OneOf(Focus(G))=F;l:Line;IsParallel(l,Asymptote(G));Intersection(l, yAxis) = P;P:Point;O:Origin;MidPoint(LineSegmentOf(O,P))=Endpoint(ImageinaryAxis(G));PointOnCurve(F,l)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 47], [97, 100], [118, 121], [56, 57]], [[1, 47]], [[4, 47]], [[4, 47]], [[52, 55]], [[1, 55]], [[64, 67], [68, 73]], [[56, 67]], [[68, 84]], [[80, 84]], [[107, 110]], [[86, 105]], [[0, 67]]]", "query_spans": "[[[118, 127]]]", "process": "From $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the asymptotes are given by $y=\\pm\\frac{b}{a}x$. Without loss of generality, take $F(c,0)$ and the asymptote $y=\\frac{b}{a}x$, then the equation of line $l$ is: $y=\\frac{b}{a}(x-c)$. Letting $x=0$, we get $y=-\\frac{bc}{a}$, so $P(0,-\\frac{bc}{a})$. The endpoint of the imaginary axis is $B_{1}(0,-b)$. Since $B_{1}(0,-b)$ is the midpoint of segment $OP$, we have $\\frac{-\\frac{bc}{a}+0}{2}=-b$, which gives $c=2a$. Therefore, the eccentricity of the hyperbola is $e=\\frac{c}{a}=2$." }, { "text": "If point $P$ lies on the curve $C_{1}$: $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, point $Q$ lies on the curve $C_{2}$: $(x-5)^{2}+y^{2}=1$, and point $R$ lies on the curve $C_{3}$: $(x+5)^{2}+y^{2}=1$, then the maximum value of $|P Q|-|P R|$ is?", "fact_expressions": "C1: Curve;C2: Curve;C3: Curve;P: Point;Q: Point;R: Point;Expression(C1) = (x^2/16 - y^2/9 = 1);Expression(C2) = (y^2 + (x - 5)^2 = 1);Expression(C3) = (y^2 + (x + 5)^2 = 1);PointOnCurve(P, C1);PointOnCurve(Q, C2);PointOnCurve(R, C3)", "query_expressions": "Max(Abs(LineSegmentOf(P, Q)) - Abs(LineSegmentOf(P, R)))", "answer_expressions": "10", "fact_spans": "[[[6, 53]], [[60, 89]], [[96, 125]], [[1, 5]], [[55, 59]], [[91, 95]], [[6, 53]], [[60, 89]], [[96, 125]], [[1, 54]], [[55, 90]], [[91, 126]]]", "query_spans": "[[[128, 147]]]", "process": "" }, { "text": "Given the ellipse $\\Gamma$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, a line $l$ passing through the point $P(1,1)$ intersects the ellipse $\\Gamma$ at points $A$ and $B$. If the chord $AB$ has point $P$ as its midpoint, then what is the equation of line $l$?", "fact_expressions": "Gamma: Ellipse;Expression(Gamma) = (x^2/4 + y^2/3 = 1);P: Point;Coordinate(P) = (1, 1);l: Line;PointOnCurve(P, l);A: Point;B: Point;Intersection(l, Gamma) = {A, B};IsChordOf(LineSegmentOf(A, B), Gamma);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(l)", "answer_expressions": "3*x+4*y-7=0", "fact_spans": "[[[2, 49], [67, 77]], [[2, 49]], [[51, 60], [100, 104]], [[51, 60]], [[61, 66], [109, 114]], [[50, 66]], [[80, 83]], [[84, 87]], [[61, 89]], [[67, 97]], [[92, 107]]]", "query_spans": "[[[109, 119]]]", "process": "" }, { "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and one of its foci is $(\\sqrt{10}, 0)$, then what is the equation of this hyperbola?", "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], [24, 25], [50, 53]], [[1, 22]], [25, 46]]", "query_spans": "[[49, 57]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, a line $l$ passes through the left focus and has an inclination angle of $\\frac{\\pi}{3}$. The circle with the major axis of the ellipse as its diameter intercepts a chord on $l$ whose length equals the focal distance of the ellipse. Find the eccentricity of the ellipse.", "fact_expressions": "l: Line;G: Ellipse;b: Number;a: Number;H: Circle;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(LeftFocus(G), l);Inclination(l) = pi/3;IsDiameter(MajorAxis(G), H);Length(InterceptChord(l, H)) = FocalLength(G)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(7)/7", "fact_spans": "[[[55, 60], [97, 100]], [[2, 54], [86, 88], [107, 109], [114, 116]], [[4, 54]], [[4, 54]], [[95, 96]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 64]], [[55, 84]], [[85, 96]], [[95, 112]]]", "query_spans": "[[[114, 122]]]", "process": "By the given condition, the left focus of the ellipse is (-c,0), the length of the major axis is 2a, and the focal distance is 2c. Let the equation of line l be y=\\sqrt{3}(x+c), that is, \\sqrt{3}x-y+\\sqrt{3}c=0. The circle with the major axis of the ellipse as diameter has center (0,0) and radius a. Therefore, the distance from the center to line l is d=\\frac{|\\sqrt{3}c|}{2}=\\frac{\\sqrt{3}}{2}c. Then 2c=2\\sqrt{a^{2}-d^{2}}}=2\\sqrt{a^{2}-\\frac{3}{4}c^{2}}}, simplifying yields c^{2}=\\frac{4}{7}a^{2}. Hence, the eccentricity of the ellipse is \\frac{c}{a}=\\sqrt{\\frac{4}{7}}=\\frac{2\\sqrt{7}}{7}" }, { "text": "If the circle $x^{2}+y^{2}=a^{2} (a>0)$ and the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ have common points, then the range of real values for $a$ is?", "fact_expressions": "H: Circle;Expression(H) = (x^2 + y^2 = a^2);a: Real;a>0;G: Ellipse;Expression(G) = (x^2/9 + y^2/4 = 1);IsIntersect(H, G)", "query_expressions": "Range(a)", "answer_expressions": "[2, 3]", "fact_spans": "[[[1, 28]], [[1, 28]], [[72, 77]], [[2, 28]], [[29, 66]], [[29, 66]], [[1, 70]]]", "query_spans": "[[[72, 84]]]", "process": "" }, { "text": "A point $P$ on the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ 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 PF_{1} F_{2}$ is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2:Point;Expression(G) = (x^2/49 + y^2/24 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);IsPerpendicular(LineSegmentOf(P,F1),LineSegmentOf(P,F2))", "query_expressions": "Area(TriangleOf(P,F1,F2))", "answer_expressions": "24", "fact_spans": "[[[0, 39], [46, 48]], [[42, 45]], [[53, 60]], [[62, 69]], [[0, 39]], [[46, 69]], [[0, 45]], [[42, 76]]]", "query_spans": "[[[78, 107]]]", "process": "" }, { "text": "If the directrix of the parabola $y^{2}=2 p x (p>0)$ passes through the point $(-1,1)$, then the coordinates of the focus of the parabola are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;H: Point;Coordinate(H) = (-1, 1);PointOnCurve(H, Directrix(G))", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(1, 0)", "fact_spans": "[[[1, 24], [41, 44]], [[1, 24]], [[4, 24]], [[4, 24]], [[29, 38]], [[29, 38]], [[1, 38]]]", "query_spans": "[[[41, 50]]]", "process": "\\because the directrix of the parabola y^{2}=2px (p>0) passes through the point (-1,1), \\because \\frac{p}{2}=1, \\therefore the focus of this parabola has coordinates (1,0)" }, { "text": "Let $P$ be an arbitrary point on the hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$. Draw lines through $P$ parallel to the asymptotes $l_{1}$: $y=\\frac{b}{a} x$ and $l_{2}$: $y=-\\frac{b}{a} x$, intersecting them at points $R$ and $Q$, respectively. Then $|P R| \\cdot|P Q|$=?", "fact_expressions": "P: Point;PointOnCurve(P, E);E: Hyperbola;Expression(E) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;l1: Line;Expression(l1) = (y = (b/a)*x);l2: Line;Expression(l2) = (y = (-b/a)*x);Asymptote(E) = {l1, l2};Z1: Line;Z2: Line;PointOnCurve(P, Z1);PointOnCurve(P, Z2);IsParallel(Z1, l1);IsParallel(Z2, l2);R: Point;Q: Point;Intersection(Z1, l2) = R;Intersection(Z2, l1) = Q", "query_expressions": "Abs(LineSegmentOf(P, Q))*Abs(LineSegmentOf(P, R))", "answer_expressions": "(a^2+b^2)/4", "fact_spans": "[[[1, 4], [63, 66]], [[1, 61]], [[5, 56]], [[5, 56]], [[13, 56]], [[13, 56]], [[70, 96]], [[70, 96]], [[97, 124]], [[97, 124]], [[5, 124]], [], [], [[62, 128]], [[62, 128]], [[62, 128]], [[62, 128]], [[132, 136]], [[137, 140]], [[62, 140]], [[62, 140]]]", "query_spans": "[[[143, 163]]]", "process": "" }, { "text": "The standard equation of a parabola with directrix $x=1$ is?", "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (x = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "y^2=-4*x", "fact_spans": "[[[11, 14]], [[0, 14]]]", "query_spans": "[[[11, 20]]]", "process": "Since the equation of the directrix of the parabola is x=1, we have \\frac{p}{2}=1, and since the focus of the parabola lies on the negative x-axis, we get p=2. Therefore, the standard equation of the parabola is y^{2}=-4x." }, { "text": "Given that $M$ and $N$ are two distinct points on the parabola $y^{2}=4x$, and $F$ is the focus of the parabola. If the $y$-coordinate of the midpoint of segment $MN$ is $2$, and $|MF|+|NF|=10$, then the equation of line $MN$ is?", "fact_expressions": "G: Parabola;M: Point;N: Point;F:Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M,G);PointOnCurve(N,G);Negation(M=N);Focus(G)=F;YCoordinate(MidPoint(LineSegmentOf(M,N)))=2;Abs(LineSegmentOf(M,F))+Abs(LineSegmentOf(N,F))=10", "query_expressions": "Expression(LineOf(M,N))", "answer_expressions": "x-y-2=0", "fact_spans": "[[[10, 24], [36, 39]], [[2, 5]], [[6, 9]], [[32, 35]], [[10, 24]], [[2, 31]], [[2, 31]], [[2, 31]], [[32, 42]], [[44, 62]], [[63, 79]]]", "query_spans": "[[[81, 93]]]", "process": "Let M(x_{1},y_{1}), N(x_{2},y_{2}), then |MF| + |NF| = x_{1} + x_{2} + 2 = 10. Therefore, x_{1} + x_{2} = 8, ∴ the midpoint of segment MN has coordinates (4,2). From y_{1}^{2} = 4x_{1}, y_{2}^{2} = 4x_{2}, subtracting the two equations gives (y_{1} + y_{2})(y_{1} - y_{2}) = 4(x_{1} - x_{2}), and since \\frac{y_{1} + y_{2}}{2} = 2, ∴ k = \\frac{y_{1} - y_{2}}{x_{1} - x_{2}} = 1. ∴ the equation of line MN is y - 2 = x - 4, that is, x - y - 2 = 0." }, { "text": "Given that the upper vertex of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$ is $A$, and the left vertex is $B$, then the slope of the line $AB$ is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;Expression(G) = (x^2/3 + y^2 = 1);UpperVertex(G) = A;LeftVertex(G) = B", "query_expressions": "Slope(LineOf(A, B))", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 29]], [[34, 37]], [[42, 45]], [[2, 29]], [[2, 37]], [[2, 45]]]", "query_spans": "[[[47, 59]]]", "process": "Since the equation of the ellipse is $\\frac{x^{2}}{3}+y^{2}=1$, it follows that $a^{2}=3$, $b^{2}=1$, i.e., $a=\\sqrt{3}$, $b=1$, so the upper vertex of the ellipse is $A(0,1)$, the left vertex is $B(-\\sqrt{3},0)$, thus $k_{AB}=\\frac{1}{5}=\\frac{\\sqrt{3}}{3}$" }, { "text": "The hyperbola $C$ and the standard ellipse $C^{\\prime}$ share common foci. The length of the real axis of $C$ is half the length of the major axis of $C^{\\prime}$. The eccentricity of $C^{\\prime}$ is smaller than the eccentricity of $C$ by $\\frac{\\sqrt{11}}{6}$, and the focal distance of $C^{\\prime}$ is $2 \\sqrt{11}$. Then, what is the equation of this hyperbola?", "fact_expressions": "C: Hyperbola;C1: Ellipse;Focus(C) = Focus(C1);Length(RealAxis(C)) = (1/2)*Length(MajorAxis(C1));Eccentricity(C1) = Eccentricity(C) - sqrt(11)/6;FocalLength(C1) = 2*sqrt(11)", "query_expressions": "Expression(C)", "answer_expressions": "{x^2/9-y^2/2=1, y^2/9-x^2/2=1}", "fact_spans": "[[[0, 6], [32, 35], [76, 79], [141, 144]], [[11, 25], [40, 52], [59, 71], [108, 120]], [[0, 31]], [[32, 58]], [[59, 105]], [[108, 137]]]", "query_spans": "[[[141, 149]]]", "process": "" }, { "text": "$AB$ is a chord of $C$: $y^{2}=4x$ passing through the focus, and $|AB|=10$, then the horizontal coordinate of the midpoint of $AB$ is?", "fact_expressions": "A: Point;B: Point;C:Curve;Expression(C)=(y^2=4*x);IsChordOf(LineSegmentOf(A,B),C);PointOnCurve(Focus(C),LineSegmentOf(A,B));Abs(LineSegmentOf(A, B)) = 10", "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "4", "fact_spans": "[[[0, 4]], [[0, 4]], [[6, 21]], [[6, 21]], [[0, 25]], [[0, 25]], [[27, 36]]]", "query_spans": "[[[38, 50]]]", "process": "" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right focus is $F$, the left and right vertices are $A$, $B$ respectively, and point $P$ lies on the hyperbola. If $|O P|=|O F|$, and $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}+\\overrightarrow{O F} \\cdot \\overrightarrow{O A}=0$, 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 > b;b > 0;F: Point;RightFocus(G) = F;A: Point;B: Point;LeftVertex(G) = A;RightVertex(G) = B;P: Point;PointOnCurve(P, G);O: Origin;Abs(LineSegmentOf(O, P)) = Abs(LineSegmentOf(O, F));DotProduct(VectorOf(O, F), VectorOf(O, A)) + DotProduct(VectorOf(P, A), VectorOf(P, B)) = 0", "query_expressions": "Eccentricity(G)", "answer_expressions": "(1+sqrt(5))/2", "fact_spans": "[[[2, 55], [85, 88], [209, 212]], [[2, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[5, 55]], [[60, 63]], [[2, 63]], [[71, 74]], [[75, 78]], [[2, 78]], [[2, 78]], [[79, 83]], [[79, 91]], [[93, 106]], [[93, 106]], [[108, 207]]]", "query_spans": "[[[209, 218]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus is $F$, the left vertex is $A$, and the upper vertex is $B$. If the distance from point $F$ to the line $AB$ is $\\frac{2b}{\\sqrt{17}}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;B: Point;A: Point;F: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F;LeftVertex(G) = A;UpperVertex(G) = B;Distance(F, LineOf(A,B)) = (2*b)/sqrt(17)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/3", "fact_spans": "[[[2, 54], [123, 125]], [[4, 54]], [[4, 54]], [[75, 78]], [[67, 70]], [[59, 62], [81, 85]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 62]], [[2, 70]], [[2, 78]], [[81, 120]]]", "query_spans": "[[[123, 131]]]", "process": "(Analysis) Find the equation of line AB, set up an equation using the distance from the center of the ellipse to the line, obtain a relation among a, b, c, then solve for the eccentricity of the ellipse to get the answer. According to the problem, given the ellipse equation \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), we have F(-c,0), A(-a,0), B(0,b). Then the equation of line AB is bx-ay+ab=0. Since the distance from point F to line AB is \\frac{2b}{\\sqrt{17}}, it follows that \\frac{|-bc+ab|}{\\sqrt{a^{2}+b^{2}}}=\\frac{2b}{\\sqrt{17}}. Simplifying yields 9a^{2}-34ac+2lc^{2}=0, i.e., 2le^{2}-34e+9=0, or (3e-1)(7e-9)=0. Solving gives e=\\frac{1}{2} or e=\\frac{9}{7} (discarded). Hence the answer is e=\\frac{1}{3}." }, { "text": "Given the parabola $y^{2}=8x$ with focus $F$, and a point $P$ on the parabola such that $|PF|=5$, find the area of $\\triangle POF$.", "fact_expressions": "G:Parabola;P: Point;O: Origin;F: Point;Expression(G) = (y^2 = 8*x);Focus(G) = F;PointOnCurve(P,G);Abs(LineSegmentOf(P, F)) = 5", "query_expressions": "Area(TriangleOf(P, O, F))", "answer_expressions": "2*sqrt(6)", "fact_spans": "[[[2, 16], [24, 27]], [[30, 33]], [[46, 63]], [[20, 23]], [[2, 16]], [[2, 23]], [[24, 33]], [[35, 44]]]", "query_spans": "[[[46, 68]]]", "process": "Since |PF| = 5, it follows that x_{P} + 2 = 5, therefore x_{P} = 3, y_{p}^{2} = 24, |y_{p}| = 2\\sqrt{6}. Hence, the area of APOF is \\frac{1}{2}|y_{P}||OF| = \\frac{1}{2} \\times 2\\sqrt{6} \\times 2 = 2\\sqrt{6}. [Note: This question examines the application of the definition of a parabola, and tests basic analytical transformation and problem-solving skills; it is a fundamental problem." }, { "text": "Given that $A$ and $B$ are the left and right vertices of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and point $P$ lies on $E$. In $\\triangle A P B$, $\\tan \\angle P A B=\\frac{1}{2}$, $\\tan \\angle P B A=\\frac{2}{9}$. Find the eccentricity of the ellipse $E$.", "fact_expressions": "E: Ellipse;b: Number;a: Number;A: Point;P: Point;B: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(E) = A;RightVertex(E) = B;PointOnCurve(P, E);Tan(AngleOf(P, A, B)) = 1/2;Tan(AngleOf(P, B, A)) = 2/9", "query_expressions": "Eccentricity(E)", "answer_expressions": "2*sqrt(2)/3", "fact_spans": "[[[10, 67], [79, 82], [171, 176]], [[17, 67]], [[17, 67]], [[2, 5]], [[74, 78]], [[6, 9]], [[17, 67]], [[17, 67]], [[10, 67]], [[2, 73]], [[2, 73]], [[74, 83]], [[104, 135]], [[138, 169]]]", "query_spans": "[[[171, 182]]]", "process": "By the symmetry of the ellipse, assume point P lies above the x-axis, let P(m,n) (n>0). From tan∠PAB = 1/2 ⇒ n/(m+a) = 1/2, tan∠PBA = 2/9 ⇒ n/(a−m) = 2/9. Solving these equations simultaneously gives: \n{ n = (4/13)a, m = −(5/13)a }, \nsubstituting into the ellipse equation yields: \n(−(5/13)a)²/a² + ((4/13)a)²/b² = 1 ⇒ a² = 9b², so a² = 9(a²−c²) ⇒ e = c/a = (2√2)/3" }, { "text": "Let $F_{1}$ and $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)$, respectively. If there exists a point $A$ on the ellipse such that $\\angle F_{1} A F_{2}=90^{\\circ}$ and $|A F_{1}|=3|A F_{2}|$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;A: 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(A, G);AngleOf(F1, A, F2) = ApplyUnit(90, degree);Abs(LineSegmentOf(A, F1)) = 3*Abs(LineSegmentOf(A, F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(10)/4", "fact_spans": "[[[19, 71], [79, 81], [148, 150]], [[21, 71]], [[21, 71]], [[1, 8]], [[84, 88]], [[9, 16]], [[21, 71]], [[21, 71]], [[19, 71]], [[1, 77]], [[1, 77]], [[79, 88]], [[90, 123]], [[124, 146]]]", "query_spans": "[[[148, 156]]]", "process": "Problem Analysis: According to the definition of an ellipse, |AF₁| + |AF₂| = 2a. Since AF₁ = 3|AF₂|, it follows that |AF₂| = a/2, |AF₁| = 3a/2. Because ∠F₁AF₂ = 90°, by the Pythagorean theorem we have (3a/2)² + (a/2)² = (2c)². Simplifying gives 5a² = 8c², i.e., c²/a² = 5/8. Therefore, the eccentricity e = c/a = √(c²/a²) = √10/4." }, { "text": "Through the point $P(1,3)$, draw two perpendicular lines $l_{1}$ and $l_{2}$, which intersect the $x$-axis and $y$-axis at points $A$ and $B$ respectively. Then the equation of the locus of the midpoint of segment $AB$ is?", "fact_expressions": "l1:Line;l2:Line;A: Point;B: Point;P: Point;Coordinate(P) = (1, 3);PointOnCurve(P,l1);PointOnCurve(P,l2);Intersection(l1,xAxis)=A;Intersection(l2,yAxis)=B;IsPerpendicular(l1,l2)", "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(A,B)))", "answer_expressions": "x+3*y-5=0", "fact_spans": "[[[18, 27]], [[28, 35]], [[52, 55]], [[56, 59]], [[1, 10]], [[1, 10]], [[0, 35]], [[0, 35]], [[36, 61]], [[36, 61]], [[13, 35]]]", "query_spans": "[[[63, 79]]]", "process": "Let the coordinates of the midpoint of segment AB be (x, y), then A(2x, 0), B(0, 2y), where PA ⊥ PB, so $\\overrightarrow{PA} \\cdot \\overrightarrow{PB} = (2x - 1, -3) \\cdot (-1, 2y - 3) = 1 - 2x - 6y + 9 = 0$. Hence, the trajectory equation of the midpoint of segment AB is $x + 3y - 5 = 0$. Upon verification, all points on this line satisfy the given condition." }, { "text": "The line $ l $ passing through the point $ M(4, 0) $ intersects the two asymptotes of the hyperbola $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ $ (a > b > 0) $ at two distinct points $ A $ and $ B $, and $ O $ is the origin. If the eccentricity of the hyperbola is $ 2 $, then the range of values of $ \\overrightarrow{O A} \\cdot \\overrightarrow{O B} $ is?", "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;M: Point;O: Origin;A: Point;B: Point;a > b;b > 0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (4, 0);PointOnCurve(M,l);L1:Line;L2:Line;Asymptote(G)={L1,L2};Intersection(l,L1)=A;Intersection(l,L2)=B;Negation(A=B);Eccentricity(G) = 2", "query_expressions": "Range(DotProduct(VectorOf(O, A), VectorOf(O, B)))", "answer_expressions": "(-oo,-32]+(0,oo)", "fact_spans": "[[[13, 18]], [[19, 72], [101, 104]], [[22, 72]], [[22, 72]], [[1, 12]], [[92, 95]], [[84, 87]], [[88, 91]], [[22, 72]], [[22, 72]], [[19, 72]], [[1, 12]], [[0, 18]], [], [], [[19, 76]], [[13, 91]], [[13, 91]], [[80, 91]], [[101, 112]]]", "query_spans": "[[[114, 170]]]", "process": "Since the eccentricity of the hyperbola is 2, we find $\\frac{b^{2}}{a^{2}}=3$. Let the equation of line $l$ be $x=ty+4$, solve simultaneously with the asymptotes' equations to obtain the coordinates of $A$ and $B$, express $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}$, and use the range of the function to determine the range of $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}$. Solution: Since the eccentricity of the hyperbola is 2, we have $e=\\frac{c}{a}=2$, so $\\frac{c^{2}}{a^{2}}=4$, that is, $\\frac{a^{2}+b^{2}}{a^{2}}=4$, hence $\\frac{b^{2}}{a^{2}}=3$. Let the equation of line $l$ be $x=ty+4$, where $t\\neq\\pm\\frac{a}{b}$, i.e., $t^{2}\\neq\\frac{1}{3}$. Substituting $x=ty+4$ into $y=\\frac{b}{a}x$ and $y=-\\frac{b}{a}x$ respectively, the coordinates of $A$ and $B$ are $(\\frac{4a}{a-bt},\\frac{4b}{a-bt})$ and $(\\frac{4a}{a+bt},-\\frac{4b}{a+bt})$ respectively. Therefore, $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=\\frac{16(a^{2}-b^{2})}{a^{2}-b^{2}t^{2}}$. Since $t^{2}\\neq\\frac{1}{3}$, the range of $\\overrightarrow{OA}\\cdot\\overrightarrow{OB}$ is $(-\\infty,-32]\\cup(0,+\\infty)$." }, { "text": "Given that the hyperbola passes through the point $(4, \\sqrt{3})$ and has asymptotes with equations $y = \\pm \\frac{1}{2} x$, then the standard equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (4, sqrt(3));Expression(Asymptote(G)) = (y = (pm*1/2)*x);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4 - y^2 = 1", "fact_spans": "[[[2, 5], [54, 57]], [[6, 22]], [[6, 22]], [[2, 51]], [[2, 22]]]", "query_spans": "[[[54, 64]]]", "process": "According to the problem, let the equation of the desired hyperbola be $ x^{2} - 4y^{2} = \\lambda $. Since the point $ M(4, \\sqrt{3}) $ lies on this hyperbola, $ \\lambda = 16 - 12 = 4 $. Therefore, the equation of the hyperbola is $ x^{2} - 4y^{2} = 4 $, that is, $ \\frac{x^{2}}{4} - y^{2} = 1 $. Hence, the correct answer is $ \\frac{x^{2}}{4} - y^{2} = 1 $." }, { "text": "If the equation $\\frac{x^{2}}{5-k}+\\frac{y^{2}}{k-3}=1$ represents a hyperbola, then what is the range of values for $k$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(5 - k) + y^2/(k - 3) = 1);k: Number", "query_expressions": "Range(k)", "answer_expressions": "(-oo, 3)+(5, +oo)", "fact_spans": "[[[44, 47]], [[1, 47]], [[49, 52]]]", "query_spans": "[[[49, 59]]]", "process": "Test analysis: From the given condition, (5-k)(k-3)<0, ∴ k<3 or k>5, so the range of k is (-∞,3)∪(5,+∞)." }, { "text": "The eccentricity of $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ is? The equation of the directrix is?", "fact_expressions": "G:Curve;Expression(G)=(x^2/25+y^2/9=1)", "query_expressions": "Eccentricity(G);Expression(Directrix(G))", "answer_expressions": "4/5\nx=pm*(25/4)", "fact_spans": "[[[0, 36]], [[0, 36]]]", "query_spans": "[[[0, 42]], [[0, 48]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $E$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$, the distance from a point $P$ on $E$ to the origin is $b$, and $\\sin \\angle P F_{2} F_{1}=3 \\sin \\angle P F_{1} F_{2}$, then the equations of the asymptotes of the hyperbola $E$ are?", "fact_expressions": "F1: Point;F2: Point;Focus(E) = {F1, F2};E: Hyperbola;Expression(E) = (-x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a>0;b>0;P: Point;PointOnCurve(P, E);O: Origin;Distance(P, O) = b;Sin(AngleOf(P, F2, F1)) = 3*Sin(AngleOf(P, F1, F2))", "query_expressions": "Expression(Asymptote(E))", "answer_expressions": "y=pm*(sqrt(2)/2)*x", "fact_spans": "[[[2, 9]], [[10, 17]], [[2, 86]], [[20, 81], [87, 90], [165, 171]], [[20, 81]], [[103, 106]], [[28, 81]], [[28, 81]], [[28, 81]], [[92, 96]], [[87, 96]], [[97, 99]], [[92, 106]], [[108, 163]]]", "query_spans": "[[[165, 179]]]", "process": "From $\\sin\\angle PF_{2}F_{1}=3\\sin\\angle PF_{1}F_{2}$ and the law of sines, we obtain $|PF_{1}|=3|PF_{2}|$. According to the definition of a hyperbola, $|PF_{1}|-|PF_{2}|=2a$, so $|PF_{2}|=a$, $|PF_{1}|=3a$. Also, $|OF_{2}|=c$, $|OP|=b$, so $\\angle OPF_{2}=\\frac{\\pi}{2}$, hence $\\cos\\angle OF_{2}P=\\frac{a}{c}$. In $\\triangle F_{1}F_{2}P$, by the law of cosines, $\\cos\\angle F_{1}F_{2}P=\\frac{a^{2}+(2c)^{2}-(3a)^{2}}{2a\\cdot2c}=\\cos\\angle OF_{2}P=\\frac{a}{c}$. Simplifying yields $c=\\sqrt{3}a$, so $b=\\sqrt{2}a$. Therefore, the asymptotes of hyperbola $E$ are $y=\\pm\\frac{a}{b}=\\pm\\frac{\\sqrt{2}}{2}x$." }, { "text": "Given that $P$ is a point on the ellipse, $F_{1}$ and $F_{2}$ are the two foci of the ellipse, $\\angle P F_{1} F_{2}=90^{\\circ}$, $\\angle P F_{2} F_{1}=30^{\\circ}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;PointOnCurve(P, G);Focus(G) = {F1, F2};AngleOf(P, F1, F2) = ApplyUnit(90, degree);AngleOf(P, F2, F1) = ApplyUnit(30, degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[6, 8], [6, 8], [6, 8]], [[2, 5]], [[13, 20]], [[21, 28]], [[2, 12]], [[13, 36]], [[37, 70]], [[72, 105]]]", "query_spans": "[[[108, 116]]]", "process": "" }, { "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 perpendicular to the $x$-axis is drawn through $F_{2}$, intersecting the ellipse at a point $P$. If $\\angle F_{1}PF_{2}=45^{\\circ}$, then the eccentricity $e$ of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;F1: Point;P: Point;F2: Point;e: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Z: Line;PointOnCurve(F2, Z);IsPerpendicular(Z, xAxis);OneOf(Intersection(Z, G)) = P;AngleOf(F1, P, F2) = ApplyUnit(45, degree);Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "sqrt(2) - 1", "fact_spans": "[[[0, 52], [97, 99], [144, 146]], [[2, 52]], [[2, 52]], [[61, 68]], [[105, 108]], [[71, 78], [81, 88]], [[150, 153]], [[2, 52]], [[2, 52]], [[0, 52]], [[0, 78]], [[0, 78]], [], [[80, 96]], [[80, 96]], [[80, 108]], [[110, 142]], [[144, 153]]]", "query_spans": "[[[150, 155]]]", "process": "" }, { "text": "Given that the equation $\\frac{x^{2}}{2-k}+\\frac{y^{2}}{2k-1}=1$ represents an ellipse with foci on the $y$-axis, what is the range of real values for $k$?", "fact_expressions": "G: Ellipse;k: Real;Expression(G) = (x^2/(2 - k) + y^2/(2*k - 1) = 1);PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(k)", "answer_expressions": "(1,2)", "fact_spans": "[[[56, 58]], [[60, 65]], [[2, 58]], [[47, 58]]]", "query_spans": "[[[60, 72]]]", "process": "Test analysis: From the given conditions, we have \n\\begin{cases}2-k>0\\\\2k-1>0\\\\2-k<2\\end{cases}11)$ is $2$ times the length of the minor axis. Then the real number $m$=?", "fact_expressions": "G: Ellipse;m: Real;m>1;Expression(G) = (y^2 + x^2/m = 1);Length(MajorAxis(G)) = 2*Length(MinorAxis(G))", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[0, 34]], [[49, 54]], [[2, 34]], [[0, 34]], [[0, 47]]]", "query_spans": "[[[49, 56]]]", "process": "From the equation of the ellipse, we know that $ a^{2} = m $, $ b^{2} = 1 $, and since the length of the major axis is twice the length of the minor axis, $ 2\\sqrt{m} = 2 \\times 2b = 4 $. Solving gives $ m = 4 $. This problem mainly examines the equation of an ellipse and its simple geometric properties, and belongs to the medium difficulty level." }, { "text": "The right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ is $F$. Then, what is the standard equation of the parabola with focus $F$?", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/5 + y^2/4 = 1);F: Point;RightFocus(H) = F;G: Parabola;Focus(G) = F", "query_expressions": "Expression(G)", "answer_expressions": "y^2=4*x", "fact_spans": "[[[0, 37]], [[0, 37]], [[42, 45], [48, 51]], [[0, 45]], [[55, 58]], [[47, 58]]]", "query_spans": "[[[55, 65]]]", "process": "The right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ is $F(1,0)$. The standard equation of the parabola with focus $F$ is $y^{2}=4x$." }, { "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 what is the value of the real number $p$?", "fact_expressions": "G: Parabola;p: Real;H: Ellipse;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": "[[[1, 17]], [[66, 71]], [[21, 58]], [[1, 17]], [[21, 58]], [[1, 64]]]", "query_spans": "[[[66, 75]]]", "process": "" }, { "text": "Given that the length of segment $AB$ is $10$, with endpoints $A$ and $B$ moving on the $x$-axis and $y$-axis respectively, if point $M$ lies on segment $AB$ and $\\overrightarrow{A M}+4 \\overrightarrow{B M}=0$, then the equation of the trajectory of point $M$ is?", "fact_expressions": "A: Point;B: Point;M: Point;Length(LineSegmentOf(A,B))=10;Endpoint(LineSegmentOf(A,B))={A,B};PointOnCurve(A,xAxis);PointOnCurve(B,yAxis);PointOnCurve(M,LineSegmentOf(A,B));VectorOf(A, M) + 4*VectorOf(B, M) = 0", "query_expressions": "LocusEquation(M)", "answer_expressions": "16*x^2+y^2=64", "fact_spans": "[[[20, 23]], [[24, 27]], [[44, 48], [108, 112]], [[2, 16]], [[2, 27]], [[20, 42]], [[20, 42]], [[44, 57]], [[59, 106]]]", "query_spans": "[[[108, 119]]]", "process": "Let M(x,y), A(a,0), B(0,b), so \\overrightarrow{AM}=(x-a,y), \\overrightarrow{MB}=(-x,b-y). Since |AB|=10, we have \\sqrt{a^{2}+b^{2}}=10, that is, a^{2}+b^{2}=100. Because \\overrightarrow{AM}+4\\overrightarrow{BM}=0, it follows that \\overrightarrow{AM}=4\\overrightarrow{MB}, so \\begin{cases}x-a=-4x\\\\y=4(b-y)\\end{cases}. Solving gives \\begin{cases}a=5x\\\\b=\\frac{5}{4}y\\end{cases}. Substituting into a^{2}+b^{2}=100 yields 25x^{2}+\\frac{25y^{2}}{16}=100, that is, 16x^{2}+y^{2}=64." }, { "text": "The equation of a circle centered at the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ and intercepted by its asymptotes to form a chord of length $6$ is?", "fact_expressions": "G: Hyperbola;H: Circle;Expression(G) = (x^2/4 - y^2/16 = 1);RightFocus(G) = Center(H);Length(InterceptChord(Asymptote(G),H))=6", "query_expressions": "Expression(H)", "answer_expressions": "(x - 2*sqrt(5)) + y^2 = 25", "fact_spans": "[[[1, 40], [50, 51]], [[64, 65]], [[1, 40]], [[0, 65]], [[49, 65]]]", "query_spans": "[[[64, 70]]]", "process": "According to the problem, the coordinates of the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{16}=1$ are $(2\\sqrt{5},0)$, and the equations of the asymptotes are $2x\\pm y=0$. The distance from the right focus of the hyperbola to the asymptote is $d=\\frac{2\\times2\\sqrt{5}}{\\sqrt{2^{2}+(\\pm1)^{2}}}=4$. Therefore, the radius of the circle is $r=\\sqrt{d^{2}+3^{2}}=\\sqrt{4^{2}+3^{2}}=5$. Thus, the equation of the required circle is $(x-2\\sqrt{5})^{2}+y^{2}=25$." }, { "text": "Given a point $M$ on the parabola $y^{2}=4x$ such that the distance from $M$ to its focus is $5$, find the distance from point $M$ to the $y$-axis.", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);M: Point;PointOnCurve(M, G) = True;Distance(M, Focus(G)) = 5", "query_expressions": "Distance(M, yAxis)", "answer_expressions": "4", "fact_spans": "[[[2, 16], [23, 24]], [[2, 16]], [[19, 22], [35, 39]], [[2, 22]], [[19, 33]]]", "query_spans": "[[[35, 49]]]", "process": "The focus of the parabola y^{2}=4x has coordinates (1,0). A point M on the parabola y^{2}=4x is at a distance |MF|=5 from the focus F of the parabola. The distance from M to the directrix is 5, and the distance from point M to the y-axis is 4. The answer is 4. This question examines the simple application of the definition of a parabola and is a basic problem." }, { "text": "An ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{m}=1$ has a focus at $(0,2)$. Then $m$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/6 + y^2/m = 1);m: Number;Coordinate(OneOf(Focus(G))) = (0,2)", "query_expressions": "m", "answer_expressions": "10", "fact_spans": "[[[0, 37]], [[0, 37]], [[52, 55]], [[0, 50]]]", "query_spans": "[[[52, 57]]]", "process": "From the problem, we have m - 6 = 4, solve for m. From the problem, we know the foci of the ellipse lie on the y-axis, ∴ m - 6 = 4, solving gives m = 10." }, { "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 of the midpoint $M$ of segment $AB$ is $2$. What is the length of segment $AB$?", "fact_expressions": "G: Parabola;H: Line;B: Point;A: Point;M:Point;Expression(G) = (x^2 = y/8);PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(A,B))=M;YCoordinate(M)=2", "query_expressions": "Length(LineSegmentOf(A,B))", "answer_expressions": "65/16", "fact_spans": "[[[1, 25], [32, 35]], [[29, 31]], [[40, 43]], [[36, 39]], [[56, 59]], [[1, 25]], [[0, 31]], [[29, 45]], [[46, 59]], [[56, 67]]]", "query_spans": "[[[69, 79]]]", "process": "" }, { "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ has two foci $F_{1}$, $F_{2}$. A line passing through $F_{1}$ and perpendicular to the $x$-axis intersects the ellipse, with one intersection point being $P$. Then $|P F_{2}|$=?", "fact_expressions": "G: Ellipse;H: Line;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Focus(G) = {F1,F2};PointOnCurve(F1,H);IsPerpendicular(H,xAxis);OneOf(Intersection(H,G)) = P;IsIntersect(G, H)", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "5/2", "fact_spans": "[[[0, 37], [79, 81]], [[76, 78]], [[89, 92]], [[51, 58]], [[43, 50], [60, 67]], [[0, 37]], [[0, 58]], [[59, 78]], [[68, 78]], [[76, 92]], [[76, 83]]]", "query_spans": "[[[94, 107]]]", "process": "Given $ a^{2}=4 $, $ b^{2}=3 $, $ \\therefore c^{2}=1 $, let $ x=1 $, we get $ \\frac{1}{4}+\\frac{y^{2}}{3}=1 $, solving gives $ y=\\pm\\frac{3}{2} $, that is $ |PF_{1}|=\\frac{3}{2} $. Also $ |PF_{1}|+|PF_{2}|=2a=4 $, $ \\therefore |PF_{2}|=4-\\frac{3}{2}=\\frac{5}{2} $" }, { "text": "The line $ l $ intersects the ellipse $ \\frac{x^{2}}{4} + y^{2} = 1 $ at points $ A $ and $ B $, with the midpoint of segment $ AB $ being $ M(1, t) $ ($ t \\in \\mathbb{R} $). The line $ l $ is the perpendicular bisector of segment $ AB $. If $ OD \\perp l $ and $ D $ is the foot of the perpendicular, then the equation of the locus of point $ D $ is?", "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;M: Point;O: Origin;D: Point;Expression(G) = (x^2/4 + y^2 = 1);t: Real;Coordinate(M) = (1, t);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M;PerpendicularBisector(LineSegmentOf(A, B)) = l;IsPerpendicular(LineSegmentOf(O, D), l);FootPoint(LineSegmentOf(O, D), l) = D", "query_expressions": "LocusEquation(D)", "answer_expressions": "(x-3/8)^2 + y^2 = 9/64", "fact_spans": "[[[0, 5], [74, 79]], [[6, 33]], [[34, 37]], [[38, 41]], [[55, 73]], [[95, 108]], [[109, 112], [117, 121]], [[6, 33]], [[55, 73]], [[55, 73]], [[0, 43]], [[44, 73]], [[74, 93]], [[95, 108]], [[95, 115]]]", "query_spans": "[[[117, 128]]]", "process": "Let the equation of the line be y = k(x - 1) + t. Substituting into the ellipse equation and simplifying yields (1 + 4k^{2})x^{2} + (8kt - 8k^{2})x + 4k^{2} - 8kt + 4t^{2} - 4 = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then \\frac{8kt - 8k^{2}}{1 + 4k^{2}} = 2, solving gives k = -\\frac{1}{4t}. Line l is the perpendicular bisector of segment AB, hence the line t: y - t = 4t(x - 1), that is: y = t(4x - 3). Let 4x - 3 = 0, then x = \\frac{3}{4}, y = 0, thus line l passes through the fixed point E(\\frac{3}{4}, 0). When the slope of line l does not exist, t = 0, line t also passes through the fixed point (\\frac{3}{4}, 0). Point D lies on the circle with OE as diameter, so the center is (\\frac{3}{8}, 0), radius r = \\frac{1}{2}|OE| = \\frac{3}{8}, therefore the trajectory equation of point D is: (x - \\frac{3}{8})^{2} + y^{2} = \\frac{9}{64}." }, { "text": "The parabola $C$: $y^{2}=4 x$ has focus $F$. A point $N$ lies on the directrix. A perpendicular to $N F$ is drawn through $N$, intersecting the $y$-axis at point $M$. If there exists a point $E$ on the parabola $C$ such that $2 \\overrightarrow{N E}=\\overrightarrow{N M}+\\overrightarrow{N F}$, then the area of $\\Delta M N F$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;PointOnCurve(N, Directrix(C));N: Point;L: Line;PointOnCurve(N, L) ;IsPerpendicular(LineSegmentOf(N, F), L);Intersection(L, yAxis) = M;M: Point;PointOnCurve(E, C);E: Point;2*VectorOf(N, E) = VectorOf(N, F) + VectorOf(N, M)", "query_expressions": "Area(TriangleOf(M, N, F))", "answer_expressions": "3*sqrt(2)/2", "fact_spans": "[[[0, 19], [57, 63]], [[0, 19]], [[23, 26]], [[0, 26]], [[0, 36]], [[33, 36]], [], [[27, 45]], [[27, 45]], [[27, 55]], [[51, 55]], [[57, 70]], [[66, 70]], [[73, 139]]]", "query_spans": "[[[141, 160]]]", "process": "From $ 2\\overrightarrow{NE} = \\overrightarrow{NM} + \\overrightarrow{NF} $ it follows that $ E $ is the midpoint of $ MF $, the directrix equation is $ x = -1 $, the focus is $ F(1,0) $. Without loss of generality, assume point $ N $ is in the third quadrant. Since $ \\angle MNF $ is a right angle, $ |NE| = \\frac{1}{2}|MF| = |EF| $. By the definition of a parabola, $ NE \\parallel x $-axis. Then we can find $ E\\left(\\frac{1}{2}, -\\sqrt{2}\\right) $, $ M(0, -2\\sqrt{2}) $, $ N(-1, -\\sqrt{2}) $, so $ |NF| = \\sqrt{6} $, $ |MN| = \\sqrt{3} $, thus $ s_{AMNF} = \\frac{3\\sqrt{2}}{2} $." }, { "text": "Points $A$ and $B$ are the endpoints of the major axis of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and points $C$ and $D$ are the endpoints of the minor axis of ellipse $E$. A moving point $M$ satisfies $\\frac{|M A|}{|M B|}=2$. If the maximum area of $\\triangle M A B$ is $8$ and the minimum area of $\\Delta M C D$ is $1$, then the eccentricity of the ellipse is?", "fact_expressions": "E: Ellipse;b: Number;a: Number;M: Point;A: Point;B: Point;C: Point;D: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);Endpoint(MajorAxis(E)) = {A, B};Endpoint(MinorAxis(E)) = {C, D};Abs(LineSegmentOf(M, A))/Abs(LineSegmentOf(M, B)) = 2;Max(TriangleOf(M, A, B)) = 8;Min(TriangleOf(M, C, D)) = 1", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[9, 66], [80, 85], [177, 179]], [[16, 66]], [[16, 66]], [[93, 96]], [[0, 4]], [[5, 8]], [[72, 75]], [[76, 79]], [[16, 66]], [[16, 66]], [[9, 66]], [[0, 71]], [[72, 90]], [[98, 121]], [[123, 150]], [[151, 175]]]", "query_spans": "[[[177, 185]]]", "process": "" }, { "text": "The shortest distance from a point on the parabola $y = x^{2}$ to the line $x - y - 2 = 0$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (x - y - 2 = 0)", "query_expressions": "Min(Distance(P, H))", "answer_expressions": "7*sqrt(2)/8", "fact_spans": "[[[0, 12]], [[0, 12]], [[14, 15]], [[0, 15]], [[16, 27]], [[16, 27]]]", "query_spans": "[[[14, 34]]]", "process": "Problem Analysis: To find the shortest distance from a parabola to a line, there are two possibilities: 1. If the parabola and the line intersect, the shortest distance is definitely 0; 2. If they do not intersect, another method must be used to find the shortest distance. We solve the parabola and the line simultaneously to obtain the equation x^{2}-x+2=0. Since the discriminant is less than 0, they do not intersect. Now assume a point on the parabola is P(a,a^{2}). The problem then becomes finding the shortest distance from the moving point P to the line. Substituting into the point-to-line distance formula gives d=\\frac{|a^{2}-a+2|}{\\sqrt{2}}, which means we need to find the minimum value of |a^{2}-a+2|. Completing the square yields \\left|(a-\\frac{1}{2})^{2}+\\frac{7}{4}\\right|, so the shortest distance is \\frac{7\\sqrt{2}}{8}." }, { "text": "If a point $M$ on the parabola $y^{2}=4x$ is at a distance of $3$ from the focus, then what is the distance from point $M$ to the $y$-axis?", "fact_expressions": "G: Parabola;M: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(M, G);Distance(M, Focus(G)) = 3", "query_expressions": "Distance(M, yAxis)", "answer_expressions": "2", "fact_spans": "[[[1, 15]], [[18, 21], [33, 37]], [[1, 15]], [[1, 21]], [[1, 31]]]", "query_spans": "[[[33, 47]]]", "process": "On the parabola y^{2}=4x, the distance from a point M to the focus is 3. Then the distance from a point M on the parabola y^{2}=4x to the directrix x=-1 is 3, so the distance from point M to the y-axis is 3-1=2." }, { "text": "Let the focus of the parabola $y^{2}=2 p x(p>0)$ be $F$, and the directrix be $l$, and let point $A(0,2)$. If the midpoint $B$ of segment $F A$ lies on the parabola, then what is the distance from $F$ to $l$? What is $|F B|$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;l: Line;Directrix(G) = l;A: Point;Coordinate(A) = (0, 2);B: Point;MidPoint(LineSegmentOf(F, A)) = B;PointOnCurve(B, G)", "query_expressions": "Distance(F, l);Abs(LineSegmentOf(F, B))", "answer_expressions": "sqrt(2)\n(3/4)*sqrt(2)", "fact_spans": "[[[1, 22], [63, 66]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 29], [69, 72]], [[1, 29]], [[33, 36], [73, 76]], [[1, 36]], [[37, 46]], [[37, 46]], [[59, 62]], [[49, 62]], [[59, 67]]]", "query_spans": "[[[69, 81]], [[81, 90]]]", "process": "Test analysis: According to the problem, the midpoint of segment FA is B(\\frac{p}{4},1), so we have 1=2p\\cdot\\frac{p}{4}. Solving gives p=\\sqrt{2}. Therefore, the distance from F to l is p=\\sqrt{2}, and |FB|=\\frac{p}{4}+\\frac{p}{2}=\\frac{3}{4}\\sqrt{2}." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus of $C$ is $F$. A line passing through the origin intersects $C$ at points $A$ and $B$. If $\\angle A F B \\geq 150^{\\circ}$, then the range of the eccentricity of $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;A: Point;F: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O,G);Intersection(G, C) = {A, B};AngleOf(A, F, B) >= ApplyUnit(150, degree);O:Origin", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(0,(sqrt(6)-sqrt(2))/4)", "fact_spans": "[[[2, 59], [76, 79], [125, 128]], [[9, 59]], [[9, 59]], [[73, 75]], [[81, 84]], [[64, 67]], [[85, 88]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 67]], [[68, 75]], [[73, 90]], [[92, 123]], [[70, 72]]]", "query_spans": "[[[125, 139]]]", "process": "As shown in the figure, let the right focus of the ellipse be F, connect AF, BF. Since AB and FF' bisect each other, quadrilateral AFBF is a parallelogram. Therefore, ∠AFB + ∠FBF' = 180°. Since ∠AFB ≥ 150°, it follows that ∠FBF' ≤ 30°. From the given condition, when B is at the endpoint B₂ of the minor axis, ∠FBF' is maximized. At this time, in the right triangle △B₂OF', ∠OB₂F' = 15°, so sin15° = (√6 - √2)/4 ≤ (√6 - √2)/4, i.e., e ∈ (0, (√6 - √2)/4]." }, { "text": "Given that point $P$ lies on the right branch of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, $A(0,2)$, and moving point $B$ satisfies $|A B|=2$, $F$ is the right focus of the hyperbola, then the maximum value of $|P F|-|P B|$ is?", "fact_expressions": "G: Hyperbola;A: Point;B: Point;P: Point;F: Point;Expression(G) = (x^2/4 - y^2/5 = 1);Coordinate(A) = (0, 2);PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(A, B)) = 2;RightFocus(G) = F", "query_expressions": "Max(-Abs(LineSegmentOf(P, B)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "sqrt(13) - 2", "fact_spans": "[[[7, 45], [82, 85]], [[50, 58]], [[61, 64]], [[2, 6]], [[78, 81]], [[7, 45]], [[50, 58]], [[2, 49]], [[66, 75]], [[77, 89]]]", "query_spans": "[[[91, 110]]]", "process": "Moving point B satisfies |AB| = 2, then the locus of point B is a circle with center A and radius 2. Let the left focus of the hyperbola be F_{1}. By the given condition, |PF_{1}| - |PF| = 4, |PF| = |PF_{1}| - 4, then |PF| - |PA| = |PF_{1}| - |PA| - 4 \\leqslant |AF_{1}| - 4 = \\sqrt{13} - 4, therefore the maximum value of |PF| - |PB| is \\sqrt{13} - 2." }, { "text": "A 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 $M$ and $N$, and $F_{2}$ is its right focus. Then the value of $|MF_{2}|+|NF_{2}|-|MN|$ is?", "fact_expressions": "G: Hyperbola;H: Line;M: Point;F2: Point;N: Point;Expression(G) = (x^2/4 - y^2/3 = 1);RightFocus(G) = F2;F1: Point;LeftFocus(G) = F1;PointOnCurve(F1, H);Intersection(H, LeftPart(G)) = {M, N}", "query_expressions": "Abs(LineSegmentOf(M, F2)) + Abs(LineSegmentOf(N, F2)) - Abs(LineSegmentOf(M, N))", "answer_expressions": "8", "fact_spans": "[[[1, 39], [53, 56], [78, 79]], [[50, 52]], [[60, 63]], [[70, 77]], [[64, 67]], [[1, 39]], [[70, 82]], [[42, 49]], [[1, 49]], [[0, 52]], [[50, 69]]]", "query_spans": "[[[84, 113]]]", "process": "" }, { "text": "Through the left focus $F(-c, 0)$ $(c>0)$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw a tangent line to the circle $x^{2}+y^{2}=\\frac{a^{2}}{4}$, with the point of tangency $E$. Extend $FE$ to intersect the right branch of the hyperbola at point $P$. If $\\overrightarrow{O E}=\\frac{1}{2}(\\overrightarrow{O F}+\\overrightarrow{O P})$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;F: Point;E: Point;O: Origin;P: Point;l1: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2/4);Coordinate(F) = (-c, 0);LeftFocus(G) = F;TangentOfPoint(F, H) = l1;TangentPoint(l1,H)=E;Intersection(OverlappingLine(LineSegmentOf(F,E)),RightPart(G))=P;VectorOf(O,E)=(VectorOf(O,F)+VectorOf(O,P))/2;c:Number;c>0", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[1, 58], [127, 130], [218, 221]], [[4, 58]], [[4, 58]], [[78, 108]], [[62, 77]], [[115, 118]], [[139, 216]], [[133, 137]], [], [[4, 58]], [[4, 58]], [[1, 58]], [[78, 108]], [[62, 77]], [[1, 77]], [[0, 111]], [[0, 118]], [[119, 137]], [[139, 216]], [[62, 77]], [[62, 77]]]", "query_spans": "[[[218, 227]]]", "process": "As shown in the figure, since $\\overrightarrow{OE}=\\frac{1}{2}(\\overrightarrow{OF}+\\overrightarrow{OP})$, $E$ is the midpoint of $FP$, and since $O$ is the midpoint of $FF'$, $OE$ is the midline of triangle $PFF'$, therefore $OE \\parallel PF$, $OE = \\frac{1}{2}PF'$. Since $OE = \\frac{1}{2}a$, $PF' = a$. $OE \\perp PF$, so $PF \\perp PF'$, $FF' = 2c$, $PF - PF' = 2a$, so $PF = 2a$. Therefore, by the Pythagorean theorem, $a^{2} + 9a^{2} = 4c^{2}$, so $10a^{2} = 4c^{2}$, thus $e = \\frac{c}{a} = \\frac{\\sqrt{10}}{2}$." }, { "text": "A line with slope $2$ passes through the right focus of the ellipse $\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$ and intersects the ellipse at points $A$ and $B$. Let $O$ be the origin. Then the area of $\\triangle O A B$ is?", "fact_expressions": "G: Ellipse;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (x^2/5 + y^2/4 = 1);PointOnCurve(RightFocus(G), H);Slope(H) = 2;Intersection(H, G) = {A, B}", "query_expressions": "Area(TriangleOf(O, A, B))", "answer_expressions": "5/3", "fact_spans": "[[[1, 38], [55, 57]], [[52, 54]], [[69, 72]], [[59, 62]], [[63, 66]], [[1, 38]], [[0, 54]], [[45, 54]], [[52, 68]]]", "query_spans": "[[[80, 102]]]", "process": "The right focus of the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{4} = 1$ is $(1,0)$, the equation of line $AB$ is $y - 0 = 2(x - 1)$, that is, $y = 2x - 2$. Substituting into the ellipse $\\frac{x^{2}}{5} + \\frac{y^{2}}{4} = 1$, simplifying yields $3x^{2} - 5x = 0$, hence $x_{1} + x_{2} = \\frac{5}{3}$, $x_{1} \\cdot x_{2} = 0$, $\\therefore AL\\frac{y^{2}}{4} = 1$ simplifies to $3x^{2} - 5x = 0$, the distance from $O$ to line $AB$ is $d = \\frac{|0 - 0 - 2|}{\\sqrt{4 + 1}} = \\frac{2}{\\sqrt{5}}$, thus the area of $\\triangle OAB$ is $\\frac{1}{2} \\cdot AB \\cdot d = \\frac{1}{2} \\cdot \\frac{5\\sqrt{5}}{3} \\cdot \\frac{2}{\\sqrt{5}} = \\frac{5}{3}$." }, { "text": "Given the parabola $C$: $x^{2}=8 y$, draw lines $M A$ and $M B$ from point $M(x_{0} , y_{0})$ tangent to the parabola $C$ at points $A$ and $B$ respectively, such that the circle with diameter $A B$ passes through point $M$. Then the value of $y_{0}$ is?", "fact_expressions": "C: Parabola;G: Circle;M: Point;A: Point;B: Point;x0:Number;y0:Number;Expression(C) = (x^2 = 8*y);Coordinate(M) = (x0, y0);PointOnCurve(M,LineOf(M,A));PointOnCurve(M,LineOf(M,B));TangentPoint(LineOf(M,A),C)=A;TangentPoint(LineOf(M,B),C)=B;IsDiameter(LineSegmentOf(A,B),G);PointOnCurve(M,G)", "query_expressions": "y0", "answer_expressions": "-2", "fact_spans": "[[[2, 21], [59, 65]], [[89, 90]], [[23, 42], [91, 95]], [[69, 73]], [[74, 77]], [[24, 42]], [[97, 104]], [[2, 21]], [[23, 42]], [[22, 51]], [[22, 58]], [[43, 77]], [[43, 77]], [[79, 90]], [[89, 95]]]", "query_spans": "[[[97, 108]]]", "process": "Let the tangent points be A(x_{1},y_{1}), B(x_{2},y_{2}). Parabola: y = \\frac{x^{2}}{8} \\therefore y' = \\frac{x}{4}. Thus: k_{MA} = \\frac{x_{1}}{4}, line MA: y = \\frac{x_{1}}{4}(x - x_{1}) + \\frac{x_{1}^{2}}{8}; k_{MB} = \\frac{x_{2}}{4}, line MB: y = \\frac{x_{2}}{4}(x - x_{2}) + \\frac{x_{2}^{2}}{8}. Hence: x_{0} = \\frac{x_{1} + x_{2}}{2}, y_{0} = \\frac{x_{1}x_{2}}{8}. Also, since the circle with AB as diameter passes through point M, MA \\bot MB, i.e., k_{MA} \\cdot k_{MB} = -1. \\therefore x_{1}x_{2} = -16. Hence y_{0} = \\frac{x_{1}x_{2}}{8} = -2." }, { "text": "Given that $P$ and $Q$ are two points on the parabola $x^{2}=2 y$, with the horizontal coordinates of points $P$ and $Q$ being $4$ and $-2$ respectively, tangents to the parabola are drawn at $P$ and $Q$, and these two tangents intersect at point $A$. Then the vertical coordinate of point $A$ is?", "fact_expressions": "G: Parabola;P: Point;Q: Point;A: Point;l1:Line;l2:Line;Expression(G) = (x^2 = 2*y);PointOnCurve(P,G);PointOnCurve(Q,G);XCoordinate(P)=4;XCoordinate(Q)=-2;TangentOnPoint(P,G)=l1;TangentOnPoint(Q,G)=l2;Intersection(l1,l2)=A", "query_expressions": "YCoordinate(A)", "answer_expressions": "-4", "fact_spans": "[[[10, 24], [64, 67]], [[2, 5], [28, 32], [54, 57]], [[6, 9], [33, 36], [58, 61]], [[76, 79], [81, 85]], [], [], [[10, 24]], [[2, 27]], [[2, 27]], [[28, 52]], [[28, 52]], [[53, 70]], [[53, 70]], [[53, 79]]]", "query_spans": "[[[81, 91]]]", "process": "From the given, let P(4,y₁), Q(-2,y₂), ∴\\begin{cases}4^{2}=2y_{1}\\\\(-2)^{2}=2y_{2}\\end{cases}∴P(4,8),Q(-2,2); ∵ the parabola can be written as y=\\frac{1}{2}x^{2}, ∴y'=x, ∴ the tangent line equation at P is y=4x-8, the tangent line equation at Q is y=-2x-2. Solving these two tangent equations simultaneously gives the coordinates of point A as (1,-4). Hence, the y-coordinate of point A is -4. Topic location: This question examines tangent equations of a parabola and the geometric meaning of derivatives, testing students' ability to transform problems and compute." }, { "text": "Let the focus of the parabola $x^{2}=p y$ coincide with the upper focus of the hyperbola $\\frac{y^{2}}{3}-x^{2}=1$, then the value of $p$ is?", "fact_expressions": "H: Parabola;Expression(H) = (x^2 = p*y);p: Number;G: Hyperbola;Expression(G) = (-x^2 + y^2/3 = 1);Focus(H) = UpperFocus(G)", "query_expressions": "p", "answer_expressions": "8", "fact_spans": "[[[1, 15]], [[1, 15]], [[55, 58]], [[19, 47]], [[19, 47]], [[1, 53]]]", "query_spans": "[[[55, 62]]]", "process": "" }, { "text": "The line $ l $ passing through the origin intersects the hyperbola $ \\frac{x^{2}}{4}-\\frac{y^{2}}{3}=-1 $ at two points. What is the range of possible values for the slope of line $ l $?", "fact_expressions": "l: Line;G: Hyperbola;Expression(G) = (x^2/4 - y^2/3 = -1);O: Origin;PointOnCurve(O, l);NumIntersection(l, G) = 2", "query_expressions": "Range(Slope(l))", "answer_expressions": "(-oo, -sqrt(3)/2) + (sqrt(3)/2, +oo)", "fact_spans": "[[[4, 9], [55, 60]], [[10, 49]], [[10, 49]], [[1, 3]], [[0, 9]], [[4, 53]]]", "query_spans": "[[[55, 70]]]", "process": "" }, { "text": "Given that the product of the distances from a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to the two foci is $m$, then when $m$ takes its maximum value, what are the coordinates of point $P$?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);F1:Point;F2:Point;Distance(P,F1)*Distance(P,F2) = m;WhenMax(m);m:Number;Focus(G)={F1,F2}", "query_expressions": "Coordinate(P)", "answer_expressions": "(0,pm*3)", "fact_spans": "[[[2, 40]], [[43, 46], [71, 75]], [[2, 40]], [[2, 46]], [], [], [[2, 59]], [[61, 70]], [[56, 59], [62, 65]], [[2, 50]]]", "query_spans": "[[[71, 80]]]", "process": "\\because \\frac{x^{2}}{25} + \\frac{y^{2}}{9} = 1, \\therefore for the ellipse, a = 5, b = 3. Let the left and right foci of the ellipse be F_{1}, F_{2}, respectively, then |PF_{1}| + |PF_{2}| = 2a = 10. \\because |PF_{1}| + |PF_{2}| \\geqslant 2\\sqrt{|PF_{1}| \\cdot |PF_{2}|}, \\therefore the product m of the distances from point P to the two foci satisfies: m = |PF_{1}| \\times |PF_{2}| \\leqslant \\left( \\frac{|PF_{1}| + |PF_{2}|}{2} \\right)^{2} = 25. The equality holds if and only if |PF_{1}| = |PF_{2}| = 5, at which point m attains its maximum value 25. At this moment, point P lies at the vertices on the minor axis of the ellipse, giving P(0,3) or (0,-3)." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2/4 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 43]]]", "process": "From the standard equation of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$, we know $a=2$, $b=\\sqrt{3}$, $\\therefore c=\\sqrt{a^{2}-b^{2}}=1$, $\\therefore e=\\frac{c}{a}=\\frac{1}{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)$, point $Q$ lies on the line $l_{1}$: $y=-\\frac{b}{a} x$, point $P$ lies on the line $l_{2}$: $y=\\frac{b}{a} x$, $O$ is the origin, $2 \\overrightarrow{F Q}=\\overrightarrow{F P}$, $\\overrightarrow{F P} \\cdot \\overrightarrow{O Q}=0$, and the area of $\\Delta F O P$ is $\\sqrt{3}$, then the standard equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;l1: Line;l2: Line;F: Point;O: Origin;P: Point;Q: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(Q, l1);Expression(l1)=(y = -x*b/a);PointOnCurve(P, l2);Expression(l2)=(y = x*(b/a));2*VectorOf(F, Q) = VectorOf(F, P);DotProduct(VectorOf(F, P), VectorOf(O, Q)) = 0;Area(TriangleOf(F, O, P)) = sqrt(3)", "query_expressions": "Expression(C)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[6, 70], [283, 289]], [[13, 70]], [[13, 70]], [[80, 109]], [[118, 144]], [[2, 5]], [[146, 149]], [[111, 115]], [[75, 79]], [[13, 70]], [[13, 70]], [[6, 70]], [[2, 74]], [[75, 110]], [[80, 109]], [[111, 145]], [[118, 144]], [[155, 200]], [[201, 252]], [[253, 281]]]", "query_spans": "[[[283, 296]]]", "process": "2\\overrightarrow{FQ}=\\overrightarrow{FP},\\overrightarrow{FP}\\cdot\\overrightarrow{OQ}=0 implies that OQ is the perpendicular bisector of segment FP, thus \\frac{b}{a}=\\sqrt{3}. Then from the area of the triangle we obtain c, and subsequently determine a and b to find the standard equation of the hyperbola. From 2\\overrightarrow{FQ}=\\overrightarrow{FP},\\overrightarrow{FP}\\cdot\\overrightarrow{OQ}=0, it follows that OQ is the perpendicular bisector of segment FP, so \\angle FOQ = \\angle POQ = \\angle POx = 60^{\\circ}, hence \\frac{b}{a}=\\sqrt{3}, |OQ|=\\frac{1}{2}|OF|=\\frac{1}{2}c, |FP|=\\sqrt{3}c, so the area of \\triangle FOP is \\frac{1}{4}\\times\\sqrt{3}c^{2}=\\sqrt{3}, solving gives c=2. Also, a^{2}+b^{2}=c^{2}, \\frac{b}{a}=\\sqrt{3}, so a=1, b=\\sqrt{3}, therefore the standard equation of hyperbola C is x^{2}-\\frac{y^{2}}{3}=1" }, { "text": "$O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=4 \\sqrt{2} x$, and $P$ is a point on $C$. If $|P F|=4 \\sqrt{2}$, then what is the area of $\\Delta P O F$?", "fact_expressions": "C: Parabola;P: Point;O: Origin;F: Point;Expression(C) = (y^2 = 4*sqrt(2)*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 4*sqrt(2)", "query_expressions": "Area(TriangleOf(P, O, F))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[13, 41], [49, 52]], [[45, 48]], [[0, 3]], [[9, 12]], [[13, 41]], [[9, 44]], [[45, 55]], [[57, 75]]]", "query_spans": "[[[77, 95]]]", "process": "\\because the equation of parabola C is y^{2}=4\\sqrt{2}x, \\therefore 2p=4\\sqrt{2}, it follows that \\frac{p}{2}=\\sqrt{2}, yielding the focus F(\\sqrt{2},0). Let P(m,n), according to the definition of a parabola, |PF|=m+\\frac{p}{2}=4\\sqrt{2}, so m+\\sqrt{2}=4\\sqrt{2}, solving gives m=3\\sqrt{2}. \\because point P lies on parabola C, we have n^{2}=4\\sqrt{2}\\times3\\sqrt{2}=24, \\therefore n=\\pm2\\sqrt{6}, x|oF|=\\sqrt{2}, \\therefore the area of \\triangle POF is S=\\frac{1}{2}|OF|\\times|n|=2\\sqrt{3}" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, and $A$, $B$ are two points on $C$ such that the midpoint of segment $AB$ is $M(2,2)$, then the area of $\\triangle ABF$ equals?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;A: Point;B: Point;PointOnCurve(A, C);PointOnCurve(B, C);M: Point;Coordinate(M) = (2, 2);MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Area(TriangleOf(A, B, F))", "answer_expressions": "2", "fact_spans": "[[[6, 25], [37, 40]], [[6, 25]], [[2, 5]], [[2, 28]], [[29, 32]], [[33, 36]], [[29, 45]], [[29, 45]], [[57, 65]], [[57, 65]], [[46, 65]]]", "query_spans": "[[[67, 90]]]", "process": "Let the equation of the line passing through M be y−2=k(x−2). From \\begin{cases}y−2=k(x−2)\\\\y^{2}=4x\\end{cases} \\Rightarrow k^{2}x^{2}−4kx+4(k−1)^{2}=0, \\therefore x_{1}+x_{2}=\\frac{4}{k}, x_{1}x_{2}=\\frac{4(k−1)^{2}}{k^{2}}. According to the condition, \\frac{4}{k}=4 \\Rightarrow k=1, then the line equation is y=x. x_{1}+x_{2}=4, x_{1}x_{2}=0, \\therefore |AB|=4\\sqrt{2}. The distance d from the focus F(1,0) to the line y=x is \\frac{1}{\\sqrt{2}}. \\because the area of \\triangle ABF is 2." }, { "text": "Given that the distance from a point $M(1, m)$ 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 the focal length of the hyperbola $C$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;m: Number;Coordinate(M) = (1, m);PointOnCurve(M, G);Distance(M, Focus(G)) = 5;C: Hyperbola;Expression(C) = (x^2 - y^2/b^2 = 1);b: Number;b>0;A: Point;LeftVertex(C) = A;IsPerpendicular(OneOf(Asymptote(C)), LineOf(A, M))", "query_expressions": "FocalLength(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 23], [36, 37]], [[2, 23]], [[5, 23]], [[5, 23]], [[26, 35]], [[26, 35]], [[26, 35]], [[2, 35]], [[26, 46]], [[47, 89], [99, 105], [123, 129]], [[47, 89]], [[55, 89]], [[55, 89]], [[94, 97]], [[47, 97]], [[99, 121]]]", "query_spans": "[[[123, 134]]]", "process": "From the definition of the parabola, we have 1+\\frac{p}{2}=5, so p=8. Thus, the equation of the parabola is y^{2}=16x, and therefore M(1,4) or M(1,-4). Let M(1,4). From the given conditions, A(-1,0), then k_{AM}=\\frac{4-0}{1-(-1)}=2. Since the asymptotes of the hyperbola are y=\\pm bx, and one asymptote of hyperbola C is perpendicular to line AM, we have 2\\times(-b)=-1, so b=\\frac{1}{2}. Then c=\\sqrt{a^{2}+b^{2}}=\\sqrt{1+\\frac{1}{4}}=\\frac{\\sqrt{5}}{2}. Therefore, the focal length of the hyperbola is 2c=\\sqrt{5}." }, { "text": "The left and right foci of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ are $F_{1}$ and $F_{2}$, respectively. There exists a point $P$ on $C$ such that $\\angle F_{1} P F_{2}=120^{\\circ}$. What is the range of 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;P: Point;F2: Point;LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P,C);AngleOf(F1,P,F2)=ApplyUnit(120,degree)", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[sqrt(3)/2,1)", "fact_spans": "[[[0, 57], [82, 85], [131, 133]], [[0, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[7, 57]], [[66, 73]], [[90, 93]], [[74, 81]], [[0, 81]], [[0, 81]], [[82, 93]], [[95, 129]]]", "query_spans": "[[[131, 141]]]", "process": "First, according to the definition of an ellipse, obtain |PF_{1}| = a + ex_{1}, |PF_{2}| = a - ex_{1}; then using the law of cosines, find x_{1}^{2} = \\frac{4c^{2} - 3a^{2}}{e^{2}}; use the range of the ellipse to set up inequalities and find the range of the eccentricity. Let P(x_{1}, y_{1}), F_{1}(-c, 0), F_{2}(c, 0), c > 0, then |PF_{1}| = a + ex_{1}, |PF_{2}| = a - ex_{1}. In \\triangle PF_{1}F_{2}, by the law of cosines: since x_{1}^{2} \\in [0, a^{2}], so 0 \\leqslant \\frac{4c^{2} - 3a^{2}}{e^{2}} \\leqslant a^{2}, thus 4c^{2} - 3a^{2} \\geqslant 0, and e^{2} < 1, therefore e = \\frac{c}{a} \\geqslant \\frac{\\sqrt{3}}{2}. Hence, the range of the eccentricity of the ellipse is [\\frac{\\sqrt{3}}{2}, 1)." }, { "text": "If the equation $\\frac{x^{2}}{m+3}-\\frac{y^{2}}{1-m}=1$ represents a hyperbola with foci on the $x$-axis, then the range of real values for $m$ is?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/(m + 3) - y^2/(1 - m) = 1);PointOnCurve(Focus(G), xAxis)", "query_expressions": "Range(m)", "answer_expressions": "(-3,1)", "fact_spans": "[[[53, 56]], [[58, 63]], [[1, 56]], [[44, 56]]]", "query_spans": "[[[58, 70]]]", "process": "The equation $\\frac{x^{2}}{m+3}-\\frac{y^{2}}{1-m}=1$ represents a hyperbola with foci on the $x$-axis, then $\\begin{cases}m+3>0\\\\1-m>0\\end{cases}$, solving it yields $-3b>0)$, respectively. Given that the ellipse $C$ passes through the point $P(2 , 1)$, what is the eccentricity of the ellipse $C$ when the length of segment $AB$ is minimized?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;A: Point;RightVertex(C) = A;B: Point;UpperVertex(C) = B;P: Point;Coordinate(P) = (2, 1);PointOnCurve(P, C);WhenMin(Length(LineSegmentOf(A, B)))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[12, 69], [80, 85], [109, 114]], [[12, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[19, 69]], [[1, 4]], [[1, 77]], [[6, 9]], [[1, 77]], [[86, 97]], [[86, 97]], [[80, 97]], [[98, 109]]]", "query_spans": "[[[109, 120]]]", "process": "First, from the given conditions we obtain A(a,0), B(0,b); then, since the ellipse passes through point P(2,1), we get \\frac{4}{a^{2}}+\\frac{1}{b^{2}}=1. Using the basic inequality, we determine the condition under which |AB|=\\sqrt{a^{2}+b^{2}} takes its minimum value, and thus obtain the result. Solution: Since A and B are respectively the right vertex and upper vertex of the ellipse C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), we have A(a,0), B(0,b). Also, since the ellipse C passes through point P(2,1), it follows that \\frac{4}{a^{2}}+\\frac{1}{b^{2}}=1. Therefore, |AB|=\\sqrt{a^{2}+b^{2}}=\\sqrt{(a^{2}+b^{2})(\\frac{4}{a^{2}}+\\frac{1}{b^{2}})}=\\sqrt{4+\\frac{a^{2}}{b^{2}}+\\frac{4b^{2}}{a^{2}}+1}\\geqslant\\sqrt{9}=3. Equality holds if and only if \\frac{a^{2}}{b^{2}}=\\frac{4b^{2}}{a^{2}}, that is, when a^{2}=2b^{2}. At this time, a^{2}=2c^{2}, so the eccentricity is e=\\frac{c}{a}=\\sqrt{\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}" }, { "text": "The length of the imaginary axis of the hyperbola $y^{2}-\\frac{x^{2}}{m}=1$ is twice the length of the real axis, then $m=$?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (y^2 - x^2/m = 1);Length(ImageinaryAxis(G)) = 2*Length(RealAxis(G))", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[0, 28]], [[43, 46]], [[0, 28]], [[0, 41]]]", "query_spans": "[[[43, 48]]]", "process": "The hyperbola $ y^{2} - \\frac{x^{2}}{m} = 1 $ has a conjugate axis length of $ 2\\sqrt{m} $ and a transverse axis length of $ 2 $. According to the problem, $ 2\\sqrt{m} = 4 $, solving gives $ m = 4 $." }, { "text": "If point $P$ is a moving point on the parabola $x^{2}=4 y$, $F$ is the focus of the parabola, and point $A(2,3)$, then the minimum value of $|P A|+|P F|$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);F: Point;Focus(G) = F;A: Point;Coordinate(A) = (2, 3)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "4", "fact_spans": "[[[6, 20], [29, 32]], [[6, 20]], [[1, 5]], [[1, 24]], [[25, 28]], [[25, 35]], [[36, 45]], [[36, 45]]]", "query_spans": "[[[47, 66]]]", "process": "The focus of the parabola $ x^{2} = 4y $ is $ F(0,1) $, and the directrix is $ y = -1 $. Draw a perpendicular from point $ A $ to the directrix, intersecting the directrix at point $ C $. By the definition of a parabola, $ |PF| $ equals the distance from point $ P $ to the directrix of the parabola, $ |PB| $. Therefore, $ |PA| + |PF| = |PA| + |PB| \\geqslant |AC| = 4 $. Hence, the minimum value of $ |PA| + |PF| $ is 4." }, { "text": "Let $F$ be the left focus of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$, and let $P$ be a point on $C$ in the first quadrant. If $\\angle F P O = \\frac{\\pi}{6}$, $|P F| = \\sqrt{3}|O F|$, then the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;F: Point;P: Point;O: Origin;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(P, C);Quadrant(P)=1;AngleOf(F, P, O) = pi/6;Abs(LineSegmentOf(P, F)) = sqrt(3)*Abs(LineSegmentOf(O, F))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[5, 55], [64, 67], [130, 135]], [[12, 55]], [[12, 55]], [[1, 4]], [[60, 63]], [[107, 128]], [[5, 55]], [[1, 59]], [[60, 75]], [[60, 75]], [[77, 106]], [[107, 128]]]", "query_spans": "[[[130, 141]]]", "process": "Connect PF₁, and by the law of cosines combined with plane geometry knowledge, we get |PF₁| = |OF₁|. Then, using the definition of the ellipse and the formula for eccentricity, we can solve. Let F(-c, 0), the right focus of the ellipse be F₁(c, 0), connect PF₁. As shown in the figure, since ∠FPO = π/6, |PF| = √3|OF|, so cos∠FPO = (|PF|² + |OP|² - |OF|²) / (2|PF|·|OP|) = (|OP|² + 2|OF|²) / (2√3|OP|·|OF|) = √3/2. Therefore, |OP| = |OF|, so |OP| = |OF₁|, ∠POF₁ = π/3, hence △POF₁ is an equilateral triangle, |PF₁| = |OF₁|, so |PF| + |PF₁| = √3|OF| + |OF₁| = (√3 + 1)c = 2a. Therefore, the eccentricity e = c/a = 1/(√3 + 1) = √3 - 1" }, { "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{9}=1$ $(a>3)$ has two foci $F_{1}$ and $F_{2}$, and $|F_{1} F_{2}|=8$. The chord $A B$ passes through point $F_{1}$. What is the perimeter of $\\triangle A B F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/9 + x^2/a^2 = 1);a: Number;a>3;F1: Point;F2: Point;Focus(G) = {F1, F2};Abs(LineSegmentOf(F1, F2)) = 8;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) ;PointOnCurve(F1, LineSegmentOf(A, B))", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "20", "fact_spans": "[[[0, 48]], [[0, 48]], [[2, 48]], [[2, 48]], [[54, 61], [96, 104]], [[62, 69]], [[0, 69]], [[71, 88]], [[90, 95]], [[90, 95]], [[0, 95]], [[90, 104]]]", "query_spans": "[[[106, 132]]]", "process": "From the given conditions we have c=4, using a^{2}=b^{2}+c^{2}, find a, then apply the ellipse definition to obtain the perimeter of \\triangleABF_{2} as 4a. As shown in the figure, from |F_{1}F_{2}|=8, we get 2c=8, so c=4, the ellipse is \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{9}=1 (a>3), \\therefore a^{2}=b^{2}+c^{2}=9+16=25, thus a=5, chord AB passes through point F_{1}, according to the ellipse definition the perimeter of \\triangleABF_{2} is 4a=20" }, { "text": "The ellipse $a x^{2}+b y^{2}=1$ $(a>0, b>0, a \\neq b)$ intersects the line $y=1-2 x$ at points $A$ and $B$. The slope of the line passing through the origin and the midpoint of segment $A B$ is $\\frac{\\sqrt{3}}{2}$. Find the value of $\\frac{a}{b}$.", "fact_expressions": "G: Ellipse;Expression(G) = (a*x^2 + b*y^2 = 1);b: Number;a: Number;a>0;b>0;Negation(a=b);H: Line;Expression(H) = (y = 1 - 2*x);A: Point;B: Point;Intersection(G, H) = {A, B};O: Origin;Z: Line;PointOnCurve(O, Z);PointOnCurve(MidPoint(LineSegmentOf(A, B)), Z);Slope(Z) = sqrt(3)/2", "query_expressions": "a/b", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 41]], [[0, 41]], [[2, 41]], [[2, 41]], [[2, 41]], [[2, 41]], [[2, 41]], [[42, 53]], [[42, 53]], [[55, 58]], [[59, 62]], [[0, 64]], [[66, 68]], [[79, 81]], [[65, 81]], [[65, 81]], [[79, 105]]]", "query_spans": "[[[107, 124]]]", "process": "Problem Analysis: Let point A(x_{1},y_{1}), B(x_{2},y_{2}). Substitute y = 1 - 2x into the ellipse ax^{2} + by^{2} = 1, yielding (a + 4b)x^{2} - 4bx + b - 1 = 0. The discriminant is (-4b)^{2} - 4(a + 4b)(b - 1) = 4a + 16b - 4ab\\textcircled{1}. Then, x_{1} + x_{2} = \\frac{4b}{a + 4b}, x_{1}x_{2} = \\frac{b - 1}{a + 4b}. The y-coordinate of the midpoint is \\frac{-2(x_{1} + x_{2})}{2} = 1 - (x_{1} + x_{2}) = 1 - \\frac{4b}{a + 4b} = \\frac{a}{a + 4b}. Let M be the midpoint of segment AB, so M(\\frac{2b}{a + 4b}, \\frac{a}{a + 4b}). Therefore, the slope of line OM is \\frac{\\frac{a}{a + 4b}}{\\frac{2b}{a + 4b}} = \\frac{a}{2b} = \\frac{\\sqrt{3}}{2}. Hence, \\frac{a}{b} = \\sqrt{3}, which when substituted into \\textcircled{1} satisfies \\triangle > 0 (a > 0, b > 0)." }, { "text": "Points $A$ and $B$ are on the parabola $C$: $y^{2}=2 p x$ ($p>0$), $F$ is the focus of parabola $C$. If $\\angle A F B=120^{\\circ}$, and the distance from the midpoint $D$ of $AB$ to the directrix of parabola $C$ is $d$, then the maximum value of $\\frac{d}{|A B|}$ 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/Abs(LineSegmentOf(A, B)))", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[9, 34], [43, 49], [94, 100]], [[16, 34]], [[0, 4]], [[5, 8]], [[39, 42]], [[90, 93]], [[16, 34]], [[9, 34]], [[0, 38]], [[0, 38]], [[39, 52]], [[54, 81]], [[83, 93]], [[90, 110]], [[107, 110]]]", "query_spans": "[[[112, 135]]]", "process": "" }, { "text": "The equation of the directrix of the parabola $y^{2}=-8 x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -8*x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x = 2", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "From the given, $ y^{2} = -8x $ opens to the left, and $ 2p = 8 \\Rightarrow p = 4 $, so the equation of the directrix is $ x = \\frac{p}{2} = 2 $, that is, $ x = 2 $." }, { "text": "The circle centered at the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ and passing through the right vertex $A$ of this ellipse intersects the line $3x+4y-21=0$. What is the length of the chord intercepted?", "fact_expressions": "G: Ellipse;H: Line;F1: Point;A:Point;C:Circle;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H) = (3*x + 4*y - 21 = 0);LeftFocus(G)=F1;Center(C)=F1;RightVertex(G)=A;PointOnCurve(A,C)", "query_expressions": "Length(InterceptChord(H,C))", "answer_expressions": "4*sqrt(7)", "fact_spans": "[[[1, 40], [57, 59]], [[68, 84]], [[44, 51]], [[62, 65]], [[66, 67]], [[1, 40]], [[68, 84]], [[1, 51]], [[0, 67]], [[57, 65]], [[55, 67]]]", "query_spans": "[[[66, 91]]]", "process": "Calculate the left focus and right vertex to obtain the circle equation (x+3)^{2}+y^{2}=64, then compute the distance from the center of the circle to the line as 6, and finally apply the perpendicular chord theorem to obtain the answer. [Detailed solution] For the ellipse \\frac{x^{2}}{2}+\\frac{y^{2}}{16}=1, the left focus is F_{1}(-3,0), the right vertex is A(5,0), the circle equation is (x+3)^{2}+y^{2}=64, the distance from the center to the line is: d=\\frac{|-9-21|}{x}=6, the chord length is 2\\sqrt{64-36}=4\\sqrt{7}" }, { "text": "The eccentricity of an ellipse is $e=\\frac{1}{2}$. Its foci coincide with the foci of the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$. Then, what is the equation of this ellipse?", "fact_expressions": "e: Number;Eccentricity(H) = e;e = 1/2;G: Hyperbola;Expression(G) = (x^2/3 - y^2 = 1);Focus(H) = Focus(G);H: Ellipse", "query_expressions": "Expression(H)", "answer_expressions": "x^2/16+y^2/12=1", "fact_spans": "[[[3, 18]], [[0, 21]], [[3, 18]], [[27, 55]], [[27, 55]], [[22, 60]], [[19, 21], [22, 23], [63, 65]]]", "query_spans": "[[[63, 70]]]", "process": "From the given conditions, the coordinates of the foci of this ellipse are $F_{1}(-2,0)$, $F_{2}(2,0)$. Since the eccentricity $e=\\frac{1}{2}$, $\\therefore a=4$, $b^{2}=12$, $\\therefore$ the equation of this ellipse is $\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$." }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$ is equal to?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/3 + y^2/4 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 44]]]", "process": "According to the problem, a=2, b=\\sqrt{3}, so c=\\sqrt{4-3}=1, and the eccentricity is e=\\frac{c}{a}=\\frac{1}{2}" }, { "text": "Given that the line $l$ passes through points $M(2,0)$ and $N(3, 1)$, and intersects the parabola $y^{2}=8x$ at points $A$ and $B$, then $|AB|=$?", "fact_expressions": "l: Line;G: Parabola;M: Point;N: Point;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Coordinate(M) = (2, 0);Coordinate(N) = (3, 1);PointOnCurve(M, l);PointOnCurve(N, l);Intersection(l,G)={A,B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "16", "fact_spans": "[[[2, 7]], [[33, 47]], [[8, 17]], [[20, 30]], [[49, 52]], [[53, 56]], [[33, 47]], [[8, 17]], [[20, 30]], [[2, 18]], [[2, 30]], [[2, 58]]]", "query_spans": "[[[60, 69]]]", "process": "" }, { "text": "A moving circle is externally tangent to the fixed circle $F$: $(x-3)^{2}+y^{2}=4$ and tangent to the line $l$: $x=-1$. Find the equation of the locus of the center of the moving circle.", "fact_expressions": "l: Line;F: Circle;G:Circle;Expression(F) = (y^2 + (x - 3)^2 = 4);Expression(l)=(x=-1);IsOutTangent(F,G);IsTangent(l,G)", "query_expressions": "LocusEquation(Center(G))", "answer_expressions": "y^2=12*x", "fact_spans": "[[[36, 49]], [[7, 30]], [[2, 4], [53, 55]], [[7, 30]], [[36, 49]], [[2, 33]], [[2, 51]]]", "query_spans": "[[[53, 64]]]", "process": "From the given conditions, the center of circle F is at F(3,0) and its radius is 2. Since the moving circle is externally tangent to the fixed circle F: (x-3)^{2}+y^{2}=4 and tangent to the line l: x=-1, the distance from the center of the moving circle to point F is 2 greater than its distance to the line l. Therefore, the distance from the center of the moving circle to point F equals its distance to the line x=-3. Hence, the locus of the center of the moving circle is a parabola with focus at point F(3,0) and directrix the line x=-3. Let the equation of the locus be y^{2}=2px. Then \\frac{p}{2}=3, so p=6. Thus, the equation of the locus of the center of the moving circle is y^{2}=12x." }, { "text": "The distance from point $M$ to point $F(0,-2)$ is 1 less than its distance to the line $l$: $y - 3 = 0$. Then, what is the equation of the trajectory of point $M$?", "fact_expressions": "l: Line;F: Point;M: Point;Expression(l) = (y - 3 = 0);Coordinate(F) = (0, -2);Distance(M, F) = Distance(M, l) - 1", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2=-8*y", "fact_spans": "[[[21, 35]], [[5, 15]], [[0, 4], [44, 48], [19, 20]], [[21, 35]], [[5, 15]], [[0, 42]]]", "query_spans": "[[[44, 55]]]", "process": "Let M(x, y). According to the problem, since the distance from point M to point F(0, -2) is 1\\frac{1}{2} less than its distance to the line l: y - 3 = 0, by geometric knowledge we get: y < 3. The original equation becomes \\sqrt{(x-0)^{2}+(y+2)^{2}} = 2 - y. Squaring both sides, we obtain x^{2}+(y+2)^{2}=(2-y)^{2}. Simplifying gives x^{2} = -8y. Thus, the trajectory equation of point M is x^{2} = -8y." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the right vertex and upper vertex are denoted as $A$ and $B$ respectively. Point $P$ lies on the ellipse, line $AP$ intersects the $y$-axis at point $C$, and line $BP$ intersects the $x$-axis at point $D$. If $|AD| \\cdot |BC| = a^{2}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;P: Point;B: Point;D: Point;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightVertex(G) = A;UpperVertex(G) = B;PointOnCurve(P, G);Intersection(LineSegmentOf(A, P), yAxis) = C;Intersection(LineSegmentOf(B, P), xAxis) = D;Abs(LineSegmentOf(A, D))*Abs(LineSegmentOf(B, C)) = a^2", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[2, 54], [78, 80], [144, 146]], [[4, 54]], [[4, 54]], [[65, 68]], [[73, 77]], [[69, 72]], [[111, 115]], [[93, 97]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 72]], [[2, 72]], [[73, 81]], [[82, 98]], [[100, 115]], [[117, 141]]]", "query_spans": "[[[144, 152]]]", "process": "Let P(x₀,y₀), write the equations of lines BP and AP, and find D(bx₀/(b−y₀),0), C(0,ay₀/(a−x₀)), then express |BC|, |AD|. From |AD|⋅|BC|=a², simplifying yields a=2b, thus the eccentricity can be found. Let P(x₀,y₀); then BP: y=((y₀−b)/x₀)x+b; let y=0, D(bx₀/(b−y₀),0). Since x₀²/a² + y₀²/b² =1, it follows that D. \n\\begin{matrix}b^{-}\\\\BC&=(2a²b²−2ba²y₀−2ab²x₀+2abx₀y₀)/(ab−ay₀−bx₀+x₀y₀)=2ab=a², so a=2b, hence e=c/a=√(1−b²/a²)=√(1−1/4)=√3/2." }, { "text": "Given that an 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})$, and one focus of the hyperbola is $F(-\\sqrt{7}, 0)$, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Point;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(H) = (2, sqrt(3));Coordinate(F) = (-sqrt(7), 0);PointOnCurve(H, OneOf(Asymptote(G)));OneOf(Focus(G)) = F", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2/3=1", "fact_spans": "[[[2, 59], [84, 87], [112, 115]], [[5, 59]], [[5, 59]], [[66, 82]], [[93, 110]], [[5, 59]], [[5, 59]], [[2, 59]], [[66, 82]], [[93, 110]], [[2, 82]], [[84, 110]]]", "query_spans": "[[[112, 120]]]", "process": "From the given conditions, the asymptotes of the hyperbola are: $ y = \\pm\\frac{b}{a}x $. Substituting the point $ (2,\\sqrt{3}) $ into $ y = \\frac{b}{a}x $ gives $ \\sqrt{3}a = 2b $. Given $ c = \\sqrt{7} $, it follows that $ a^{2} = 4 $, $ b^{2} = 3 $. The equation of the hyperbola is: $ \\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1 $." }, { "text": "The ellipse $\\frac{x^{2}}{81}+\\frac{y^{2}}{27}=1$ has two foci $F_{1}$, $F_{2}$. A line passes through $F_{1}$ and intersects the ellipse at points $P$, $Q$. Find the perimeter of $\\triangle P Q F_{2}$.", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/81 + y^2/27 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};H: Line;PointOnCurve(F1, H);P: Point;Q: Point;Intersection(H, G) = {P, Q}", "query_expressions": "Perimeter(TriangleOf(P, Q, F2))", "answer_expressions": "36", "fact_spans": "[[[0, 39], [72, 74]], [[0, 39]], [[44, 51], [64, 71]], [[52, 59]], [[0, 59]], [[61, 63]], [[61, 71]], [[75, 78]], [[79, 82]], [[61, 82]]]", "query_spans": "[[[84, 110]]]", "process": "From the equation of the ellipse, we know that a=9. By the definition of the ellipse, the perimeter of \\trianglePQF_{2} is 4a. \\therefore the perimeter of \\trianglePQF_{2} is 36." }, { "text": "The line $ l $ passes through the focus of the parabola $ y^{2} = 4x $ and has an inclination angle of $ \\frac{\\pi}{3} $. If $ l $ intersects the parabola at points $ A $ and $ B $, and $ O $ is the origin, then the area of $ \\triangle AOB $ is?", "fact_expressions": "l: Line;G: Parabola;A: Point;O: Origin;B: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), l);Inclination(l) = pi/3;Intersection(l, G) = {A, B}", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[18, 23], [45, 48]], [[1, 15], [49, 52]], [[55, 58]], [[65, 68]], [[59, 62]], [[1, 15]], [[0, 23]], [[18, 43]], [[45, 64]]]", "query_spans": "[[[74, 96]]]", "process": "Problem Analysis: The focus of the parabola y^{2}=4x is (1,0). The equation of line l is y=\\sqrt{3}(x-1). Combining this with y^{2}=4x and eliminating x gives:" }, { "text": "If point $M(1,1)$ is the midpoint of chord $AB$ of the parabola $y^{2}=4x$, then what is the length of chord $AB$?", "fact_expressions": "G: Parabola;A: Point;B: Point;M: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (1, 1);IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "sqrt(15)", "fact_spans": "[[[11, 25]], [[27, 32]], [[27, 32]], [[1, 10]], [[11, 25]], [[1, 10]], [[11, 32]], [[1, 35]]]", "query_spans": "[[[38, 47]]]", "process": "Let the coordinates of points A and B be A(x_{1},y_{1}), B(x_{2},y_{2}), substitute into the parabola equation y^{2}=4x, we get y_{1}^{2}=4x_{1}, y_{2}^{2}=4x_{2}. Subtracting these two equations, we obtain k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{4}{y_{1}+y_{2}}=2. Therefore, the equation of line AB is y-1=2(x-1), or y=2x-1. Substituting into the parabola equation yields 4x^{2}-8x+1=0, then x_{1}+x_{2}=2, x_{1}x_{2}=\\frac{1}{4}. Then |AB|=\\sqrt{1+k^{2}}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\sqrt{5\\times(2^{2}-4\\times\\frac{1}{4})}=\\sqrt{15}. Thus, the length of chord AB is \\sqrt{15}." }, { "text": "If $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)$, and the circle with diameter $F_{1}F_{2}$ intersects the right branch of the hyperbola at point $P$, and if $\\angle P F_{1} F_{2}=\\alpha$, then the eccentricity of the hyperbola is (expressed in terms of $\\alpha$)?", "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;IsDiameter(LineSegmentOf(F1,F2),H);H: Circle;Intersection(H,RightPart(G)) = P;P: Point;AngleOf(P, F1, F2) = alpha;alpha: Number", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/(Cos(alpha)-Sin(alpha))", "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 81]], [[1, 81]], [[19, 75], [104, 107], [147, 150]], [[19, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[82, 103]], [[102, 103]], [[102, 114]], [[110, 114]], [[116, 145]], [[160, 168]]]", "query_spans": "[[[147, 172]]]", "process": "According to the problem, PF_{1}\\bot PF_{2}, so |PF_{1}|=2c\\cdot\\cos\\alpha, |PF_{2}|=2c\\cdot\\sin\\alpha, therefore e=\\frac{2c}{2a}=\\frac{2c}{|PF_{1}|-|PF_{2}|}=\\frac{1}{\\cos\\alpha-\\sin\\alpha}." }, { "text": "The distance from the focus to the directrix of the parabola ${y}^{2}=8 x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x)", "query_expressions": "Distance(Focus(G), Directrix(G))", "answer_expressions": "4", "fact_spans": "[[[0, 16]], [[0, 16]]]", "query_spans": "[[[0, 27]]]", "process": "" }, { "text": "Let the left and right foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ be $F_{1}$ and $F_{2}$, respectively. A line passing through focus $F_{1}$ intersects the ellipse at points $M$ and $N$. If the area of the incircle of $\\triangle M N F_{2}$ is $\\pi$, then $S_{\\Delta M N F_{2}}$=?", "fact_expressions": "G: Ellipse;H: Line;M: Point;N: Point;F2: Point;F1: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1, H);Intersection(H, G) = {M, N};Area(InscribedCircle(TriangleOf(M, N, F2))) = pi", "query_expressions": "Area(TriangleOf(M, N, F2))", "answer_expressions": "4", "fact_spans": "[[[1, 38], [77, 79]], [[74, 76]], [[80, 83]], [[84, 87]], [[55, 62]], [[47, 54], [66, 73]], [[1, 38]], [[1, 62]], [[1, 62]], [[63, 76]], [[74, 89]], [[91, 125]]]", "query_spans": "[[[127, 151]]]", "process": "As shown in the figure, for the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, $a=2$. A line passing through focus $F_{1}$ intersects the ellipse at points $M(x_{1},y_{1})$ and $N(x_{2},y_{2})$. The area of the incircle of $\\triangle MNF_{2}$ is $\\pi$, therefore the inradius of $\\triangle MNF_{2}$ is $1$, hence the area $S$ of $\\triangle MNF_{2}$ is $\\frac{1}{2}\\times1\\times(|MN|+|MF_{2}|+|NF_{2}|)=2a=4$." }, { "text": "$O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=4x$, and $P$ is a point on $C$. If $|PF|=3$, then the area of triangle $POF$ is?", "fact_expressions": "C: Parabola;P: Point;F: Point;O: Origin;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(P, C);Abs(LineSegmentOf(P, F)) = 3", "query_expressions": "Area(TriangleOf(P,O,F))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[13, 32], [40, 43]], [[36, 39]], [[9, 12]], [[0, 3]], [[13, 32]], [[9, 35]], [[36, 47]], [[49, 58]]]", "query_spans": "[[[60, 75]]]", "process": "According to the problem, the focus of the parabola C is F(1,0), and the equation of the directrix is x = -1. Given |PF| = 3, let P(x,y), then x + 1 = 3, so x = 2, hence y = \\pm2\\sqrt{2}, that is, the coordinates of point P are (2,\\pm2\\sqrt{2}). Then the area of \\triangle POF is S = \\frac{1}{2} \\times 1 \\times 2\\sqrt{2} = \\sqrt{2}." }, { "text": "The ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ has eccentricity $\\frac{\\sqrt{3}}{3}$. The line $x-2 y+b=0$ intersects the ellipse at points $P$ and $Q$, and $E$ is the midpoint of $P Q$, $O$ is the origin. Then the slope of line $O E$ is?", "fact_expressions": "G: Ellipse;a: Number;b: Number;H: Line;O: Origin;E: Point;P: Point;Q: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (b + x - 2*y = 0);Eccentricity(G)= sqrt(3)/3;Intersection(H, G) = {P, Q};MidPoint(LineSegmentOf(P,Q))=E", "query_expressions": "Slope(LineOf(O,E))", "answer_expressions": "-4/3", "fact_spans": "[[[0, 52], [92, 94]], [[2, 52]], [[2, 52]], [[78, 91]], [[119, 122]], [[115, 118]], [[96, 99]], [[100, 103]], [[2, 52]], [[2, 52]], [[0, 52]], [[78, 91]], [[0, 76]], [[78, 105]], [[107, 118]]]", "query_spans": "[[[127, 139]]]", "process": "Since the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{3}}{3}$, it follows that $e=\\frac{c}{a}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\frac{\\sqrt{3}}{3}$, so $\\frac{b^{2}}{a^{2}}=\\frac{2}{3}$. Let $P(x_{1},y_{1})$, $Q(x_{2},y_{2})$, so $k_{PQ}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{1}{2}$, $E(\\frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2})$. Since $P$, $Q$ lie on the ellipse, we have $\\begin{cases}\\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\\end{cases}$, subtracting the two equations gives $\\frac{x_{1}^{2}-x_{2}^{2}}{a^{2}}+\\frac{y_{1}^{2}-y_{2}^{2}}{b^{2}}=0$, that is, $\\frac{y_{1}^{2}-y_{2}^{2}}{x_{1}^{2}-x_{2}^{2}}=-\\frac{b^{2}}{a^{2}}$, i.e., $\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{(x_{1}-x_{2})(x_{1}+x_{2})}=-\\frac{2}{3}$, thus $k_{PQ}\\cdot k_{OE}=-\\frac{2}{3}$, so $k_{OE}=-\\frac{4}{3}$." }, { "text": "It is known that the center of an ellipse is at the origin, its right focus coincides with the center of the circle $x^{2}+m y^{2}-2 m x-3=0$, and the length of the major axis equals the diameter of the circle. What is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;H: Circle;m: Number;O:Origin;Center(G)=O;Expression(H) = (-2*m*x + m*y^2 + x^2 - 3 = 0);RightFocus(G) = Center(H);Length(MajorAxis(G))=Diameter(H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/2", "fact_spans": "[[[2, 4], [61, 63]], [[17, 43], [54, 55]], [[18, 43]], [[8, 12]], [[2, 12]], [[17, 43]], [[2, 48]], [[2, 58]]]", "query_spans": "[[[61, 69]]]", "process": "Since the equation $ x^{2} + my^{2} - 2mx - 3 = 0 $ represents a circle, we have $ m = 1 $, so the equation of the circle is $ x^{2} + y^{2} - 2x - 3 = 0 $, or $ (x - 1)^{2} + y^{2} = 2^{2} $, with center $ (1, 0) $ and radius $ 2 $. Therefore, $ a = 2 $, $ c = 1 $, $ e = \\frac{c}{a} = \\frac{1}{2} $." }, { "text": "The line $y = x - 2$ intersects the parabola $y^{2} = 2x$ at points $A$ and $B$, and $O$ is the origin. Find $\\angle AOB$.", "fact_expressions": "G: Parabola;H: Line;A: Point;O: Origin;B: Point;Expression(G) = (y^2 = 2*x);Expression(H) = (y = x - 2);Intersection(H, G) = {A, B}", "query_expressions": "AngleOf(A, O, B)", "answer_expressions": "ApplyUnit(90,degree)", "fact_spans": "[[[10, 24]], [[0, 9]], [[27, 30]], [[37, 40]], [[31, 34]], [[10, 24]], [[0, 9]], [[0, 36]]]", "query_spans": "[[[47, 63]]]", "process": "" }, { "text": "Given a point $P$ on the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$. Draw lines $l_{1}$, $l_{2}$ through point $P$ parallel to the two asymptotes of the hyperbola. The lines $l_{1}$, $l_{2}$ intersect the $x$-axis at points $M$, $N$ respectively. Then $|O M| \\cdot|O N|$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2 = 1);P: Point;PointOnCurve(P, G);L1: Line;L2: Line;Asymptote(G) = {L1, L2};l1: Line;l2: Line;PointOnCurve(P, l1);PointOnCurve(P, l2);IsParallel(L1, l1);IsParallel(L2, l2);M: Point;N: Point;Intersection(l1, xAxis) = M;Intersection(l2, xAxis) = N;O: Origin", "query_expressions": "Abs(LineSegmentOf(O, M))*Abs(LineSegmentOf(O, N))", "answer_expressions": "4", "fact_spans": "[[[2, 30], [43, 46]], [[2, 30]], [[33, 36], [38, 42]], [[2, 36]], [], [], [[43, 50]], [[54, 61], [71, 80]], [[62, 70], [82, 89]], [[37, 70]], [[37, 70]], [[43, 70]], [[43, 70]], [[97, 100]], [[101, 104]], [[71, 106]], [[71, 106]], [[108, 126]]]", "query_spans": "[[[108, 128]]]", "process": "The hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ has two asymptotes with slopes $\\pm\\frac{1}{2}$. Let point $P(x_{0},y_{0})$, then the equations of lines $l_{1}$, $l_{2}$ are respectively $y-y_{0}=\\frac{1}{2}(x-x_{0})$, $y-y_{0}=-\\frac{1}{2}(x-x_{0})$, so the coordinates of points $M$, $N$ are $M(x_{0}-2y_{0},0)$, $N(x_{0}+2y_{0},0)$. Therefore, $|OM|\\cdot|ON|=|x_{0}-2y_{0}||x_{0}+2y_{0}|=|x_{0}^{2}-4y_{0}^{2}|$. Since point $P$ lies on the hyperbola, $\\frac{x_{0}^{2}}{4}-y_{0}^{2}=1$, thus $|OM|\\cdot|ON|=4$." }, { "text": "Given the parabola $y=x^{2}$ and the point $P(0 , 1)$, if from a certain point $C$ two tangents can be drawn to the parabola with points of tangency $A$ and $B$, respectively, such that $\\overrightarrow{C P}=\\frac{1}{3} \\overrightarrow{C A}+\\frac{2}{3} \\overrightarrow{C B}$, then what is the area of $\\triangle A B C$?", "fact_expressions": "G: Parabola;P: Point;A: Point;B: Point;C: Point;l1: Line;l2: Line;Expression(G) = (y = x^2);Coordinate(P) = (0, 1);TangentOfPoint(C, G) = {l1, l2};TangentPoint(l1, G) = A;TangentPoint(l2, G) = B;VectorOf(C, P) = VectorOf(C, A)/3 + (2/3)*VectorOf(C, B)", "query_expressions": "Area(TriangleOf(A, B, C))", "answer_expressions": "sqrt(2)*27/16", "fact_spans": "[[[2, 14], [36, 39]], [[15, 26]], [[50, 53]], [[54, 57]], [[31, 34]], [], [], [[2, 14]], [[15, 26]], [[28, 44]], [[28, 57]], [[28, 57]], [[61, 149]]]", "query_spans": "[[[151, 173]]]", "process": "From $\\overrightarrow{CP}=\\frac{1}{3}\\overrightarrow{CA}+\\frac{2}{3}\\overrightarrow{CB}$ it follows that $\\overrightarrow{PA}=-2\\overrightarrow{PB}$, then the line $AB$ always passes through the fixed point $P$. Let the equation of line $AB$ be solved together with the parabola equation, thus the length of chord $AB$ can be obtained. Differentiate the parabola equation to find the slope of the tangent lines, then obtain the tangent line equations, and further solve for the coordinates of point $C$. Using the point-to-line distance formula and the triangle area formula, the required result can be solved. [Solution] $\\because \\overrightarrow{CP}=\\frac{1}{3}\\overrightarrow{CA}+\\frac{2}{3}\\overrightarrow{CB}$, then $3\\overrightarrow{CP}=(\\overrightarrow{CP}+\\overrightarrow{PA})+2(\\overrightarrow{CP}+\\overrightarrow{PB})$ $\\therefore \\overrightarrow{PA}=-2\\overrightarrow{PB}$, hence line $AB$ passes through point $P$, and $AP=2PB$. Let line $AB: y=kx+1$, $A(x_{1},y_{1})$, $B(x_{2},y_{2})$. Solving the system $\\begin{cases} y=kx+1 \\\\ y=x^{2} \\end{cases}$ gives $x^{2}-kx-1=0$, then $x_{1}x_{2}=-1$, $x_{1}+x_{2}=k$. From $AP=2PB$, we get $x_{1}+2x_{2}=0$, so $k=\\pm\\frac{\\sqrt{2}}{2}$, $AB=\\sqrt{1+k^{2}}\\cdot\\sqrt{(x_{1}+x_{2})^{2}-4x_{1}x_{2}}=\\frac{3\\sqrt{3}}{2}$. From derivative $y'=2x$, the tangents at $A,B$ are $y+y_{1}=2x_{1}x$, $y+y_{2}=2x_{2}x$. Solving the tangent equations gives $C(\\frac{k}{2},-1)$. The distance from $C$ to $y=kx+1$ is $d=\\frac{2}{\\sqrt{6}}$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $\\frac{\\sqrt{3}}{3}$. If the maximum radius of a circle centered at $M(0,-1)$ that has common points with the ellipse $C$ is $\\sqrt{15}$, then what is the equation of the ellipse $C$?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Circle;M: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(M) = (0, -1);Eccentricity(C) = sqrt(3)/3;Center(G)=M;IsIntersect(G,C);Max(Radius(G))=sqrt(15)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/18+y^2/8=1", "fact_spans": "[[[2, 59], [101, 106], [132, 137]], [[9, 59]], [[9, 59]], [[111, 112]], [[87, 96]], [[9, 59]], [[9, 59]], [[2, 59]], [[87, 96]], [[2, 84]], [[86, 112]], [[100, 112]], [[111, 129]]]", "query_spans": "[[[132, 142]]]", "process": "Given that $ e = \\frac{c}{a} = \\frac{\\sqrt{3}}{3} $, so $ a = \\sqrt{3}c $, then $ b = \\sqrt{a^{2} - c^{2}} = \\sqrt{2}c $. Let the coordinates of an arbitrary point $ P $ on the ellipse be $ (\\sqrt{3}c\\cos\\theta, \\sqrt{2}c\\sin\\theta) $. Then \n$$\n|MP|^{2} = (\\sqrt{3}c\\cos\\theta)^{2} + (\\sqrt{2}c\\sin\\theta + 1)^{2} = 3c^{2}\\cos^{2}\\theta + 2c^{2}\\sin^{2}\\theta + 2\\sqrt{2}c\\sin\\theta + 1 = -c^{2}\\sin^{2}\\theta + 2\\sqrt{2}c\\sin\\theta + 3c^{2} + 1 = -c^{2}(\\sin\\theta - \\frac{\\sqrt{2}}{c})^{2} + 3c^{2} + 3.\n$$\nIf $ c \\geqslant \\sqrt{2} $, then when $ \\sin\\theta = \\frac{\\sqrt{2}}{c} $, $ (|MP|^{2})_{\\max} = 3c^{2} + 3 $. From $ 3c^{2} + 3 = 15 $, we get $ c = 2 $, which satisfies the condition. At this time, $ a = 2\\sqrt{3} $, $ b = 2\\sqrt{2} $, and the equation of the ellipse is $ \\frac{x^{2}}{18} + \\frac{y^{2}}{8} = 1 $. If $ 0 < c < \\sqrt{2} $, then when $ \\sin\\theta = 1 $, $ (|MP|^{2})_{\\max} = 2c^{2} + 2\\sqrt{2}c + 1 $, so $ 2c^{2} + 2\\sqrt{2}c + 1 = 15 $, i.e., $ c^{2} + \\sqrt{2}c = 7 $. But when $ 0 < c < \\sqrt{2} $, $ c^{2} + \\sqrt{2}c < 4 $, thus $ c^{2} + \\sqrt{2}c = 7 $ has no solution. In conclusion, the equation of the ellipse is $ \\frac{x^{2}}{18} + \\frac{y^{2}}{8} = 1 $." }, { "text": "The equation of the directrix of the parabola $y^{2}=2 x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-1/2", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "The equation of the directrix of the parabola y^{2}=2x is x=-\\frac{1}{2}." }, { "text": "Given the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ is a moving point on the ellipse. The maximum 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);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G)", "query_expressions": "Max(Area(TriangleOf(P, F1, F2)))", "answer_expressions": "12", "fact_spans": "[[[2, 40], [69, 71]], [[64, 68]], [[48, 55]], [[56, 63]], [[2, 40]], [[2, 63]], [[2, 63]], [[64, 75]]]", "query_spans": "[[[76, 108]]]", "process": "In the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, $a^{2}=25$, $b^{2}=9$, $\\therefore c^{2}=a^{2}-b^{2}=16$, $\\therefore b=3$, $c=4$. By the geometric properties of the ellipse, when point $P$ is an endpoint of the minor axis of the ellipse, the area of $\\triangle PF_{1}F_{2}$ is maximized, hence $\\frac{1}{2}|F_{1}F_{2}|b=bc=12$." }, { "text": "Given that point $P(x, y)$ lies on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, find the range of values for $2x+3y$.", "fact_expressions": "G: Ellipse;P: Point;x1:Number;y1:Number;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(P) = (x1, y1);PointOnCurve(P, G)", "query_expressions": "Range(2*x1 + 3*y1)", "answer_expressions": "[-5,5]", "fact_spans": "[[[13, 40]], [[2, 12]], [[3, 12]], [[3, 12]], [[13, 40]], [[2, 12]], [[2, 44]]]", "query_spans": "[[[46, 62]]]", "process": "Since point P(x,y) is a moving point on the ellipse \\frac{x^{2}}{4}+y^{2}=1, we set the parametric equations of the ellipse as \\begin{cases}x=2\\cos\\theta\\\\y=\\sin\\theta\\end{cases} (\\theta is the parameter). Then z=2x+3y=4\\cos\\theta+3\\sin\\theta=5(\\frac{4}{5}\\cos\\theta+\\frac{3}{5}\\sin\\theta)=5\\sin(\\theta+\\phi). Since \\theta\\in[0,2\\pi), it follows that z\\in[-5,5], so the range of z=2x+3y is [-5,5]." }, { "text": "The point $M(x, y)$ moves in such a way that the equation $\\sqrt{x^{2}+(y+1)^{2}}+\\sqrt{x^{2}+(y-1)^{2}}=6$ is always satisfied. What is the equation of the trajectory of point $M$?", "fact_expressions": "M: Point;Coordinate(M) = (x1, y1);x1:Number;y1:Number;sqrt(x1^2+(y1+1)^2)+sqrt(x1^2+(y1-1)^2)=6", "query_expressions": "LocusEquation(M)", "answer_expressions": "y^2/9+x^2/8=1", "fact_spans": "[[[0, 11], [74, 78]], [[0, 11]], [[1, 11]], [[1, 11]], [[24, 73]]]", "query_spans": "[[[74, 85]]]", "process": "The sum of the distances from point M to (0,-1) and (0,1) is 6, and the distance between the points (0,-1) and (0,1) is 2, so its trajectory is an ellipse with foci at (0,-1) and (0,1), where c=1, a=3, b^{2}=8. Therefore, the equation of the trajectory of point M is \\frac{y^{2}}{9}+\\frac{x^{2}}{8}=1." }, { "text": "Through the focus $F$ of the parabola $y^{2}=8x$, draw a line $l$ with slope $2$ intersecting the parabola at points $A$ and $B$. Then the standard equation of the circle with $AB$ as diameter is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);F: Point;Focus(G) = F;l: Line;PointOnCurve(F, l);Slope(l) = 2;A: Point;B: Point;Intersection(l, G) = {A, B};H: Circle;IsDiameter(LineSegmentOf(A, B), H)", "query_expressions": "Expression(H)", "answer_expressions": "(x-3)^2+(y-2)^2=25", "fact_spans": "[[[1, 15], [35, 38]], [[1, 15]], [[18, 21]], [[1, 21]], [[29, 34]], [[0, 34]], [[22, 34]], [[39, 42]], [[43, 46]], [[29, 48]], [[60, 61]], [[50, 61]]]", "query_spans": "[[[60, 68]]]", "process": "From y^{2}=8x we get p=4, so F(2,0), thus the equation of line AB is: y=2(x-2)=2x-4. Solving the system \\begin{cases}y=2x-4\\\\y^{2}=8x\\end{cases}, eliminating y and simplifying yields x^{2}-6x+4=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=6, x_{1}x_{2}=4, so y_{1}+y_{2}=2(x_{1}+x_{2})-8=4. Therefore, the center of the circle with AB as diameter is (3,2), the radius of the circle with AB as diameter is 5, hence the standard equation of the circle with AB as diameter is: (x-3)^{2}+(y-2)^{2}=25" }, { "text": "A line $l$ passing through the focus of the parabola $y^{2}=4x$ intersects the parabola at points $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$. If the horizontal coordinate of the midpoint of $PQ$ is $3$, then what is the length of $PQ$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);l: Line;PointOnCurve(Focus(G), l);P: Point;Q: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);Intersection(l, G) = {P, Q};XCoordinate(MidPoint(LineSegmentOf(P, Q))) = 3", "query_expressions": "Length(LineSegmentOf(P, Q))", "answer_expressions": "8", "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[19, 24]], [[0, 24]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[29, 46]], [[48, 65]], [[19, 67]], [[69, 84]]]", "query_spans": "[[[86, 95]]]", "process": "From the parabola $ y^{2} = 4x $, we have $ 2p = 4 $, so the focus is $ F(1,0) $, and the equation of the directrix is $ x = -1 $. According to the definition of the parabola: $ |PF| = x_{1} + 1 $, $ |QF| = x_{2} + 1 $, so $ |PQ| = x_{1} + x_{2} + 2 $. Since the horizontal coordinate of the midpoint of $ PQ $ is 3, we have $ x_{1} + x_{2} = 6 $, so $ |PQ| = x_{1} + x_{2} + 2 = 6 + 2 = 8 $." }, { "text": "The number of intersection points between the line $y=2k$ and the curve $9k^{2}x^{2}+y^{2}=18k^{2}|x|$ $(k \\in \\mathbb{R}, k \\neq 0)$ is?", "fact_expressions": "G: Line;k: Real;H: Curve;Expression(G) = (y = 2*k);Expression(H) = (9*(k^2*x^2) + y^2 = 18*k^2*Abs(x));Negation(k=0)", "query_expressions": "NumIntersection(G, H)", "answer_expressions": "4", "fact_spans": "[[[0, 9]], [[12, 64]], [[10, 64]], [[0, 9]], [[10, 64]], [[12, 64]]]", "query_spans": "[[[0, 73]]]", "process": "Solving the system of the line equation y=2k and the curve equation 9k^{2}x^{2}+y^{2}=18k^{2}|x|, we obtain 9x^{2}-18|x|+4=0, and determine based on the number of roots of the equation. Solving 9x^{2}-18|x|+4=0 yields |x|=1-\\frac{\\sqrt{5}}{3} or |x|=1+\\frac{\\sqrt{5}}{3}, so x=1-\\frac{\\sqrt{5}}{3} or x=\\frac{\\sqrt{5}}{3}-1, and x=1+\\frac{\\sqrt{5}}{3} or x=-1-\\frac{\\sqrt{5}}{3}. Therefore, there are 4 common points between the line and the curve. The final answer is: 4" }, { "text": "If a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{8}=1$ is $(4 , 0)$, then what is the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;Expression(G) = (-y^2/8 + x^2/a^2 = 1);Coordinate(OneOf(Focus(G))) = (4, 0)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*x", "fact_spans": "[[[1, 43], [60, 63]], [[4, 43]], [[1, 43]], [[1, 58]]]", "query_spans": "[[[60, 71]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $y^{2}=-2 x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -2*x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(-1/2,0)", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "" }, { "text": "Let $F_{1}$ and $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)$, respectively. A line passing through $F_{1}$ intersects the left branch of hyperbola $C$ at points $A$ and $B$, with $|A F_{2}|=6$, $|B F_{2}|=10$, $|A B|=8$. Then the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;A: Point;F2: Point;B: 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(F1, G);Intersection(G, LeftPart(C)) = {A, B};Abs(LineSegmentOf(A, F2)) = 6;Abs(LineSegmentOf(B, F2)) = 10;Abs(LineSegmentOf(A, B)) = 8", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(10)/2", "fact_spans": "[[[19, 80], [98, 104], [158, 164]], [[27, 80]], [[27, 80]], [[95, 97]], [[107, 110]], [[9, 16]], [[111, 114]], [[1, 8], [1, 8]], [[27, 80]], [[27, 80]], [[19, 80]], [[1, 85]], [[1, 85]], [[86, 97]], [[95, 116]], [[118, 131]], [[132, 146]], [[147, 156]]]", "query_spans": "[[[158, 170]]]", "process": "By the definition of a hyperbola, |AF₂| - |AF₁| = |BF₂| - |BF₁|. Given |AF₁| + |BF₁| = |AB|, we obtain |AF₁| = 2, |BF₁| = 6, so a = (6 - 2)/2 = 2, thus a₁, a₁ = 1/2·|2| = 6, |BF₂| = 10, |AB| = 8, hence ∠A is a right angle. Therefore, the eccentricity of hyperbola C is √10." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{m}-y^{2}=1$ $(m>0)$ has an asymptote $\\sqrt{3} x+m y=0$, then the focal distance of $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2 + x^2/m = 1);m: Number;m>0;Expression(OneOf(Asymptote(C))) = (m*y + sqrt(3)*x = 0)", "query_expressions": "FocalLength(C)", "answer_expressions": "4", "fact_spans": "[[[2, 40], [67, 70]], [[2, 40]], [[10, 40]], [[10, 40]], [[2, 65]]]", "query_spans": "[[[67, 75]]]", "process": "From the asymptote equation $\\sqrt{3}x + my = 0$, simplifying yields $y = -\\frac{\\sqrt{3}}{m}x$, that is, $\\frac{b}{a} = \\frac{\\sqrt{3}}{m}$; squaring both sides gives $\\frac{b^{2}}{a^{2}} = \\frac{3}{m^{2}}$. Also, in the hyperbola, $a^{2} = m$, $b^{2} = 1$, hence $\\frac{3}{m^{2}} = \\frac{1}{m}$, solving yields $m = 3$, $m = 0$ (discarded), $c^{2} = a^{2} + b^{2} = 3 + 1 = 4 \\Rightarrow c = 2$, thus the focal length $2c = 4$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. If a circle centered at the right focus $F_{2}(c, 0)$ $(c>0)$ with radius $c$ intersects the right branch of the hyperbola at a point $M$, and the line $F_{1} M$ is tangent to the circle, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;c:Number;H: Circle;F1: Point;M: Point;F2: Point;a>0;b>0;c>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F2) = (c, 0);LeftFocus(G) = F1;RightFocus(G) = F2;Center(H)=F2;Radius(H)=c;OneOf(Intersection(H,RightPart(G)))=M;IsTangent(LineOf(F1,M),H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)+1", "fact_spans": "[[[2, 58], [59, 60], [120, 123], [156, 159]], [[5, 58]], [[5, 58]], [[114, 117]], [[118, 119], [151, 152]], [[68, 75]], [[132, 135]], [[89, 107], [76, 83]], [[5, 58]], [[5, 58]], [[89, 107]], [[2, 58]], [[89, 107]], [[59, 83]], [[59, 83]], [[85, 119]], [[111, 119]], [[118, 135]], [[137, 154]]]", "query_spans": "[[[156, 165]]]", "process": "Problem Analysis: Since the right focus $ F_{2}(c,0) $ ($ c>0 $) is taken as the center to draw a circle with radius $ c $, intersecting the right branch of the hyperbola at a point $ M $, and the line $ F_{1}M $ is exactly tangent to the circle, it follows that $ MF_{1} \\perp MF_{2} $, $ |MF_{2}| = c $. By the Pythagorean theorem, $ |MF_{1}| = \\sqrt{3}c $. From the definition of the hyperbola, $ |MF_{1}| - |MF_{2}| = 2a = \\sqrt{3}c - c $. The eccentricity is $ e = \\frac{c}{a} = \\frac{2}{\\sqrt{3}-1} = \\sqrt{3}+1 $." }, { "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 $R$ and $Q$, respectively. Then the value of $\\overrightarrow{P R} \\cdot \\overrightarrow{P Q}$ is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;R: Point;Q: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);PointOnCurve(P, H);IsParallel(H, RealAxis(G));L1: Line;L2: Line;Asymptote(G) = {L1, L2};Intersection(H, L1) = R;Intersection(H, L2) = Q", "query_expressions": "DotProduct(VectorOf(P, R), VectorOf(P, Q))", "answer_expressions": "a^2", "fact_spans": "[[[1, 57]], [[4, 57]], [[4, 57]], [[73, 75]], [[62, 65]], [[82, 85]], [[86, 89]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[0, 75]], [[0, 75]], [], [], [[0, 81]], [[0, 91]], [[0, 91]]]", "query_spans": "[[[93, 146]]]", "process": "Let p(x,y), then M(\\frac{a}{b}y,y), N(-\\frac{a}{b}y,y), thus \\overrightarrow{PM}\\cdot\\overrightarrow{PN}=(\\frac{a}{b}y-x,0)\\cdot(-\\frac{a}{b}y-x,0)=(\\frac{a}{b}y-x)(-\\frac{a}{b}y-x)=x^{2}-\\frac{a^{2}}{b^{2}}y^{2}=\\frac{1}{b^{2}}(b^{2}x^{2}-a^{2}y^{2})=\\frac{a2b^{2}}{b^{2}}=a^{2}," }, { "text": "Construct a circle centered at the right focus $F_{2}$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ such that the circle passes through the center of the ellipse and intersects the ellipse at points $M$ and $N$. If the line $M F_{1}$ passing through the left focus $F_{1}$ of the ellipse is tangent to the circle $F_{2}$, then what is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Circle;F2: Point;M: Point;F1: Point;N: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F2;Center(H) = F2;PointOnCurve(Center(G), H);Intersection(H, G) = {M, N};LeftFocus(G) = F1;PointOnCurve(F1, LineOf(M, F1));IsTangent(LineOf(M, F1), H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3) - 1", "fact_spans": "[[[1, 53], [76, 78], [82, 84], [97, 99], [136, 138]], [[3, 53]], [[3, 53]], [[70, 71], [74, 75], [122, 130]], [[57, 64]], [[85, 88]], [[102, 109]], [[89, 92]], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 64]], [[0, 71]], [[74, 80]], [[74, 94]], [[97, 109]], [[96, 121]], [[110, 133]]]", "query_spans": "[[[136, 144]]]", "process": "By the given condition, point M is the point of tangency, $ MF_{2} = c $, $ MF_{1} \\perp MF_{2} $. Also, $ \\because F_{1}F_{2} = 2c $, $ \\therefore \\angle MF_{1}F_{2} = 30^{\\circ} $, $ \\therefore |MF_{1}| = \\sqrt{3}c $. According to the definition of ellipse: $ |MF_{1}| + |MF_{2}| = 2a $, $ \\therefore |MF_{2}| = 2a - c $. $ \\therefore 2a - c = \\sqrt{3}c $, i.e., $ a = \\frac{\\sqrt{3}+1}{2}c $. $ \\therefore e = \\frac{c}{a} = \\frac{2}{\\sqrt{3}+1} = \\sqrt{3} - 1 $, hence option C is selected." }, { "text": "If the line $ l $ with slope $ 1 $ passing through the point $ M(2,0) $ intersects the parabola $ y^{2}=4x $ at points $ A $ and $ B $, then $ |AB| = $?", "fact_expressions": "l: Line;G: Parabola;M: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (2, 0);PointOnCurve(M, l);Slope(l) = 1;Intersection(l, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "4*sqrt(6)", "fact_spans": "[[[20, 25]], [[26, 40]], [[3, 12]], [[42, 45]], [[46, 49]], [[26, 40]], [[3, 12]], [[1, 25]], [[13, 25]], [[20, 51]]]", "query_spans": "[[[53, 62]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then the equation of line l is y = x - 2. From \\begin{cases} y = x - 2 \\\\ y^{2} = 4x \\end{cases}, we obtain x^{2} - 8x + 4 = 0. Therefore, x_{1} + x_{2} = 8, x_{1} \\cdot x_{2} = 4. Hence, |AB| = \\sqrt{2} \\times \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1} \\cdot x_{2}} = \\sqrt{2} \\times \\sqrt{64 - 16} = 4\\sqrt{6}." }, { "text": "The equation of the locus of the midpoint of the line segment joining point $P(4,-2)$ and an arbitrary point on the circle $x^{2}+y^{2}=4$?", "fact_expressions": "P: Point;Coordinate(P) = (4, -2);G: Circle;Expression(G) = (x^2 + y^2 = 4);Z: Point;PointOnCurve(Z, G)", "query_expressions": "LocusEquation(MidPoint(LineSegmentOf(P, Z)))", "answer_expressions": "(x - 2)^2 + (y + 1)^2 = 1", "fact_spans": "[[[0, 10]], [[0, 10]], [[11, 27]], [[11, 27]], [], [[11, 31]]]", "query_spans": "[[[0, 45]]]", "process": "" }, { "text": "Let $O$ be the origin, the line $x=a$ intersects the two asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) at points $D$ and $E$, respectively. If the area of $\\triangle O D E$ is $4$, then the minimum value of the focal distance of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Line;O: Origin;D: Point;E: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x = a);L1:Line;L2:Line;Asymptote(C)={L1,L2};Intersection(G,L1)=D;Intersection(G,L2)=E;Area(TriangleOf(O, D, E)) = 4", "query_expressions": "Min(FocalLength(C))", "answer_expressions": "4*sqrt(2)", "fact_spans": "[[[18, 79], [126, 129]], [[26, 79]], [[26, 79]], [[10, 17]], [[1, 4]], [[89, 92]], [[93, 96]], [[26, 79]], [[26, 79]], [[18, 79]], [[10, 17]], [], [], [[18, 85]], [[10, 98]], [[10, 98]], [[100, 124]]]", "query_spans": "[[[126, 138]]]", "process": "Because $ C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) $, the asymptotes of the hyperbola are $ y=\\pm\\frac{b}{a}x $. Solving simultaneously with the line $ x=a $ gives the coordinates of points $ D $ and $ E $, so $ |ED| $ can be found. Given that the area of $ \\triangle ODE $ is 4, the value of $ ab $ can be obtained. Using $ 2c=2\\sqrt{a^{2}+b^{2}} $ and the AM-GM inequality, the answer can be found.\n\n$ \\because C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) $\n\n$ \\therefore $ The asymptotes of the hyperbola are $ y=\\pm\\frac{b}{a}x $\n\n$ \\because $ The line $ x=a $ intersects the two asymptotes of the hyperbola $ C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) $ at points $ D $ and $ E $ respectively. Without loss of generality, assume $ D $ is in the first quadrant and $ E $ is in the fourth quadrant. Solving the system\n$$\n\\begin{cases}\nx=a \\\\\ny=\\frac{b}{a}x\n\\end{cases}\n$$\ngives\n$$\n\\begin{cases}\nx=a \\\\\ny=b\n\\end{cases}\n$$\nso $ D(a,b) $\n\n$ \\therefore |ED|=2b $\n\n$ \\therefore $ The area of $ \\triangle ODE $ is: $ S_{\\triangle ODE}=\\frac{1}{2}a\\times2b=ab=4 $\n\n$ \\because $ Hyperbola $ C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0,b>0) $\n\n$ \\therefore $ Its focal distance is $ 2c=2\\sqrt{a^{2}+b^{2}}\\geqslant2\\sqrt{2ab}=2\\sqrt{8}=4\\sqrt{2} $\n\nEquality holds if and only if $ a=b=2 $\n\n$ \\therefore $ The minimum focal distance of $ C $ is $ 4\\sqrt{2} $" }, { "text": "Let $P$ be a moving point on the curve $y^{2}=4(x-1)$. What is 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 $y$-axis?", "fact_expressions": "G: Curve;A: Point;P: Point;Expression(G) = (y^2 = 4*(x - 1));Coordinate(A) = (0, 1);PointOnCurve(P, G)", "query_expressions": "Min(Distance(P, A) + Distance(P, yAxis))", "answer_expressions": "sqrt(5)", "fact_spans": "[[[5, 21]], [[34, 45]], [[1, 4], [29, 33], [49, 53]], [[5, 21]], [[34, 45]], [[1, 27]]]", "query_spans": "[[[29, 69]]]", "process": "Since the equation of the parabola is y^{2}=4(x-1), the focus of the parabola has coordinates F(2,0) and the directrix has equation: x=0, as shown in the figure. By the definition of the parabola, PD=PF, so PA+PD=PA+PF\\geqslantAF, with equality when A, P, F are collinear. Therefore, 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 y-axis is \\sqrt{5}." }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{100}+\\frac{y^{2}}{36}=1$. If the distance from $P$ to the right directrix of the ellipse is $\\frac{17}{2}$, then what is the distance from point $P$ to the left focus?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/100 + y^2/36 = 1);P: Point;PointOnCurve(P, G);Distance(P, RightDirectrix(G)) = 17/2", "query_expressions": "Distance(P, LeftFocus(G))", "answer_expressions": "66/5", "fact_spans": "[[[6, 46], [55, 57]], [[6, 46]], [[2, 5], [51, 54], [81, 85]], [[2, 50]], [[51, 78]]]", "query_spans": "[[[55, 94]]]", "process": "P is a point on the ellipse \\frac{x^2}{100} + \\frac{y^{2}}{36} = 1. Let the two foci of the ellipse be F_{1}, F_{2}. If the distance from P to the right directrix of the ellipse is \\frac{17}{2}, then \\frac{|PF_{1}|}{\\frac{17}{2}} = e = \\frac{c}{a} = \\frac{\\sqrt{100-36}}{10} = \\frac{4}{5}. Solving gives |PF_{1}| = \\frac{34}{5}. By the definition of the ellipse, we get: |PF_{2}| = 2a - |PF_{1}| = 20 - \\frac{34}{5} = \\frac{66}{5}." }, { "text": "Given the ellipse $C$: $\\frac{y^{2}}{5}+x^{2}=1$ with right vertex $A$, and let $P$ be a point on $C$, then the maximum value of $P A$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2 + y^2/5 = 1);A: Point;RightVertex(C) = A;P: Point;PointOnCurve(P, C) = True", "query_expressions": "Max(LineSegmentOf(P, A))", "answer_expressions": "5/2", "fact_spans": "[[[2, 33], [46, 49]], [[2, 33]], [[38, 41]], [[2, 41]], [[42, 45]], [[42, 52]]]", "query_spans": "[[[54, 65]]]", "process": "The ellipse $ C: \\frac{y^{2}}{5} + x^{2} = 1 $ has its right vertex at $ A(1,0) $. Let point $ P(x_{0},y_{0}) $, then $ \\frac{y_{0}^{2}}{5} + x_{0}^{2} = 1 $, i.e., $ y_{0}^{2} = 5 - 5x_{0}^{2} $, and $ -1 \\leqslant x_{0} \\leqslant 1 $. Thus, $ |PA| = \\sqrt{(x_{0}-1)^{2} + y_{0}^{2}} = \\sqrt{(x_{0}-1)^{2} + 5 - 5x_{0}^{2}} = \\sqrt{-4(x_{0} + \\frac{1}{4})^{2} + \\frac{25}{4}} $. Since $ -1 \\leqslant x_{0} \\leqslant 1 $, when $ x_{0} = -\\frac{1}{4} $, $ |PA|_{\\max} = \\frac{5}{2} $. Therefore, the maximum value of $ PA $ is $ \\frac{5}{2} $." }, { "text": "The foci of a hyperbola lie on the $x$-axis, its center is at the origin, one asymptote is $y = \\sqrt{2}x$, and the point $P(1, -2)$ lies on the hyperbola. What is the standard equation of the hyperbola?", "fact_expressions": "G: Hyperbola;P: Point;O: Origin;Center(G)=O;Coordinate(P) = (1, -2);PointOnCurve(Focus(G), xAxis);PointOnCurve(P, G);Expression(OneOf(Asymptote(G)))=(y=sqrt(2)*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/2-y^2/4=1", "fact_spans": "[[[0, 3], [51, 54], [57, 60]], [[40, 50]], [[16, 18]], [[0, 18]], [[40, 50]], [[0, 12]], [[40, 55]], [[0, 39]]]", "query_spans": "[[[57, 67]]]", "process": "" }, { "text": "The coordinates of the point on the parabola $y=x^{2}$ that is closest to the line $2x-y=4$ are?", "fact_expressions": "G: Parabola;H: Line;Expression(G) = (y = x^2);Expression(H) = (2*x - y = 4);Z: Point;PointOnCurve(Z, G);WhenMin(Distance(Z, H))", "query_expressions": "Coordinate(Z)", "answer_expressions": "(1, 1)", "fact_spans": "[[[0, 12]], [[13, 24]], [[0, 12]], [[13, 24]], [[29, 30]], [[0, 30]], [[0, 30]]]", "query_spans": "[[[29, 35]]]", "process": "Test Analysis: Let the line parallel to the line 2x - y = 4 be 2x - y = b. Solving 2x - y = b and y = x^{2} simultaneously by eliminating y gives x^{2} - 2x + b = 0. When 4 = (-2)^{2} - 4b = 0, that is, b = 1, the line 2x - y = b is tangent to y = x^{2}. Substituting b = 1 into x^{2} - 2x + b = 0 yields x = 1. At this time, the point of tangency is (1,1). By combining numerical and geometric analysis, it can be seen that the point (1,1) on the parabola y = x^{2} is the closest point to the line 2x - y = 4. The line and the conic section are tangent." }, { "text": "If the equation $\\frac{x^{2}}{m}+\\frac{y^{2}}{m^{2}-2}=1$ represents an ellipse with foci on the $x$-axis, then the range of real values for $m$ is?", "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (y^2/(m^2 - 2) + x^2/m = 1);PointOnCurve(Focus(G), xAxis)", "query_expressions": "Range(m)", "answer_expressions": "(sqrt(2),2)", "fact_spans": "[[[55, 57]], [[59, 64]], [[1, 57]], [[46, 57]]]", "query_spans": "[[[59, 71]]]", "process": "\\because the equation \\frac{x^{2}}{m}+\\frac{y^{2}}{m^{2}-2}=1 represents an ellipse with foci on the x-axis, \\therefore m>m^{2}-2>0, solving yields \\sqrt{2}0$ and $\\lambda<0$, and find the eccentricity. Transform $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=\\lambda$ into standard form: $\\frac{x^{2}}{9\\lambda}-\\frac{y^{2}}{16\\lambda}=1$. \n① When $\\lambda>0$, the hyperbola represents a hyperbola with foci on the $x$-axis. At this time, $a^{2}=9\\lambda$, $b^{2}=16\\lambda$, so $c^{2}=a^{2}+b^{2}=25\\lambda$. The eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{\\frac{25\\lambda}{9\\lambda}}=\\frac{5}{3}$. \n② When $\\lambda<0$, the hyperbola represents a hyperbola with foci on the $y$-axis. At this time, $a^{2}=-16\\lambda$, $b^{2}=-9\\lambda$, so $c^{2}=a^{2}+b^{2}=-25\\lambda$. The eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{\\frac{-25\\lambda}{-16\\lambda}}=\\frac{5}{4}$. Therefore, the eccentricity of the hyperbola is $\\frac{5}{3}$ or $\\frac{5}{4}$." }, { "text": "If the line $y = kx + 2$ and the parabola $y^2 = x$ have only one intersection point, then the value of the real number $k$ is?", "fact_expressions": "H: Line;Expression(H) = (y = k*x + 2);G: Parabola;Expression(G) = (y^2 = x);NumIntersection(H, G) = 1;k: Real", "query_expressions": "k", "answer_expressions": "{0,1/8}", "fact_spans": "[[[1, 12]], [[1, 12]], [[13, 25]], [[13, 25]], [[1, 31]], [[33, 38]]]", "query_spans": "[[[33, 42]]]", "process": "Solving the system of the line equation and the parabola equation gives: k^{2}x^{2}+(4k-1)x+4=0\\textcircled{1} If k=0, then x=4, which satisfies the condition; \\textcircled{2} If k\\neq0, then a=(4k-1)^{2}-16k^{2}=0, solving yields k=\\frac{1}{8}. In conclusion, k=0 or \\frac{1}{8}." }, { "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{12}=1(a>0)$ has an inclination angle of $60^{\\circ}$, then $a$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/12 + x^2/a^2 = 1);a: Number;a>0;Inclination(OneOf(Asymptote(G))) = ApplyUnit(60, degree)", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[1, 49]], [[1, 49]], [[74, 77]], [[4, 49]], [[1, 72]]]", "query_spans": "[[[74, 79]]]", "process": "" }, { "text": "The focal distance of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ is? The equations of the asymptotes are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/2 - y^2 = 1)", "query_expressions": "FocalLength(G);Expression(Asymptote(G))", "answer_expressions": "2*sqrt(3)\ny=pm*(sqrt(2)/2)*x", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 33]], [[0, 40]]]", "process": "From the problem, we have: $ a = \\sqrt{2}, b = 1, c = \\sqrt{a^{2} + b^{2}} = \\sqrt{2 + 1} = \\sqrt{3} $, $ \\therefore $ the focal distance is $ 2c = 2\\sqrt{3} $, and the asymptotes are $ y = \\pm \\frac{b}{a}x = \\pm \\frac{\\sqrt{2}}{2}x $." }, { "text": "On the hyperbola $4 x^{2}-y^{2}+64=0$, the distance from a point $P$ to one of its foci is equal to $1$. Then, the distance from point $P$ to the other focus is equal to?", "fact_expressions": "G: Hyperbola;P: Point;F1:Point;F2:Point;Expression(G) = (4*x^2 - y^2 + 64 = 0);PointOnCurve(P, G);Distance(P,F1) = 1;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)", "query_expressions": "Distance(P,F2)", "answer_expressions": "17", "fact_spans": "[[[0, 23], [30, 31]], [[26, 29], [47, 51]], [], [], [[0, 23]], [[0, 29]], [[26, 44]], [[30, 36]], [[30, 57]], [[30, 57]]]", "query_spans": "[[[30, 63]]]", "process": "" }, { "text": "The hyperbola $E$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. Let $P$ be a point on the left branch of $E$ such that $|P F_{1}|=|F_{1} F_{2}|$. The line $P F_{2}$ is tangent to the circle $x^{2}+y^{2}=a^{2}$. Find the eccentricity of $E$?", "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;P: Point;PointOnCurve(P, LeftPart(E));Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(F1, F2));G: Circle;Expression(G) = (x^2 + y^2 = a^2);IsTangent(LineOf(P, F2), G)", "query_expressions": "Eccentricity(E)", "answer_expressions": "5/3", "fact_spans": "[[[0, 61], [90, 93], [162, 165]], [[0, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[8, 61]], [[70, 77]], [[78, 85]], [[0, 85]], [[0, 85]], [[86, 89]], [[86, 98]], [[100, 125]], [[138, 158]], [[138, 158]], [[126, 160]]]", "query_spans": "[[[162, 171]]]", "process": "Let the line $ PF_{1} $ be tangent to the circle $ x^{2} + y^{2} = a^{2} $ at point $ M $, then $ |OM| = a $, $ OM \\perp PF_{1} $. Take the midpoint $ N $ of $ PF_{1} $, and connect $ NF_{2} $. Since $ |PF_{1}| = |F_{1}F_{2}| = 2c $, then $ NF_{2} \\perp PF_{1} $, $ |NP| = |NF_{1}| $. Given $ |NF_{2}| = 2|OM| = 2a $, then $ |NP| = 2b $, so $ |PF_{1}| = 4b $. By the definition of the hyperbola, we have $ |PF_{1}| - |PF_{2}| = 2a $. Thus $ 4b - 2c = 2a $, i.e., $ 2b = c + a $. Then $ 4b^{2} = (c + a)^{2} $, i.e., $ 4(c^{2} - a^{2}) = (c + a)^{2} $, $ 4(c - a) = c + a $, so $ 3c = 5a $. Hence $ e = \\frac{5}{3} $." }, { "text": "Let the focus of the parabola $x^{2}=4 y$ be $F$. A line $l$ passing through the point $P(1,4)$ intersects the parabola at points $A$ and $B$, and point $P$ is exactly the midpoint of $AB$. Then $|\\overrightarrow{A F}|+| \\overrightarrow{B F} |$=?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);F: Point;Focus(G) = F;P: Point;Coordinate(P) = (1, 4);l: Line;PointOnCurve(P,l) = True;Intersection(l,G) = {A,B};A: Point;B: Point;MidPoint(LineSegmentOf(A,B)) = P", "query_expressions": "Abs(VectorOf(A,F))+Abs(VectorOf(B,F))", "answer_expressions": "10", "fact_spans": "[[[1, 15], [41, 44]], [[1, 15]], [[19, 22]], [[1, 22]], [[25, 34], [58, 62]], [[25, 34]], [[35, 40]], [[23, 40]], [[35, 56]], [[47, 50]], [[51, 54]], [[58, 72]]]", "query_spans": "[[[75, 126]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). Since point P(1,4) is exactly the midpoint of AB, we have y_{1}+y_{2}=8. By the definition of the parabola, |\\overrightarrow{AF}|+|\\overrightarrow{BF}|=y_{1}+y_{2}+2=10." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, a line $l$ passing through the point $P(-1, \\frac{1}{2})$ intersects the ellipse $C$ at points $A$ and $B$. If point $P$ is exactly the midpoint of chord $AB$, then what is the slope of line $l$?", "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;P: Point;Expression(C) = (x^2/4 + y^2/3 = 1);Coordinate(P) = (-1, 1/2);PointOnCurve(P, l);Intersection(l, C) = {A, B};IsChordOf(LineSegmentOf(A, B), C);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Slope(l)", "answer_expressions": "3/2", "fact_spans": "[[[68, 73], [108, 113]], [[2, 44], [74, 79]], [[81, 84]], [[85, 88]], [[46, 67], [92, 96]], [[2, 44]], [[46, 67]], [[45, 73]], [[68, 90]], [[74, 104]], [[92, 106]]]", "query_spans": "[[[108, 117]]]", "process": "Let the coordinates of A and B be (x_{1},y_{1}), (x_{2},y_{2}) respectively. Then \\frac{x_{1}2}{4}\\frac{2}{3}=1, \\frac{x_{2}2}{4}. Subtracting these two equations gives: \\underline{3}. Rearranging yields: \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{3(x_{1}+x_{2})}{4(y_{1}+y_{2})}. Since point P is exactly the midpoint of chord AB, we know x_{1}+_{2}=-2, y_{1}+y\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{3\\times(-2)}{4}=\\frac{3}{2}, that is, the slope of line l is \\frac{3}{2}." }, { "text": "The center of a circle lies on the parabola $y^{2}=16 x$, and the circle passes through the vertex and the focus of the parabola. If the center is in the first quadrant, then the standard equation of the circle is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 16*x);H: Circle;PointOnCurve(Center(H), G);PointOnCurve(Vertex(G), H);PointOnCurve(Focus(G), H);Quadrant(Center(H)) = 1", "query_expressions": "Expression(H)", "answer_expressions": "(x-2)^2 + (y - 4*sqrt(2))^2 = 36", "fact_spans": "[[[7, 22], [29, 32]], [[7, 22]], [[2, 3], [26, 27], [50, 51]], [[2, 23]], [[26, 35]], [[26, 38]], [[26, 47]]]", "query_spans": "[[[50, 58]]]", "process": "Problem Analysis: The vertex and focus of the parabola y^{2}=16x are (0,0) and (4,0), respectively. Therefore, the center of the circle lies on the line x=2. Substituting x=2 into y^{2}=16x gives y=4\\sqrt{2}, so the center is (2,4\\sqrt{2}). Thus, the standard equation of the circle is (x-2)^{2}+(y-4\\sqrt{2})^{2}=36." }, { "text": "Given the line $l$: $x - y + 1 = 0$ and the parabola $C$: $x^{2} = 2y$ intersect at points $A$ and $B$. Point $P$ is a moving point on line $l$, and points $M$, $N$ are two moving points on parabola $C$. If $\\overrightarrow{MN} \\parallel \\overrightarrow{AB}$, $|\\overrightarrow{MN}| < |\\overrightarrow{AB}|$, then the maximum area of $\\Delta PMN$ is?", "fact_expressions": "l: Line;Expression(l) = (x - y + 1 = 0);C: Parabola;Expression(C) = (x^2 = 2*y);A: Point;B: Point;Intersection(l, C) = {A, B};P: Point;PointOnCurve(P, l);M: Point;N: Point;PointOnCurve(M, C);PointOnCurve(N, C);IsParallel(VectorOf(M, N), VectorOf(A, B));Abs(VectorOf(M, N)) < Abs(VectorOf(A, B))", "query_expressions": "Max(Area(TriangleOf(P, M, N)))", "answer_expressions": "1", "fact_spans": "[[[2, 18], [55, 60]], [[2, 18]], [[19, 38], [75, 81]], [[19, 38]], [[40, 43]], [[44, 47]], [[2, 49]], [[50, 54]], [[50, 64]], [[65, 68]], [[71, 74]], [[65, 86]], [[65, 86]], [[88, 133]], [[134, 179]]]", "query_spans": "[[[181, 202]]]", "process": "Problem Analysis: From the given conditions, when line MN passes through the origin, the area of APMN is maximized. Therefore, the equation of line MN is x - y = 0. The distance d from point P to line MN is d = \\frac{|1-0|}{\\sqrt{1^{2}+(-1)^{2}}} = \\frac{\\sqrt{2}}{2}. From \\begin{cases}x-y=0\\\\x^{2}=2y\\end{cases}, we obtain: \\begin{cases}x=0\\\\y=0\\end{cases} or \\begin{cases}x=2\\\\y=2\\end{cases}, so N(0,0), M(2,2). Thus |MN| = \\sqrt{(2-0)^{2}+(2-0)^{2}} = 2\\sqrt{2}. Therefore, the maximum area of APMN is \\frac{1}{2}\\cdot|MN|\\cdot d = \\frac{1}{2}\\times2\\sqrt{2}\\times\\frac{\\sqrt{2}}{2} = 1. So the answer should be: 1" }, { "text": "Let the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ have a length of $2$ for its imaginary axis and a focal distance of $2 \\sqrt{3}$. Then the equations of the asymptotes of the hyperbola are?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Length(ImageinaryAxis(G)) = 2;FocalLength(G) = 2*sqrt(3)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(sqrt(2)/2)*x", "fact_spans": "[[[1, 57], [83, 86]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[1, 81]]]", "query_spans": "[[[83, 94]]]", "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) has an imaginary axis length of $2$ and a focal distance of $2\\sqrt{3}$, so $b=1$, $c=\\sqrt{3}$, $\\therefore a^{2}=c^{2}-b^{2}=2$, thus the hyperbola equation is $\\frac{x^{2}}{2}-y^{2}=1$, and the asymptotes are $y=\\pm\\frac{\\sqrt{2}}{2}x$." }, { "text": "From the left focus $F$ of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$, draw a tangent line $F P$ to the circle $x^{2}+y^{2}=4$, intersecting the right branch of the hyperbola at point $P$, $T$ being the point of tangency, $N$ the midpoint of segment $F P$, and $O$ the origin. Then $|N O|-|N T|$=?", "fact_expressions": "G: Hyperbola;H: Circle;P: Point;F: Point;N: Point;O: Origin;T: Point;Expression(G) = (x^2/4 - y^2/12 = 1);Expression(H) = (x^2 + y^2 = 4);LeftFocus(G)=F;TangentOfPoint(F,H)=LineOf(F,P);Intersection(LineOf(F,P),RightPart(G))=P;TangentPoint(LineOf(F,P),H)=T;MidPoint(LineSegmentOf(F,P)) = N", "query_expressions": "Abs(LineSegmentOf(N, O)) - Abs(LineSegmentOf(N, T))", "answer_expressions": "2*sqrt(3)-2", "fact_spans": "[[[1, 40], [73, 76]], [[48, 64]], [[79, 83]], [[44, 47]], [[91, 94]], [[106, 109]], [[84, 87]], [[1, 40]], [[48, 64]], [[1, 47]], [[0, 72]], [[67, 83]], [[0, 90]], [[91, 105]]]", "query_spans": "[[[116, 131]]]", "process": "Let the right focus be F', then NO-NT=\\frac{1}{2}PF-NT=\\frac{1}{2}(PF-2a)-NT=-a+NF-NT=-a+FT=-2+\\sqrt{16-4}=-2+2\\sqrt{3}" }, { "text": "Given that $F$ is the right focus of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$, and the line $y=k x+1$ intersects the ellipse $C$ at points $A$ and $B$. If $A F \\perp B F$, then the value of the real number $k$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);F: Point;RightFocus(C) = F;G: Line;Expression(G) = (y = k*x + 1);k: Real;A: Point;B: Point;Intersection(G, C) = {A, B};IsPerpendicular(LineSegmentOf(A, F), LineSegmentOf(B, F))", "query_expressions": "k", "answer_expressions": "-1/2", "fact_spans": "[[[6, 38], [55, 60]], [[6, 38]], [[2, 5]], [[2, 42]], [[43, 54]], [[43, 54]], [[89, 94]], [[62, 65]], [[66, 69]], [[43, 71]], [[72, 87]]]", "query_spans": "[[[89, 98]]]", "process": "According to the problem, solve the system of equations consisting of the line and the ellipse: \n\\begin{cases}y=kx+1\\\\\\frac{x^{2}}{2}+y^{2}=1\\end{cases}, \neliminate $ y $ and simplify to obtain $ (2k^{2}+1)x^{2}+4kx=0 $, \nsolve to get $ x=0 $ or $ x=\\frac{-4k}{2k^{2}+1} $, \nwithout loss of generality, take $ x_{A}=0 $, then $ y_{A}=1 $, $ x_{B}=\\frac{-4k}{2k^{2}+1} $, \n$ y_{B}=k\\cdot\\frac{-4k}{2k^{2}+1}+1=\\frac{1-2k^{2}}{2k^{2}+1} $, \nso $ A(0,1) $, $ B\\left(\\frac{-4k}{2k^{2}+1},\\frac{1-2k^{2}}{2k^{2}+1}\\right) $, \nalso $ F(1,0) $, so $ k_{AF}=-1 $, \nsince $ AF\\bot BF $, then $ k_{BF}=1 $, \ni.e. $ \\frac{\\frac{1-2k^{2}}{2k^{2}+1}}{\\frac{-4k}{2k^{2}+1}-1}=1 $, \nthus $ \\frac{1-2k^{2}}{2k^{2}+1}=\\frac{-4k}{2k^{2}+1}-1 $, \nso $ 1-2k^{2}=-4k-(2k^{2}+1) $, \nsolve to get $ k=-\\frac{1}{2} $. \nThe answer is: $ \\frac{1}{2} $" }, { "text": "$P$ is a point on the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{9}=1$, $F_{1}$, $F_{2}$ are its left and right foci, and $|P F_{1}|=5$, then $|P F_{2}|$=?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/4 - y^2/9 = 1);PointOnCurve(P, G);LeftFocus(G)=F1;RightFocus(G)=F2;Abs(LineSegmentOf(P, F1)) = 5", "query_expressions": "Abs(LineSegmentOf(P, F2))", "answer_expressions": "9", "fact_spans": "[[[4, 42], [63, 64]], [[0, 3]], [[47, 54]], [[55, 62]], [[4, 42]], [[0, 46]], [[47, 70]], [[47, 70]], [[72, 85]]]", "query_spans": "[[[88, 101]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a+t}+\\frac{y^{2}}{a}=1$ ($a>0$, $t>0$), the distance between the two foci is $2$. What is the sum of the distances from any point on the ellipse to the two foci? (Express in terms of $a$)", "fact_expressions": "C: Ellipse;a: Number;t: Number;a>0;t>0;Expression(C) = (x^2/(a + t) + y^2/a = 1);F1:Point;F2:Point;Focus(C)={F1,F2};Distance(F1,F2)=2;P:Point;PointOnCurve(P,C)", "query_expressions": "Distance(P,F1)+Distance(P,F2)", "answer_expressions": "2*sqrt(a+1)", "fact_spans": "[[[2, 56], [71, 73]], [[9, 56]], [[9, 56]], [[9, 56]], [[9, 56]], [[2, 56]], [], [], [[2, 61]], [[2, 69]], [], [[71, 79]]]", "query_spans": "[[[71, 91]]]", "process": "Since the distance between the two foci of $\\frac{x^{2}}{a+t}+\\frac{y^{2}}{a}=1$ $(a>0,t>0)$ is 2, we have $2\\sqrt{(a+t)-a}=2$, so $t=1$. Therefore, the sum of the distances from any point on the ellipse to the two foci is $2\\sqrt{a+1}$." }, { "text": "Given the line $ l_{1} $: $ x - \\sqrt{3} y + b = 0 $ ($ b > 3 $), the parabola $ C $: $ y^{2} = 4x $ has focus $ F $ and directrix $ l_{2} $. Let $ A $ be a point on the parabola $ C $, and let the distances from $ A $ to $ l_{1} $ and $ l_{2} $ be $ d_{1} $ and $ d_{2} $, respectively. When $ d_{1} + d_{2} $ takes the minimum value, $ d_{1} = 2d_{2} $. Then $ b = $?", "fact_expressions": "l1: Line;b: Number;b>3;Expression(l1) = (x - sqrt(3)*y + b = 0);C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l2: Line;Directrix(C) = l2;A: Point;PointOnCurve(A, C);d1: Number;d2: Number;Distance(A, l1) = d1;Distance(A, l2) = d2;WhenMin(d1+d2);d1 = 2*d2", "query_expressions": "b", "answer_expressions": "7", "fact_spans": "[[[2, 36], [96, 103]], [[13, 36]], [[13, 36]], [[2, 36]], [[37, 56], [81, 87]], [[37, 56]], [[60, 63]], [[37, 63]], [[67, 74], [106, 113]], [[37, 74]], [[77, 80], [92, 95]], [[77, 91]], [[119, 126]], [[129, 136]], [[92, 136]], [[92, 136]], [[137, 156]], [[157, 172]]]", "query_spans": "[[[174, 179]]]", "process": "" }, { "text": "Given that the line $l$ intersects the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$, and the midpoint of $AB$ has coordinates $(\\frac{1}{2}, \\frac{1}{2})$, then the equation of line $l$ is?", "fact_expressions": "l: Line;C: Ellipse;A: Point;B: Point;Expression(C) = (x^2/2 + y^2 = 1);Coordinate(MidPoint(LineSegmentOf(A,B))) = (1/2, 1/2);Intersection(l, C) = {A, B}", "query_expressions": "Expression(l)", "answer_expressions": "2*x+4*y-3=0", "fact_spans": "[[[2, 7], [93, 98]], [[8, 40]], [[41, 44]], [[45, 48]], [[8, 40]], [[52, 91]], [[2, 50]]]", "query_spans": "[[[93, 103]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\frac{x_{1}^{2}}{2}+y_{1}^{2}=1, \\frac{x_{2}^{2}}{2}+y_{2}^{2}=1. Subtracting these two equations and simplifying yields -\\frac{1}{2}=\\frac{y_{1}+y_{2}}{x_{1}+x_{2}}\\cdot\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, -\\frac{1}{2}=\\frac{\\frac{1}{2}}{\\frac{1}{2}}\\cdot\\frac{y_{1}-y_{2}}{x_{1}}-x_{2}, \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{1}{2}. Therefore, the equation of line l is y-\\frac{1}{2}=-\\frac{1}{2}(x-\\frac{1}{2}), 2x+4y-3=0." }, { "text": "If the focal distance of the ellipse $x^{2}+y^{2} \\cos \\theta=1$ $(0<\\theta<\\frac{\\pi}{2})$ is $2$, then what is the eccentricity of this ellipse?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + y^2*Cos(theta) = 1);theta: Number;01$. Thus, $a^{2}=\\frac{1}{\\cos\\theta}$, $b^{2}=1$, $c^{2}=a^{2}-b^{2}=\\frac{1}{\\cos\\theta}-1$. From the given condition, $2c=2$, so $c=1$. From $c^{2}=a^{2}-b^{2}=\\frac{1}{\\cos\\theta}-1$, we obtain $a^{2}=\\frac{1}{\\cos\\theta}=2$. The eccentricity is $e=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}$." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, respectively, and let $M$ be a point on $C$ in the first quadrant. If $|M F_{1}|=|F_{1} F_{2}|$, then the coordinates of point $M$ are?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/9 + y^2/5 = 1);F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;M: Point;PointOnCurve(M, C);Quadrant(M) = 1;Abs(LineSegmentOf(M, F1)) = Abs(LineSegmentOf(F1, F2))", "query_expressions": "Coordinate(M)", "answer_expressions": "(3/2, sqrt(15)/2)", "fact_spans": "[[[19, 61], [72, 75]], [[19, 61]], [[1, 8]], [[9, 16]], [[1, 67]], [[1, 67]], [[68, 71], [114, 118]], [[68, 78]], [[68, 84]], [[87, 112]]]", "query_spans": "[[[114, 123]]]", "process": "" }, { "text": "Let the left and right foci of the ellipse be $F_{1}$ and $F_{2}$, respectively, and the upper vertex be $B_{1}$. If $|B_{1} F_{2}| = |F_{1} F_{2}| = 4$, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;B1: Point;F2: Point;F1: Point;LeftFocus(G) = F1;RightFocus(G) = F2;UpperVertex(G) = B1;Abs(LineSegmentOf(B1,F2))=Abs(LineSegmentOf(F1,F2));Abs(LineSegmentOf(F1,F2))=4", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16+y^2/12=1", "fact_spans": "[[[1, 3], [75, 77]], [[32, 39]], [[20, 27]], [[12, 19]], [[1, 27]], [[1, 27]], [[1, 39]], [[41, 72]], [[41, 72]]]", "query_spans": "[[[75, 84]]]", "process": "According to the Pythagorean theorem, we obtain |B₁F₂| = a. Combining this with c = 2, we can find b and then obtain the standard equation of the ellipse. Let F₂(c, 0), and let the standard equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 (a > b > 0). Since |B₁F₂|² = |OB₁|² + |OF₂|² = b² + c² = a², it follows that |B₁F₂| = a, so a = 4. While |F₁F₂| = 2c = 4, hence c = 2. Therefore, b = \\sqrt{a^{2}-c^{2}} = 2\\sqrt{3}. Thus, the standard equation of the ellipse is \\frac{x^{2}}{16}+\\frac{x^{2}}{12}=1." }, { "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{2}=1$ $(a>0)$ is $F(2,0)$, what is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;F: Point;OneOf(Focus(G)) = F;a>0;Expression(G) = (+ x^2/a^2-y^2/2 = 1);Coordinate(F) = (2, 0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 49], [65, 68]], [[5, 49]], [[55, 63]], [[2, 63]], [[5, 49]], [[2, 49]], [[55, 63]]]", "query_spans": "[[[65, 74]]]", "process": "From the given conditions: c=2, b^{2}=2, from c^{2}=a^{2}+b^{2} \\Rightarrow a=\\sqrt{2}, hence the eccentricity e=\\frac{c}{a}=\\frac{2}{\\sqrt{2}}=\\sqrt{2}" }, { "text": "Given that the three vertices of triangle $A B C$ lie on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $A B \\perp x$-axis, $A C \\| x$-axis, then the maximum value of $\\frac{|A C| \\cdot|A B|}{|B C|^{2}}$ is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: Point;C: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(Vertex(TriangleOf(A,B,C)),G);IsPerpendicular(LineSegmentOf(A,B),xAxis);IsParallel(LineSegmentOf(A,C),xAxis)", "query_expressions": "Max((Abs(LineSegmentOf(A, B))*Abs(LineSegmentOf(A, C)))/Abs(LineSegmentOf(B, C))^2)", "answer_expressions": "1/2", "fact_spans": "[[[19, 71]], [[21, 71]], [[21, 71]], [[74, 87]], [[74, 87]], [[89, 99]], [[21, 71]], [[21, 71]], [[19, 71]], [[2, 72]], [[74, 88]], [[89, 100]]]", "query_spans": "[[[102, 144]]]", "process": "Problem Analysis: Let the coordinates of point A be (x_{0},y_{0}), (x_{0}>0, y_{0}>0). Then, by the symmetry of the ellipse, \\frac{|aC|\\cdot|AB|}{|BC|^{2}} = \\frac{2x_{0}\\cdot2y_{0}}{(2\\sqrt{x_{0}^{2}+y_{0}^{2}})^{2}} = \\frac{x_{0}y_{0}}{x_{0}^{2}} \\leqslant \\frac{1}{2}, with equality holding if and only if x_{0} = y_{0} = \\frac{ab}{\\sqrt{a^{2}+b^{2}}}. Therefore, the answer should be: \\frac{1}{2}" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has eccentricity $e=\\frac{1}{2}$, $A$, $B$ are the left and right vertices of the ellipse, and $P$ is a point on the ellipse different from $A$ and $B$. The angles of inclination of lines $PA$ and $PB$ are $\\alpha$ and $\\beta$ respectively. Then $\\frac{\\cos (\\alpha-\\beta)}{\\cos (\\alpha+\\beta)}$=?", "fact_expressions": "P: Point;A: Point;G: Ellipse;b: Number;a: Number;B: Point;a > b;b > 0;e: Number;alpha: Number;beta: Number;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = e ;e = 1/2;LeftVertex(G) = A;RightVertex(G) = B;PointOnCurve(P, G);Negation(P = A);Negation(P = B);Inclination(LineOf(P, A)) = alpha;Inclination(LineOf(P, B)) = beta", "query_expressions": "Cos(alpha - beta)/Cos(alpha + beta)", "answer_expressions": "1/7", "fact_spans": "[[[94, 97]], [[104, 107], [104, 107]], [[2, 55], [85, 87], [98, 100]], [[4, 54]], [[4, 54]], [[81, 84], [108, 111]], [[4, 54]], [[4, 54]], [[58, 73]], [[132, 140]], [[141, 149]], [[2, 54]], [[2, 73]], [[58, 73]], [[76, 93]], [[76, 93]], [[94, 114]], [[94, 114]], [[94, 114]], [[115, 149]], [[115, 149]]]", "query_spans": "[[[151, 202]]]", "process": "Problem Analysis: From the given conditions, A(-a,0), B(a,0), let P(x,y), then $\\tan\\alpha=\\frac{y}{x+a}$, $\\tan\\beta=\\frac{y}{x-a}$, $\\therefore\\tan\\alpha\\tan\\beta=\\frac{y}{x+a}\\cdot\\frac{y}{x-a}=\\frac{y^{2}}{x^{2}-a^{2}}$, $\\because$ the eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ is $e$, $\\therefore\\frac{a^{2}-b^{2}}{a^{2}}=\\frac{1}{4}$, $\\therefore a^{2}=\\frac{4}{3}b^{2}$, $\\therefore \\frac{x^{2}}{\\frac{4}{3}b^{2}}+\\frac{y^{2}}{b^{2}}=1$, $\\therefore y^{2}=b^{2}-\\frac{3x^{2}}{4}$, $\\frac{y^{2}}{x^{2}-a^{2}}=-\\frac{3}{4}$, $\\tan\\alpha\\tan\\beta=-\\frac{3}{4}$, $\\frac{\\cos(\\alpha-\\beta)}{\\cos(\\alpha+\\beta)}=\\frac{\\cos\\alpha\\cos\\beta+\\sin\\alpha\\sin\\beta}{\\cos\\alpha\\cos\\beta-\\sin\\alpha\\sin\\beta}=\\frac{1+\\tan\\alpha\\tan\\beta}{1-\\tan\\alpha\\tan\\beta}=\\frac{1-\\frac{3}{4}}{1+\\frac{3}{4}}=\\frac{1}{7}$" }, { "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}$ and $F_{2}$ respectively. Point $M$ lies on the left branch of $C$. A perpendicular is drawn from point $M$ to one asymptote of $C$, with foot of the perpendicular at $N$. When $|M F_{2}|+|M N|$ attains its minimum value of $10$, the maximum area of $\\Delta F_{1} N F_{2}$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;F1: Point;N: Point;F2: Point;M: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, LeftPart(C));L:Line;PointOnCurve(M, L);IsPerpendicular(L,OneOf(Asymptote(C)));FootPoint(L,OneOf(Asymptote(C)))=N;Min(Abs(LineSegmentOf(M,F2))+Abs(LineSegmentOf(M,N)))=10", "query_expressions": "Max(Area(TriangleOf(F1, N, F2)))", "answer_expressions": "25/2", "fact_spans": "[[[2, 63], [93, 96], [107, 110]], [[10, 63]], [[10, 63]], [[72, 79]], [[123, 126]], [[80, 87]], [[88, 92], [102, 106]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 100]], [], [[101, 119]], [[101, 119]], [[101, 126]], [[129, 154]]]", "query_spans": "[[[156, 186]]]", "process": "By the given condition, |MF₂| - |MF₁| = 2a, hence |MF₂| = |MF₁| + 2a. As shown in the figure: \\frac{,}{b^{2}} = \\frac{bc}{c} = b, then |MF₂| + |MN| = |MF₁| + 2a + |MN| \\geqslant |F₁N| + 2a = b + 2a, with equality if and only if points M, F₁, N are collinear. ∴ The minimum value of |MF₂| + |MN| is b + 2a = 10. ∴ 10 \\geqslant 2\\sqrt{2ab}, i.e., ab \\leqslant \\frac{25}{2}, with equality if and only if b = 2a = 5. Also, |OF₁| = c, hence |ON| = \\sqrt{c^{2} - b^{2}} = a. ∴ S_{ΔF₁NF₂} = 2S_{ΔF₁NO} = 2 × \\frac{1}{2}|NF₁|·|NO| = ab \\leqslant \\frac{25}{2}, i.e., the maximum area of ΔF₁NF₂ is \\frac{25}{2}." }, { "text": "The line $ l $ passing through the point $ P(-2,0) $ intersects the parabola $ C: y^2 = 8x $ at two points $ A $ and $ B $. If $ A $ and $ B $ lie in the first quadrant and point $ A $ is the midpoint of segment $ PB $, then what is the slope of line $ l $?", "fact_expressions": "l: Line;C: Parabola;B: Point;P: Point;A: Point;Expression(C) = (y^2 = 8*x);Coordinate(P) = (-2, 0);PointOnCurve(P, l);Intersection(l, C) = {A, B};Quadrant(A) = 1;Quadrant(B) = 1;MidPoint(LineSegmentOf(P,B)) = A", "query_expressions": "Slope(l)", "answer_expressions": "2*sqrt(2)/3", "fact_spans": "[[[12, 17], [82, 87]], [[18, 37]], [[44, 47], [55, 58]], [[1, 11]], [[40, 43], [51, 54], [65, 69]], [[18, 37]], [[1, 11]], [[0, 17]], [[12, 49]], [[51, 63]], [[55, 63]], [[65, 80]]]", "query_spans": "[[[82, 92]]]", "process": "According to the problem, the slope of line $ l $ exists and is positive. Thus, we can set the equation of line $ l $ as $ x = my - 2 $ ($ m > 0 $). Let points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system of equations of line $ l $ and parabola $ C $, listing Vieta's formulas, we can obtain $ y_{2} = 2y_{1} $, and substitute into Vieta's formulas to find the value of $ m $, then the slope of line $ l $ is $ \\frac{1}{m} $. Since line $ l $ passing through point $ P(-2, 0) $ intersects the parabola $ C: y^{2} = 8x $ at two points $ A $, $ B $, and $ A $, $ B $ are in the first quadrant, the slope of line $ l $ exists and is positive. Set the equation of line $ l $ as $ x = my - 2 $ ($ m > 0 $), let points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Solving the system\n$$\n\\begin{cases}\nx = my - 2 \\\\\ny^{2} = 8x\n\\end{cases}\n$$\nwe get $ y^{2} - 8my + 16 = 0 $, $ \\Delta = 64m^{2} - 64 > 0 $, since $ m > 0 $, solving gives $ m > 1 $. By Vieta's formulas, $ y_{1} + y_{2} = 8m $, $ y_{1}y_{2} = 16 $. Since point $ A $ is the midpoint of segment $ PB $, then $ y_{2} = 2y_{1} $, $ \\therefore 8m = $, $ \\therefore y_{1} = \\frac{8m}{3} $, $ 16 = y_{1}y_{2} = 2y_{1}^{2} = 2 \\times \\left( \\frac{8m}{3} \\right)^{2} $, we get $ m^{2} = \\frac{9}{8} $, since $ m > 0 $, solving gives $ m = \\frac{3\\sqrt{}}{4} $. Therefore, the slope of line $ l $ is $ k = \\frac{1}{m} = \\frac{4}{3\\sqrt{2}} = \\frac{2\\sqrt{2}}{3} $." }, { "text": "The hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1 (m n \\neq 0)$ has eccentricity $2$. What is the value of $\\frac{m}{n}$?", "fact_expressions": "G: Hyperbola;m: Number;n: Number;Expression(G) = (-y^2/n + x^2/m = 1);Eccentricity(G) = 2;Negation(m*n=0)", "query_expressions": "m/n", "answer_expressions": "{3,1/3}", "fact_spans": "[[[0, 52]], [[3, 52]], [[3, 52]], [[0, 52]], [[0, 60]], [[3, 52]]]", "query_spans": "[[[62, 79]]]", "process": "" }, { "text": "The equation $\\frac{x^{2}}{16-m}+\\frac{y^{2}}{m+4}=1$ represents an ellipse with foci on the $y$-axis. Find the range of real values for $m$.", "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(16 - m) + y^2/(m + 4) = 1);PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(m)", "answer_expressions": "(6,16)", "fact_spans": "[[[53, 55]], [[57, 62]], [[0, 55]], [[44, 55]]]", "query_spans": "[[[57, 69]]]", "process": "\\because the equation \\frac{x^{2}}{16-m}+\\frac{y^{2}}{m+4}=1 represents an ellipse with foci on the y-axis, \\begin{cases}\\\\m+4>0\\end{cases}, solving gives: 616-m)" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ with endpoints of the minor axis at points $A$ and $B$. Point $P$ is any point on the ellipse other than $A$ and $B$, then the product of the slopes of lines $PA$ and $PB$ is?", "fact_expressions": "P: Point;A: Point;G: Ellipse;a: Number;b: Number;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Endpoint(MinorAxis(G)) = {A, B};PointOnCurve(P, G);Negation(P = A);Negation(P = B)", "query_expressions": "Slope(LineOf(P, A))*Slope(LineOf(P, B))", "answer_expressions": "-b^2/a^2", "fact_spans": "[[[67, 72]], [[59, 62], [77, 81]], [[2, 54], [73, 75]], [[4, 54]], [[4, 54]], [[63, 66], [83, 86]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 66]], [[68, 91]], [[68, 91]], [[68, 91]]]", "query_spans": "[[[93, 111]]]", "process": "" }, { "text": "The distance from a moving point $P$ to the point $F(2,0)$ is equal to its distance to the line $x+2=0$. Then, what is the equation of the trajectory of point $P$?", "fact_expressions": "G: Line;F: Point;P: Point;Expression(G) = (x + 2 = 0);Coordinate(F) = (2, 0);Distance(P, F) = Distance(P, G)", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2 = 8*x", "fact_spans": "[[[21, 30]], [[6, 15]], [[37, 41], [2, 5], [19, 20]], [[21, 30]], [[6, 15]], [[2, 35]]]", "query_spans": "[[[37, 48]]]", "process": "" }, { "text": "The line $x - my + m = 0$ passing through the focus of the parabola $y^2 = 2px$ ($p > 0$) intersects the parabola at points $A$ and $B$, and the area of $\\triangle OAB$ ($O$ being the origin) is $2\\sqrt{2}$. Then $m^6 + m^4 = $?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;H: Line;Expression(H) = (m - m*y + x = 0);PointOnCurve(Focus(G), H);m: Number;Intersection(H, G) = {A, B};O: Origin;A: Point;B: Point;Area(TriangleOf(O, A, B)) = 2*sqrt(2)", "query_expressions": "m^6 + m^4", "answer_expressions": "2", "fact_spans": "[[[1, 22], [39, 42]], [[1, 22]], [[4, 22]], [[4, 22]], [[26, 38]], [[26, 38]], [[0, 38]], [[28, 38]], [[26, 55]], [[73, 76]], [[44, 47]], [[50, 53]], [[57, 98]]]", "query_spans": "[[[100, 115]]]", "process": "Problem Analysis: Let A(x_{1},y_{1}), B(x_{2},y_{2}), and consider the system \n\\begin{cases}y^{2}=2px\\\\x=my-m\\end{cases}, \neliminating x gives y^{2}-2mpy+2pm=0, y_{1}+y_{2}=2pm, y_{1}y_{2}=2pm, (y_{1}-y_{2})^{2}=(y_{1}+y_{2})^{2}-4y_{1}y_{2}=4p^{2}m^{2}-8pm. Since the focus (\\frac{p}{2},0) lies on x-my+m=0, \\therefore p=-2m, \\therefore |y_{1}-y_{2}|=4\\sqrt{m^{4}+m^{2}}, \\therefore S_{\\triangle OAB}=\\frac{1}{2}\\times\\frac{p}{2}\\times|y_{1}-y_{2}|=2\\sqrt{2}-m\\sqrt{m^{4}+m^{2}}=\\sqrt{2}, squaring both sides yields m^{6}+m^{4}=2" }, { "text": "Through the left focus $F_{1}$ of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$, draw a line $AB$ with inclination angle $\\frac{\\pi}{6}$, intersecting the left and right branches of the hyperbola at points $A$ and $B$, respectively. Find the chord length $AB$=?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/3 = 1);F1: Point;LeftFocus(G) = F1;PointOnCurve(F1, LineOf(A, B)) = True;Inclination(LineOf(A, B)) = pi/6;A: Point;B: Point;Intersection(LineOf(A, B), LeftPart(G)) = A;Intersection(LineOf(A, B), RightPart(G)) = B;IsChordOf(LineSegmentOf(A, B), G) = True", "query_expressions": "LineSegmentOf(A, B)", "answer_expressions": "3", "fact_spans": "[[[2, 30], [73, 76]], [[2, 30]], [[34, 41]], [[2, 41]], [[0, 69]], [[42, 69]], [[82, 86]], [[87, 90]], [[62, 90]], [[62, 90]], [[73, 100]]]", "query_spans": "[[[95, 102]]]", "process": "Since the left focus of the hyperbola is F_{1}(-2,0), let A(x_{1},y_{1}), B(x_{2},y_{2}). The equation of line AB can be set as y=\\frac{\\sqrt{3}}{13}(x+2). Substituting into the equation x^{2}-\\frac{y^{2}}{3}=1 gives 8x^{2}\\cdot4x-13=0. \\therefore x_{1}+x_{2}=\\frac{1}{2}, x_{1}x_{2}=-\\frac{13}{8}. \\therefore AB=\\sqrt{1+k^{2}}\\cdot|x_{1}-x_{2}|=\\sqrt{1+\\frac{1}{3}}\\cdot\\sqrt{(\\frac{1}{2})^{2}+4\\times\\frac{13}{8}}=3" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of ellipse $C$, point $P$ lies on the ellipse and satisfies $P F_{1}=2 P F_{2}$, $\\angle PF_{1} F_{2}=30^{\\circ}$, find the eccentricity of the ellipse.", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P, C);LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2);AngleOf(P,F1,F2)=ApplyUnit(30,degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[18, 23], [35, 37], [98, 100]], [[30, 34]], [[2, 9]], [[10, 17]], [[2, 29]], [[2, 29]], [[30, 38]], [[42, 62]], [[64, 96]]]", "query_spans": "[[[98, 106]]]", "process": "" }, { "text": "Let a point $P$ on the parabola $y^{2}=-8x$ be at a distance of $4$ from the $y$-axis. Then the distance from point $P$ to the focus of the parabola is?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (y^2 = -8*x);PointOnCurve(P, G);Distance(P, yAxis) = 4", "query_expressions": "Distance(P, Focus(G))", "answer_expressions": "6", "fact_spans": "[[[1, 16], [42, 45]], [[19, 22], [36, 40]], [[1, 16]], [[1, 22]], [[19, 34]]]", "query_spans": "[[[36, 52]]]", "process": "From $ y^{2} = -8x \\geqslant 0 $, we get $ x \\leqslant 0 $. From the given parabolic equation, we can obtain its directrix equation $ x = 2 $. Since the distance from point $ P $ to the $ y $-axis is 4, the horizontal coordinate of point $ P $ is $ x_{p} = -4 $. By the definition of a parabola, the distance from point $ P $ to the focus equals the distance from point $ P $ to the directrix, that is, $ |PF| = -x_{p} + 2 = 4 + 2 = 6 $." }, { "text": "Given that $A$ and $B$ are two fixed points on the coordinate plane with $|AB| = 2$, and the sum of distances from a moving point $P$ to points $A$ and $B$ is a constant $2$, then the trajectory of point $P$ is?", "fact_expressions": "A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 2;P: Point;Distance(P, A) + Distance(P, B) = 2", "query_expressions": "Locus(P)", "answer_expressions": "Line segment", "fact_spans": "[[[2, 5], [40, 43]], [[6, 9], [44, 47]], [[22, 33]], [[36, 39], [61, 65]], [[36, 59]]]", "query_spans": "[[[61, 70]]]", "process": "\\because|AB|=2, and the sum of the distances from moving point P to points A and B is also 2, \\because2=2, therefore the locus of point P is not an ellipse, \\therefore it is the line segment AB." }, { "text": "A hyperbola $2 m x^{2}-m y^{2}=2$ has a directrix $y=1$. Find the value of $m$.", "fact_expressions": "G: Hyperbola;Expression(G) = (2*(m*x^2) - m*y^2 = 2);m: Number;Expression(OneOf(Directrix(G))) = (y = 1)", "query_expressions": "m", "answer_expressions": "-4/3", "fact_spans": "[[[0, 24]], [[0, 24]], [[37, 40]], [[0, 35]]]", "query_spans": "[[[37, 44]]]", "process": "It is known that the foci of the hyperbola lie on the y-axis, ∴ m < 0, the equation of the hyperbola can be rewritten as \\frac{y^{2}}{-\\frac{2}{m}} - \\frac{x^{2}}{-\\frac{1}{m}} = 1, thus a^{2} = -\\frac{2}{m}, b^{2} = -\\frac{1}{m}, c^{2} = -\\frac{3}{m}. Since the directrix is y = 1, ∴ a^{2} = c, that is, -\\frac{2}{m} = \\sqrt{-\\frac{3}{m}}, solving gives m = -\\frac{4}{3}." }, { "text": "Given that point $F$ is the focus of the parabola $y^{2}=2 p x(p>0)$, and points $A(2 , y_{1})$, $B(\\frac{1}{2} , y_{2})$ are on the parabola located in the first and fourth quadrants respectively, if $|A F|=10$, then $|y_{1}-y_{2}|$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;F: Point;Focus(G) = F;A: Point;B: Point;y1: Number;y2: Number;Coordinate(A) = (2, y1);Coordinate(B) = (1/2, y2);PointOnCurve(A, G);PointOnCurve(B, G);Quadrant(A) = 1;Quadrant(B) = 4;Abs(LineSegmentOf(A, F)) = 10", "query_expressions": "Abs(y1 - y2)", "answer_expressions": "12", "fact_spans": "[[[7, 28], [75, 78]], [[7, 28]], [[10, 28]], [[10, 28]], [[2, 6]], [[2, 31]], [[32, 47]], [[48, 72]], [[33, 47]], [[48, 72]], [[32, 47]], [[48, 72]], [[32, 89]], [[32, 89]], [[32, 89]], [[32, 89]], [[91, 101]]]", "query_spans": "[[[103, 120]]]", "process": "\\because|AF|=2+\\frac{p}{2}=10,\\therefore p=16, then the equation of the parabola is y^{2}=32x, substitute x=\\frac{1}{2} into the equation, get y=-4 (y=4 is discarded), so B(\\frac{1}{2},-4); substitute x=2 into the equation, get y=8 (y=-8 is discarded), so A(2,8), then |y_{1}-y_{2}|=|8-(-4)|=12," }, { "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. Point $P$ lies on the right branch of the hyperbola such that $|P F_{2}|=|F_{1} F_{2}|$, and the distance from $F_{2}$ to the line $P F_{1}$ equals the length of the real axis of the hyperbola. What is the equation of the asymptotes 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;P: Point;PointOnCurve(P, RightPart(G));Abs(LineSegmentOf(P, F2)) = Abs(LineSegmentOf(F1, F2));Distance(F2, LineOf(P, F1)) = Length(RealAxis(G))", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*((4/3)*x)", "fact_spans": "[[[19, 75], [87, 90], [147, 150], [156, 159]], [[19, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[22, 75]], [[1, 8]], [[9, 16], [123, 130]], [[1, 81]], [[1, 81]], [[82, 86]], [[82, 94]], [[96, 121]], [[123, 154]]]", "query_spans": "[[[156, 167]]]", "process": "" }, { "text": "Draw a line through the focus of the ellipse $x^{2}+2 y^{2}=2$ with an inclination angle of $45^{\\circ}$, intersecting the ellipse at points $A$ and $B$. The center of the ellipse is $O$. Then the area of $\\triangle A O B$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 2*y^2 = 2);H: Line;PointOnCurve(Focus(G), H) = True;Inclination(H) = ApplyUnit(45, degree);A: Point;B: Point;Intersection(H,G) = {A,B};O: Origin;Center(G) = O", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "2/3", "fact_spans": "[[[1, 20], [46, 48], [60, 62]], [[1, 20]], [[43, 45]], [[0, 45]], [[26, 45]], [[50, 53]], [[54, 57]], [[43, 59]], [[66, 69]], [[60, 69]]]", "query_spans": "[[[71, 93]]]", "process": "Convert $x^{2}+2y^{2}=2$ into $\\frac{x^{2}}{2}+y^{2}=1$. Since the line passes through the focus $F(1,0)$ of the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ and has an inclination angle of $45^{\\circ}$, the equation of the line is $y=x-1$. Solving the system \n\\[\n\\begin{cases}\ny=x-1 \\\\\nx^{2}-2y^{2}=2\n\\end{cases}\n\\]\nyields $3y^{2}+2y-1=0$, giving $y=\\frac{1}{3}$, $y_{2}=-1$. Then $S_{AAOB}=S_{AAOF}+S_{ABOF}=\\frac{1}{2}\\times\\left(\\frac{1}{3}+1\\right)\\times1=\\frac{2}{3}$." }, { "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;F2: Point;B: Point;F1: Point;A: 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": "", "fact_spans": "[[[19, 46], [62, 64]], [[9, 16]], [[58, 61]], [[1, 8]], [[53, 57], [123, 127]], [[19, 46]], [[1, 52]], [[1, 52]], [[53, 65]], [[53, 65]], [[67, 121]]]", "query_spans": "[[[123, 132]]]", "process": "" }, { "text": "5. The line $l$ passes through the focus $F$ of the parabola $y^{2}=4x$ and intersects the parabola at two points $A$ and $B$. Then $\\frac{1}{|F A|}+\\frac{1}{|F B|}=$?", "fact_expressions": "l: Line;E: Parabola;F: Point;A: Point;B: Point;Expression(E) = (y^2 = 4*x);Focus(E) = F;PointOnCurve(F, l);Intersection(l, E) = {A, B}", "query_expressions": "1/Abs(LineSegmentOf(F, A)) + 1/Abs(LineSegmentOf(F,B))", "answer_expressions": "1", "fact_spans": "[[[2, 9]], [[11, 26], [36, 39]], [[11, 35]], [[41, 44]], [[45, 48]], [[11, 27]], [[11, 34]], [[3, 35]], [[3, 52]]]", "query_spans": "[[[57, 92]]]", "process": "" }, { "text": "Given the curve $C$: $|x|+|y|=m$ ($m>0$), and the curve $C$ intersects the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ at four distinct points, find the range of real values for $m$.", "fact_expressions": "G: Ellipse;C: Curve;m: Real;Expression(G) = (x^2/9 + y^2/4 = 1);m>0;Expression(C) = (Abs(x) + Abs(y) = m);NumIntersection(C, G) = 4", "query_expressions": "Range(m)", "answer_expressions": "{(2, 3), sqrt(13)}", "fact_spans": "[[[32, 69]], [[2, 24], [26, 31]], [[79, 84]], [[32, 69]], [[8, 24]], [[2, 24]], [[26, 77]]]", "query_spans": "[[[79, 91]]]", "process": "Problem Analysis: (1) If m=1, the curve C: |x|+|y|=1 represents a square with diagonal length 2, and the area is 2; (2) The ellipse has semi-major axis 3 and semi-minor axis 2. When 20, y>0, combining x+y-m=0 with the ellipse equation yields 13x^{2}-18mx+9m^{2}-36=0, thus \\triangle=(-18m)^{2}-52(9m^{2}\\cdot36)=0, since m>0, therefore m=\\sqrt{13}. At this time, the curve C and the ellipse \\frac{x^{2}}{4}+\\frac{y^{2}}{4}=1 have four distinct intersection points." }, { "text": "Given $A(1,2)$, $B(-1,2)$, a moving point $P$ satisfies $\\overrightarrow{A P} \\perp \\overrightarrow{B P}$. If 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 trajectory of the moving point $P$, then the range of the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;A: Point;B: Point;P: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (1, 2);Coordinate(B) = (-1, 2);IsPerpendicular(VectorOf(A,P),VectorOf(B,P));NumIntersection(Asymptote(G),Locus(P))=0", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,2)", "fact_spans": "[[[81, 137], [157, 160]], [[84, 137]], [[84, 137]], [[2, 10]], [[11, 20]], [[144, 147], [24, 27]], [[84, 137]], [[84, 137]], [[81, 137]], [[2, 10]], [[11, 20]], [[29, 78]], [[81, 155]]]", "query_spans": "[[[157, 170]]]", "process": "Let P(x, y). Since \\overrightarrow{AP} \\bot \\overrightarrow{BP}, A(1,2), B(-1,2), \\therefore (x-1)(x+1) + (y-2)^{2} = 0, that is, x^{2} + (y-2)^{2} = 1. Let an asymptote of the hyperbola be y = \\frac{b}{a}x, then \\frac{|2a|}{\\sqrt{a^{2}+b^{2}}} > 1 \\therefore 3a^{2} > b^{2}, and since b^{2} = c^{2} - a^{2}, it follows that c^{2} < 4a^{2}, \\therefore e = \\frac{c}{a} < 2. Also, the eccentricity of the hyperbola satisfies e > 1, so 1 < e < 2." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and point $P$ lies on the hyperbola such that $F P_{1} \\perp F P_{2}$. Then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - y^2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(F, P1), LineSegmentOf(F, P2))", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "1", "fact_spans": "[[[17, 45], [56, 59]], [[1, 8]], [[51, 55]], [[9, 16]], [[17, 45]], [[1, 50]], [[51, 60]], [[62, 85]]]", "query_spans": "[[[87, 114]]]", "process": "According to the hyperbola equation, calculate the foci $ F_{1}(-\\sqrt{5},0) $, $ F_{2}(\\sqrt{5},0) $. Using the Pythagorean theorem, compute $ |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=20 $. By the definition of a hyperbola, $ ||PF_{1}|-|PF_{2}||=2a=4 $. Solving these equations simultaneously gives $ |PF_{1}|\\cdot|PF_{2}|=2 $, thus obtaining the area of $ \\triangle F_{1}PF_{2} $. [Solution] $ \\because $ For the hyperbola $ \\frac{x^{2}}{4}-y^{2}=1 $, $ a=2 $, $ b=1 $. $ \\because $ Point $ P $ lies on the hyperbola and $ \\angle F_{1}PF_{2}=90^{\\circ} $, $ \\therefore |PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}=20 $. According to the definition of the hyperbola, $ ||PF_{1}|-|PF_{2}||=2a=4 $. $ \\therefore $ Solving the two equations simultaneously yields $ |PF_{1}|\\cdot|PF_{2}|=2 $. Therefore, the area of $ \\triangle F_{1}PF_{2} $ is $ S=\\frac{1}{2}|PF_{1}|\\cdot|PF_{2}|=1 $." }, { "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{20}-\\frac{y^{2}}{5}=1$, what is its focal distance?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/20 - y^2/5 = 1)", "query_expressions": "FocalLength(G)", "answer_expressions": "10", "fact_spans": "[[[2, 5], [48, 49]], [[2, 45]]]", "query_spans": "[[[48, 54]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $\\Gamma$: $\\frac{x^{2}}{2}-\\frac{y^{2}}{14}=10$, respectively. A line $l$ passing through $F_{1}$ intersects the left and right branches of the hyperbola $\\Gamma$ at points $P$ and $Q$, respectively, and $|P Q|=|Q F_{2}|$. Then $\\cos \\angle P F_{1} F_{2}$=?", "fact_expressions": "l: Line;Gamma: Hyperbola;P: Point;Q: Point;F2: Point;F1: Point;Expression(Gamma) = (x^2/2 - y^2/14 = 10);LeftFocus(Gamma) = F1;RightFocus(Gamma) = F2;PointOnCurve(F1, l);Intersection(l,LeftPart(Gamma))=P;Intersection(l,RightPart(Gamma))=Q;Abs(LineSegmentOf(P, Q)) = Abs(LineSegmentOf(Q, F2))", "query_expressions": "Cos(AngleOf(P, F1, F2))", "answer_expressions": "5*sqrt(2)/8", "fact_spans": "[[[85, 90]], [[20, 70], [91, 102]], [[110, 113]], [[114, 117]], [[10, 17]], [[2, 9], [77, 84]], [[20, 70]], [[2, 75]], [[2, 75]], [[76, 90]], [[85, 119]], [[85, 119]], [[121, 138]]]", "query_spans": "[[[140, 169]]]", "process": "Hyperbola $\\Gamma: \\frac{x^{2}}{2} - \\frac{y^{2}}{14} = 10$, the real semi-axis length $a = \\sqrt{2}$, the imaginary semi-axis length $b = \\sqrt{14}$, the semi-focal distance $c = 4$. As shown in the figure, by the definition of hyperbola, $|PF_{1}| = |QF_{1}| - |QP| = |QF_{1}| - |QF_{2}| = 2a = 2\\sqrt{2}$, then $|PF_{2}| = 2a + |PF_{1}| = 4\\sqrt{2}$, $|F_{1}F_{2}| = 2c = 8$. In $\\triangle PF_{1}F_{2}$, by the cosine theorem, $\\cos\\angle PF_{1}F_{2} = \\frac{|PF_{1}|^{2} + |F_{1}F_{2}|^{2} - |PF_{2}|^{2}}{2|PF_{1}|\\cdot|F_{1}F_{2}|} = \\frac{(2\\sqrt{2})^{2} + 8^{2} - (4\\sqrt{2})^{2}}{2\\cdot2\\sqrt{2}\\cdot8} = \\frac{5\\sqrt{2}}{8}$" }, { "text": "Given the parabola $y^{2}=4 x$, a line passing through the point $N(2,0)$ intersects the parabola at points $A$ and $B$. If $\\frac{|A N|}{|B N|}=2$, then the length of segment $A B$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);N: Point;Coordinate(N) = (2, 0);H: Line;PointOnCurve(N, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, N))/Abs(LineSegmentOf(B, N)) = 2", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "3*sqrt(5)", "fact_spans": "[[[2, 16], [31, 34]], [[2, 16]], [[18, 27]], [[18, 27]], [[28, 30]], [[17, 30]], [[35, 38]], [[39, 42]], [[28, 44]], [[45, 68]]]", "query_spans": "[[[70, 80]]]", "process": "" }, { "text": "Given the ellipse $\\Gamma$: $\\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. Points $A$ and $B$ lie on the ellipse $\\Gamma$, satisfying $\\overrightarrow{A F_{1}} \\cdot \\overrightarrow{F_{1} F_{2}}=0$ and $\\overrightarrow{A F_{2}}=\\lambda \\overrightarrow{F_{2} B}$. When $\\lambda \\in[2,3]$, what is the range of values for the eccentricity of the ellipse?", "fact_expressions": "Gamma: Ellipse;a: Number;b: Number;A: Point;F1: Point;F2: Point;B: Point;a > b;b > 0;lambda:Number;In(lambda,[2,3]);Expression(Gamma) = (x^2/a^2 + y^2/b^2 = 1);LeftFocus(Gamma) = F1;RightFocus(Gamma) = F2;PointOnCurve(A, Gamma);PointOnCurve(B, Gamma);DotProduct(VectorOf(A,F1),VectorOf(F1,F2))=0;VectorOf(A,F2)=lambda*VectorOf(F2,B)", "query_expressions": "Range(Eccentricity(Gamma))", "answer_expressions": "[sqrt(5)/5,sqrt(3)/3]", "fact_spans": "[[[2, 63], [97, 107], [255, 257]], [[13, 63]], [[13, 63]], [[88, 92]], [[72, 79]], [[80, 87]], [[93, 96]], [[13, 63]], [[13, 63]], [[235, 253]], [[235, 253]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 108]], [[93, 108]], [[109, 172]], [[173, 232]]]", "query_spans": "[[[255, 268]]]", "process": "Since $\\overrightarrow{AF}_{1}\\cdot\\overrightarrow{F_{1}F_{2}}=0$, we can set $A(-c,\\frac{b^{2}}{a})$, $F_{2}(c,0)$, $B(x,y)$, so we can set $A(-c,\\frac{b}{a})$, $F_{2}(c,0)$, $B(x,y)$, so $\\frac{(1+\\frac{2}{2})^{2}}{a^{2}}+\\frac{(-\\frac{b^{2}}{2a})^{2}}{b^{2}}=1$, that is $(\\lambda+2)^{2}c^{2}+b^{2}=\\lambda^{2}a^{2}$, that is $(\\lambda+2)^{2}c^{2}+b^{2}=\\lambda^{2}a^{2}$, that is $(\\lambda^{2}+4\\lambda+3)c^{2}=(\\lambda^{2}-1)a^{2}$, that is $\\frac{c}{a}=\\sqrt{\\frac{2^{2}-1}{2+41+3}}=\\sqrt{\\frac{\\lambda-1}{2+3}}==\\sqrt{1-\\frac{-2}{2}}\\frac{4}{+3}$, which is an increasing function on the interval $[2,3]$, so $\\frac{c}{a}\\in\\left[\\frac{\\sqrt{5}}{5},\\frac{\\sqrt{3}}{3}\\right]$, therefore the range of the ellipse's eccentricity is $\\left[\\frac{\\sqrt{5}}{5},\\frac{\\sqrt{3}}{3}\\right]$" }, { "text": "The line with slope $1$ intersects the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ at points $A$ and $B$. Then the maximum value of $|AB|$ is?", "fact_expressions": "G: Ellipse;H: Line;A: Point;B: Point;Expression(G) = (x^2/2 + y^2 = 1);Slope(H)=1;Intersection(H, G) = {A, B}", "query_expressions": "Max(Abs(LineSegmentOf(A, B)))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[10, 37]], [[7, 9]], [[40, 43]], [[44, 47]], [[10, 37]], [[0, 9]], [[7, 49]]]", "query_spans": "[[[51, 64]]]", "process": "Substitute the line L: y = x + b with slope 1 into \\frac{x^{2}}{2} + y^{2} = 1, simplify to obtain 3x^{2} + 4bx + 2b^{2} - 2 = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then x_{1} + x_{2} = -\\frac{4b}{3}, x_{1}x_{2} = \\frac{2b^{2} - 2}{3}, and a = 16b^{2} - 12(2b^{2} - 2) > 0, solving gives b^{2} < 3. |AB| = \\sqrt{2} \\times \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{2} \\times \\sqrt{\\frac{16b^{2}}{9} - \\frac{8b^{2} - 8}{3}} = \\sqrt{2} \\times \\sqrt{\\frac{-8b^{2} + 24}{9}} \\leqslant \\frac{4\\sqrt{3}}{3}. Therefore, when b = 0, the maximum value of |AB| is \\frac{4\\sqrt{3}}{3}." }, { "text": "The equation of the directrix of the parabola $y^{2}=x$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x = -1/4", "fact_spans": "[[[0, 12]], [[0, 12]]]", "query_spans": "[[[0, 19]]]", "process": "The equation of the directrix of the parabola y^{2}=x is x=-\\frac{1}{4}; therefore, fill in x=-\\frac{1}{4}." }, { "text": "The equation of the directrix of the parabola $x^{2}=-8 y$ is?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -8*y)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=2", "fact_spans": "[[[0, 15]], [[0, 15]]]", "query_spans": "[[[0, 22]]]", "process": "" }, { "text": "The line $x + y + 2 = 0$ intersects the $x$-axis and $y$-axis at points $A$ and $B$ respectively. Point $P$ lies on the parabola $y^2 = 4x$. Then the minimum area of $\\triangle ABP$ is?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (x + y + 2 = 0);Intersection(H,xAxis) = A;Intersection(H,yAxis) = B;PointOnCurve(P, G)", "query_expressions": "Min(Area(TriangleOf(A, B, P)))", "answer_expressions": "1", "fact_spans": "[[[40, 54]], [[0, 11]], [[25, 28]], [[29, 32]], [[35, 39]], [[40, 54]], [[0, 11]], [[0, 34]], [[0, 34]], [[35, 55]]]", "query_spans": "[[[57, 82]]]", "process": "According to the problem, A(-2,0), B(0,-2). Let the line l, which is parallel to the line x+y+2=0 and tangent to the parabola, have the equation: x+y+t=0. Combining the equation of line l and the parabola, eliminating y gives: y^{2}+4y+4t=0. Then \\triangle=16-16t=0, so t=1. Since the distance d between the line x+y+2=0 and line l is d=\\frac{|2-1|}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2}, then \\frac{1}{2}|AB|d=\\frac{1}{2}\\times2\\sqrt{2}\\times\\frac{\\sqrt{2}}{2}=1. [Analysis] This problem examines the relationship between a line and a conic section, tests computational and problem-solving abilities; combining algebra and geometry is key to solving this problem, and it is a medium-difficulty problem." }, { "text": "The standard equation of a parabola with the focus at the intersection point of the line $3x - 4y - 12 = 0$ and the $y$-axis is?", "fact_expressions": "C:Parabola;H: Line;Expression(H) = (3*x - 4*y - 12 = 0);Intersection(H,yAxis)=Focus(C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2=-12*y", "fact_spans": "[[[29, 32]], [[1, 17]], [[1, 17]], [[0, 32]]]", "query_spans": "[[[29, 38]]]", "process": "According to the problem, the focus of the parabola is (0, -3), so 2p = 12. Since the focus lies on the negative y-axis, the equation of the parabola is x^{2} = -12y. Hence, fill in x^{2} = -12y. [Note] This problem mainly examines the equation of a parabola and its simple geometric properties, and is a medium-difficulty question." }, { "text": "The eccentricity of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{7}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/7 + y^2/4 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(11)/2", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 44]]]", "process": "Since $ a^{2} = 4 $, $ b^{2} = 7 $, so $ c^{2} = a^{2} + b^{2} = 4 + 7 = 11 $. Therefore, the eccentricity is $ \\frac{c}{a} = \\frac{\\sqrt{11}}{2} $." }, { "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 left and right foci $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}$. Then, what is the slope of the asymptotes 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;P: Point;PointOnCurve(P, RightPart(G));F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;O: Origin;H: Circle;A: Point;TangentPoint(LineOf(P, F1), H) = A;Center(H) = O;Radius(H) = a;PointOnCurve(F2, PerpendicularBisector(LineSegmentOf(P, F1)))", "query_expressions": "Slope(Asymptote(G))", "answer_expressions": "pm*(4/3)", "fact_spans": "[[[5, 61], [66, 67], [164, 167]], [[5, 61]], [[8, 61]], [[117, 120]], [[8, 61]], [[8, 61]], [[0, 4]], [[0, 65]], [[75, 82]], [[84, 91], [153, 161]], [[66, 91]], [[66, 91]], [[106, 113]], [[124, 125]], [[128, 132]], [[93, 132]], [[105, 125]], [[117, 125]], [[133, 161]]]", "query_spans": "[[[164, 176]]]", "process": "As shown in the figure, A is the point of tangency, B is the midpoint of PF_{1}, since |OA| = a, so |BF_{2}| = 2a, and |F_{1}F_{2}| = 2c, thus |BF_{1}| = 2b, |PF_{2}| = 4b, also |PF_{2}| = |F_{1}F_{2}| = 2c, according to the definition of hyperbola, |PF_{1}| - |PF_{2}| = 2a, that is, 4b - 2c = 2a, squaring both sides and simplifying yields 3c^{2} - 2ac - 5a^{2} = 0, so \\frac{c}{a} = \\frac{5}{3}, therefore the slope of asymptotes is \\pm\\frac{b}{a} = \\pm\\sqrt{(\\frac{c}{a})^{2}-1} = \\pm\\frac{4}{3}" }, { "text": "The left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1$ are $F_{1}$ and $F_{2}$ respectively, and $P$ is a point on the ellipse. If $P F_{1}=2 P F_{2}$, then what is the distance from $P$ to the left directrix?", "fact_expressions": "G: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/3 = 1);LeftFocus(G)=F1;RightFocus(G)=F2;PointOnCurve(P,G);LineSegmentOf(P, F1) = 2*LineSegmentOf(P, F2)", "query_expressions": "Distance(P,LeftDirectrix(G))", "answer_expressions": "2*sqrt(6)", "fact_spans": "[[[0, 37], [66, 68]], [[62, 65], [94, 97]], [[46, 53]], [[54, 61]], [[0, 37]], [[0, 61]], [[0, 61]], [[62, 71]], [[73, 92]]]", "query_spans": "[[[66, 106]]]", "process": "\\because \\text{ellipse} \\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1, \\therefore a=3, b^{2}=3, \\because |PF_{1}|+|PF_{2}|=2a=6, |PF_{1}|=2|PF_{2}| \\therefore |PF_{1}|=4, |PF_{2}|=2. \\text{Find the eccentricity of the ellipse} e=\\frac{c}{a}=\\frac{\\sqrt{6}}{3}, \\text{let } d \\text{ be the distance from } P \\text{ to the left directrix, according to the unified definition of conic sections, we get: } \\frac{|PF_{1}|}{d}=e=\\frac{\\sqrt{6}}{3}, \\therefore d=\\frac{4\\times3}{\\sqrt{6}}=2\\sqrt{6}, \\text{so the distance from } P \\text{ to the left directrix is } 2\\sqrt{6}. \\text{Hence fill in: } \\frac{\\sqrt{6}}{2\\sqrt{6}}. \\text{This problem examines the application of the definition of an ellipse, geometric properties of ellipses, and the unified definition of conic sections; the ratio of the distance from a point on the ellipse to the left focus and to the corresponding directrix is the eccentricity of the ellipse.}" }, { "text": "Given that $F$ is the left focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a, b>0)$, $B_{1} B_{2}$ is the imaginary axis of the hyperbola, $M$ is the midpoint of $O B_{1}$, and the line passing through $F$ and $M$ intersects the hyperbola $C$ at point $A$, with $F M=2 M A$. What is the eccentricity of the hyperbola $C$?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(C) = ( -y^2/b^2 + x^2/a^2 = 1 );F: Point ;LeftFocus(C) = F;B1: Point;B2: Point;O: Origin;LineSegmentOf(B1,B2) = ImageinaryAxis(C);M: Point;MidPoint(LineSegmentOf(O,B1)) = M;Line_1: Line ;A: Point;PointOnCurve(F, Line_1) = True;PointOnCurve(M,Line_1) = True;Intersection(Line_1, C) = A;LineSegmentOf(F,M) = 2*LineSegmentOf(M,A)", "query_expressions": "Eccentricity(C)", "answer_expressions": "5/2", "fact_spans": "[[[6, 65], [84, 87], [121, 127], [147, 153]], [[14, 65]], [[14, 65]], [[14, 65]], [[14, 65]], [[6, 65]], [[2, 5], [110, 113]], [[2, 69]], [[70, 83]], [[70, 83]], [[95, 104]], [[70, 90]], [[91, 94], [114, 117]], [[91, 108]], [[118, 120]], [[128, 131]], [[109, 120]], [[109, 120]], [[118, 131]], [[133, 144]]]", "query_spans": "[[[147, 158]]]", "process": "" }, { "text": "If the line $x + y + a = 0$ and the circle $x^{2} + (y + \\sqrt{2})^{2} = 1$ have a common point, then the range of real values for $a$ is?", "fact_expressions": "G: Circle;H: Line;a: Real;Expression(G) = (x^2 + (y + sqrt(2))^2 = 1);Expression(H) = (a + x + y = 0);IsIntersect(H, G)", "query_expressions": "Range(a)", "answer_expressions": "[0,2*sqrt(2)]", "fact_spans": "[[[14, 45]], [[2, 13]], [[51, 56]], [[14, 45]], [[2, 13]], [[2, 49]]]", "query_spans": "[[[51, 63]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $y^{2}=100 x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 100*x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(25, 0)", "fact_spans": "[[[0, 16]], [[0, 16]]]", "query_spans": "[[[0, 23]]]", "process": "Since 2p = 100, we have p = 50, so the coordinates of the focus are (25, 0)." }, { "text": "Given that $P$ is a moving point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{8}=1$, and $O$ is the origin, then the equation of the trajectory of the midpoint $Q$ of the segment $OP$ is?", "fact_expressions": "G: Ellipse;O: Origin;P: Point;Q:Point;Expression(G) = (x^2/4 + y^2/8 = 1);PointOnCurve(P, G);MidPoint(LineSegmentOf(O,P))=Q", "query_expressions": "LocusEquation(Q)", "answer_expressions": "x^2+y^2/2=1", "fact_spans": "[[[6, 43]], [[48, 51]], [[2, 5]], [[67, 70]], [[6, 43]], [[2, 47]], [[58, 70]]]", "query_spans": "[[[67, 77]]]", "process": "Let Q(x, y). Since Q is the midpoint of OP, then P(2x, 2y). Substituting into the ellipse equation gives \\frac{(2x)^{2}}{4}+\\frac{(2y)^{2}}{8}=1, simplifying yields x^{2}+\\frac{y^{2}}{2}=1. Thus, the trajectory equation of point Q is x^{2}+\\frac{y^{2}}{2}=1. This question mainly examines the substitution method for finding the trajectory equation of a moving point and the midpoint coordinates, and belongs to a basic problem." }, { "text": "The line $ l $ intersects the parabola $ y^{2} = 4x $ at points $ A $ and $ B $, $ |AB| = 6 $, then the minimum value of the distance from the midpoint of $ AB $ to the $ y $-axis is?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, B)) = 6", "query_expressions": "Min(Distance(MidPoint(LineSegmentOf(A, B)),yAxis))", "answer_expressions": "2", "fact_spans": "[[[0, 5]], [[6, 20]], [[23, 26]], [[27, 30]], [[6, 20]], [[0, 32]], [[33, 42]]]", "query_spans": "[[[44, 64]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), and the intersection point be F. According to the problem, the distance from the midpoint of AB to the y-axis is d = \\frac{x_{1}+x_{2}}{2} = \\frac{x_{1}+1+x_{2}+1-2}{2} = \\frac{|AF|+|BF|}{2}-1. Since |AF|+|BF| \\geqslant |AB| = 6, the minimum value of d is 2." }, { "text": "The left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are $F_{1}$ and $F_{2}$, respectively. A line passing through $F_{1}$ with an inclination angle of $45^{\\circ}$ intersects the right branch of the hyperbola at point $M$. If $M F_{2}$ is perpendicular to the $x$-axis, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;M: 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(F1, H);Inclination(H)=ApplyUnit(45,degree);Intersection(H, RightPart(G)) = M;IsPerpendicular(LineSegmentOf(M,F2),xAxis)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1+sqrt(2)", "fact_spans": "[[[0, 56], [110, 113], [140, 143]], [[3, 56]], [[3, 56]], [[107, 109]], [[116, 120]], [[73, 80]], [[65, 72], [82, 89]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 80]], [[0, 80]], [[81, 109]], [[90, 109]], [[107, 120]], [[122, 138]]]", "query_spans": "[[[140, 149]]]", "process": "Problem Analysis: From the given conditions, $\\frac{c^{2}}{a^{2}}-\\frac{(2c)^{2}}{b^{2}}=1\\Rightarrow\\frac{(2c)^{2}}{b^{2}}=\\frac{b^{2}}{a^{2}}\\Rightarrow2ac=c^{2}-a^{2}\\Rightarrow e^{2}-2e-1=0$, since $e>1$, therefore $e=1+\\sqrt{2}$" }, { "text": "Let the parabola $C$: $y^{2}=4x$ have focus $F$ and directrix $l$. Point $M$ lies on $C$, point $N$ lies on $l$, and $\\overrightarrow{F N}=\\lambda \\overrightarrow{F M}(\\lambda>0)$. If $|M F|=\\frac{4}{3}$, then the value of $\\lambda$ is?", "fact_expressions": "C: Parabola;F: Point;N: Point;M: Point;l: Line;lambda:Number;lambda>0;Expression(C) = (y^2 = 4*x);Focus(C) = F;Directrix(C) = l;PointOnCurve(M, C);PointOnCurve(N, l);VectorOf(F, N) = lambda*VectorOf(F, M);Abs(LineSegmentOf(M, F)) = 4/3", "query_expressions": "lambda", "answer_expressions": "3", "fact_spans": "[[[1, 20], [39, 42]], [[23, 26]], [[44, 48]], [[34, 38]], [[30, 33], [49, 52]], [[141, 150]], [[55, 117]], [[1, 20]], [[1, 26]], [[1, 33]], [[34, 43]], [[44, 53]], [[55, 117]], [[120, 139]]]", "query_spans": "[[[141, 154]]]", "process": "The parabola $ C: y^2 = 4x $ has focus $ F(1,0) $, and directrix $ l: x = -1 $. Let $ M(m,n) $, where $ n < 0 $, and $ |MF| = \\frac{4}{3} $. Then $ m + 1 = \\frac{4}{3} $, solving gives $ m = \\frac{1}{3} $, $ n = -\\frac{2\\sqrt{3}}{3} $, so $ M\\left(\\frac{1}{3}, -\\frac{2\\sqrt{3}}{3}\\right) $. The equation of line $ MF $ is $ y = \\sqrt{3}(x - 1) $. Letting $ x = -1 $, we get $ y = -2\\sqrt{3} $, so $ N(-1, -2\\sqrt{3}) $. Therefore, $ \\lambda = \\frac{|FN|}{|FM|} = \\frac{4}{\\frac{4}{3}} = 3 $." }, { "text": "If the asymptotes of a hyperbola are given by $y = \\pm 3x$, and one of its foci coincides with the focus of the parabola $y^2 = 4\\sqrt{10}x$, then the standard equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(H) = (y^2 = 4*sqrt(10)*x);Expression(Asymptote(G)) = (y = pm*(3*x));OneOf(Focus(G)) = Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/9=1", "fact_spans": "[[[1, 4], [23, 24], [61, 64]], [[30, 54]], [[30, 54]], [[1, 22]], [[23, 59]]]", "query_spans": "[[[61, 71]]]", "process": "Test analysis: The focus of the parabola y^{2}=4\\sqrt{10}x is (\\sqrt{10},0) \\therefore c=10, asymptotes y=\\pm3x \\therefore \\frac{b}{a}=3 \\because a^{2}+b^{2}=c^{2} \\therefore a^{2}=1,b^{2}=9 \\therefore x^{2}-\\frac{y^{2}}{9}=1" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, and a line passing through $F$ with slope $1$ intersects the parabola $C$ at points $A$ and $B$. Then the value of $|AB|$ equals?", "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) = 1;Intersection(G, C) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[6, 25], [44, 50]], [[41, 43]], [[51, 54]], [[55, 58]], [[2, 5], [30, 33]], [[6, 25]], [[2, 28]], [[29, 43]], [[34, 43]], [[41, 60]]]", "query_spans": "[[[63, 75]]]", "process": "The focus of the parabola is (1,0), and the slope is 1. Then the equation of the line is y = x - 1. Substituting into the parabola equation y^{2} = 4x gives x^{2} - 6x + 1 = 0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), \\therefore x_{1} + x_{2} = 6. According to the definition of the parabola, |AB| = x_{1} + \\frac{p}{2} + x_{2} + \\frac{p}{2} = x_{1} + x_{2} + p = 6 + 2 = 8" }, { "text": "The distance from the focus of the parabola $x^{2}=8 y$ to the asymptotes of the hyperbola $\\frac{y^{2}}{3}-x^{2}=1$ is?", "fact_expressions": "G: Hyperbola;H: Parabola;Expression(G) = (-x^2 + y^2/3 = 1);Expression(H) = (x^2 = 8*y)", "query_expressions": "Distance(Focus(H),Asymptote(G))", "answer_expressions": "1", "fact_spans": "[[[18, 46]], [[0, 14]], [[18, 46]], [[0, 14]]]", "query_spans": "[[[0, 55]]]", "process": "The focus of the parabola \\( x^{2} = 8y \\) is at \\( (0, 2) \\), and the asymptotes of the hyperbola \\( \\frac{y^{2}}{3} - x^{2} = 1 \\) are given by \\( y = \\pm\\sqrt{3}x \\). Therefore, the distance from the focus to the asymptote is \\( d = \\frac{| \\pm\\sqrt{3} \\times 0 + 2 |}{\\sqrt{1 + (\\sqrt{3})^{2}}} = 1 \\)." }, { "text": "The line $ l $ has the equation $ y = x + 3 $. Take any point $ P $ on $ l $. If a point $ P $ is given and an ellipse is formed having foci at the foci of the hyperbola $ 12x^{2} - 4y^{2} = 3 $, 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]], [[38, 61]], [[65, 67], [80, 82]], [[26, 29], [32, 36]], [[38, 61]], [[0, 16]], [[17, 29]], [[31, 67]], [[37, 70]], [[75, 82]]]", "query_spans": "[[[80, 86]]]", "process": "" }, { "text": "If $A$ and $B$ are the two vertices on the minor axis of the ellipse $E$: $\\frac{x^{2}}{m}+y^{2}=1$ ($m>1$), and point $P$ is any point on the ellipse distinct from $A$ and $B$, and if the product of the slopes of lines $AP$ and $BP$ is $-\\frac{m}{4}$, then what is the eccentricity of ellipse $E$?", "fact_expressions": "A: Point;B: Point;E: Ellipse;Expression(E) = (y^2 + x^2/m = 1);m: Number;m > 1;Vertex(MinorAxis(E)) = {A, B};P: Point;PointOnCurve(P, E) = True;Negation(P = A);Negation(P = B);Slope(LineOf(A, P))*Slope(LineOf(B, P)) = -m/4", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[1, 4], [67, 70]], [[5, 8], [71, 74]], [[11, 48], [62, 64], [119, 124]], [[11, 48]], [[18, 48]], [[18, 48]], [[1, 56]], [[57, 61]], [[57, 79]], [[57, 79]], [[57, 79]], [[81, 116]]]", "query_spans": "[[[119, 130]]]", "process": "Let the equations of lines AP and BP be y−1=k_{A}x, y+1=k_{B}x, point P(x_{0},y_{0}), k_{A}=\\frac{y_{0}-1}{x_{0}}, k_{B}=\\frac{y_{0}+1}{x_{0}} then k_{A}\\cdot k_{B}=\\frac{y_{0}^{2}-1}{x_{0}^{2}}=-\\frac{m}{4}\\textcircled{1}, also point P lies on the ellipse \\frac{x^{2}}{m}+y^{2}=1, y_{0}^{2}-1=-\\frac{x_{0}^{2}}{m}\\textcircled{2}, from \\textcircled{1}\\textcircled{2} we get m^{2}=4, \\because m>1, m=2. Thus the eccentricity e=\\frac{c}{a}=\\frac{1}{\\sqrt{2}}=\\frac{\\sqrt{2}}{2}, this problem is of medium difficulty" }, { "text": "Given that the focus of the parabola $x^{2}=4 y$ is $F$, if a point $P$ on the parabola is at a distance of $2$ from the $x$-axis, then the value of $P F$ is?", "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (x^2 = 4*y);Focus(G) = F;PointOnCurve(P, G);Distance(P, xAxis) = 2", "query_expressions": "LineSegmentOf(P, F)", "answer_expressions": "3", "fact_spans": "[[[2, 16], [25, 28]], [[31, 34]], [[20, 23]], [[2, 16]], [[2, 23]], [[25, 34]], [[31, 46]]]", "query_spans": "[[[48, 57]]]", "process": "The focus of the parabola $ x^{2} = 4y $ is $ F(0,1) $, and the directrix is $ y = -1 $. Since the distance from a point $ P $ on the parabola to the $ x $-axis is 2, by the definition of a parabola, we have $ |PF| = 2 + 1 = 3 $. The answer is: $ 3 $." }, { "text": "The line $ l $: $ y = kx + 2 $ intersects the hyperbola $ C $: $ \\frac{x^{2}}{2} - y^{2} = 1 $ at two points. What is the range of values for $ k $?", "fact_expressions": "l: Line;C: Hyperbola;Expression(C) = (x^2/2 - y^2 = 1);Expression(l)=(y=k*x+2);NumIntersection(l,C)=2;k:Number", "query_expressions": "Range(k)", "answer_expressions": "(-sqrt(10)/2,-sqrt(2)/2)+(-sqrt(2)/2,sqrt(2)/2)+(sqrt(2)/2,sqrt(10)/2)", "fact_spans": "[[[0, 16]], [[17, 50]], [[17, 50]], [[0, 16]], [[0, 56]], [[57, 60]]]", "query_spans": "[[[57, 67]]]", "process": "Solve the system of equations of line $ l $ and hyperbola $ C $: \n$$\n\\begin{cases}\n\\frac{x^{2}}{2} - y^{2} = 1 \\\\\ny = kx + 2\n\\end{cases}\n$$\nEliminating $ y $, we obtain $ (1 - 2k^{2})x^{2} - 8kx - 10 = 0 $. Since line $ l $ and hyperbola $ C $ have common points, we can find the answer. \nSolve the system of equations of line $ l $ and hyperbola $ C $: \n$$\n\\begin{cases}\n\\frac{x^{2}}{2} - y^{2} = 1 \\\\\ny = kx + 2\n\\end{cases}\n$$\nEliminating $ y $, we get $ (1 - 2k^{2})x^{2} - 8kx - 10 = 0 $. \n$ \\because $ line $ l $ and hyperbola $ C $ have common points, \n$ \\therefore $\n$$\n\\begin{cases}\n1 - 2k^{2} \\neq 0 \\\\\n\\Delta = (8k)^{2} + 40(1 - 2k^{2}) \\geq 0\n\\end{cases}\n$$\nSolving gives $ -\\frac{\\sqrt{10}}{2} < k < -\\frac{\\sqrt{2}}{2} $ or $ -\\frac{\\sqrt{2}}{2} < k < \\frac{\\sqrt{2}}{2} $ and $ \\sqrt{2} $" }, { "text": "Given that the foci and the endpoints of the real axis of hyperbola $ C $ are exactly the endpoints of the major axis and the foci of the ellipse $ \\frac{x^{2}}{25} + \\frac{y^{2}}{16} = 1 $, respectively, then the equation of the asymptotes of hyperbola $ C $ is?", "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(G)=(x^2/25+y^2/16=1);Focus(C)=Endpoint(MajorAxis(G));Endpoint(RealAxis(C))=Focus(G)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "pm*3*y+4*x=0", "fact_spans": "[[[2, 8], [68, 74]], [[19, 58]], [[19, 58]], [[2, 66]], [[2, 66]]]", "query_spans": "[[[68, 82]]]", "process": "" }, { "text": "If the focus of the parabola $y^{2}=2 px$ coincides with the right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, then the equation of the directrix of this parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;H: Ellipse;Expression(H) = (x^2/9 + y^2/5 = 1);Focus(G) = RightFocus(H)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x=-2", "fact_spans": "[[[1, 16], [66, 69]], [[1, 16]], [[4, 16]], [[20, 57]], [[20, 57]], [[1, 63]]]", "query_spans": "[[[66, 76]]]", "process": "The right focus of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$ is $(2,0)$, therefore $\\frac{p}{2}=2$, $p=4$, and the equation of the directrix is $x=-2$. 【Topic】Geometric properties of ellipses and parabolas" }, { "text": "Given that the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ ($a>0$, $b>0$) intersect the circle $x^{2}+y^{2}-4x+2=0$, then the range of the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Circle;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (-4*x + x^2 + y^2 + 2 = 0);IsIntersect(Asymptote(G),H)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,sqrt(2)]", "fact_spans": "[[[2, 59], [92, 95]], [[5, 59]], [[5, 59]], [[64, 86]], [[5, 59]], [[5, 59]], [[2, 59]], [[64, 86]], [[2, 89]]]", "query_spans": "[[[92, 106]]]", "process": "Problem Analysis: The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}\\cdot\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ and the circle $x^{2}+y^{2}-4x+2=0$ have intersection points; the distance from the point $(2,0)$ to the asymptotes is $\\leqslant$ the radius $r$. Solve accordingly. From the circle $x^{2}+y^{2}-4x+2=0$, rewrite as $(x-2)^{2}+y^{2}=2$, yielding center $(2,0)$ and radius $r=\\sqrt{2}$. $\\because$ the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}\\cdot\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$, given by $y=\\pm\\frac{b}{a}x$, intersect the circle $x^{2}+y^{2}-4x+2=0$, we have $\\frac{|2b|}{\\sqrt{a^{2}+b^{2}}}\\leqslant\\sqrt{2}$, simplifying to $b^{2}\\leqslanta^{2}$. $\\therefore 10)$ has a focus at $(4 , 0)$. What is the eccentricity of this ellipse?", "fact_expressions": "G: Ellipse;m: Number;m>0;Expression(G)=(y^2/9 + x^2/m^2 = 1);Coordinate(OneOf(Focus(G)))=(4,0)", "query_expressions": "Eccentricity(G)", "answer_expressions": "4/5", "fact_spans": "[[[0, 48], [66, 68]], [[2, 48]], [[2, 48]], [[0, 48]], [0, 60]]", "query_spans": "[[[66, 74]]]", "process": "" }, { "text": "Given that $F$ is the right focus of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{8}=1$, $P$ is a point on the left branch of $C$, and $A(0,6 \\sqrt{6})$, when the perimeter of $\\triangle A P F$ is minimized, the area of the triangle is?", "fact_expressions": "C: Hyperbola;A: Point;P: Point;F: Point;Expression(C) = (x^2 - y^2/8 = 1);Coordinate(A) = (0, 6*sqrt(6));RightFocus(C) = F;PointOnCurve(P, LeftPart(C));WhenMin(Perimeter(TriangleOf(A,P,F)))", "query_expressions": "Area(TriangleOf(A,P,F))", "answer_expressions": "12*sqrt(6)", "fact_spans": "[[[6, 39], [48, 51]], [[57, 74]], [[44, 47]], [[2, 5]], [[6, 39]], [[57, 74]], [[2, 43]], [[44, 56]], [[75, 98]]]", "query_spans": "[[[76, 108]]]", "process": "Let the left focus of the hyperbola be F_{1}. By the definition of a hyperbola, |PF| = 2a + |PF_{1}|. Therefore, the perimeter of \\triangle APF is |PA| + |PF| + |AF| = |PA| + 2a + |PF_{1}| + |AF| = |PA| + |PF_{1}| + |AF| + 2a. Since 2a + |AF| is constant, to minimize the perimeter of \\triangle APF, |PA| + |PF_{1}| must be minimized, which occurs when P, A, and F_{1} are collinear. Given A(0, 6\\sqrt{6}), F_{1}(-3, 0), the equation of line AF_{1} is \\frac{x}{-3} + \\frac{y}{6\\sqrt{6}} = 1, or x = \\frac{y}{2\\sqrt{6}} - 3. Substituting into x^{2} - \\frac{y^{2}}{8} = 1 and simplifying yields y^{2} + 6\\sqrt{6}y - 96 = 0, solving gives y = 2\\sqrt{6} or y = -8\\sqrt{6} (discarded). Thus, the y-coordinate of point P is 2\\sqrt{6}. Therefore, S_{\\triangle APF} = S_{\\triangle AFF_{1}} - S_{\\triangle PF_{1}F} = \\frac{1}{2} \\times 6 \\times 6\\sqrt{6} - \\frac{1}{2} \\times 6 \\times 2\\sqrt{6} = 12\\sqrt{6}. The answer is: 12\\sqrt{6}. [Note] This problem examines the area of a triangle in a hyperbola, involving the definition of a hyperbola, and is a moderate-level comprehensive problem." }, { "text": "Given that the focus of the parabola $x^{2}=-2 p y(p>0)$ has coordinates $F(0,-3)$, find the coordinates of the midpoint of the chord cut by the line $y=x$ on the parabola.", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = -2*p*y);p: Number;p>0;F: Point;Coordinate(F) = (0, -3);Focus(G) = F;H: Line;Expression(H) = (y = x)", "query_expressions": "Coordinate(MidPoint(InterceptChord(H, G)))", "answer_expressions": "(-6, -6)", "fact_spans": "[[[2, 24], [49, 52]], [[2, 24]], [[5, 24]], [[5, 24]], [[30, 39]], [[30, 39]], [[2, 39]], [[41, 48]], [[41, 48]]]", "query_spans": "[[[41, 63]]]", "process": "According to the coordinates of the focus of the parabola, the standard equation of the parabola can be determined. By solving simultaneously the equations of the parabola and the line y = x, the coordinates of the two intersection points can be obtained, and then the midpoint coordinates of the chord can be found using the midpoint formula. From the focus coordinates of the parabola, we get p = 6, hence the equation of the parabola is x^{2} = -12y. Solving the system of equations \\begin{cases} y = x \\\\ x^{2} = -12y \\end{cases}, rearranging gives x^{2} + 12x = 0, solving yields x = 0 or x = -12, so the coordinates of the two intersection points are \\begin{cases} x = 0 \\\\ y = 0 \\end{cases} and \\begin{cases} x = -12 \\\\ y = -12 \\end{cases}. Therefore, by the midpoint formula, the midpoint coordinates of the chord are (-6, -6)." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $e=2$, what is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = e ;e: Number;e = 2", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*sqrt(3)*x", "fact_spans": "[[[2, 58], [69, 70]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 67]], [[62, 67]], [[62, 67]]]", "query_spans": "[[[69, 78]]]", "process": "" }, { "text": "Let $O$ be the coordinate origin, $F$ be the focus of the parabola $y^{2}=4x$, and $P$ be a point on the parabola. If $|P F|=3$, then the area of $\\triangle O P F$ is?", "fact_expressions": "G: Parabola;O: Origin;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(P,G);Abs(LineSegmentOf(P, F)) = 3", "query_expressions": "Area(TriangleOf(O, P, F))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[36, 39], [10, 24]], [[1, 4]], [[32, 35]], [[28, 31]], [[10, 24]], [[10, 31]], [[32, 42]], [[44, 53]]]", "query_spans": "[[[55, 77]]]", "process": "From the problem, we have F(1,0). Let P(x_{0},y_{0}). Then by the definition of a parabola, |PF| = x_{0} + 1 = 3. Solving gives x_{0} = 2. Substituting into the parabola equation yields |y_{0}| = 2\\sqrt{2}. Therefore, S_{\\triangleOPF} = \\frac{1}{2} \\times |OF| \\times |y_{1}| = \\frac{1}{2} \\times 1 \\times 2\\sqrt{2} = \\sqrt{2}. The answer is ^{^{2}}" }, { "text": "Let $A B$ be the major axis of ellipse $\\Gamma$, point $C$ lies on $\\Gamma$, and $\\angle C B A = \\frac{\\pi}{4}$. If $A B = 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": "[[[7, 17], [26, 34], [93, 101]], [[1, 6]], [[1, 6]], [[21, 25]], [], [], [[1, 20]], [[21, 35]], [[37, 65]], [[67, 74]], [[76, 91]], [[93, 106]]]", "query_spans": "[[[93, 113]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $x=\\frac{1}{4 m} y^{2}$ are?", "fact_expressions": "G: Parabola;m: Number;Expression(G) = (x = y^2/(4*m))", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(m,0)", "fact_spans": "[[[0, 26]], [[3, 26]], [[0, 26]]]", "query_spans": "[[[0, 33]]]", "process": "The standard equation of the parabola is y^{2}=4mx, and the coordinates of the focus are (m,0)." }, { "text": "Given circle $M$: $(x-4)^{2}+y^{2}=169$, circle $N$: $(x+4)^{2}+y^{2}=9$, and a moving circle $P$ that is internally tangent to circle $M$ and externally tangent to circle $N$, what is the equation of the locus of the center $P$ of the moving circle?", "fact_expressions": "M: Circle;Expression(M) = (y^2 + (x - 4)^2 = 169);N: Circle;Expression(N) = (y^2 + (x + 4)^2 = 9);P: Circle;IsInTangent(P,M) = True;IsOutTangent(P,N) = True;P1: Point;Center(P) = P1", "query_expressions": "LocusEquation(P1)", "answer_expressions": "x^2/64+y^2/48=1", "fact_spans": "[[[2, 28], [60, 64]], [[2, 28]], [[29, 53], [69, 73]], [[29, 53]], [[56, 59]], [[56, 67]], [[56, 76]], [[82, 85]], [[78, 85]]]", "query_spans": "[[[82, 92]]]", "process": "From M: (x-4)^{2}+y^{2}=169, we obtain the center of the circle M(4,0) and radius r_{1}=13. From N: (x+4)^{2}+y^{2}=9, we obtain the center of the circle N(-4,0) and radius r_{2}=3. Let the radius of the moving circle be r. Since the moving circle P is internally tangent to circle M and externally tangent to circle N, we have |PM|=13-r and |PN|=3+r. Also, |MN|=8. Adding the two equations, we get |PM|+|PN|=16>|MN|=8. By the definition of an ellipse, the locus of point P is an ellipse with foci M and N, where 2a=16, 2c=8, so a=8, c=4, and thus b^{2}=a^{2}-c^{2}=48. Therefore, the equation of the ellipse is \\frac{x^{2}}{64}+\\frac{y^{2}}{48}=1. That is, the trajectory equation of the center P of the moving circle is \\frac{x^{2}}{64}+\\frac{y^{2}}{48}=1" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{3}=1$, and $A$, $B$ are the left vertex and the upper vertex of the ellipse respectively. Point $P$ lies on the line segment $AB$. Then the minimum value of $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}$ is?", "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (x^2/6 + y^2/3 = 1);Focus(G) = {F1, F2};A: Point;B: Point;LeftVertex(G) = A;UpperVertex(G) = B;P: Point;PointOnCurve(P, LineSegmentOf(A, B)) = True", "query_expressions": "Min(DotProduct(VectorOf(P, F1), VectorOf(P, F2)))", "answer_expressions": "-1", "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 55], [72, 74]], [[18, 55]], [[2, 60]], [[61, 64]], [[65, 68]], [[61, 82]], [[61, 82]], [[83, 87]], [[83, 96]]]", "query_spans": "[[[98, 161]]]", "process": "From the problem, we have $ F_{1}(-\\sqrt{3},0) $, $ F_{2}(\\sqrt{3},0) $, $ A(-\\sqrt{6},0) $, $ B(0,\\sqrt{3}) $, let $ P(x,y) $. Since point $ P $ lies on segment $ AB $, it follows that $ y = \\frac{\\sqrt{2}}{2}x + \\sqrt{3} $, $ -\\sqrt{6} \\leqslant x \\leqslant 0 $. Therefore, $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = (x+\\sqrt{3},y) \\cdot (x-\\sqrt{3},y) = x^{2} + y^{2} - 3 = \\frac{3}{2}x^{2} + \\sqrt{6}x = \\frac{3}{2}\\left(x + \\frac{\\sqrt{6}}{3}\\right)^{2} - 1 $. Hence, when $ x = -\\frac{\\sqrt{6}}{3} $, the minimum value of $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} $ is $ -1 $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $C$: $\\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1$, a line passing through $F_{1}$ intersects the ellipse $C$ at points $M$ and $N$. Then, what is the perimeter of $\\Delta F_{2} M N$?", "fact_expressions": "C: Ellipse;G: Line;F2: Point;M: Point;N: Point;F1: Point;Expression(C) = (x^2/3 + y^2/4 = 1);Focus(C) = {F1, F2};PointOnCurve(F1,G);Intersection(G,C) = {M,N}", "query_expressions": "Perimeter(TriangleOf(F2, M, N))", "answer_expressions": "8", "fact_spans": "[[[18, 60], [78, 83]], [[75, 77]], [[10, 17]], [[85, 88]], [[89, 92]], [[2, 9], [67, 74]], [[18, 60]], [[2, 65]], [[66, 77]], [[75, 94]]]", "query_spans": "[[[96, 119]]]", "process": "Since the ellipse C: \\frac{x^{2}}{3}+\\frac{y^{2}}{4}=1, we have a^{2}=4, so a=2. Therefore, C_{\\DeltaF_{2}MN}=|MN|+|F_{2}M|+|F_{2}N|=|F_{1}M|+|F_{1}N|+|F_{2}M|+|F_{2}N|=4a=8" }, { "text": "Given point $A(-1,1)$ and ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, $M$ is a moving point on ellipse $C$, $F$ is the right focus of ellipse $C$, then the minimum value of $|M A|+2|M F|$ is?", "fact_expressions": "A: Point;Coordinate(A) = (-1, 1);C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);M: Point;PointOnCurve(M, C) = True;F: Point;RightFocus(C) = F", "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + 2*Abs(LineSegmentOf(M, F)))", "answer_expressions": "5", "fact_spans": "[[[2, 12]], [[2, 12]], [[13, 55], [62, 67], [76, 81]], [[13, 55]], [[58, 61]], [[57, 71]], [[72, 75]], [[72, 85]]]", "query_spans": "[[[87, 107]]]", "process": "From the ellipse equation, we have $ e = \\frac{c}{a} = \\frac{1}{2} $. Therefore, $ \\frac{1}{2}|MA| + |MF| = \\frac{1}{2}(|MA| + 2|MF|) $. According to the second definition of the ellipse: draw a perpendicular line from A to the right directrix, intersecting at point N, where the equation of the right directrix is $ x = 4 $. Then $ |MA| + 2|MF| = |MA| + |MN| \\geqslant |AN| $. Since $ |AN| = 4 + 1 = 5 $, the answer is 5." }, { "text": "The equation $16 x^{2}+k y^{2}=16$ represents an ellipse. What is the range of values for $k$?", "fact_expressions": "G: Ellipse;Expression(G)=(k*y^2 + 16*x^2 = 16);k:Number", "query_expressions": "Range(k)", "answer_expressions": "{(0,1),(1,+oo)}", "fact_spans": "[[[25, 27]], [[0, 27]], [[29, 32]]]", "query_spans": "[[[29, 39]]]", "process": "" }, { "text": "The hyperbola $C$ shares common foci with the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and one asymptote of $C$ has the equation $x+\\sqrt{3} y=0$. Then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);Focus(C) = Focus(G);Expression(OneOf(Asymptote(C))) = (x + sqrt(3)*y = 0)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/3-y^2=1", "fact_spans": "[[[0, 6], [51, 54], [81, 84]], [[7, 44]], [[7, 44]], [[0, 49]], [[51, 79]]]", "query_spans": "[[[81, 89]]]", "process": "From the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, we get $a^{2}=9$, $b^{2}=5$, $c=\\sqrt{a^{2}-b^{2}}=2$. $\\therefore$ the foci of hyperbola $C$ are $F_{1}(-2,0)$, $F_{2}(2,0)$. Let the real semi-axis of the hyperbola be $a_{1}$, and the imaginary semi-axis be $b_{1}$. $\\because$ the asymptote equation is $x+\\sqrt{3}y=0$, i.e., $y=-\\frac{\\sqrt{3}}{3}x$, $\\therefore \\frac{b_{1}}{a_{1}}=\\frac{\\sqrt{3}}{3}$. Also $c_{1}=c=2$, and $a_{1}^{2}+b_{1}^{2}=c_{1}^{2}$. Solving gives $a_{1}^{2}=3$, $b_{1}^{2}=1$. $\\therefore$ the equation of hyperbola $C$ is $\\frac{x^{2}}{3}-y^{2}=1$." }, { "text": "If the asymptotes of the hyperbola $x^{2}-\\frac{y^{2}}{m}=1$ are given by $y=\\pm 2 x$, then the real number $m$=?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2 - y^2/m = 1);Expression(Asymptote(G)) = (y = pm*(2*x))", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[1, 29]], [[49, 54]], [[1, 29]], [[1, 47]]]", "query_spans": "[[[49, 56]]]", "process": "The hyperbola \\( x^{2} - \\frac{y^{2}}{m} = 1 \\) has its foci on the x-axis, \\( \\therefore \\) the asymptotes are \\( y = \\pm\\sqrt{m}x \\), \\( \\therefore \\sqrt{m} = 2 \\Rightarrow m = 4 \\)" }, { "text": "A line passing through the focus of the parabola $y^{2}=2 p x$, $(p>0)$, intersects the parabola at points $A$ and $B$, with $|A B|=3$, and the ordinate of the midpoint of $A B$ is $\\frac{1}{2}$. Find the value of $p$.", "fact_expressions": "G: Parabola;p: Number;H: Line;A: Point;B: Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(Focus(G), H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 3;YCoordinate(MidPoint(LineSegmentOf(A, B))) = 1/2", "query_expressions": "p", "answer_expressions": "(3 + pm*sqrt(5))/4", "fact_spans": "[[[1, 23], [29, 32]], [[83, 86]], [[26, 28]], [[34, 37]], [[38, 41]], [[4, 23]], [[1, 23]], [[0, 28]], [[26, 43]], [[44, 53]], [[55, 81]]]", "query_spans": "[[[83, 90]]]", "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}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the ellipse at points $A$ and $B$. If $|A B|=|F_{1} F_{2}|$ and $|F_{1} A|=\\frac{1}{2}|F_{1} B|$, 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 > b;b > 0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;A: Point;B: Point;PointOnCurve(F2, G);Intersection(G, C) = {A, B};Abs(LineSegmentOf(A, B)) = Abs(LineSegmentOf(F1, F2));Abs(LineSegmentOf(F1, A)) = 1/2*Abs(LineSegmentOf(F1, B))", "query_expressions": "Eccentricity(C)", "answer_expressions": "(1+sqrt(145))/18", "fact_spans": "[[[2, 59], [95, 97], [168, 173]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[67, 74]], [[75, 82], [84, 91]], [[2, 82]], [[2, 82]], [[92, 94]], [[99, 102]], [[104, 107]], [[83, 94]], [[92, 109]], [[111, 132]], [[134, 166]]]", "query_spans": "[[[168, 179]]]", "process": "As shown in the figure, the perimeter of $\\triangle ABF$ is $4a$, $|AB|=|F_{1}F_{2}|$, $|F_{1}A|=\\frac{1}{2}|F_{1}B|$, so $|BF_{1}|=\\frac{8a-4c}{3}$, $|AF_{1}|=\\frac{4a-2c}{3}$, $|BF_{2}|=\\frac{4c-2a}{3}$, $|AF_{2}|=\\frac{2c+2}{3}$. Also, $\\cos\\angle BF_{2}F_{1}=-\\cos\\angle AF_{2}F_{1}$, so $\\frac{|BF_{2}|^{2}+|F_{1}F_{2}|^{2}-|BF_{1}|^{2}}{2|BF_{2}||F_{1}F_{2}|}=-\\frac{|AF_{2}|^{2}+|F_{1}F_{2}|^{2}-|AF_{1}|^{2}}{2|AF_{2}||F_{1}F_{2}|}$. Then simplifying gives $9c^{2}-4a^{2}-ac=0$, so $9e^{2}-e-4=0$, hence $e=\\frac{1\\pm\\sqrt{145}}{18}$. Since $e>1$, we have $e=\\frac{1+\\sqrt{145}}{18}$." }, { "text": "The coordinates of the focus of the parabola $y=\\frac{1}{2} x^{2}$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/2)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,1/2)", "fact_spans": "[[[0, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "Rewriting the parabola equation in standard form gives, from y = \\frac{1}{2}x^2, we obtain x^{2} = 2y. Therefore, 2p = 2, p = 1, so the focus of the parabola y = \\frac{1}{2}x^2 has coordinates (0, \\frac{1}{2}) as the answer." }, { "text": "Points $M(-3,0)$, $N(3,0)$, a moving point $P$ satisfies $|P M|=10-|P N|$, then the trajectory equation of point $P$ is?", "fact_expressions": "M: Point;N: Point;P: Point;Coordinate(M) = (-3, 0);Coordinate(N) = (3, 0);Abs(LineSegmentOf(P, M)) = 10 - Abs(LineSegmentOf(P, N))", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/24+y^2/16=1", "fact_spans": "[[[0, 10]], [[11, 20]], [[23, 26], [46, 50]], [[0, 10]], [[11, 20]], [[28, 44]]]", "query_spans": "[[[46, 57]]]", "process": "" }, { "text": "The ellipse $E$: $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ has its left focus at $F_{1}$. The line $x=m$ intersects the ellipse $E$ at points $A$ and $B$. When the perimeter of $\\Delta F_{1} A B$ is maximized, what is the value of $m$?", "fact_expressions": "E: Ellipse;G: Line;m: Number;F1: Point;A: Point;B: Point;Expression(E) = (x^2/25 + y^2/9 = 1);Expression(G) = (x = m);LeftFocus(E) = F1;Intersection(G, E) = {A, B};WhenMax(Perimeter(TriangleOf(F1,A,B)))", "query_expressions": "m", "answer_expressions": "4", "fact_spans": "[[[0, 43], [64, 69]], [[56, 63]], [[108, 111]], [[48, 55]], [[71, 74]], [[75, 78]], [[0, 43]], [[56, 63]], [[0, 55]], [[56, 80]], [[81, 106]]]", "query_spans": "[[[108, 116]]]", "process": "Let the right focus of the ellipse be E. By the definition of the ellipse, the perimeter of $\\triangle FAB$ is: $AB + AF + BF = AB + (2a - AE) + (2a - BE) = 4a + AB - AE - BE$. Using the triangle inequality relating the sum of two sides to the third side, the result follows. [Detailed Solution] Let the right focus of the ellipse be E. By the definition of the ellipse, the perimeter of $\\triangle FAB$ is: $AB + AF + BF = AB + (2a - AE) + (2a - BE) = 4a + AB - AE - BE$. Since $AE + BE > AB$, it follows that $AB - AE - BE \\leqslant 0$, with equality when AB passes through point E. Therefore, $AB + AF + BF = 4a + AB - AE - BE \\leqslant 4a$; that is, the perimeter of $\\triangle FAB$ is maximized when the line $x = m$ passes through the right focus E of the ellipse; hence the line $x = m = c = 4$." }, { "text": "Given that the equation of an ellipse with foci on the $x$-axis is $(m-1) x^{2}+(3 m-4) y^{2}=m$, what is the range of $m$?", "fact_expressions": "G: Ellipse;PointOnCurve(Focus(G), xAxis);m: Number;Expression(G) = (x^2*(m - 1) + y^2*(3*m - 4) = m)", "query_expressions": "Range(m)", "answer_expressions": "(-oo, 0) + (3/2, +oo)", "fact_spans": "[[[11, 13]], [[2, 13]], [[47, 50]], [[11, 45]]]", "query_spans": "[[[47, 55]]]", "process": "When m=0, the original equation is not an ellipse, so (m-1)x^{2}+(3m-4)y^{2}=m \\Rightarrow \\frac{x^{2}}{m-1}+\\frac{y^{2}}{3m-4}=1. Since the foci lie on the x-axis, we have \\begin{cases}\\frac{m}{m-1}>0\\\\\\frac{m}{3m-4}>0\\\\\\frac{m}{m-1}>\\frac{m}{3m-4}\\end{cases}. Solving gives m<0 or m>\\frac{3}{2}." }, { "text": "Given an ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ with eccentricity $e=\\frac{1}{2}$. A line passing through the center of the ellipse intersects the ellipse at points $A$ and $B$ ($A$ in the first quadrant). Draw a perpendicular from $A$ to the $x$-axis intersecting the ellipse at point $C$. Draw a line $AP$ perpendicular to $AB$ intersecting the ellipse at point $P$. Connect $BP$ intersecting $AC$ at point $Q$. Then $\\frac{AQ}{QC}$=?", "fact_expressions": "G: Ellipse;b: Number;a: Number;e: Number;H: Line;A: Point;P: Point;B: Point;C: Point;Q: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Eccentricity(G) = e ;e = 1/2;PointOnCurve(Center(G), H);Intersection(H, G) = {A, B};Quadrant(A) = 1;Z: Line;PointOnCurve(A, Z);IsPerpendicular(Z, xAxis);Intersection(Z, G) = C;PointOnCurve(A, LineOf(A, P));IsPerpendicular(LineOf(A, P), LineSegmentOf(A, B));Intersection(LineOf(A, P), G) = P;Intersection(LineSegmentOf(B, P), LineSegmentOf(A, C)) = Q", "query_expressions": "LineSegmentOf(A, Q)/LineSegmentOf(Q, C)", "answer_expressions": "1/3", "fact_spans": "[[[2, 47], [67, 69], [75, 77], [111, 113], [139, 141]], [[4, 47]], [[4, 47]], [[50, 65]], [[72, 74]], [[78, 81], [89, 92], [100, 103], [120, 123]], [[142, 146]], [[82, 85]], [[114, 118]], [[161, 165]], [[2, 47]], [[2, 65]], [[50, 65]], [[66, 74]], [[72, 87]], [[89, 97]], [], [[99, 110]], [[99, 110]], [[99, 118]], [[119, 131]], [[124, 138]], [[124, 146]], [[149, 165]]]", "query_spans": "[[[167, 186]]]", "process": "" }, { "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/25 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0,pm*3)", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 46]]]", "process": "Converting the ellipse into standard form gives \\frac{y^{2}}{25}+\\frac{x^{2}}{16}=1, so a^{2}=25, b^{2}=16, the foci lie on the y-axis, thus c^{2}=9, c=3, therefore the coordinates of the foci of the ellipse \\frac{x^{2}}{16}+\\frac{y^{2}}{25}=1 are (0,\\pm3)." }, { "text": "Given $A(2,1)$, $B(2,-1)$, and $O$ as the origin, a moving point $P(x, y)$ satisfies $\\overrightarrow{O P}=m \\overrightarrow{O A}+n \\overrightarrow{O B}$, where $m, n \\in \\mathbf{R}$ and $m^{2}+n^{2}=\\frac{1}{2}$. Then the trajectory equation of the moving point $P$ is?", "fact_expressions": "A: Point;B: Point;P: Point;O: Origin;x1:Number;y1:Number;m:Real;n:Real;Coordinate(A) = (2, 1);Coordinate(B) = (2, -1);Coordinate(P) = (x1, y1);VectorOf(O, P) = m*VectorOf(O, A) + n*VectorOf(O, B);m^2 + n^2 = 1/2", "query_expressions": "LocusEquation(P)", "answer_expressions": "", "fact_spans": "[[[2, 10]], [[12, 22]], [[35, 44], [169, 172]], [[23, 27]], [[35, 44]], [[35, 44]], [[117, 138]], [[117, 138]], [[2, 10]], [[12, 21]], [[35, 44]], [[46, 114]], [[140, 165]]]", "query_spans": "[[[169, 179]]]", "process": "Let a moving point be P(x, y). According to the relationship between vectors, we get x = 2m + 2n, y = m - n. Substituting into m^{2} + n^{2} = \\frac{1}{2}, we can simplify to obtain the trajectory equation of the moving point P. Let the moving point be P(x, y), then since point P satisfies \\overrightarrow{OP} = m\\overrightarrow{OA} + n\\overrightarrow{OB}, where m, n \\in \\mathbb{R}, \\therefore (x, y) = (2m + 2n, m - n), \\therefore x = 2m + 2n, y = m - n, \\therefore m = \\frac{x + 2y}{4}, n = \\frac{x - 2y}{4}, \\because m^{2} + n^{2} = \\frac{1}{2}, \\therefore (\\frac{x + 2y}{4})^{2} + (\\frac{x - 2y}{4})^{2} = \\frac{1}{2}, which simplifies to \\frac{x^{2}}{4} + y^{2} = 1." }, { "text": "Given the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, its two foci are $F_{1}$ and $F_{2}$ respectively. Point $P$ lies on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$. Find the area of $\\triangle F_{1} P F_{2}$.", "fact_expressions": "G: Hyperbola;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/4 - 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)", "fact_spans": "[[[2, 30], [60, 63]], [[38, 46]], [[55, 59]], [[47, 54]], [[2, 30]], [[2, 54]], [[55, 64]], [[67, 100]]]", "query_spans": "[[[102, 131]]]", "process": "Because |F_{1}F_{2}^{2}|=|PF_{1}^{2}|+|PF_{2}|-2|PF_{1}||PF_{2}|\\cos\\angleF_{1}PF_{2}=(|PF_{1}|-|PF_{2}|)^{2}+2|PF_{1}||PF_{2}|-2|PF_{1}||PF_{2}|\\cos\\angleF_{1}PF_{2}, so 4(4+1)=(2\\times2)^{2}+2|PF_{1}|_{PF_{2}}|-2|PF_{1}||PF_{2}|\\cos\\frac{\\pi}{3} \\therefore |PF_{1}|PF_{2}|=4, S_{\\DeltaPF_{1}}F_{2}=\\frac{1}{2}\\times4\\times\\sin\\frac{\\pi}{3}=\\sqrt{3}. This problem examines the definition of a hyperbola and the focal triangle, testing basic analytical and solving abilities, and is of medium difficulty." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ shares the same foci with the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, and its left directrix is given by $x=-1$. Let $F_{1}$, $F_{2}$ be 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^2/25 + y^2/9 = 1);Focus(C)=Focus(G);Expression(LeftDirectrix(C))=(x=-1);P: Point;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(P,RightPart(C))", "query_expressions": "Min(LineSegmentOf(P, F1)^2/LineSegmentOf(P, F2))", "answer_expressions": "16", "fact_spans": "[[[2, 63], [140, 146]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[64, 102]], [[64, 102]], [[2, 107]], [[2, 120]], [[155, 158]], [[122, 129]], [[130, 137]], [[122, 154]], [[122, 154]], [[140, 166]]]", "query_spans": "[[[168, 203]]]", "process": "" }, { "text": "Given the equation of circle $C$ is $(x-1)^{2}+y^{2}=1$, and $P$ is a point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$. Two tangents are drawn from $P$ to the circle, touching the circle at points $A$ and $B$. Then the range of $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ is?", "fact_expressions": "C: Circle;Expression(C) = (y^2 + (x - 1)^2 = 1);G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);P: Point;PointOnCurve(P, G);L1: Line;L2: Line;TangentPoint(L1, C) = A;TangentPoint(L2, C) = B;A: Point;B: Point;TangentOfPoint(P, C) = {L1, L2}", "query_expressions": "Range(DotProduct(VectorOf(P, A), VectorOf(P, B)))", "answer_expressions": "[2*sqrt(2) - 3, 56/9]", "fact_spans": "[[[2, 6], [79, 80]], [[2, 28]], [[33, 70]], [[33, 70]], [[75, 78], [29, 32]], [[29, 73]], [], [], [[74, 96]], [[74, 96]], [[89, 92]], [[93, 96]], [72, 82]]", "query_spans": "[[[98, 154]]]", "process": "" }, { "text": "Given that the focal distance of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $2 \\sqrt{5}$, and the length of the real axis of the hyperbola is twice the length of the imaginary axis, 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;FocalLength(G) = 2*sqrt(5);Length(RealAxis(G)) = 2*Length(ImageinaryAxis(G))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2=1", "fact_spans": "[[[2, 58], [76, 79], [94, 97]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 74]], [[76, 92]]]", "query_spans": "[[[94, 102]]]", "process": "" }, { "text": "The equations of the asymptotes of the hyperbola $\\frac{y^{2}}{4}-\\frac{x^{2}}{16}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/16 + y^2/4 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(x/2)", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 47]]]", "process": "" }, { "text": "Given that the distance from one vertex of the major axis of the ellipse $\\frac{x^{2}}{a}+\\frac{y^{2}}{9}=1$ $(a>9)$ to the line $x-\\sqrt{3} y=0$ is at least $2$, then the range of the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/9 + x^2/a = 1);a: Number;a>9;H: Line;Expression(H) = (x - sqrt(3)*y = 0);Negation(Distance(OneOf(Vertex(MajorAxis(G))),H) < 2)", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[sqrt(7)/4, 1)", "fact_spans": "[[[2, 44], [81, 83]], [[2, 44]], [[4, 44]], [[4, 44]], [[52, 70]], [[52, 70]], [[2, 79]]]", "query_spans": "[[[81, 94]]]", "process": "The vertices of the major axis are $(\\pm\\sqrt{a},0)$, hence $\\underline{|\\sqrt{a}-0|}\\geqslant2$, i.e., $a\\geqslant16$. Thus, the eccentricity $e=\\sqrt{\\frac{a-9}{a}}=\\sqrt{1-\\frac{9}{a}}\\in\\left[\\frac{\\sqrt{7}}{4},1\\right)$." }, { "text": "Through the focus $F$ of the parabola $y^{2}=2x$, a line is drawn intersecting the parabola at points $A$ and $B$. If $|AB|=\\frac{25}{12}$, $|AF|<|BF|$, then $|AF|=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*x);F: Point;Focus(G) = F;H: Line;A: Point;B: Point;PointOnCurve(F, H);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 25/12;Abs(LineSegmentOf(A, F)) < Abs(LineSegmentOf(B, F))", "query_expressions": "Abs(LineSegmentOf(A, F))", "answer_expressions": "5/6", "fact_spans": "[[[1, 15], [25, 28]], [[1, 15]], [[18, 21]], [[1, 21]], [[22, 24]], [[29, 32]], [[33, 36]], [[0, 24]], [[22, 38]], [[40, 61]], [[64, 80]]]", "query_spans": "[[[83, 92]]]", "process": "" }, { "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 the point $P(m, 2m)$ lies on the hyperbola, find 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);Coordinate(OneOf(Focus(G))) = (m, 0);Coordinate(P) = (m, 2*m);PointOnCurve(P, G);m:Number;m>0", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)+1", "fact_spans": "[[[2, 61], [98, 101], [104, 107]], [[5, 61]], [[5, 61]], [[85, 96]], [[5, 61]], [[5, 61]], [[2, 61]], [[2, 83]], [[85, 96]], [[85, 102]], [[68, 83]], [[68, 83]]]", "query_spans": "[[[104, 113]]]", "process": "" }, { "text": "If the foci of the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{m}=1$ lie on the $x$-axis and the focal distance is $8$, then the real number $m$=?", "fact_expressions": "G: Hyperbola;m: Real;Expression(G) = (x^2/7 - y^2/m = 1);PointOnCurve(Focus(G), xAxis);FocalLength(G) = 8", "query_expressions": "m", "answer_expressions": "9", "fact_spans": "[[[2, 40]], [[59, 64]], [[2, 40]], [[2, 49]], [[2, 56]]]", "query_spans": "[[[59, 66]]]", "process": "According to the problem, for the hyperbola $\\frac{x^{2}}{7}-\\frac{y^{2}}{m}=1$, the foci lie on the $x$-axis and the focal distance is 8, so we have $m>0$ and $7+m=4^{2}$, solving gives $m=9$." }, { "text": "A point $M$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ has a distance of $\\frac{20}{3}$ to the right directrix. Then, what is the distance from point $M$ to the left focus?", "fact_expressions": "G: Ellipse;M: Point;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(M, G);Distance(M, RightDirectrix(G)) = 20/3", "query_expressions": "Distance(M, LeftFocus(G))", "answer_expressions": "6", "fact_spans": "[[[0, 39]], [[43, 46], [70, 74]], [[0, 39]], [[0, 46]], [[0, 68]]]", "query_spans": "[[[0, 83]]]", "process": "" }, { "text": "Given that the length of the imaginary axis of a hyperbola with foci on the $x$-axis is equal to the semi-focal distance, then the equations of the asymptotes of the hyperbola are?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G),xAxis) = True;Length(ImageinaryAxis(G)) = HalfFocalLength(G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(3)*x", "fact_spans": "[[[11, 14], [25, 28]], [[2, 14]], [[11, 23]]]", "query_spans": "[[[25, 36]]]", "process": "" }, { "text": "Let a line passing through an arbitrary point $P$ (distinct from the origin $O$) on the parabola $y^{2}=2 p x$ ($p>0$) intersect the parabola $y^{2}=8 p x$ ($p>0$) at points $A$ and $B$. The line $OP$ intersects the parabola $y^{2}=8 p x$ ($p>0$) again at point $Q$. Then $\\frac{S_{\\triangle A B Q}}{S_{\\triangle A B O}}$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*p*x);p: Number;p>0;P: Point;O: Origin;Negation(P=O);PointOnCurve(P, G);H: Line;PointOnCurve(P, H);Z: Parabola;Expression(Z) = (y^2 = 8*p*x);A: Point;B: Point;Intersection(H, Z)={A, B};Q: Point;Intersection(LineOf(O,P), Z)={P, Q}", "query_expressions": "Area(TriangleOf(A,B,Q))/Area(TriangleOf(A,B,O))", "answer_expressions": "3", "fact_spans": "[[[2, 23]], [[2, 23]], [[5, 23]], [[5, 23]], [[28, 31]], [[34, 39]], [[28, 41]], [[2, 31]], [[42, 44]], [[1, 44]], [[45, 66], [86, 107]], [[45, 66]], [[68, 71]], [[72, 75]], [[42, 77]], [[114, 117]], [[28, 117]]]", "query_spans": "[[[119, 170]]]", "process": "Analysis: Draw the figure, transform the ratio of triangle areas into a ratio of segment lengths, then into a ratio of coordinates. Set the coordinates of the points, write the equation of the line, solve the system of equations to find the coordinates of the intersection point, and finally substitute the coordinates to obtain the ratio. By drawing the corresponding figure, it can be observed that $\\frac{S_{\\triangle ABQ}}{S_{\\triangle ABO}} = \\frac{PQ}{OP} = \\frac{x_{Q}-x_{P}}{x_{P}} = \\frac{y_{Q}}{y_{P}} - 1$. Let $P\\left(\\frac{y_{1}^{2}}{2p}, y_{1}\\right)$, then the line $OP: y = \\frac{y_{1}}{\\frac{y_{1}^{2}}{2p}}x$, i.e., $y = \\frac{2p}{y_{1}}x$. Solving this together with $y^{2} = 8px$, we obtain $y_{O} = 4y_{1}$. Thus, the area ratio is $\\frac{4y_{1}}{y_{1}} - 1 = 3$. Hence, the answer is 3." }, { "text": "$P$ is a point on the right branch of the hyperbola $C$: $\\frac{x^{2}}{4}-y^{2}=1$, line $l$ is an asymptote of the hyperbola $C$, the projection of $P$ onto $l$ is $Q$, and $F_{1}$ is the left focus of the hyperbola $C$. Then the minimum value of $|P F_{1}|+|P Q|$ is?", "fact_expressions": "l: Line;C: Hyperbola;P: Point;F1: Point;Q: Point;Expression(C) = (x^2/4 - y^2 = 1);PointOnCurve(P, RightPart(C));OneOf(Asymptote(C))=l;Projection(P,l)=Q;LeftFocus(C)=F1", "query_expressions": "Min(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))", "answer_expressions": "5", "fact_spans": "[[[43, 48], [66, 69]], [[4, 37], [49, 55], [86, 92]], [[0, 3], [62, 65]], [[78, 85]], [[74, 77]], [[4, 37]], [[0, 42]], [[43, 61]], [[62, 77]], [[78, 96]]]", "query_spans": "[[[98, 121]]]", "process": "According to the problem, |PF₂| + |PQ| is minimized if and only if points Q, P, F₂ are collinear and P lies between F₂ and Q, and the minimum value is the distance from F₂ to line l. Thus, the minimum value of |PF₁| + |PQ| can be found. Let F₂ be the right focus of hyperbola C. Since |PF₁| - |PF₂| = 4, it follows that |PF₁| + |PQ| = 4 + |PF₂| + |PQ|. Clearly, |PF₂| + |PQ| is minimized if and only if points Q, P, F₂ are collinear and P lies between F₂ and Q, and the minimum value is the distance from F₂ to line l. The equation of line l is y = (1/2)x or y = -(1/2)x, and F₂(√5, 0). The distance from F₂ to line l is 1. Therefore, the minimum value of |PF₁| + |PQ| is 5. Answer: 5." }, { "text": "The hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$ has an asymptote with equation $y=\\sqrt{3} x$. Then the foci of the hyperbola are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/b^2 = 1);b: Number;b>0;Expression(OneOf(Asymptote(G))) = (y = sqrt(3)*x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(pm*2, 0)", "fact_spans": "[[[0, 37], [62, 65]], [[0, 37]], [[3, 37]], [[3, 37]], [[0, 60]]]", "query_spans": "[[[62, 70]]]", "process": "\\because the hyperbola x^{2}-\\frac{y^{2}}{b^{2}}=1 (b>0) has an asymptote with equation y=\\sqrt{3}x, \\therefore b=\\sqrt{3}, \\therefore c=2. \\therefore the foci of the hyperbola are (\\pm2,0)" }, { "text": "The right vertex of the hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ is $A$, and the right focus is $F$. A line passing through point $F$ and parallel to one asymptote of the hyperbola intersects the hyperbola at point $B$. Find the area of $\\triangle A F B$.", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1);A: Point;RightVertex(G) = A;RightFocus(G) = F;F: Point;PointOnCurve(F, H) = True;IsParallel(H, OneOf(Asymptote(G))) = True;H: Line;Intersection(H, G) = B;B: Point", "query_expressions": "Area(TriangleOf(A, F, B))", "answer_expressions": "32/15", "fact_spans": "[[[0, 39], [64, 67], [77, 80]], [[0, 39]], [[44, 47]], [[0, 47]], [[0, 55]], [[52, 55], [57, 61]], [[56, 76]], [[61, 76]], [[74, 76]], [[74, 86]], [[82, 86]]]", "query_spans": "[[[88, 110]]]", "process": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has right vertex $A(3,0)$, right focus $F(5,0)$, $a=3$, $b=4$, $c=5$, so the asymptotes are $y=\\pm\\frac{4}{3}x$. Without loss of generality, assume the equation of line $FB$ is $y=\\frac{4}{3}(x-5)$. Substituting $y=\\frac{4}{3}(x-5)$ into the hyperbola equation and simplifying yields $x^{2}-(x-5)^{2}=9$. Solving gives $x=\\frac{17}{5}=-\\frac{32}{15}$, so $B(\\frac{17}{5},-\\frac{32}{15})$. Therefore, $S_{\\triangle AFB}=\\frac{15}{2}|AF||y_{B}|=\\frac{1}{2}(c-a)|y_{B}|=\\frac{1}{2}\\times(5-3)\\times\\frac{32}{15}=\\frac{32}{15}$." }, { "text": "The center of the ellipse is at the origin, and it passes through the fixed point $(2,-3)$. One of its foci coincides with the focus of the parabola $y^{2}=8x$. Then the equation of the ellipse is?", "fact_expressions": "O: Origin;Center(H) = O;H: Ellipse;PointOnCurve(I, H);I: Point;Coordinate(I) = (2, -3);G: Parabola;Expression(G) = (y^2 = 8*x);OneOf(Focus(H)) = Focus(G)", "query_expressions": "Expression(H)", "answer_expressions": "x^2/16+y^2/12=1", "fact_spans": "[[[5, 7]], [[0, 7]], [[22, 23], [0, 2], [50, 52]], [[0, 21]], [[13, 21]], [[13, 21]], [[28, 42]], [[28, 42]], [[22, 47]]]", "query_spans": "[[[50, 57]]]", "process": "" }, { "text": "If the eccentricity $e = \\frac{1}{3}$ of the ellipse $\\frac{x^{2}}{k+8}+\\frac{y^{2}}{9}=1$, then what is the value of $k$?", "fact_expressions": "G: Ellipse;k: Number;e: Number;Expression(G) = (x^2/(k + 8) + y^2/9 = 1);Eccentricity(G) = e ;e = 1/3", "query_expressions": "k", "answer_expressions": "{0, 17/8}", "fact_spans": "[[[1, 40]], [[61, 64]], [[44, 59]], [[1, 40]], [[1, 59]], [[44, 59]]]", "query_spans": "[[[61, 68]]]", "process": "" }, { "text": "The distance from a moving point $P(x, y)$ to the point $F(3,0)$ is greater by $1$ than its distance to the line $x+2=0$. Then, the equation of the trajectory of the moving point is?", "fact_expressions": "G: Line;P: Point;F: Point;x1:Number;y1:Number;Expression(G) = (x + 2 = 0);Coordinate(P) = (x1, y1);Coordinate(F) = (3, 0);Distance(P, F) = Distance(P, G) + 1", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2 = 12*x", "fact_spans": "[[[27, 36]], [[2, 11], [45, 47], [25, 26]], [[12, 21]], [[2, 11]], [[2, 11]], [[27, 36]], [[2, 11]], [[12, 21]], [[2, 43]]]", "query_spans": "[[[45, 54]]]", "process": "" }, { "text": "The length of the major axis of the ellipse $x^{2} + 3 y^{2}=1$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2 + 3*y^2 = 1)", "query_expressions": "Length(MajorAxis(G))", "answer_expressions": "2", "fact_spans": "[[[0, 21]], [[0, 21]]]", "query_spans": "[[[0, 27]]]", "process": "" }, { "text": "The equation of the hyperbola passing through the point $M(2 \\sqrt{3}, 2 \\sqrt{5})$ and having the same asymptotes as the hyperbola $\\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1$ is?", "fact_expressions": "G: Hyperbola;C:Hyperbola;M: Point;Expression(G) = (x^2/3 - y^2/2 = 1);Coordinate(M) = (2*sqrt(3), 2*sqrt(5));Asymptote(G) = Asymptote(C);PointOnCurve(M,C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/12-x^2/18=1", "fact_spans": "[[[33, 71]], [[78, 81]], [[2, 31]], [[33, 71]], [[2, 31]], [[32, 81]], [[0, 81]]]", "query_spans": "[[[78, 85]]]", "process": "Let the required hyperbola equation be \\frac{x^{2}}{3}-\\frac{y^{2}}{2}=\\lambda(\\lambda\\neq0). Since the hyperbola passes through the point M(2\\sqrt{3},2\\sqrt{5}), we have \\frac{12}{3}-\\frac{20}{2}=-6=\\lambda. Therefore, the hyperbola equation is \\frac{y^{2}}{12}-\\frac{x^{2}}{18}=1." }, { "text": "Given that $M$ is a point on the parabola $y^{2}=2 p x(p>0)$, $F$ is the focus of the parabola, and a perpendicular is drawn from point $M$ to the directrix $l$, with foot of perpendicular at $E$. If $|E O|=|M F|$ and the horizontal coordinate of point $M$ is $3$, then $p=$?", "fact_expressions": "G: Parabola;p: Number;E: Point;O: Origin;M: Point;F: Point;l: Line;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(M, G);Focus(G) = F;L:Line;PointOnCurve(M, L);IsPerpendicular(L, l);FootPoint(L, l)=E;Abs(LineSegmentOf(E, O)) = Abs(LineSegmentOf(M, F));XCoordinate(M) = 3;Directrix(G)=l", "query_expressions": "p", "answer_expressions": "3", "fact_spans": "[[[6, 27], [35, 38]], [[93, 96]], [[59, 62]], [[65, 78]], [[42, 46], [79, 83], [2, 5]], [[31, 34]], [[49, 52]], [[9, 27]], [[6, 27]], [[2, 30]], [[31, 40]], [], [[41, 55]], [[41, 55]], [[41, 62]], [[65, 78]], [[79, 91]], [[35, 52]]]", "query_spans": "[[[93, 98]]]", "process": "According to the problem, we have M(3,\\pm\\sqrt{6p}), E(-\\frac{p}{2},\\pm\\sqrt{6p}), \\sqrt{(-\\frac{p}{2})^{2}+(\\pm\\sqrt{6p})^{2}}=3+\\frac{p}{2}. Solving gives: p=3." }, { "text": "The two asymptotes of the hyperbola $\\frac{x^{2}}{b^{2}}-\\frac{y^{2}}{a^{2}}=1$ are perpendicular to each other. What is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/b^2 - y^2/a^2 = 1);b: Number;a: Number;l1: Line;l2: Line;Asymptote(G) = {l1, l2};IsPerpendicular(l1, l2)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 46], [60, 63]], [[0, 46]], [[3, 46]], [[3, 46]], [], [], [[0, 52]], [[0, 56]]]", "query_spans": "[[[60, 69]]]", "process": "From the hyperbola equation, the asymptotes are given by $ y = \\pm\\frac{a}{b}x $. Since the two asymptotes are perpendicular, we have $ -\\frac{a}{b} \\times \\frac{a}{a} = -1 $, solving gives $ a = b $. Using $ e = \\sqrt{1 + \\frac{a^{2}}{b^{2}}} $, the result can be obtained. [Detailed solution] From the hyperbola $ \\frac{x^{2}}{b^{2}} - \\frac{y^{2}}{a^{2}} = 1 $, the asymptotes are $ y = \\pm\\frac{a}{b}x $. Since the two asymptotes are perpendicular, $ -\\frac{a}{b} \\times \\frac{a}{a} = -1 $, solving gives $ a = b $. The eccentricity of this hyperbola is $ e = \\sqrt{1 + \\frac{a^{2}}{b^{2}}} = \\sqrt{2} $." }, { "text": "If a point $P$ on the hyperbola $5 x^{2}-4 y^{2}=20$ is at a distance of $3$ from the right focus of the hyperbola, then what is the distance from point $P$ to the left directrix?", "fact_expressions": "G: Hyperbola;Expression(G) = (5*x^2 - 4*y^2 = 20);P: Point;PointOnCurve(P, G) = True;Distance(P, RightFocus(G)) = 3", "query_expressions": "Distance(P, LeftDirectrix(G))", "answer_expressions": "14/3", "fact_spans": "[[[2, 25], [33, 36]], [[2, 25]], [[29, 32], [49, 53]], [[2, 32]], [[29, 46]]]", "query_spans": "[[[33, 62]]]", "process": "" }, { "text": "Let $F(0,-3)$ be a focus of the ellipse $\\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1$ $(a>b>0)$, and let point $A(0,2)$. If there exists a point $P$ on the ellipse such that $|P A|+|P F|=9$, then the range of the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F: Point;Coordinate(F) = (0, -3);OneOf(Focus(G)) = F;A: Point;Coordinate(A) = (0, 2);P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F)) = 9", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "[3/5, 3/4]", "fact_spans": "[[[11, 63], [80, 82], [108, 110]], [[11, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[1, 10]], [[1, 10]], [[1, 68]], [[69, 78]], [[69, 78]], [[85, 89]], [[80, 89]], [[91, 106]]]", "query_spans": "[[[108, 120]]]", "process": "Let the upper focus of the ellipse be E(0,3). By the definition of an ellipse, |PE| + |PF| = 2a. Also, |PA| + |PF| = 9, so ||PA| - |PE|| = |2a - 9|. Since |EA| = 1, it follows that |2a - 9| ≤ 1, hence 4 ≤ a ≤ 5. The eccentricity of the ellipse e = \\frac{c}{a} \\in \\left[\\frac{3}{5}, \\frac{3}{4}\\right]. This problem examines the definition and geometric properties of an ellipse. Distance problems involving points on conic sections and foci are often solved using the definition." }, { "text": "The parabola takes the center of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ as its vertex and the left directrix of the ellipse as its directrix. This parabola intersects the right directrix of the ellipse at points $A$ and $B$. Find the value of $|A B|$.", "fact_expressions": "H: Ellipse;Expression(H) = (x^2/25 + y^2/16 = 1);G: Parabola;Center(H) = Vertex(G);LeftDirectrix(H) = Directrix(G);Intersection(G,RightDirectrix(H)) = {A,B};A: Point;B: Point", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "100/3", "fact_spans": "[[[1, 40], [48, 50], [62, 64]], [[1, 39]], [[58, 61]], [[0, 61]], [[47, 61]], [[58, 78]], [[69, 72]], [[73, 76]]]", "query_spans": "[[[80, 91]]]", "process": "" }, { "text": "The eccentricity of the hyperbola $x^{2}-\\frac{y^{2}}{5}=1$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2/5 = 1)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)", "fact_spans": "[[[0, 28]], [[0, 28]]]", "query_spans": "[[[0, 34]]]", "process": "According to the simple geometric properties of hyperbolas, we can calculate: since $x^{2}-\\frac{y^{2}}{5}=1$, then $a=1$, $b^{2}=5$, $\\therefore c=\\sqrt{a^{2}+b^{2}}=\\sqrt{6}$, $\\therefore e=\\frac{c}{a}=\\sqrt{6}$." }, { "text": "The equation of a hyperbola with the same foci as the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and eccentricity $2$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);Focus(G)=Focus(H);Eccentricity(G)=2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2/12=1", "fact_spans": "[[[54, 57]], [[1, 39]], [[1, 39]], [[0, 57]], [[46, 57]]]", "query_spans": "[[[54, 61]]]", "process": "The foci of the ellipse \\frac{x^2}{25}+\\frac{y^{2}}{9}=1 are (\\pm4,0). Since the foci of the hyperbola coincide with those of the ellipse, c=4. Given that the eccentricity of the hyperbola is 2, we have e=\\frac{c}{a}=2, solving gives a=2. Thus, b^{2}=c^{2}-a^{2}=16-4=12, so b=2\\sqrt{3}. Therefore, the equation of the hyperbola is \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1." }, { "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$. If the distance from $P$ to the left focus of the ellipse is $2$, then what is the distance from $P$ to the right focus?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/25 + y^2/9 = 1);PointOnCurve(P, G);Distance(P, LeftFocus(G)) = 2", "query_expressions": "Distance(P, RightFocus(G))", "answer_expressions": "8", "fact_spans": "[[[5, 43], [52, 54]], [[1, 4], [47, 50], [65, 68]], [[5, 43]], [[1, 46]], [[47, 64]]]", "query_spans": "[[[52, 77]]]", "process": "According to the definition of an ellipse, the distance from P to the right focus is obtained. Given that a = 5 and the distance from P to the left focus of the ellipse is 2, then the distance from P to the right focus is 5 × 2 − 2 = 8." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ passes through points $P(-\\sqrt{2},-\\sqrt{3})$ and $Q(\\frac{\\sqrt{15}}{3}, \\sqrt{2})$, $F_{1}$, $F_{2}$ are its left and right foci respectively, and point $M$ lies on this hyperbola. If $\\angle F_{1} M F_{2}=120^{\\circ}$, then the area of $\\Delta F_{1} M F_{2}$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;Q: Point;F1: Point;M: Point;F2: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (-sqrt(2), -sqrt(3));Coordinate(Q) = (sqrt(15)/3, sqrt(2));PointOnCurve(P,C);PointOnCurve(Q,C);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M,C);AngleOf(F1, M, F2) = ApplyUnit(120, degree)", "query_expressions": "Area(TriangleOf(F1, M, F2))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 63], [146, 147], [162, 165]], [[10, 63]], [[10, 63]], [[64, 88]], [[91, 125]], [[128, 135]], [[157, 160]], [[136, 143]], [[10, 63]], [[10, 63]], [[2, 63]], [[64, 88]], [[91, 125]], [[2, 88]], [[2, 125]], [[128, 155]], [[128, 155]], [[157, 166]], [[168, 202]]]", "query_spans": "[[[204, 231]]]", "process": "From the hyperbola equation, we know that 2a=4, 2c=2\\sqrt{5}, \\therefore |F_{1}F_{2}|^{2}=20. By the definition of the hyperbola, ||PF_{1}|-|PF_{2}||=4. Squaring both sides gives |PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|=16 \\textcircled{1}. By the law of cosines, |F_{1}F_{2}|^{2}=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\cos60^{\\circ}, \\therefore |PF_{1}|^{2}+|PF_{2}|^{2}-|PF_{1}||PF_{2}|=20 \\textcircled{2}. From \\textcircled{1} and \\textcircled{2}, we obtain |PF_{1}||PF_{2}|=20-16=4, \\therefore S_{\\Delta F_{1}PF_{2}}=\\frac{1}{2}|PF_{1}||PF_{2}|\\sin60^{\\circ}=\\frac{1}{2}\\times4\\times\\frac{\\sqrt{3}}{2}=\\sqrt{3}" }, { "text": "The hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has left and right foci $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through $F_{1}$ intersects the right branch of the hyperbola at point $P$. Given that $PF_{2} \\perp x$-axis and $\\sin \\angle PF_{1} F_{2}=\\frac{1}{3}$, find the eccentricity of the hyperbola?", "fact_expressions": "C1: Hyperbola;Expression(C1) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C1) = F1;RightFocus(C1) = F2;PointOnCurve(F1, l) = True;Intersection(l, RightPart(C1)) = P;l: Line;P: Point;IsPerpendicular(LineSegmentOf(P, F2), xAxis) = True;Sin(AngleOf(P, F1, F2)) = 1/3", "query_expressions": "Eccentricity(C1)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 65], [109, 112], [181, 184]], [[0, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[73, 81], [93, 100]], [[83, 90]], [[0, 90]], [[0, 90]], [[92, 108]], [[103, 119]], [[103, 108]], [[115, 119]], [[122, 139]], [[141, 179]]]", "query_spans": "[[[181, 190]]]", "process": "Using the given conditions, set up an equation and solve for the eccentricity of the hyperbola. [Detailed solution] From the given condition that PF₂ ⊥ x-axis and sin∠PF₁F₂ = 1/3, it follows that tan∠PF₁F₂ = √2/4. In triangle PF₁F₂, tan∠PF₁F₂ = a/(2c) = √2/4. Thus, 2e² - √2 e - 2 = 0. Solving gives e = √2." }, { "text": "Given that point $P$ is any point on the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, what is the minimum value of the distance from point $P$ to the line $l$: $x+2 y-12=0$?", "fact_expressions": "P: Point;G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);PointOnCurve(P, G) = True;l: Line;Expression(l) = (x + 2*y - 12 = 0)", "query_expressions": "Min(Distance(P, l))", "answer_expressions": "8*sqrt(5)/5", "fact_spans": "[[[2, 6], [50, 54]], [[7, 44]], [[7, 44]], [[2, 48]], [[55, 74]], [[55, 74]]]", "query_spans": "[[[50, 83]]]", "process": "Problem Analysis: Draw a line r' parallel to l through any point on the ellipse. When r' is tangent to the ellipse, the extreme distance from point P to the line l: x+2y-12=0 equals the distance between l and l'. Let the equation of l' be x+2y+c=0. Solve the system \\begin{cases}\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1\\\\x+2y+c=0\\end{cases}. Substitute the line into the ellipse to obtain 4x^{2}-2cx+c^{2}-12=0. Setting Δ=0 gives c=\\pm4. Thus, the distance d from l to l' is \\frac{|-12-c|}{\\sqrt{1^{2}+2^{2}}}=\\frac{16\\sqrt{5}}{5} or \\frac{8\\sqrt{5}}{5}." }, { "text": "Given a fixed point $A(0,-2)$, point $B$ moves on the circle $C$: $x^{2}+y^{2}-4y-32=0$, where $C$ is the center of the circle. The perpendicular bisector of segment $AB$ intersects $BC$ at point $P$. Then the equation of the trajectory $E$ of the moving point $P$ is?", "fact_expressions": "C1: Circle;Expression(C1) = (-4*y + x^2 + y^2 - 32 = 0);B: Point;PointOnCurve(B, C1);A: Point;Coordinate(A) = (0, -2);C: Point;Center(C1) = C;P: Point;Intersection(PerpendicularBisector(LineSegmentOf(A,B)), LineSegmentOf(B, C)) = P;Locus(P) = E;E: Curve", "query_expressions": "Expression(E)", "answer_expressions": "y^2/9 + x^2/5 = 1", "fact_spans": "[[[19, 47]], [[19, 47]], [[14, 18]], [[14, 50]], [[4, 13]], [[4, 13]], [[51, 54]], [[19, 57]], [[86, 89], [78, 82]], [[58, 82]], [[86, 95]], [[92, 95]]]", "query_spans": "[[[92, 100]]]", "process": "By the given condition, |PA| = |PB|, ∴ |PA| + |PC| = |PB| + |PC| = r = 6 > |AC| = 4, ∴ the locus E of point P is an ellipse with foci at A and C, where c = 2, a = 3, ∴ b = \\sqrt{5}, ∴ the equation of the ellipse is \\frac{y^{2}}{9} + \\frac{x^{2}}{5} = 1" }, { "text": "Draw a vertical line through the left focus $F_{1}$ of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersecting the ellipse at point $P$, and let $F_{2}$ be the right focus. If $\\angle F_{1}PF_{2}=60^{\\circ}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: Point;F2: Point;F1: Point;Z: Line;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;PointOnCurve(F1, Z);IsPerpendicular(Z, xAxis);Intersection(Z, G) = P;RightFocus(G) = F2;AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[1, 53], [73, 75], [128, 130]], [[3, 53]], [[3, 53]], [[76, 80]], [[81, 88]], [[57, 64]], [], [[3, 53]], [[3, 53]], [[1, 53]], [[1, 64]], [[0, 72]], [[0, 72]], [[0, 80]], [[73, 92]], [[94, 126]]]", "query_spans": "[[[128, 136]]]", "process": "" }, { "text": "Given the parabola $E$: $y^{2}=8x$ with focus $F$, a line $l$ passing through point $F$ intersects $E$ at points $A$ and $B$. If $\\overrightarrow{A B}=3 \\overrightarrow{F B}$, then $\\frac{1}{|A F|}+\\frac{1}{|B F|}$=? What is the equation of $l$?", "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 8*x);F: Point;Focus(E) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, E) = {A, B};VectorOf(A,B) = 3*VectorOf(F,B)", "query_expressions": "1/Abs(LineSegmentOf(B, F)) + 1/Abs(LineSegmentOf(A, F));Expression(l)", "answer_expressions": "1/2\n{((2*sqrt(2))*x-y-4*sqrt(2)=0), ((2*sqrt(2))*x+y-4*sqrt(2)=0)}", "fact_spans": "[[[2, 21], [41, 44]], [[2, 21]], [[25, 28], [30, 34]], [[2, 28]], [[35, 40], [139, 144]], [[29, 40]], [[46, 49]], [[50, 53]], [[35, 55]], [[57, 102]]]", "query_spans": "[[[105, 140]], [[140, 148]]]", "process": "① According to the problem, in parabola E, p=4, line l passes through focus F(2,0). Let points A(x_{1},y_{1}), B(x_{2},y_{2}). From the property of a focal chord in a parabola, x_{1}x_{2}=\\frac{p^{2}}{4}=4, |AF|=x_{1}+2, |BF|=x_{2}+2, \\frac{1}{|AF|}+\\frac{1}{|BF|}=\\frac{1}{x_{1}+2}+\\frac{1}{x_{2}+2}=\\frac{x_{1}+2+x_{2}+2}{x_{1}x_{2}+2(x_{1}+x_{2})+4}=\\frac{x_{1}+x_{2}+4}{2(x_{1}+x_{2})+8}=\\frac{1}{2} ② As shown in the figure, draw perpendiculars from points A and B to the directrix l of the parabola, with feet M and N respectively. By the definition of a parabola and \\overrightarrow{AB}=3\\overrightarrow{FB}, we have |AM|=|AF|=2|FB|=2|BN|. In right trapezoid AMNB, draw BC\\bot AM at C, |MC|=|BN|, |BF|\\cos\\angle MAB=\\frac{|AC|}{|AB|}=\\frac{|AM|-|MC|}{|AB|}=\\frac{|BF|}{3|BF|}=\\frac{1}{3}. The inclination angle \\alpha of line l satisfies \\cos\\alpha=\\pm\\frac{1}{2}, so the slope k=\\tan\\alpha=\\pm2\\sqrt{2}. Thus, the equation of l is y=\\pm2\\sqrt{2}(x-2), or equivalently, 2\\sqrt{2}x-y-4\\sqrt{2}=0 and 2\\sqrt{2}x+y-4\\sqrt{2}=0." }, { "text": "Pass a line $l$ through the point $M(1,2)$ intersecting the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$ at points $A$ and $B$. If point $M$ is exactly the midpoint of segment $AB$, then the equation of line $l$ is?", "fact_expressions": "M: Point;Coordinate(M) = (1, 2);l: Line;PointOnCurve(M, l);G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);A: Point;B: Point;Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Expression(l)", "answer_expressions": "8*x+25*y-58=0", "fact_spans": "[[[1, 10], [68, 72]], [[1, 10]], [[11, 16], [86, 91]], [[0, 16]], [[17, 56]], [[17, 56]], [[57, 60]], [[61, 64]], [[11, 66]], [[68, 84]]]", "query_spans": "[[[86, 96]]]", "process": "Problem Analysis: Let A(x₁, y₁), B(x₂, y₂), substitute into the equations \n\\begin{cases}\\frac{x^{2}}{25}+\\frac{y_{1}^{2}}{16}=1\\\\\\frac{x_{2}}{25}+\\frac{y_{2}^{2}}{16}=1\\end{cases} \nSubtracting the two equations gives: \n\\frac{x_{1}^{2}-x_{2}^{2}}{25}+\\frac{y_{1}^{2}-y_{2}^{2}}{16}=0 \n\\Leftrightarrow \\frac{(x_{1}+x_{2})(x_{1}-x_{2})}{25}+\\frac{(y_{1}+y_{2})(y_{1}-y_{2})}{16}=0 \nx_{1}+x_{2}=2, y_{1}+y_{2}=4 \nWhen x_{1} \\neq x_{2}, rearrange to: \n\\frac{2}{25}+\\frac{4}{16}\\times\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0, \nand k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{8}{25}, \nso the equation of the line is y-2=-\\frac{8}{25}(x-1), \nrearranged as: 8x+25y-58=0, \nthus fill in: 8x+25y-58=0" }, { "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=2$, then the value of $a$ is?", "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 2)", "query_expressions": "a", "answer_expressions": "-1/8", "fact_spans": "[[[0, 14]], [[27, 30]], [[0, 14]], [[0, 25]]]", "query_spans": "[[[27, 34]]]", "process": "" }, { "text": "Given that $M$ is a point on the ellipse $\\frac{x^{2}}{7}+\\frac{y^{2}}{3}=1$ with foci $F_{1}$ and $F_{2}$, $O$ is the origin, and $2 M O = F_{1} F_{2}$, then the area of $\\triangle F_{1} M F_{2}$ is?", "fact_expressions": "G: Ellipse;F1: Point;M: Point;F2: Point;O: Origin;Expression(G) = (x^2/7 + y^2/3 = 1);PointOnCurve(M, G);Focus(G) = {F1, F2};2*LineSegmentOf(M, O) = LineSegmentOf(F1, F2)", "query_expressions": "Area(TriangleOf(F1, M, F2))", "answer_expressions": "3", "fact_spans": "[[[26, 63]], [[7, 14]], [[2, 5]], [[15, 22]], [[68, 71]], [[26, 63]], [[2, 67]], [[6, 63]], [[78, 97]]]", "query_spans": "[[[99, 128]]]", "process": "" }, { "text": "Let $P$ be a point on the curve represented by the equation $\\sqrt{(x+4)^{2}+y^{2}}+\\sqrt{(x-4)^{2}+y^{2}}=12$, and let $M$, $N$ be points on the circles $(x+4)^{2}+y^{2}=4$ and $(x-4)^{2}+y^{2}=1$, respectively. Then the minimum value of $|P M|+|P N|$ is?", "fact_expressions": "G: Circle;C:Circle;H: Curve;P: Point;M: Point;N: Point;Expression(H)=(sqrt((x+4)^2+y^2)+sqrt((x-4)^2+y^2)=12);Expression(G) = (y^2 + (x + 4)^2 = 4);Expression(C) = (y^2 + (x - 4)^2 = 1);PointOnCurve(P,H);PointOnCurve(M, G);PointOnCurve(N, C)", "query_expressions": "Min(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, N)))", "answer_expressions": "9", "fact_spans": "[[[76, 96]], [[97, 117]], [[60, 62]], [[1, 4]], [[66, 69]], [[70, 73]], [[5, 62]], [[76, 96]], [[97, 117]], [[1, 65]], [[66, 120]], [[66, 120]]]", "query_spans": "[[[122, 141]]]", "process": "The equation $\\sqrt{(x+4)^{2}+y^{2}}+\\sqrt{(x-4)^{2}+y^{2}}=12$ represents an ellipse, with foci $F_{1}(-4,0)$, $F_{2}(4,0)$. The circle centered at $F_{1}$ has radius $R=2$, and the circle centered at $F_{2}$ has radius $r=1$. Then $|PM|+|PN|\\geqslant|PF_{1}|-R+|PF_{2}|-r=(|PF_{1}|+|PF_{2}|)-(2+1)=12-3=9$, with equality if and only if $P$, $M$, $F_{1}$ are collinear, $P$, $N$, $F_{2}$ are collinear, and $M$ lies between $P$ and $F_{1}$." }, { "text": "Given that the distance from a point $M(1 , m)$ on the parabola $y^{2}=2 p x(p>0)$ to its focus is $5$, and the left vertex of the hyperbola $x^{2}-\\frac{y^{2}}{a}=1$ is $A$. If one asymptote of the hyperbola is perpendicular to the line $A M$, then the real number $a$=?", "fact_expressions": "H: Parabola;Expression(H) = (y^2 = 2*p*x);p: Number;p>0;M: Point;Coordinate(M) = (1, m);m: Number;PointOnCurve(M, H);Distance(M, Focus(H)) = 5;G: Hyperbola;Expression(G) = (x^2 - y^2/a = 1);a: Real;A: Point;LeftVertex(G) = A;IsPerpendicular(OneOf(Asymptote(G)), LineOf(A, M))", "query_expressions": "a", "answer_expressions": "1/4", "fact_spans": "[[[2, 23], [37, 38]], [[2, 23]], [[5, 23]], [[5, 23]], [[26, 36]], [[26, 36]], [[26, 36]], [[2, 36]], [[26, 47]], [[48, 76], [86, 89]], [[48, 76]], [[107, 112]], [[81, 84]], [[48, 84]], [[86, 105]]]", "query_spans": "[[[107, 114]]]", "process": "According to the definition of the parabola, we have 1+\\frac{p}{2}=5. Substituting and solving gives p=8, hence y^{2}=16x. Substituting M(1,m) yields m=\\pm4; without loss of generality, take M(1,4). Given A(-1,0), the slope of line AM is 2. From the hyperbola x^{2}-\\frac{y^{2}}{a}=1, its asymptotes are y=\\pm\\sqrt{a}x. Using the given condition, -\\sqrt{a}\\times2=-1, solving gives a=\\frac{1}{4}." }, { "text": "Given that the line $x - y = 2$ intersects the parabola $y^2 = 4x$ at points $A$ and $B$, what are the coordinates of the midpoint of segment $AB$?", "fact_expressions": "H: Line;Expression(H) = (x - y = 2);G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;Intersection(H, G) = {A, B}", "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A, B)))", "answer_expressions": "(4,2)", "fact_spans": "[[[2, 11]], [[2, 11]], [[12, 26]], [[12, 26]], [[29, 32]], [[33, 36]], [[2, 38]]]", "query_spans": "[[[41, 54]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}). From \\begin{cases}y^{2}=4x\\\\x-y=2\\end{cases} we obtain y^{2}=4(y+2), that is, y^{2}-4y-8=0. Therefore, \\frac{y_{1}+y_{2}}{2}=2, hence \\frac{x_{1}+x_{2}}{2}=4. Thus, the midpoint coordinates are (4,2)." }, { "text": "The distance from the right focus to the right directrix of the hyperbola $8 x^{2}-y^{2}=8$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (8*x^2 - y^2 = 8)", "query_expressions": "Distance(RightFocus(G),RightDirectrix(G))", "answer_expressions": "8/3", "fact_spans": "[[[0, 20]], [[0, 20]]]", "query_spans": "[[[0, 33]]]", "process": "" }, { "text": "If the directrix of the parabola $y^{2}=2 a x (a \\neq 0)$ is tangent to the circle $(x-3)^{2}+y^{2}=16$, then the value of the real number $a$ is?", "fact_expressions": "G: Parabola;a: Real;H: Circle;Expression(G) = (y^2 = 2*(a*x));Expression(H) = (y^2 + (x - 3)^2 = 16);IsTangent(Directrix(G),H);Negation(a=0)", "query_expressions": "a", "answer_expressions": "{2,-14}", "fact_spans": "[[[1, 29]], [[58, 63]], [[33, 54]], [[1, 29]], [[33, 54]], [[1, 56]], [[4, 29]]]", "query_spans": "[[[58, 67]]]", "process": "" }, { "text": "Through the right vertex of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, draw a perpendicular line to the $x$-axis, intersecting an asymptote of $C$ at point $A$. If a circle centered at the right focus of $C$ with radius $4$ passes through points $A$ and $O$ ($O$ being the origin), then the equation of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;A: Point;O:Origin;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);L:Line;PointOnCurve(RightVertex(C),L);IsPerpendicular(xAxis,L);Intersection(L,OneOf(Asymptote(C)))=A;RightFocus(C)=Center(G);Radius(G)=4;PointOnCurve(A,G);PointOnCurve(O,G)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4-y^2=1", "fact_spans": "[[[1, 52], [65, 68], [84, 87], [126, 132]], [[9, 52]], [[9, 52]], [[102, 103]], [[77, 80], [105, 108]], [[115, 118], [109, 112]], [[1, 52]], [], [[0, 64]], [[0, 64]], [[0, 80]], [[83, 103]], [[95, 103]], [[102, 114]], [[102, 114]]]", "query_spans": "[[[126, 137]]]", "process": "\\because the circle with center at the right focus of C and radius 4 passes through points A and O (O is the origin), \\therefore the radius is r=4, the center is (4,0), thus the standard equation of the circle is (x-4)^{2}+y^{2}=16, B(a,0), y=\\frac{b}{a}\\cdota=b, we obtain A(a,b), so a^{2}-8a+16+b^{2}=16, i.e., c^{2}-8a=0, i.e., 8a=16, then a=2, b^{2}=16-4=12, then the equation of hyperbola C is \\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1," }, { "text": "If the equation $\\frac{x^{2}}{a-6}+\\frac{y^{2}}{a^{2}}=1$ represents a hyperbola, then what is the range of real values for $a$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(a-6) + y^2/a^2 = 1);a: Real", "query_expressions": "Range(a)", "answer_expressions": "(-oo,0)+(0,6)", "fact_spans": "[[[47, 50]], [[2, 50]], [[52, 57]]]", "query_spans": "[[[52, 64]]]", "process": "From the equation representing a hyperbola, we know: \n\\begin{cases}a-6<0\\\\a\\neq0\\end{cases} \n\\therefore a\\in(-\\infty,0)\\cup(0,6)." }, { "text": "A point $P$ lies on the parabola $y^{2}=4x$ with focus $F$. From $P$, a perpendicular is drawn to the directrix, meeting it at $Q$. If $\\angle QPF=120^{\\circ}$, then $|PF|=$?", "fact_expressions": "G: Parabola;Q: Point;P: Point;F: Point;H:Line;Expression(G) = (y^2 = 4*x);Focus(G)=F;PointOnCurve(P,G);PointOnCurve(P,H);IsPerpendicular(H,Directrix(G));FootPoint(H,Directrix(G))=Q;AngleOf(Q, P, F) = ApplyUnit(120, degree)", "query_expressions": "Abs(LineSegmentOf(P,F))", "answer_expressions": "4/3", "fact_spans": "[[[8, 22], [29, 30]], [[39, 42]], [[25, 28]], [[4, 7]], [], [[8, 22]], [[1, 22]], [[8, 28]], [[0, 35]], [[0, 35]], [[0, 42]], [[44, 68]]]", "query_spans": "[[[70, 78]]]", "process": "Problem Analysis: From the given conditions, the focus of the parabola y^{2}=4x is F(1,0), and the directrix of the parabola y^{2}=4x is x=-1. Without loss of generality, let P(m,n), (where both m and n are positive), then 4m=n^{2}, so |PQ|=1+m, |FQ|=\\sqrt{2^{2}+n^{2}}. By the definition of a parabola, |PF|=|PQ|=1+m, so AQPF is an isosceles triangle. Since \\angle QPF=120^{\\circ}, it follows that \\frac{1}{2}|PQ|=|PF|\\sin60^{\\circ}, that is, \\frac{1}{2}\\sqrt{2^{2}+n^{2}}=\\frac{\\sqrt{3}}{2}(1+m). Simplifying gives 3m+2n-1=0, solving yields m=\\frac{1}{3} or m=0 (discarded), so |PF|=\\frac{4}{3}." }, { "text": "Let $A$ and $B$ be two points on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(b>a>0)$, and let $O$ be the origin. If $OA \\perp OB$, then the minimum area of $\\triangle AOB$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;b > a;a > 0;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);O: Origin;IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B))", "query_expressions": "Min(Area(TriangleOf(A, O, B)))", "answer_expressions": "a^2*b^2/(b^2-a^2)", "fact_spans": "[[[10, 63]], [[10, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[13, 63]], [[1, 4]], [[5, 8]], [[1, 66]], [[1, 66]], [[67, 70]], [[78, 93]]]", "query_spans": "[[[95, 120]]]", "process": "" }, { "text": "Let $O$ and $F$ be the vertex and focus of the parabola $y=\\frac{1}{2} x^{2}$, respectively. A variable line passing through $F$ intersects the parabola at points $A$ and $B$. Then the minimum value of $S_{\\triangle O A B}$ is?", "fact_expressions": "O: Point;Vertex(G)=O;F: Point;Focus(G) =F;G: Parabola;Expression(G) = (y = x^2/2);H: Line;PointOnCurve(F,H);A: Point;B: Point;Intersection(H,G) = {A, B}", "query_expressions": "Min(Area(TriangleOf(O,A,B)))", "answer_expressions": "1/2", "fact_spans": "[[[1, 4]], [[1, 41]], [[5, 8], [43, 46]], [[1, 41]], [[11, 35], [51, 54]], [[11, 35]], [[47, 50]], [[42, 50]], [[56, 59]], [[60, 63]], [[47, 65]]]", "query_spans": "[[[68, 95]]]", "process": "From the given conditions, we have $ F(0,\\frac{1}{2}) $ and the slope of line $ l $ must exist. Let the equation of the line passing through point $ F $ be $ y = kx + \\frac{1}{2} $, $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. Solving the system of equations\n\\[\n\\begin{cases}\ny = \\frac{1}{2}x^{2} \\\\\ny = kx + \\frac{1}{2}\n\\end{cases}\n\\]\neliminating $ y $ gives $ x^{2} - 2kx - 1 = 0 $. Thus, $ x_{1} + x_{2} = 2k $, $ x_{1} \\cdot x_{2} = -1 $, hence\n\\[\n|AB| = \\sqrt{1+k^{2}} \\cdot \\sqrt{4k^{2}+4} = 2+2k^{2}\n\\]\nThe distance from the origin $ O $ to the line $ y = kx + \\frac{1}{2} $ is $ d = \\frac{\\frac{1}{2}}{\\sqrt{1+}} $. Therefore,\n\\[\nS_{\\triangle OAB} = \\frac{1}{2}|AB| \\cdot d = \\frac{\\sqrt{1+k^{2}}}{2}\n\\]\nThus, when $ k = 0 $, $ S_{\\triangle OAB} $ has a minimum value of $ \\frac{1}{2} $." }, { "text": "What is the equation of the hyperbola that has the same foci as the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1$ and has $y=\\pm \\frac{4}{3} x$ as asymptotes?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/49 + y^2/24 = 1);Focus(G)=Focus(H);Expression(Asymptote(G))=(y=pm*(4/3)*x)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9-y^2/16=1", "fact_spans": "[[[74, 77]], [[1, 40]], [[1, 40]], [[0, 77]], [[47, 77]]]", "query_spans": "[[[74, 80]]]", "process": "Test analysis: \\because the foci of the ellipse \\frac{x^{2}}{49}+\\frac{y^{2}}{24}=1 are (5,0) and (-5,0), therefore in the hyperbola, c=5, and \\frac{b}{a}=\\frac{4}{3}, a^{2}+b^{2}=25, \\therefore a^{2}=9, b^{2}=16, the required hyperbola equation is \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1." }, { "text": "The line $ l $ intersects the ellipse $ \\Gamma $: $ \\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1 $ at points $ P $ and $ Q $. If $ OP \\perp OQ $ ($ O $ being the origin), then what is the equation of the circle centered at $ O $ that is tangent to the line $ l $?", "fact_expressions": "l: Line;Gamma: Ellipse;H: Circle;O: Origin;P: Point;Q: Point;Expression(Gamma)=(x^2/4+y^2/2=1);Intersection(l,Gamma) = {P, Q};IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(O, Q));Center(H)=O;IsTangent(l,H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2+y^2=4/3", "fact_spans": "[[[0, 5], [104, 109]], [[6, 53]], [[112, 113]], [[83, 86], [95, 99]], [[56, 59]], [[60, 63]], [[6, 53]], [[0, 65]], [[67, 82]], [[94, 113]], [[103, 113]]]", "query_spans": "[[[112, 117]]]", "process": "Analysis: Let the line $ l $ be a special case: $ x = t $, thus obtaining the coordinates of $ P $ and $ Q $, and solve using $ OP \\perp OQ $. Detailed solution: The line $ l $ intersects the ellipse $ r: \\frac{x^{2}}{4} + \\frac{y^{2}}{2} = 1 $ at points $ P $ and $ Q $. If $ OP \\perp OQ $ ($ O $ being the origin), assume the line $ l $ is: $ x = t $. Then we have: $ P(t, \\sqrt{2(1 - \\frac{t^{2}}{4})}) $. From $ OP \\perp OQ $, we get $ t^{2} - 2(1 - \\frac{t^{2}}{4}) = 0 $, solving gives $ t^{2} = \\frac{4}{3} $. So in this case $ l $ is: $ x = \\pm\\sqrt{\\frac{4}{3}} $. Then the circle centered at $ O $ and tangent to the line $ l $ is $ x^{2} + y^{2} = \\frac{4}{3} $." }, { "text": "Given the parabola $y^{2}=8x$ with focus $F$ and point $A(1,2)$, let point $P$ be a moving point on the parabola. When the perimeter of $\\triangle PAF$ is minimized, what are the coordinates of point $P$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);Focus(G) = F;F: Point;A: Point;Coordinate(A) = (1, 2);P: Point;PointOnCurve(P, G) = True;WhenMin(Perimeter(TriangleOf(P,A,F))) = True", "query_expressions": "Coordinate(P)", "answer_expressions": "(1/2,2)", "fact_spans": "[[[2, 16], [37, 40]], [[2, 16]], [[2, 22]], [[19, 22]], [[23, 31]], [[23, 31]], [[32, 36], [72, 76]], [[32, 44]], [[46, 72]]]", "query_spans": "[[[72, 81]]]", "process": "The focus of the parabola y^{2}=8x is F(2,0), point A(1,2). Find the minimum perimeter of \\triangle PAF, which is to find the minimum value of |PA|+|PF|. Let D be the projection of point P on the directrix. According to the definition of a parabola, |PF|=|PD|. Therefore, the minimum value of |PA|+|PF| is the minimum value of |PA|+|PD|. By plane geometry knowledge, |PA|+|PD| is minimized when points D, P, A are collinear. The y-coordinate of P is 2, so 4=8x, solving gives x=\\frac{1}{2}. Then the coordinates of point P when the perimeter of \\triangle PAF reaches its minimum value are (\\frac{1}{2},2)." }, { "text": "The line $y = x + m$ intersects the hyperbola $x^{2} - \\frac{y^{2}}{4} = 1$ at points $A$ and $B$. If $|AB| = \\frac{8}{3} \\sqrt{2}$, then $m = $?", "fact_expressions": "G: Hyperbola;H: Line;m: Number;A: Point;B: Point;Expression(G) = (x^2 - y^2/4 = 1);Expression(H) = (y = m + x);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = sqrt(2)*(8/3)", "query_expressions": "m", "answer_expressions": "pm*1", "fact_spans": "[[[10, 38]], [[0, 9]], [[81, 84]], [[40, 43]], [[44, 47]], [[10, 38]], [[0, 9]], [[0, 49]], [[51, 79]]]", "query_spans": "[[[81, 86]]]", "process": "Solving the line and hyperbola simultaneously gives: $3x^{2}-2mx-m^{2}-4=0$, then $\\triangle=16m^{2}+48>0$, so $x_{A}+x_{B}=\\frac{2m}{3}$, $x_{A}x_{B}=-\\frac{m^{2}+4}{3}$, and $|AB|=\\sqrt{1+k^{2}}\\cdot\\sqrt{(x_{A}+x_{B})^{2}-4x_{A}x_{B}}=\\sqrt{2}\\times\\frac{4\\sqrt{m^{2}+3}}{3}=\\frac{8}{3}\\sqrt{2}$, which gives $m=\\pm1$." }, { "text": "The difference between the distance from a moving point $P$ to the line $x+4=0$ and the distance from $P$ to the point $M(2,0)$ is equal to $2$. Then what is the trajectory of point $P$?", "fact_expressions": "G: Line;M: Point;P: Point;Expression(G) = (x + 4 = 0);Coordinate(M) = (2, 0);Distance(P, G)-Distance(P, M)=2", "query_expressions": "Locus(P)", "answer_expressions": "Parabola", "fact_spans": "[[[6, 15]], [[22, 31]], [[2, 5], [20, 21], [44, 48]], [[6, 15]], [[22, 31]], [[2, 41]]]", "query_spans": "[[[44, 53]]]", "process": "" }, { "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3}=1$ $(a>0)$ has eccentricity $\\sqrt{2}$, then the distance from the focus of the parabola $y^{2}=8x$ to an asymptote of $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/3 + x^2/a^2 = 1);a: Number;a>0;Eccentricity(C) = sqrt(2);G: Parabola;Expression(G) = (y^2 = 8*x)", "query_expressions": "Distance(Focus(G), Asymptote(C))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 53], [88, 91]], [[1, 53]], [[9, 53]], [[9, 53]], [[1, 68]], [[70, 84]], [[70, 84]]]", "query_spans": "[[[70, 99]]]", "process": "" }, { "text": "Given that the hyperbola $x^{2}+n y^{2}=1$ $(n \\in \\mathbb{R})$ has the same foci as the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$, what is the equation of the asymptotes of this hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (n*y^2 + x^2 = 1);n: Real;H: Ellipse;Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G) = Focus(H)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(3)*x", "fact_spans": "[[[2, 31], [78, 81]], [[2, 31]], [[5, 31]], [[32, 69]], [[32, 69]], [[2, 75]]]", "query_spans": "[[[78, 89]]]", "process": "" }, { "text": "If the eccentricity of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{m+9}=1$ is $\\frac{1}{2}$, then what is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/9 + y^2/(m + 9) = 1);Eccentricity(G) = 1/2", "query_expressions": "m", "answer_expressions": "{3,-9/4}", "fact_spans": "[[[1, 40]], [[60, 63]], [[1, 40]], [[1, 58]]]", "query_spans": "[[[60, 68]]]", "process": "" }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ are given by the equations $y=\\pm 2 \\sqrt{2} x$. Then, the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(Asymptote(G)) = (y = pm*2*sqrt(2)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "3", "fact_spans": "[[[0, 56], [87, 90]], [[3, 56]], [[3, 56]], [[3, 56]], [[3, 56]], [[0, 56]], [[0, 84]]]", "query_spans": "[[[87, 96]]]", "process": "The hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0, b>0) has eccentricity \\frac{c}{a}=e. Since \\frac{b}{a}=2\\sqrt{2}, then \\frac{c^{2}}{a^{2}}=\\frac{a^{2}+b^{2}}{a^{2}}=9, \\therefore \\frac{c}{a}=e=3" }, { "text": "Let the hyperbola $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ have a length of $2$ for its imaginary axis and a focal distance of $2 \\sqrt{3}$. Then the equations of the asymptotes of the hyperbola are?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;a>0;b>0;Expression(G) = (-x^2/b^2 + y^2/a^2 = 1);Length(ImageinaryAxis(G))=2;FocalLength(G) = 2*sqrt(3)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(2)*x", "fact_spans": "[[[1, 57], [83, 86]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[1, 81]]]", "query_spans": "[[[83, 94]]]", "process": "Analysis: From the given conditions, first find the equation of the hyperbola, then solve for the asymptotes. According to the given conditions: $ 2b = 2 $, $ \\therefore b = 1 $; from the focal distance: $ 2c = 2\\sqrt{3} $, $ \\therefore c = \\sqrt{3} $. Thus, $ a = \\sqrt{c^{2} - b^{2}} = \\sqrt{2} $. The equation of the hyperbola is: $ \\frac{y^{2}}{2} - \\frac{x^{2}}{1} = 1 $. Then the asymptotes satisfy: $ \\frac{y^{2}}{2} - \\frac{x^{2}}{1} = 0 $. Solving this yields the asymptote equations: $ y = \\pm\\sqrt{2}x $." }, { "text": "Given that the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ intersects the line $x+2 y-2=0$ at points $A$ and $B$, $|A B|=\\sqrt{5}$, and the midpoint of $A B$ has coordinates $\\left(m, \\frac{1}{2}\\right)$, find the equation of this ellipse?", "fact_expressions": "G: Ellipse;b: Number;a: Number;H: Line;A: Point;B: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x + 2*y - 2 = 0);Coordinate(MidPoint(LineSegmentOf(A, B))) = (m, 1/2);Intersection(G, H) = {A, B};Abs(LineSegmentOf(A, B)) = sqrt(5);m:Number", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4+y^2=1", "fact_spans": "[[[2, 54], [130, 132]], [[4, 54]], [[4, 54]], [[55, 68]], [[70, 73]], [[74, 77]], [[4, 54]], [[4, 54]], [[2, 54]], [[55, 68]], [[98, 127]], [[2, 79]], [[80, 96]], [[109, 127]]]", "query_spans": "[[[130, 137]]]", "process": "Since the midpoint coordinates of AB are (m, \\frac{1}{2}) and satisfy the line equation x + 2y - 2 = 0, we have m + 2 \\times \\frac{1}{2} - 2 = 0, solving gives m = 1, then the midpoint coordinates of AB are (1, \\frac{1}{2}). Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), from \\begin{cases} x + 2y - 2 = 0 \\\\ \\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\end{cases} we get (a^{2} + 4b^{2})x^{2} - 4a^{2}x + 4a^{2} - 4a^{2}b^{2} = 0, then x_{1} + x_{2} = \\frac{4a^{2}}{a^{2} + 4b^{2}}, x_{1}x_{2} = \\frac{4a^{2} - 4a^{2}b^{2}}{a^{2} + 4b^{2}}. Since the midpoint coordinates of AB are (1, \\frac{1}{2}), \\frac{x_{1} + x_{2}}{2} = 1, i.e., x_{1} + x_{2} = 2, then \\frac{4a^{2}}{a^{2} + 4b^{2}} = 2, so a^{2} = 4b^{2}, hence x_{1}x_{2} = \\frac{4a^{2} - 4a^{2}b^{2}}{a^{2} + 4b^{2}} = 2 - 2b^{2}. Also |AB| = \\sqrt{1 + k^{2}} \\cdot \\sqrt{(x_{1} + x_{2})^{2} - 4x_{1}x_{2}} = \\sqrt{1 + (-\\frac{1}{2})^{2}} \\cdot \\sqrt{2^{2} - 4 \\cdot (2 - 2b^{2})} = \\sqrt{5}, solving gives b^{2} = 1, thus a^{2} = 4b^{2} = 4. Therefore, the ellipse equation is \\frac{x^{2}}{4} + y^{2} = 1." }, { "text": "Write a standard equation of a hyperbola with an asymptote inclined at $60^{\\circ}$ and foci on the $y$-axis.", "fact_expressions": "G: Hyperbola;Inclination(Asymptote(G))=ApplyUnit(60,degree);PointOnCurve(Focus(G),yAxis)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/3-x^2=1", "fact_spans": "[[[34, 37]], [[4, 37]], [[25, 37]]]", "query_spans": "[[[34, 42]]]", "process": "If $\\frac{y^{2}}{3}-x^{2}=1$, the foci lie on the $y$-axis. Let $\\frac{y^{2}}{3}-x^{2}=0$, then the asymptotes are given by $y=\\pm\\sqrt{3}x$, where the inclination angle of $y=\\sqrt{3}x$ is $60^{\\circ}$." }, { "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 $T$, and $P$, $Q$ are two points on $C$ such that the line $TP$ is tangent to $C$ and $\\overrightarrow{T P}=\\lambda \\overrightarrow{F Q}$ ($\\lambda>0$), then $\\lambda=$?", "fact_expressions": "C: Parabola;p: Number;lambda:Number;T: Point;P: Point;F: Point;Q: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C) = F;Intersection(Directrix(C),xAxis)=T;PointOnCurve(P,C);PointOnCurve(Q,C);IsTangent(LineOf(T,P),C);VectorOf(T, P) = lambda*VectorOf(F, Q);lambda>0", "query_expressions": "lambda", "answer_expressions": "sqrt(2)-1", "fact_spans": "[[[6, 32], [36, 39], [62, 65], [78, 81]], [[13, 32]], [[148, 157]], [[50, 53]], [[54, 57]], [[2, 5]], [[58, 61]], [[13, 32]], [[6, 32]], [[2, 35]], [[36, 53]], [[54, 69]], [[54, 69]], [[70, 83]], [[84, 146]], [[84, 146]]]", "query_spans": "[[[148, 159]]]", "process": "From the given conditions, we have T(-\\frac{p}{2},0), F(\\frac{p}{2},0) for y^{2}=2px. Without loss of generality, assume y>0, then y=\\sqrt{2px}, and by differentiation we get y'=\\sqrt{\\frac{p}{2x}}. Let P(x_{0},y_{0}) with y_{0}>0, then the equation of line TP is y-\\sqrt{2p}x_{0}=\\sqrt{\\frac{p}{2x_{0}}}(x-x_{0}). Since point T(-\\frac{p}{2},0) lies on line TP, it follows that -\\sqrt{2px_{0}}=\\sqrt{\\frac{p}{2x_{0}}}(-\\frac{p}{2}-x_{0}), solving gives x_{0}=\\frac{p}{2}, then y_{0}=p, so P(\\frac{p}{2},p), \\overrightarrow{TP}=(p,p). Let Q(x_{1},y_{1}), then \\overrightarrow{FQ}=(x_{1}-\\frac{p}{2},y_{1}). Since \\overrightarrow{TP}=\\lambda\\overrightarrow{FQ}, we have (p,p)=\\lambda(x_{1}-\\frac{p}{2},y_{1}), thus \\begin{cases}\\lambda(x_{1}-\\frac{p}{2})=p\\\\\\lambda y_{1}=p\\end{cases}. Substituting into y^{2}=2px, we obtain \\lambda^{2}+2\\lambda-1=0, solving gives \\lambda=-\\sqrt{2}-1 (discarded) or \\lambda=\\sqrt{2}-1." }, { "text": "If the focus of the parabola $y^{2}=2 p x$ coincides with the right focus of the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, then $p$=?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;H: Ellipse;Expression(H) = (x^2/8 + y^2/4 = 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, 71]]]", "process": "From the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$, we know $a^{2}=8$, $b^{2}=4$, $c^{2}=a^{2}-b^{2}=4$, so the coordinates of the right focus are $(2,0)$. The focus of the parabola $y^{2}=2px$ is $\\left(\\frac{p}{2},0\\right)$, thus $\\frac{p}{2}=2$, solving gives $p=4$. This problem mainly examines the application of properties of parabolas and ellipses, and finding focus coordinates from standard equations." }, { "text": "Let the hyperbola $C$ pass through the point $(1,3)$, and have the same asymptotes as $\\frac{y^{2}}{3}-x^{2}=1$. Then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;G: Point;Coordinate(G) = (1, 3);PointOnCurve(G, C);D:Curve;Expression(D)=(y^2/3-x^2=1);Asymptote(D)=Asymptote(C)", "query_expressions": "Expression(C)", "answer_expressions": "y^2/6-x^2/2=1", "fact_spans": "[[[1, 7], [54, 57]], [[9, 17]], [[9, 17]], [[1, 17]], [[20, 45]], [[20, 45]], [[1, 52]]]", "query_spans": "[[[54, 62]]]", "process": "Since the hyperbola C passes through the point (1,3) and has the same asymptotes as \\frac{y^{2}}{3}-x^{2}=1, let the equation of hyperbola C be \\frac{y^{2}}{3}-x^{2}=\\lambda, (\\lambda\\neq0). Substituting the point (1,3) gives: \\frac{9}{3}-1=\\lambda, solving yields \\lambda=2. Therefore, the equation of hyperbola C is: \\frac{y^{2}}{6}-\\frac{x^{2}}{2}=1" }, { "text": "If the parabola $y^{2}=2 p x(p>0)$ passes through the point $(2 ,-4)$, then what is the value of $p$? What is the equation of the directrix of this parabola?", "fact_expressions": "G: Parabola;p: Number;H: Point;p>0;Expression(G) = (y^2 = 2*(p*x));Coordinate(H) = (2, -4);PointOnCurve(H, G)", "query_expressions": "p;Expression(Directrix(G))", "answer_expressions": "4\nx=-2", "fact_spans": "[[[1, 22], [43, 46]], [[35, 38]], [[23, 33]], [[4, 22]], [[1, 22]], [[23, 33]], [[1, 33]]]", "query_spans": "[[[35, 42]], [[43, 53]]]", "process": "" }, { "text": "If the length of the real axis of the hyperbola $\\frac{x^{2}}{m}-y^{2}=1$ is $10$, then what is the equation of the asymptotes of this hyperbola?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (-y^2 + x^2/m = 1);Length(RealAxis(G))=10", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*(1/5)*x", "fact_spans": "[[[1, 29], [41, 44]], [[4, 29]], [[1, 29]], [[1, 38]]]", "query_spans": "[[[41, 52]]]", "process": "" }, { "text": "Let the vertex of the parabola be at the origin, and the equation of the directrix be $x = -2$. Then the equation of the parabola is?", "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;Expression(Directrix(G)) = (x =-2)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=8*x", "fact_spans": "[[[1, 4], [25, 28]], [[8, 10]], [[1, 10]], [[1, 22]]]", "query_spans": "[[[25, 33]]]", "process": "" }, { "text": "Given that the distance from the point $(m, 3)$ on the parabola $x^{2}=2 p y(p>0)$ to the focus is $5$, then $p$=?", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;H: Point;m: Number;Coordinate(H) = (m, 3);PointOnCurve(H, G);Distance(H, Focus(G)) = 5", "query_expressions": "p", "answer_expressions": "4", "fact_spans": "[[[2, 23]], [[2, 23]], [[47, 50]], [[5, 23]], [[25, 34]], [[26, 34]], [[25, 34]], [[2, 34]], [[2, 44]]]", "query_spans": "[[[47, 52]]]", "process": "Since the equation of the directrix of the parabola \\( x^{2} = 2py \\) (\\( p > 0 \\)) is \\( y = -\\frac{p}{2} \\), according to the definition of the parabola: \\( 3 - \\left(-\\frac{p}{2}\\right) = 5 \\Rightarrow p = 4 \\)." }, { "text": "Given that moving circle $M$ is externally tangent to both circle $C_{1}$: $(x+2)^{2}+y^{2}=1$ and circle $C_{2}$: $(x-2)^{2}+y^{2}=4$, what is the equation of the locus of the center of moving circle $M$?", "fact_expressions": "M: Circle;C1:Circle;C2:Circle;Expression(C1)=((x+2)^2+y^2=1);Expression(C2)=((x-2)^2+y^2=4);IsOutTangent(M,C1);IsOutTangent(M,C2)", "query_expressions": "LocusEquation(Center(M))", "answer_expressions": "(4*x^2-4*y^2/15=1)&(x<=-1/2)", "fact_spans": "[[[4, 7], [72, 75]], [[8, 36]], [[37, 65]], [[8, 36]], [[37, 65]], [[4, 68]], [[4, 68]]]", "query_spans": "[[[72, 84]]]", "process": "By the given condition, circle $ C_{1}: (x+2)^{2}+y^{2}=1 $, circle $ C_{2}: (x-2)^{2}+y^{2}=4 $ are both externally tangent. The centers of the two circles are $ C_{1}(-2,0) $, $ C_{2}(2,0) $, and their radii are 1 and 2 respectively. Let the center coordinates of the required circle be $ M(x,y) $, radius $ r $. Since the required circle is externally tangent to both circles, we have $ |MC_{1}| = r+1 $, $ |MC_{2}| = r+2 $. Subtracting these two equations gives $ |MC_{2}| - |MC_{1}| = 1 $. According to the definition of a hyperbola, the locus of point $ M $ is one branch of a hyperbola with foci $ C_{1} $, $ C_{2} $. Thus, $ 2a = 1 $, $ c = 2 $, so $ a = \\frac{1}{2} $, then $ b = \\sqrt{c^{2}-a^{2}} = \\sqrt{4-\\frac{1}{4}} = \\sqrt{\\frac{15}{4}} $. Hence, the trajectory equation of center $ M $ is $ 4x^{2}-\\frac{4y^{2}}{15}=1 $ $ (x \\leqslant -\\frac{1}{2}) $." }, { "text": "The line passing through the focus of the parabola $y^{2}=4 x$ with an inclination angle of $\\frac{3}{4} \\pi$ intersects the parabola at points $P$ and $Q$, and $O$ is the origin. Then the area of $\\triangle O P Q$ equals?", "fact_expressions": "G: Parabola;H: Line;O: Origin;P: Point;Q: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(Focus(G), H);Inclination(H)=(3/4)*pi;Intersection(H, G) = {P, Q}", "query_expressions": "Area(TriangleOf(O, P, Q))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[1, 15], [45, 48]], [[42, 44]], [[59, 62]], [[49, 52]], [[53, 56]], [[1, 15]], [[0, 44]], [[20, 44]], [[42, 58]]]", "query_spans": "[[[69, 92]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $C$: $5 x^{2}-4 y^{2}=m^{2}$, respectively. A line $l$ passing through $F_{2}$ intersects the right branch of hyperbola $C$ at points $A$ and $B$, and satisfies $\\overrightarrow{A F_{2}}=7 \\overrightarrow{F_{2} B}$ ($O$ is the origin). What is the slope of line $l$?", "fact_expressions": "l: Line;C: Hyperbola;m: Number;A: Point;F2: Point;B: Point;F1: Point;O: Origin;Expression(C) = (5*x^2 - 4*y^2 = m^2);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(F2, l);Intersection(l, RightPart(C)) = {A, B};VectorOf(A, F2) = 7*VectorOf(F2, B)", "query_expressions": "Slope(l)", "answer_expressions": "pm*sqrt(3)", "fact_spans": "[[[66, 71], [161, 166]], [[19, 50], [72, 78]], [[27, 50]], [[83, 86]], [[9, 16], [58, 65]], [[87, 90]], [[1, 8]], [[150, 153]], [[19, 50]], [[1, 56]], [[1, 56]], [[57, 71]], [[66, 92]], [[96, 149]]]", "query_spans": "[[[161, 171]]]", "process": "As shown in the figure, let the left and right foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ be $F_{1}$ and $F_{2}$ respectively, and let point $M$ lie on the right branch of the hyperbola. Connect $MF_{2}$, and from point $M$ draw a perpendicular $MN$ to the right directrix $x=\\frac{a^{2}}{c}$. Denote $\\angle MF_{2}x = \\theta$. Then by the second definition of the hyperbola, $\\frac{MF_{2}}{MN} = \\frac{c}{a}$, where $e = \\frac{c}{a}$. That is, $\\frac{MF_{2}}{c - \\frac{a^{2}}{c} + MF_{2}\\cos\\theta} = e$, simplifying yields $MF_{2} = \\frac{\\frac{b^{2}}{a}}{1 - e\\cos\\theta}$. From the hyperbola $C: 5x^{2} - 4y^{2} = m^{2}$, we obtain $\\frac{x^{2}}{\\frac{m^{2}}{5}} - \\frac{y^{2}}{\\frac{m^{2}}{4}} = 1$, so $a^{2} = \\frac{m^{2}}{5}$, $b^{2} = \\frac{m^{2}}{4}$, and the eccentricity $e = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\frac{3}{2}$. Given that the inclination angle of line $l$ is $\\theta$, and from $\\overrightarrow{AF}_{2} = 7\\overrightarrow{F_{2}B}$, it follows that $AF_{2} = 7BF_{2}$. Therefore, $\\frac{b^{2}}{1 - e\\cos\\theta} = 7 \\times \\frac{b^{2}}{1 - e\\cos(\\pi - \\theta)}$, or $\\frac{b^{2}}{1 - e\\cos(\\pi - \\theta)} = 7 \\times \\frac{b^{2}}{1 - e\\cos\\theta}$. Substituting $e = \\frac{3}{2}$, we find $\\theta = 60^{\\circ}$ or $\\theta = 120^{\\circ}$. Hence, the slope of line $l$ is $\\sqrt{3}$ or $-\\sqrt{3}$." }, { "text": "Let $F$ be the right focus of the ellipse $C$: $\\frac{x^{2}}{a}+\\frac{y^{2}}{b}=1$ $(a>0 , b>0)$, and let $d$ be the maximum distance from a moving point on $C$ to $F$. If there exists a point $P$ on the right directrix of $C$ such that $PF=d$, then the range of values for the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b + x^2/a = 1);b: Number;a: Number;a>0;b>0;F: Point;RightFocus(C) = F;K: Point;PointOnCurve(K, C);d: Number;Max(Distance(K, F)) = d;P: Point;PointOnCurve(P, RightDirectrix(C));LineSegmentOf(P, F) = d", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[1/2, 1)", "fact_spans": "[[[5, 58], [63, 66], [87, 90], [112, 117]], [[5, 58]], [[12, 58]], [[12, 58]], [[12, 58]], [[12, 58]], [[1, 4], [73, 76]], [[1, 62]], [], [[63, 72]], [[82, 85]], [[63, 85]], [[97, 101]], [[87, 101]], [[104, 110]]]", "query_spans": "[[[112, 128]]]", "process": "" }, { "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ such that the distance from $P$ to the focus $F_{1}$ is $2$, $M$ is the midpoint of the segment $P F_{1}$, and $O$ is the origin. Then $|O M|$ equals?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G) = True;F1: Point;OneOf(Focus(G)) = F1;Distance(P, F1) = 2;M: Point;MidPoint(LineSegmentOf(P, F1)) = M;O: Origin", "query_expressions": "Abs(LineSegmentOf(O, M))", "answer_expressions": "4", "fact_spans": "[[[2, 40]], [[2, 40]], [[44, 47]], [[2, 47]], [[50, 57]], [[2, 57]], [[44, 64]], [[66, 69]], [[66, 84]], [[85, 88]]]", "query_spans": "[[[94, 104]]]", "process": "\\because in the ellipse \\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1, a=5, \\therefore |PF_{1}|+|PF_{2}|=2a=10, combining with |PF_{1}|=2, we get |PF_{2}|=8, \\because OM is the midline of \\triangle PF_{1}F_{2}, \\therefore |OM|=\\frac{1}{2}|PF_{2}|=\\frac{1}{2}\\times8=4" }, { "text": "Given that one asymptote of the hyperbola $\\frac{x^{2}}{m}-\\frac{y^{2}}{n}=1$ is $y=\\frac{4}{3} x$, what is the eccentricity $e$ of this hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/n + x^2/m = 1);m: Number;n: Number;Expression(OneOf(Asymptote(G))) = (y = (4/3)*x);e: Number;Eccentricity(G) = e", "query_expressions": "e", "answer_expressions": "{5/3, 5/4}", "fact_spans": "[[[2, 40], [69, 72]], [[2, 40]], [[5, 40]], [[5, 40]], [[2, 66]], [[76, 79]], [[69, 79]]]", "query_spans": "[[[69, 81]]]", "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 $F_{1}$ and $F_{2}$ respectively. On the right branch of the hyperbola, there exists a point $P$ such that $|P F_{1}|=3| P F_{2} |$. Then the range of values for 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, RightPart(G));Abs(LineSegmentOf(P, F1)) = 3*Abs(LineSegmentOf(P, F2))", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1, 2]", "fact_spans": "[[[2, 58], [86, 89], [129, 132]], [[5, 58]], [[5, 58]], [[97, 100]], [[77, 84]], [[67, 74]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 84]], [[2, 84]], [[86, 100]], [[103, 127]]]", "query_spans": "[[[129, 143]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be fixed points, $|F_{1} F_{2}|=6$, and a moving point $M$ satisfies $|M F_{1}|+|M F_{2}|=10$. Then the locus of the moving point $M$ is? (Choose from: ellipse, line, circle, line segment)", "fact_expressions": "F1: Point;F2: Point;M: Point;Abs(LineSegmentOf(F1,F2)) = 6;Abs(LineSegmentOf(M,F1)) + Abs(LineSegmentOf(M,F2)) = 10", "query_expressions": "Locus(M)", "answer_expressions": "Ellipse", "fact_spans": "[[[1, 8]], [[9, 16]], [[40, 43], [73, 76]], [[20, 37]], [[45, 69]]]", "query_spans": "[[[73, 81]]]", "process": "Directly from the definition of an ellipse, the moving point M satisfies |MF_{1}| + |MF_{2}| = 10 > 6 = |F_{1}F_{2}|, so the locus of point M is an ellipse with foci F_{1}, F_{2}." }, { "text": "Given that $M$ and $N$ are the intersection points of line $l$ passing through the focus $F$ of the parabola $C$: $y^{2}=2 p x$ ($p>0$) with the parabola $C$, $O$ is the origin, and satisfying $\\overrightarrow{M F}=4 \\overrightarrow{F N}$, $S_{\\triangle O M N}=\\sqrt{3}|M N|$, find the value of $p$.", "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;N: Point;F: Point;O: Origin;p>0;Expression(C) = (y^2 = 2*p*x);Focus(C)=F;PointOnCurve(F,l);Intersection(l,C)={M,N};VectorOf(M, F) = 4*VectorOf(F, N);Area(TriangleOf(O,M,N))=sqrt(3)*Abs(LineSegmentOf(M,N))", "query_expressions": "p", "answer_expressions": "5*sqrt(3)", "fact_spans": "[[[44, 49]], [[11, 37], [50, 56]], [[156, 159]], [[2, 5]], [[6, 9]], [[40, 43]], [[60, 63]], [[19, 37]], [[11, 37]], [[11, 43]], [[10, 49]], [[2, 59]], [[72, 117]], [[119, 154]]]", "query_spans": "[[[156, 163]]]", "process": "Without loss of generality, assume the slope $ k > 0 $ of line $ MN $. As shown in the figure, draw perpendiculars from $ M $ and $ N $ to the directrix of the parabola, with feet $ G $ and $ H $, respectively. Draw $ NK \\perp MG $ intersecting at $ K $, and let $ P $ be the intersection of $ NK $ with the $ x $-axis. Draw $ MQ \\perp x $-axis at $ Q $. From $ \\overrightarrow{MF} = 4\\overrightarrow{FN} $, we obtain $ |MF| = 4|FN| $. Let $ |FN| = m $, then $ |MF| = 4m $, $ |MN| = 5m $. By the definition of the parabola, $ |MG| = |MF| = 4m $, and $ |NH| = |FN| = m $. Therefore, $ |MK| = |MG| - |KG| = 3m $, so $ |NK| = \\sqrt{|MN|^2 - |MK|^2} = 4m $. Since $ S_{\\Delta OMN} = S_{\\Delta OMF} + S_{\\Delta ONF} = \\frac{1}{2}|OF| \\cdot (|MQ| + |NP|) = \\frac{1}{2}|OF| \\cdot |NK| = \\frac{1}{2} \\cdot \\frac{p}{2} \\cdot 4m = pm $. Also, $ S_{\\Delta OMN} = \\sqrt{3}|MN| = 5\\sqrt{3}m $, so $ pm = 5\\sqrt{3}m $, therefore $ p = 5\\sqrt{3} $." }, { "text": "The equation of the directrix of the parabola $y=\\frac{1}{4} x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y=-1", "fact_spans": "[[[0, 24]], [[0, 24]]]", "query_spans": "[[[0, 31]]]", "process": "From y = \\frac{1}{4}x^2 we get: x^{2} = 4y, so 2p = 4, that is: \\frac{p}{2} = 1. Therefore, the directrix equation of the parabola y = \\frac{1}{4}x^2 is: y = -\\frac{p}{2}. (This question mainly examines the simple properties of a parabola and is a basic problem.)" }, { "text": "A line $ l $ with slope $ k $ intersects the ellipse $ \\frac{x^{2}}{16} + \\frac{y^{2}}{8} = 1 $ at points $ A $ and $ B $. Point $ M(1,1) $ is the midpoint of segment $ AB $. Then $ k = $?", "fact_expressions": "l: Line;G: Ellipse;B: Point;A: Point;M: Point;Expression(G) = (x^2/16 + y^2/8 = 1);Coordinate(M) = (1, 1);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A,B)) = M;k:Number;Slope(l)=k", "query_expressions": "k", "answer_expressions": "-12", "fact_spans": "[[[7, 12]], [[13, 51]], [[58, 61]], [[54, 57]], [[64, 73]], [[13, 51]], [[64, 73]], [[7, 63]], [[64, 84]], [[3, 6], [86, 89]], [[0, 12]]]", "query_spans": "[[[86, 91]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}) (x_{1} \\neq x_{2}), then \\frac{y_{1}}{8}=1. Subtracting the two equations, we get \\frac{2}{8}=1. Since point M(1,1) is the midpoint of segment AB, it follows that x_{1}+x_{2}=2, y_{1}+y_{2}=2. Also, since k=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, we have \\frac{1}{2}+\\frac{2}{2}\\cdot k=0, then k=-\\frac{1}{2}." }, { "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$. If the incircle of $\\triangle P F_{2} Q$ is tangent to segment $P F_{2}$ at its midpoint and tangent to $P Q$ at point $F_{1}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;L: Line;F2: Point;P: Point;Q: Point;F1: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F1,L);Intersection(L, G) = {P, Q};TangentPoint(InscribedCircle(TriangleOf(P,F2,Q)),LineSegmentOf(P,F2))=MidPoint(LineSegmentOf(P,F2));TangentPoint(InscribedCircle(TriangleOf(P,F2,Q)),LineSegmentOf(P,Q))=F1", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 54], [91, 93], [171, 173]], [[4, 54]], [[4, 54]], [[88, 90]], [[70, 77]], [[95, 98]], [[99, 102]], [[62, 69], [79, 87], [161, 169]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 77]], [[2, 77]], [[78, 90]], [[88, 104]], [[107, 151]], [[107, 169]]]", "query_spans": "[[[171, 179]]]", "process": "Let the incenter of $\\triangle PF_{2}Q$ be $I$, and let $M$ be the point of tangency, which is also the midpoint of $PF_{2}$. Then $\\triangle PF_{2}Q$ is an isosceles triangle ($|QP|=|QF_{2}|$). Let $|PF_{1}|=m$, $|PF_{2}|=n$, so that $m+n=2a$. By the property of tangents, $m=\\frac{1}{2}n$, solving gives $m=\\frac{2a}{3}$, $n=\\frac{4a}{3}$. Let $|QF_{1}|=t$, $|QF_{2}|=2a-t$. From $|QF_{1}|=t=|QP|-|PF_{1}|=2a-t-\\frac{2a}{3}$, solving gives $t=\\frac{2a}{3}$. Then $\\triangle PF_{2}Q$ is an equilateral triangle, so that $_{2c}=\\frac{\\sqrt{3}}{2}\\times\\frac{4a}{3}$, hence $_{e}=\\frac{c}{a}=\\frac{\\sqrt{3}}{3}$." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F(\\sqrt{2}, 0)$ is its right focus. The chord length obtained by intersecting the ellipse with the line passing through $F$ and perpendicular to the $x$-axis is $2$. Then the equation 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;F: Point;Coordinate(F) = (sqrt(2), 0);RightFocus(C) = F;G: Line;PointOnCurve(F, G);IsPerpendicular(G, xAxis);Length(InterceptChord(G, C)) = 2", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4+y^2/2=1", "fact_spans": "[[[2, 59], [79, 80], [99, 101], [114, 119]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[61, 78], [85, 88]], [[61, 78]], [[61, 83]], [[96, 98]], [[84, 98]], [[88, 98]], [[96, 112]]]", "query_spans": "[[[114, 124]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, the left and right vertices are denoted as $A_{1}$ and $A_{2}$ respectively. Let $P$ be an arbitrary point on $C$, and the slope of the line $P A_{1}$ lies in the range $[-2,-1]$. Then, what is the range of the slope of the line $P A_{2}$?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/3 = 1);A1: Point;A2: Point;LeftVertex(C) = A1;RightVertex(C) = A2;P: Point;PointOnCurve(P, C) = True;Range(Slope(LineOf(P, A1))) = [-2, -1]", "query_expressions": "Range(Slope(LineOf(P, A2)))", "answer_expressions": "[3/8,3/4]", "fact_spans": "[[[2, 44], [72, 75]], [[2, 44]], [[52, 59]], [[60, 67]], [[2, 67]], [[2, 67]], [[68, 71]], [[68, 79]], [[82, 108]]]", "query_spans": "[[[110, 128]]]", "process": "From the equation of the ellipse, we have a^{2}=4, b^{2}=3, then A_{1}(-2,0), A_{2}(2,0). Let P(m,n), \\therefore \\frac{m^{2}}{4}+\\frac{n^{2}}{3}=1, that is, n^{2}=-\\frac{3}{4}(m^{2}-4). k_{pA_{1}}=\\frac{n}{m+2}, k_{pA_{2}}=\\frac{n}{m-2}, k_{pA_{1}}\\cdot k_{pA_{2}}=\\frac{n^{2}}{m^{2}-4}=-\\frac{3}{4}. The range of the slope of line PA_{1} is [-2,-1], the range of the slope of line PA_{2} is: [\\frac{3}{8},\\frac{3}{4}]" }, { "text": "Let $O$ be the coordinate origin, and let the right focus of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be $F$. Point $P$ lies on $C$ in the first quadrant, and point $Q(x_{0}, y_{0})$ satisfies $b x_{0}+a y_{0}=0$. If the line segments $O P$ and $Q F$ are perpendicular to each other and bisect each other, then the eccentricity of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;O: Origin;Q: Point;F: Point;x0: Number;y0: Number;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(Q) = (x0, y0);RightFocus(C) = F;Quadrant(P) = 1;PointOnCurve(P, C);a*y0 + b*x0 = 0;PerpendicularBisector(LineSegmentOf(O, P)) = LineSegmentOf(Q, F);PerpendicularBisector(LineSegmentOf(Q, F)) = LineSegmentOf(O, P)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2) + 1", "fact_spans": "[[[10, 71], [85, 88], [159, 162]], [[18, 71]], [[18, 71]], [[80, 84]], [[1, 4]], [[97, 115]], [[76, 79]], [[117, 136]], [[117, 136]], [[18, 71]], [[18, 71]], [[10, 71]], [[97, 115]], [[10, 79]], [[80, 96]], [[80, 96]], [[117, 136]], [[138, 157]], [[138, 157]]]", "query_spans": "[[[159, 168]]]", "process": "Since the line segments OP and QF bisect each other perpendicularly, |QO| = |FO| = c, that is, x_{0}^{2} + y_{0}^{2} = c^{2}, and bx_{0} + ay_{0} = 0. Solving gives x_{0} = -a, y_{0} = b, ∴ Q(-a, b). Hence the midpoint of QF has coordinates (\\frac{c-a}{2}, \\frac{b}{2}), so P(c-a, b). Substituting into C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 (a > 0, b > 0), we have \\frac{(c-a)^{2}}{a^{2}} - \\frac{(b)^{2}}{b^{2}} = 1, yielding c = (\\sqrt{2}+1)a, that is, e = \\frac{c}{a} = \\sqrt{2}+1." }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{9}=1$ $(a>0)$ has asymptotes with equations $3 x \\pm 2 y=0$, then the value of $a$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/9 + x^2/a^2 = 1);a: Number;a>0;Expression(Asymptote(G)) = (3*x + pm*2*y = 0)", "query_expressions": "a", "answer_expressions": "2", "fact_spans": "[[[1, 48]], [[1, 48]], [[72, 75]], [[4, 48]], [[1, 70]]]", "query_spans": "[[[72, 79]]]", "process": "From the asymptote equations 3x±2y=0, we get \\frac{b}{a}=\\frac{3}{2}\\therefore\\frac{3}{a}=\\frac{3}{2}\\therefore a=2." }, { "text": "Given that the line $l$ with slope $-1$ passing through the focus $F$ of the parabola $C$: $y^{2}=4x$ intersects the parabola $C$ at points $A$ and $B$, then $|AB|=$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;PointOnCurve(F, l) = True;Slope(l) = -1;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[3, 22], [43, 49]], [[3, 22]], [[25, 28]], [[3, 28]], [[2, 42]], [[29, 42]], [[37, 42]], [[37, 59]], [[50, 53]], [[54, 57]]]", "query_spans": "[[[61, 70]]]", "process": "The equation of line l is y = 1 - x. Substituting into y^{2} = 4x gives: x^{2} - 6x + 1 = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then x_{1} + x_{2} = 6, ∴|_{AB}| = x_{1} + 1 + x_{2} + 1 = x_{1} + x_{2} + 2 = 8b female safety as .8" }, { "text": "It is known that the focus of the parabola $y^{2}=8 x$ coincides with one focus of the hyperbola $\\frac{x^{2}}{a^{2}}-y^{2}=1$. Then, what is the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;a: Number;H: Parabola;Expression(G) = (-y^2 + x^2/a^2 = 1);Expression(H) = (y^2 = 8*x);Focus(H) = OneOf(Focus(G))", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[20, 52], [61, 64]], [[23, 52]], [[2, 16]], [[20, 52]], [[2, 16]], [[2, 59]]]", "query_spans": "[[[61, 70]]]", "process": "\\because the focus of the parabola y^{2}=8x has coordinates (2,0), and the focus of the parabola y^{2}=8x coincides with one focus of the hyperbola \\frac{x^{2}}{a^{2}}-y^{2}=1, \\therefore a^{2}+1=4, a=\\sqrt{3}, then e=\\frac{c}{a}=\\frac{2}{\\sqrt{3}}=\\frac{2\\sqrt{3}}{3}" }, { "text": "Given that $O$ is the coordinate origin, $F$ is the focus of the parabola $C$: $y^{2}=4 \\sqrt{2} x$, and $P$ is a point on $C$ such that $|P F|=3 \\sqrt{2}$, then the area of $\\triangle P O F$ is?", "fact_expressions": "O: Origin;C: Parabola;Expression(C) = (y^2 = 4*(sqrt(2)*x));F: Point;Focus(C) = F;PointOnCurve(P, C);P: Point;Abs(LineSegmentOf(P, F)) = 3*sqrt(2)", "query_expressions": "Area(TriangleOf(P, O, F))", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[2, 5]], [[15, 43], [51, 54]], [[15, 43]], [[11, 14]], [[11, 46]], [[47, 57]], [[47, 50]], [[59, 77]]]", "query_spans": "[[[79, 101]]]", "process": "\\because the equation of parabola C is y^{2}=4\\sqrt{2}x, \\therefore 2p=4\\sqrt{2}, it follows that \\frac{p}{2}=\\sqrt{2}, \\therefore the directrix of the parabola is x=-\\sqrt{2}, and the focus is F(\\sqrt{2},0). Since P is a point on C and |PF|=3\\sqrt{2}, \\therefore x_{P}=2\\sqrt{2}. Substituting into the parabola equation gives: |y_{P}|=4, \\therefore S_{\\triangle POF}=\\frac{1}{2}\\times|OF|\\times|y_{P}|=2\\sqrt{2}" }, { "text": "The standard equation of a circle with center at the focus of the parabola $y^{2}=4 x$ and radius $2$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);H: Circle;Focus(G) = Center(H);Radius(H) = 2", "query_expressions": "Expression(H)", "answer_expressions": "(x-1)^2+y^2=4", "fact_spans": "[[[1, 15]], [[1, 15]], [[29, 30]], [[0, 30]], [[22, 30]]]", "query_spans": "[[[29, 37]]]", "process": "According to the problem, the focus of the parabola has coordinates (1,0), which is also the center of the circle. Therefore, the equation of the circle is (x-1)^{2}+y^{2}=4" }, { "text": "The vertex of parabola $C$ is at the origin, and the focus is $F(1 , 0)$. A line $l$ passing through point $F$ intersects parabola $C$ at points $A$ and $B$. If the midpoint of $AB$ is $(2, m)$, then what is the length of chord $AB$?", "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;F: Point;O: Origin;Coordinate(F) = (1, 0);Vertex(C) = O;Focus(C) = F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Coordinate(MidPoint(LineSegmentOf(A, B)))= (2,m);IsChordOf(LineSegmentOf(A, B),C);m:Number", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "6", "fact_spans": "[[[35, 40]], [[0, 6], [41, 47]], [[50, 53]], [[54, 57]], [[18, 28], [30, 34]], [[10, 14]], [[18, 28]], [[0, 14]], [[0, 28]], [[29, 40]], [[35, 59]], [[61, 78]], [[41, 86]], [[70, 78]]]", "query_spans": "[[[81, 90]]]", "process": "" }, { "text": "Let $F_{1}$, $F_{2}$ be fixed points, $P F_{1}-P F_{2}=5$, $F_{1} F_{2}=8$. Then the locus of the moving point $P$ is?", "fact_expressions": "P: Point;F1: Point;F2: Point;LineSegmentOf(P, F1) - LineSegmentOf(P, F2) = 5;LineSegmentOf(F1,F2)=8", "query_expressions": "Locus(P)", "answer_expressions": "Hyperbola", "fact_spans": "[[[60, 63]], [[1, 8]], [[9, 16]], [[20, 39]], [[40, 56]]]", "query_spans": "[[[60, 68]]]", "process": "According to the definition of a hyperbola, the absolute value of the difference of the distances to two fixed points is constant. Here, PF_{1} - PF_{2} = 5 is a constant, and F_{1}, F_{2} are fixed points, so the locus of the moving point P is a hyperbola." }, { "text": "Given that point $P$ is a moving point on the parabola $y = \\frac{1}{2} x^{2}$, the projection of point $P$ onto the $x$-axis is $M$, and the coordinates of point $A$ are $(6, \\frac{17}{2})$, then the minimum value of $|PA| + |PM|$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/2);P: Point;PointOnCurve(P, G);M: Point;Projection(P, xAxis) = M;A: Point;Coordinate(A) = (6, 17/2)", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "19/2", "fact_spans": "[[[7, 31]], [[7, 31]], [[2, 6], [36, 40]], [[2, 35]], [[50, 53]], [[36, 53]], [[54, 58]], [[54, 81]]]", "query_spans": "[[[83, 102]]]", "process": "Problem Analysis: The standard equation of the parabola y = \\frac{1}{2}x^{2} is x^{2} = 2y, with focus F(0, \\frac{1}{2}). By the definition of a parabola, |PA| + |PM| = |PA| + |PF| - \\frac{1}{2} \\geqslant |AF| - \\frac{1}{2} = \\sqrt{(6-0)^{2} + \\left(\\frac{17}{2} - \\frac{1}{2}\\right)^{2}} - \\frac{1}{2} = \\frac{19}{2} (equality holds if and only if points A, P, F are collinear). Therefore, the minimum value is \\frac{19}{2}." }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has eccentricity $2$. Through a point $M$ on its left branch, draw a line parallel to the $x$-axis intersecting the asymptotes at points $P$ and $Q$. If $|P M| \\cdot|M Q|=4$, then the focal distance of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;Eccentricity(G) = 2;M: Point;PointOnCurve(M,LeftPart(G)) = True;L: Line;PointOnCurve(M,L) = True;IsParallel(L,xAxis) = True;Intersection(L,Asymptote(G)) = {P,Q};P: Point;Q: Point;Abs(LineSegmentOf(M, Q))*Abs(LineSegmentOf(P, M)) = 4", "query_expressions": "FocalLength(G)", "answer_expressions": "8", "fact_spans": "[[[0, 46], [56, 57], [115, 118]], [[0, 46]], [[3, 46]], [[3, 46]], [[0, 54]], [[62, 65]], [[56, 65]], [[74, 76]], [[55, 76]], [[66, 76]], [[56, 90]], [[81, 84]], [[85, 88]], [[92, 112]]]", "query_spans": "[[[115, 123]]]", "process": "Let M(x_{0},y_{0}), then \\frac{x_{0}2}{a^{2}}-\\frac{y_{0}^{2}}{b^{2}}=1, the asymptotes of the hyperbola are given by y=\\pm\\frac{b}{a}x. So when y=y_{0}, x=\\pm\\frac{a}{b}y_{0}, that is, P(-\\frac{a}{b}y_{0},y_{0}), Q(\\frac{a}{b}y_{0},y_{0}), since PQ//x-axis. Therefore |MP|=-\\frac{a}{b}y_{0}-x_{0}, |MQ|=\\frac{a}{b}y_{0}-x_{0}, then |PM|\\cdot|MQ|=x_{0}^{2}-\\frac{a^{2}}{b^{2}}y_{0}^{2}=4, and \\frac{x_{0}2}{a^{2}}-\\frac{y_{0}^{2}}{b^{2}}=1, which implies x_{0}2-\\frac{a^{2}y_{0}^{2}}{b^{2}}=a^{2}, so a^{2}=4, that is, a=2, then the eccentricity e=\\frac{c}{a}=\\frac{c}{2}=2, so c=4, therefore the focal length is 2c=8." }, { "text": "Given that $M$ is a point on the ellipse $C$: $\\frac{y^{2}}{9}+\\frac{x^{2}}{5}=1$, and $F_{1}$, $F_{2}$ are the foci of the ellipse $C$, then the perimeter of $\\Delta M F_{1} F_{2}$ is?", "fact_expressions": "C: Ellipse;M: Point;F1: Point;F2: Point;Expression(C) = (x^2/5 + y^2/9 = 1);PointOnCurve(M, C);Focus(C) = {F1, F2}", "query_expressions": "Perimeter(TriangleOf(M, F1, F2))", "answer_expressions": "10", "fact_spans": "[[[6, 48], [68, 73]], [[2, 5]], [[52, 59]], [[60, 67]], [[6, 48]], [[2, 51]], [[52, 76]]]", "query_spans": "[[[78, 105]]]", "process": "According to the problem, find a and c, then use the definition of an ellipse to solve. From the problem, a = 3, b = \\sqrt{5}, \\therefore c = \\sqrt{9 - 5} = 2. It follows that the perimeter of \\Delta MF_{1}F_{2} is |MF| + |MF_{2}| + |F_{1}F_{2}| = 2a + 2c = 2 \\times 3 + 2 \\times 2 = 10" }, { "text": "If the two foci of an ellipse are $F_{1}(-1,0)$, $F_{2}(1,0)$, and the major axis has length $10$, then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-1, 0);Coordinate(F2) = (1, 0);Focus(G) = {F1, F2};Length(MajorAxis(G)) = 10", "query_expressions": "Expression(G)", "answer_expressions": "x^2/25+y^2/24=1", "fact_spans": "[[[1, 3], [48, 50]], [[9, 22]], [[25, 37]], [[9, 22]], [[25, 37]], [[1, 37]], [[1, 46]]]", "query_spans": "[[[48, 55]]]", "process": "" }, { "text": "Given that $P$ is a moving point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $A(-2,1)$, $B(2,1)$, then the maximum value of $\\cos \\angle A P B$ is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;P: Point;Expression(G) = (x^2/4 + y^2 = 1);Coordinate(A) = (-2, 1);Coordinate(B) = (2, 1);PointOnCurve(P, G)", "query_expressions": "Max(Cos(AngleOf(A, P, B)))", "answer_expressions": "(sqrt(6)-sqrt(2))/4", "fact_spans": "[[[6, 33]], [[38, 47]], [[50, 58]], [[2, 5]], [[6, 33]], [[38, 47]], [[50, 58]], [[2, 37]]]", "query_spans": "[[[60, 85]]]", "process": "According to the problem, draw the ellipse as shown in the figure below: Let P(x_{0},y_{0}), draw PH\\bot AB intersecting AB at H, and since P(x_{0},y_{0}) lies on \\frac{x^{2}}{4}+y^{2}=1, substituting into the above expression yields \\tan\\angle APB=. From the properties of the ellipse, y_{0}\\in[-1,1], let t=1-y_{0}, t\\in[0,2], then \\tan\\angle APB=\\frac{4t}{t^{2}-4(1-t)^{2}}=\\frac{4t}{-3t^{2}+8t-4}, t\\in[0,2]. When t=0, \\tan\\angle APB=0, at this time \\angle APB=\\pi, \\cos\\angle APB=-1. When t\\in(0,2], by the AM-GM inequality, \\tan\\angle APB=\\frac{4}{-(3t+\\frac{4}{t})+8}\\geqslant2+\\sqrt{3}, with equality if and only if 3t=\\frac{4}{t}, i.e., t=\\frac{2\\sqrt{3}}{3}, at which point \\cos\\angle APB is maximized, thus \\begin{cases}\\frac{\\sin\\angle APB}{\\cos\\angle APB}=2+\\sqrt{3}\\\\\\sin^{2}\\angle APB+\\cos^{2}\\angle\\end{cases} simplifying gives _{\\cos^{2}}\\angle APB=\\frac{2-\\sqrt{3}}{4} B. \\begin{cases}\\sin^{2}\\angle APB+\\cos^{2}\\angle APB=1\\\\\\frac{\\sqrt{6}-\\sqrt{2}}{4},\\end{cases} in summary, the maximum value of \\cos\\angle APB is \\underline{\\sqrt{6}-\\sqrt{2}}," }, { "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}$ and $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], [90, 93], [160, 163]], [[5, 60]], [[5, 60]], [[86, 89]], [[68, 75]], [[77, 84]], [[167, 170]], [[5, 60]], [[5, 60]], [[2, 60]], [[2, 84]], [[2, 84]], [[86, 97]], [[99, 122]], [[123, 157]], [[160, 170]]]", "query_spans": "[[[167, 172]]]", "process": "\\because|PF_{1}-PF_{2}|=2a,PF_{1}^{2}+PF_{2}^{2}=F_{1}F_{2}^{2}=4c^{2}\\thereforePF_{1}\\cdot PF_{2}=\\frac{4c^{2}-4a^{2}}{2}=2c^{2}-2a^{2}=2b^{2}\\because S=\\frac{1}{2}PF_{1}\\cdot PF_{2}=2ab\\therefore b^{2}=2ab,b=2a,\\therefore c=\\sqrt{5}a,e=\\sqrt{5}. This question examines the definition of hyperbola and eccentricity, testing basic analytical and solving skills, classified as a medium-difficulty problem." }, { "text": "Given that point $M$ lies on the parabola $C$: $y^{2} = -x$, what is the shortest distance from point $M$ to the line $x + 2y - 3 = 0$?", "fact_expressions": "C: Parabola;G: Line;M: Point;Expression(C) = (y^2 = -x);Expression(G) = (x + 2*y - 3 = 0);PointOnCurve(M, C)", "query_expressions": "Min(Distance(M, G))", "answer_expressions": "2*sqrt(5)/5", "fact_spans": "[[[6, 24]], [[34, 47]], [[2, 5], [29, 33]], [[6, 24]], [[34, 47]], [[2, 27]]]", "query_spans": "[[[29, 54]]]", "process": "Assume the line $ x + 2y + c = 0 $ passing through point $ M $ is tangent to the parabola $ C: y^2 = -x $. Then $ y^2 = -2y - c $, so $ \\Delta = 4 + 4c = 0 $, $ \\therefore c = -1 $. Hence, the line is $ x + 2y - 1 = 0 $. The shortest distance from point $ M $ to the line $ x + 2y - 3 = 0 $ is the distance between the two parallel lines. $ \\therefore $ The shortest distance $ d $ from point $ M $ to the line $ x + 2y - 3 = 0 $ is $ \\frac{|-3 - (-1)|}{\\sqrt{1 + 2^2}} = \\frac{2\\sqrt{5}}{5} $." }, { "text": "Given that the foci of the ellipse lie on the $x$-axis, the focal distance is $2$, and it passes through the point $(0,2)$, then the standard equation of the ellipse is?", "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (0, 2);PointOnCurve(Focus(G),xAxis);FocalLength(G) = 2;PointOnCurve(H,G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5+y^2/4=1", "fact_spans": "[[[2, 4], [35, 37]], [[24, 32]], [[24, 32]], [[2, 13]], [[2, 20]], [[2, 32]]]", "query_spans": "[[[35, 44]]]", "process": "\\because the focal distance of the ellipse is 2, \\therefore c=1; and since the foci lie on the x-axis and the ellipse passes through the point (0,2), \\therefore b=2; \\therefore a^{2}=b^{2}+c^{2}=5; \\therefore the standard equation of the ellipse is \\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1" }, { "text": "Let the hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0, b>0 $) have two asymptotes that are perpendicular to each other, and the distance from a vertex to one asymptote is $ 1 $. Then, what is the distance from a focus of the hyperbola to the other asymptote?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Z1: Line;Z2: Line;OneOf(Asymptote(C)) = Z1;OneOf(Asymptote(C)) = Z2;Negation(Z1=Z2);IsPerpendicular(Z1, Z2);Distance(Vertex(C), Z1) = 1", "query_expressions": "Distance(OneOf(Focus(C)), Z2)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[1, 62], [90, 93]], [[1, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [[9, 62]], [], [], [[1, 81]], [[90, 105]], [[1, 105]], [[1, 72]], [[1, 88]]]", "query_spans": "[[[90, 110]]]", "process": "" }, { "text": "If the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has a focal distance of $8$, and the point $M(1, \\sqrt{3})$ lies on its asymptote, then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;FocalLength(C) = 8;M: Point;Coordinate(M) = (1, sqrt(3));PointOnCurve(M, Asymptote(C))", "query_expressions": "Expression(C)", "answer_expressions": "x^2/4-y^2/12=1", "fact_spans": "[[[1, 52], [78, 79], [85, 88]], [[1, 52]], [[9, 52]], [[9, 52]], [[1, 59]], [[60, 77]], [[60, 77]], [[60, 83]]]", "query_spans": "[[[85, 93]]]", "process": "From the given condition, $\\frac{1}{a^{2}}-\\frac{3}{b^{2}}=0$, $\\therefore b=\\sqrt{3}a \\cdot 2c=8$, $c=4$, $\\therefore b=2\\sqrt{3}$, $a$. Thus, the equation of $C$ is $\\frac{x^{2}}{4}-\\frac{y^{2}}{12}=1$." }, { "text": "Let the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ have left and right vertices $A$ and $B$, respectively. Let $P$ be a point on the ellipse different from $A$ and $B$. Let the slopes of lines $AP$ and $BP$ be $m$ and $n$, respectively. Then, when $\\frac{a}{b}(3-\\frac{2}{3 m n})+\\frac{2}{m n}+3(\\ln |m|+\\ln |n|)$ attains its minimum value, what is the eccentricity of the ellipse $C$?", "fact_expressions": "C: Ellipse;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a > b;b > 0;A: Point;B: Point;LeftVertex(C) = A;RightVertex(C) = B;P: Point;Negation(P=A);Negation(P=B);PointOnCurve(P, C);m: Number;n: Number;Slope(LineOf(A, P)) = m;Slope(LineOf(B, P)) = n;WhenMin((a/b)*(3 - 2/(3*m*n)) + 2/(m*n) + 3*(ln(Abs(m)) + ln(Abs(n))))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[1, 58], [79, 81], [203, 208]], [[1, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[8, 58]], [[67, 70], [85, 88]], [[71, 74], [89, 92]], [[1, 74]], [[1, 74]], [[75, 78]], [[75, 95]], [[75, 95]], [[75, 95]], [[118, 121]], [[124, 127]], [[97, 127]], [[97, 127]], [[130, 202]]]", "query_spans": "[[[203, 214]]]", "process": "" }, { "text": "Given the hyperbola $E$: $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ has eccentricity $2$, and point $P$ is a moving point on $E$, then the product of the distances from point $P$ to the two asymptotes of $E$ is?", "fact_expressions": "E: Hyperbola;Expression(E) = (x^2/4 - y^2/b^2 = 1);Eccentricity(E) = 2;b: Number;b>0;P: Point;PointOnCurve(P, E) = True;A1: Line;A2: Line;Asymptote(E) = {A1, A2}", "query_expressions": "Distance(P, A1)*Distance(P, A2)", "answer_expressions": "3", "fact_spans": "[[[2, 54], [68, 71], [82, 85]], [[2, 54]], [[2, 62]], [[10, 54]], [[10, 54]], [[63, 67], [77, 81]], [[63, 75]], [], [], [[82, 91]]]", "query_spans": "[[[77, 98]]]", "process": "From $\\frac{b}{a}=\\sqrt{e^{2}-1}=\\sqrt{3}$, we get $b=2\\sqrt{3}$. $\\therefore$ the asymptotes are $y=\\pm\\sqrt{3}x$, that is, $\\sqrt{3}x\\pm y=0$. Let $P(s,t)$, then $\\frac{s^{2}}{4}-\\frac{t^{2}}{12}=1$, namely $3s^{2}-t^{2}=12$. $\\therefore$ the product of the distances from point $P$ to the two asymptotes of $E$ is $\\frac{|\\sqrt{3}s+t|}{2}\\cdot\\frac{|\\sqrt{3}s-t|}{2}=\\frac{|3s^{2}-t^{2}|}{4}=3$ constant value is 3." }, { "text": "Given that the equation $\\frac{x^{2}}{m-3}+\\frac{y^{2}}{m+1}=1$ represents a hyperbola, what is the range of values for $m$?", "fact_expressions": "G: Hyperbola;Expression(G)= (x^2/(m - 3) + y^2/(m + 1) = 1);m: Number", "query_expressions": "Range(m)", "answer_expressions": "(-1,3)", "fact_spans": "[[[45, 48]], [[2, 48]], [[50, 53]]]", "query_spans": "[[[50, 60]]]", "process": "" }, { "text": "Given a point $P$ on the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{16}=1$ is at a distance of $3$ from one focus and at a distance of $7$ from the other focus, then what is the value of $m$?", "fact_expressions": "G: Ellipse;m: Number;P: Point;F1:Point;F2:Point;Expression(G) = (y^2/16 + x^2/m = 1);Distance(P, F1) = 3;Distance(P, F2) = 7;PointOnCurve(P, G);OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2)", "query_expressions": "m", "answer_expressions": "25", "fact_spans": "[[[2, 40], [48, 50]], [[75, 78]], [[44, 47]], [], [], [[2, 40]], [[44, 61]], [[44, 73]], [[2, 47]], [46, 51], [46, 64], [46, 64]]", "query_spans": "[[[75, 81]]]", "process": "From the definition of the ellipse, we have: 3 + 7 = 2a, so a = 5, then a^{2} = 25. According to the equation of the ellipse, we get: m = 25." }, { "text": "Given the parabola $y^{2}=4 x$, $P$ is a point on the parabola, $F$ is the focus, and $|P F|=5$. Then the equation of the directrix of the parabola is? The coordinates of point $P$ are?", "fact_expressions": "G: Parabola;P: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(P, G);Abs(LineSegmentOf(P, F)) = 5", "query_expressions": "Expression(Directrix(G));Coordinate(P)", "answer_expressions": "x=-1\n{(4,4),(4,-4)}", "fact_spans": "[[[2, 17], [23, 26], [49, 52]], [[19, 22], [61, 65]], [[30, 33]], [[2, 17]], [[23, 36]], [[18, 29]], [[38, 47]]]", "query_spans": "[[[49, 60]], [[61, 70]]]", "process": "Since the equation of the parabola is y^{2}=4x, we have p=2, so the equation of the directrix of the parabola is x=-1. Let P(x,y). Since |PF|=5, it follows that x+1=5, solving gives x=4, y=\\pm4. Therefore, the coordinates of point P are (4,4) or (4,-4)." }, { "text": "The line $y=kx$ ($k>0$) intersects the hyperbola $y=\\frac{4}{x}$ at two points $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$. Find the value of $2 x_{1} y_{2}-7 x_{2} y_{1}$.", "fact_expressions": "H: Line;Expression(H) = (y = k*x);k: Number;k>0;G: Hyperbola;Expression(G) = (y = 4/x);A: Point;B: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);Intersection(H, G) = {A, B}", "query_expressions": "2*(x1*y2) - 7*(x2*y1)", "answer_expressions": "20", "fact_spans": "[[[0, 14]], [[0, 14]], [[2, 14]], [[2, 14]], [[15, 33]], [[15, 33]], [[35, 52]], [[54, 71]], [[35, 52]], [[54, 71]], [[35, 52]], [[54, 71]], [[35, 52]], [[54, 71]], [[0, 73]]]", "query_spans": "[[[75, 108]]]", "process": "Method 1: Since the line $ y = kx $ intersects the hyperbola $ y = \\frac{4}{x} $ at points $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, we have $ x_{1}y_{1} = 4 $, $ x_{2}y_{2} = 4 $. Also, since both the line $ y = kx $ and the hyperbola $ y = \\frac{4}{x} $ are centrally symmetric figures with the origin as their center of symmetry, points $ A $ and $ B $ are symmetric about the origin. Therefore, $ x_{1} = -x_{2} $, $ y_{1} = -y_{2} $. Hence, $ 2x_{1}y_{2} - 7x_{2}y_{1} = -2x_{1}y_{1} + 7x_{1}y_{1} = 5x_{1}y_{1} = 5 \\times 4 = 20 $." }, { "text": "Let $a>1$, then the range of the eccentricity $e$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{(a+1)^{2}}=1$ is?", "fact_expressions": "a: Number;a > 1;G: Hyperbola;Expression(G) = (-y^2/(a + 1)^2 + x^2/a^2 = 1);e: Number;Eccentricity(G) = e", "query_expressions": "Range(e)", "answer_expressions": "(sqrt(2),sqrt(5))", "fact_spans": "[[[1, 6]], [[1, 6]], [[8, 58]], [[8, 58]], [[62, 65]], [[8, 65]]]", "query_spans": "[[[62, 72]]]", "process": "" }, { "text": "Given that line $m$ passing through point $P(-1,1)$ intersects the $x$-axis at point $A$, and there is a point $B$ on the parabola $x^{2}=y$ such that $P A \\perp P B$. If $A B$ is tangent to the parabola $x^{2}=y$, then the equation of line $m$ is?", "fact_expressions": "m: Line;G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (x^2 = y);Coordinate(P) = (-1, 1);PointOnCurve(P, m);Intersection(m, xAxis) = A;PointOnCurve(B,G);IsPerpendicular(LineSegmentOf(P,A),LineSegmentOf(P,B));IsTangent(LineSegmentOf(A, B), G)", "query_expressions": "Expression(m)", "answer_expressions": "{x+3*y-2=0,x-y+2=0}", "fact_spans": "[[[14, 19], [90, 95]], [[30, 42], [73, 85]], [[25, 29]], [[46, 49]], [[3, 13]], [[30, 42]], [[3, 13]], [[2, 19]], [[14, 29]], [[30, 49]], [[50, 65]], [[67, 88]]]", "query_spans": "[[[90, 100]]]", "process": "Analysis: Given B(t, t^{2}), differentiate to obtain the line AB: y = 2tx - t^{2}, then discuss two cases, t = 0 and t \\neq 0, to find the equation of line m. Details: Given B(t, t^{2}), differentiate 2x = y, so k_{AB} = 2t, then the line AB: y = 2tx - t^{2}. When t = 0, verify it satisfies the condition; at this time A(-2, 0), hence m: x - y + 2 = 0. When t \\neq 0, A(\\frac{t}{2}, 0), \\overrightarrow{PA} = (\\frac{t}{2} + 1, -1), \\overrightarrow{PB} = (t + 1, t^{2} - 1). \\overrightarrow{PA} \\cdot \\overrightarrow{PB} = 0 \\Rightarrow (t + 1)(\\frac{t}{2} + 1 - t + 1) = 0 \\Rightarrow t = 4 or t = -1 (discarded since B, P coincide). At this time P(-1, 1), A(2, 0), hence m: x + 3y - 2 = 0." }, { "text": "$A$, $B$ are two vertices of the hyperbola $C$. The line $l$ intersects the hyperbola $C$ at two distinct points $P$, $Q$, and is perpendicular to the transverse axis. If $\\overrightarrow{P B} \\cdot \\overrightarrow{A Q}=0$, then the eccentricity $e$ of the hyperbola $C$ is?", "fact_expressions": "B: Point;A: Point;C: Hyperbola;Vertex(C) = {A, B};l: Line;Intersection(l, C) = {P, Q};P: Point;Q: Point;Negation(P = Q);IsPerpendicular(l, RealAxis(C)) = True;DotProduct(VectorOf(P, B), VectorOf(A, Q)) = 0;e: Number;Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "sqrt(2)", "fact_spans": "[[[4, 7]], [[0, 3]], [[8, 14], [26, 32], [108, 114]], [[0, 19]], [[20, 25]], [[20, 46]], [[39, 42]], [[43, 46]], [[34, 46]], [[20, 53]], [[55, 106]], [[118, 121]], [[108, 121]]]", "query_spans": "[[[118, 123]]]", "process": "" }, { "text": "Given the left focus $F_1$ and right focus $F_2$ of ellipse $F$, with the upper vertex at $A$. The perpendicular bisector of $A F_{2}$ intersects the ellipse at point $B$. If the left focus $F_1$ lies on segment $A B$, then the eccentricity of the ellipse is?", "fact_expressions": "F: Ellipse;A: Point;B: Point;F2: Point;F1: Point;LeftFocus(F) = F1;RightFocus(F) = F2;UpperVertex(F)=A;Intersection(PerpendicularBisector(LineSegmentOf(A, F2)), F) = B;PointOnCurve(F1, LineSegmentOf(A,B))", "query_expressions": "Eccentricity(F)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 7], [54, 56], [84, 86]], [[34, 37]], [[57, 61]], [[22, 29]], [[11, 18], [66, 73]], [[2, 18]], [[2, 29]], [[2, 37]], [[39, 61]], [[66, 82]]]", "query_spans": "[[[84, 91]]]", "process": "Let |BF₂| = t. By the definition of the ellipse, |BF₁| = 2a - t. Since the perpendicular bisector of AF₂ intersects the ellipse at point B, we have |AB| = |BF₂| = t, thus |AF₁| = 2t - 2a. In △AF₁F₂, cos∠AF₁F₂ = c/a; in △BF₁F₂, by the law of cosines, cos∠BF₁F₂ = (2c)/a - a/c. From this we obtain a² = 3c². Finally, the eccentricity can be found using its definition. As shown in the figure, let |BF₂| = t. By the definition of the ellipse, |BF₁| = 2a - t. Since the perpendicular bisector of AF₂ intersects the ellipse at point B, |AB| = |BF₂| = t, so |AF₁| = |AB| - |BF₁| = 2t - 2a. Also |AF₁| = √(b² + c²) = a, so 2t - 2a = a, solving gives t = (3a)/2. Therefore, |AF₂| = |AF₁| = a, |BF₁| = a/2, |BF₂| = (3a)/2, |F₁F₂| = 2c. In △AF₁F₂, cos∠AF₁F₂ = c/a. Thus, cos∠BF₁F₂ = -cos∠AF₁F₂ = -c/a. In △BF₁F₂, by the law of cosines, cos∠BF₁F₂ = (a²/4 + 4c² - 9a²/4)/(2·(a/2)·2c) = (2c)/a - a/c. Therefore, (2c)/a - a/c = -c/a, so a² = 3c², hence e = c/a = √3/3, that is, the eccentricity of the ellipse is √3/3." }, { "text": "Given the parabola $C$: $y^{2}=6x$, a line $l$ passing through the focus $F$ of the parabola intersects the parabola at point $A$ and intersects the directrix of the parabola at point $B$. If $\\overrightarrow{FB}=3\\overrightarrow{FA}$, then what is the distance from point $A$ to the origin?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 6*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);A: Point;Intersection(l, C) = A;B: Point;Intersection(l, Directrix(C)) = B;VectorOf(F, B) = 3*VectorOf(F, A);O: Origin", "query_expressions": "Distance(A, O)", "answer_expressions": "sqrt(13)/2", "fact_spans": "[[[2, 21], [23, 26], [39, 42], [49, 52]], [[2, 21]], [[29, 32]], [[23, 32]], [[33, 38]], [[22, 38]], [[43, 47], [109, 113]], [[33, 47]], [[56, 60]], [[33, 60]], [[62, 107]], [[114, 116]]]", "query_spans": "[[[109, 121]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, with left focus $F_{1}$ and right focus $F_{2}$, and upper vertex $A$. The perpendicular bisector of $A F_{2}$ intersects the ellipse at point $B$. If the left focus $F_{1}$ lies on the line segment $A B$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;A: Point;B: 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;Intersection(PerpendicularBisector(LineSegmentOf(A, F2)), G) = B;PointOnCurve(F1,LineSegmentOf(A,B))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 54], [101, 103], [131, 133]], [[4, 54]], [[4, 54]], [[81, 84]], [[104, 108]], [[69, 76]], [[58, 65], [113, 120]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 65]], [[2, 76]], [[2, 84]], [[87, 108]], [[113, 129]]]", "query_spans": "[[[131, 138]]]", "process": "" }, { "text": "Given that the equation $(1+k) x^{2}-(1-k) y^{2}=1$ represents a hyperbola with foci on the $x$-axis, what is the range of values for $k$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2*(k + 1) - y^2*(1 - k) = 1);PointOnCurve(Focus(G), xAxis);k:Number", "query_expressions": "Range(k)", "answer_expressions": "(-1,1)", "fact_spans": "[[[42, 45]], [[2, 45]], [[33, 45]], [[47, 50]]]", "query_spans": "[[[47, 57]]]", "process": "The curve represented by the equation is a hyperbola with foci on the x-axis. Set up an inequality in terms of k and solve to find the range of k. Since (1+k)x^{2}-(1-k)y^{2}=1 represents a hyperbola with foci on the x-axis, we have \\begin{cases}(1+k)[\\\\1+k>0\\\\1-k>0\\end{cases}+k)[-(1-k)]<0, solving gives -10,b>0). From the given conditions, 2a=6, 2b=8, that is, a=3, b=4. Therefore, the required hyperbola equation is: \\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1." }, { "text": "Given that the eccentricity of the ellipse $\\frac{x^{2}}{m+9}+\\frac{y^{2}}{m}=1$ is $\\frac{\\sqrt{3}}{3}$, what is the value of the real number $m$?", "fact_expressions": "G: Ellipse;m: Real;Expression(G) = (x^2/(m + 9) + y^2/m = 1);Eccentricity(G) = sqrt(3)/3", "query_expressions": "m", "answer_expressions": "18", "fact_spans": "[[[2, 41]], [[68, 73]], [[2, 41]], [[2, 66]]]", "query_spans": "[[[68, 77]]]", "process": "In the ellipse $\\frac{x^{2}}{m+9}+\\frac{y^{2}}{m}=1$, since $m+9>m$, the foci of the ellipse lie on the x-axis, with $a=\\sqrt{m+9}$, $b=\\sqrt{m}$, then $c=3$, and its eccentricity is $e=\\frac{c}{a}=\\frac{3}{\\sqrt{m+9}}=\\frac{\\sqrt{3}}{3}$, solving gives $m=18$." }, { "text": "The standard equation of the ellipse that shares foci with the hyperbola $\\frac{x^{2}}{5}-y^{2}=1$ and passes through the point $(0,-2)$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;I: Point;Expression(G) = (x^2/5 - y^2 = 1);Coordinate(I) = (0, -2);Focus(G)=Focus(H);PointOnCurve(I, H)", "query_expressions": "Expression(H)", "answer_expressions": "x^2/10+y^2/4=1", "fact_spans": "[[[1, 29]], [[46, 48]], [[36, 45]], [[1, 29]], [[36, 45]], [[0, 48]], [[34, 48]]]", "query_spans": "[[[46, 55]]]", "process": "c=\\sqrt{6}, and b=2, so a=\\sqrt{10}, therefore the equation of the ellipse is \\frac{x^{2}}{10}+\\frac{y^{2}}{4}=1^{\\circ}" }, { "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$ and $F_{2}$. Point $P$ lies on the ellipse. If $|P F_{1}|=4$, then what is the measure of $\\angle F_{1} P F_{2}$? What is the area of $\\triangle F_{1} P F_{2}$?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/9 + y^2/2 = 1);Focus(G) = {F1, F2};PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4", "query_expressions": "AngleOf(F1,P,F2);Area(TriangleOf(F1,P,F2))", "answer_expressions": "2*pi/3\n2*sqrt(3)", "fact_spans": "[[[0, 37], [62, 64]], [[41, 48]], [[57, 61]], [[49, 56]], [[0, 37]], [[0, 56]], [[57, 65]], [[67, 80]]]", "query_spans": "[[[82, 108]], [[108, 137]]]", "process": "" }, { "text": "Given the equation of the hyperbola is $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$, then the distance from the focus of the hyperbola to the asymptote is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/9 - y^2/16 = 1)", "query_expressions": "Distance(Focus(G), Asymptote(G))", "answer_expressions": "4", "fact_spans": "[[[2, 5], [47, 50]], [[2, 45]]]", "query_spans": "[[[47, 62]]]", "process": "Problem Analysis: First, find the coordinates of the foci and the equations of the asymptotes from the given conditions, then substitute into the point-to-line distance formula to obtain the result. From the problem: the coordinates of the foci are (-5,0), (5,0). The equations of the asymptotes are y=\\pm\\frac{4}{3}x, or \\pm3y\\cdot4x=0, so the distance from a focus to an asymptote is d=\\frac{20}{5}=4." }, { "text": "The hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has eccentricity $\\sqrt{3}$. When $a=1$, the line $x-y+m=0$ intersects the hyperbola $C$ at two distinct points $A$ and $B$, and the midpoint of segment $AB$ lies on the circle $x^{2}+y^{2}=5$. Find the value of $m$?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;H: Line;m: Number;B: Point;A: Point;a>0;b>0;a=1;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (x^2 + y^2 = 5);Expression(H) = (m + x - y = 0);Eccentricity(C) = sqrt(3);Intersection(H, C) = {A, B};PointOnCurve(MidPoint(LineSegmentOf(A,B)),G);Negation(A=B)", "query_expressions": "m", "answer_expressions": "pm*1", "fact_spans": "[[[0, 61], [98, 104]], [[8, 61]], [[8, 61]], [[131, 147]], [[86, 97]], [[150, 153]], [[115, 118]], [[111, 114]], [[8, 61]], [[8, 61]], [[79, 84]], [[0, 61]], [[131, 147]], [[86, 97]], [[0, 76]], [[86, 118]], [[120, 148]], [104, 115]]", "query_spans": "[[[150, 156]]]", "process": "First, find the equation of the hyperbola. Let the coordinates of points A and B be (x_{1},y_{1}), (x_{2},y_{2}), and let the midpoint of segment AB be M(x_{0},y_{0}). By solving the system of the line and the hyperbola equations simultaneously and using Vieta's formulas, solve for m using the given conditions. When a=1, e=\\frac{c}{a}=\\sqrt{3}, so c=\\sqrt{3}. Since c^{2}=a^{2}+b^{2}, we get b^{2}=2. Thus, the equation of hyperbola C is x^{2}-\\frac{y^{2}}{2}=1. Let the coordinates of points A and B be (x_{1},y_{1}), (x_{2},y_{2}), and the midpoint of segment AB be M(x_{0},y_{0}). From \\begin{cases}x^{2}-\\frac{y^{2}}{2}=1\\\\x+y+m=0\\end{cases}, we obtain x^{2}-2mx-m^{2}-2=0 (discriminant \\Delta>0). \\therefore x_{0}=\\frac{x_{1}+x_{2}}{2}=m, y_{0}=-x_{0}-m=-2m. Since point M(x_{0},y_{0}) lies on the circle x^{2}+y^{2}=5, \\therefore m^{2}+4m^{2}=5, \\therefore m=\\pm1." }, { "text": "The equation of the directrix of the parabola $y=-\\frac{1}{8} x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = -x^2/8)", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "y = 2", "fact_spans": "[[[0, 25]], [[0, 25]]]", "query_spans": "[[[0, 32]]]", "process": "(1) First, convert to the standard equation of the parabola; based on the directrix equation $-\\frac{p}{2}$, obtain the result; (2) Find the foci of the parabola and the ellipse, then set up equations to solve; (3) Based on the directrix equation $-\\frac{p}{2}$, set up an equation to solve for $a$. [Detailed Solution] (1) The standard equation of the parabola $y = -\\frac{1}{8}x^2$ is $x^2 = -8y$, so the directrix equation is $\\boxed{}$, that is, $y = 2$; (2) The focus of the parabola $y^2 = 2px$ ($p > 0$) is , and a right focus of the ellipse is , so $B$, solving gives: $B$; (3) Since the directrix equation of the parabola is $y = 4$, so , solving gives: $B$" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a major axis of length $8$, $F_{1}$, $F_{2}$ are the two foci of $C$, $P$ is a point on $C$ such that $P F_{1} \\perp P F_{2}$, and the area of $\\Delta P F_{1} F_{2}$ is $9$. Find 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;Length(MajorAxis(C)) = 8;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2));Area(TriangleOf(P, F1, F2)) = 9", "query_expressions": "Expression(C)", "answer_expressions": "x^2/16+y^2/9=1", "fact_spans": "[[[2, 59], [87, 90], [100, 103], [165, 170]], [[2, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[8, 59]], [[2, 67]], [[70, 77]], [[78, 86]], [[70, 95]], [[96, 99]], [[96, 106]], [[108, 131]], [[134, 163]]]", "query_spans": "[[[165, 176]]]", "process": "According to the problem, a = 4. Since F_{1} and F_{2} are the two foci of ellipse C and P is a point on C, let the semi-focal distance be c. Then |PF_{1}| + |PF_{2}| = 2a = 8. Squaring both sides gives |PF_{1}|^{2} + |PF_{2}|^{2} + 2|PF_{1}| \\cdot |PF_{2}| = 64. Since PF_{1} \\perp PF_{2}, it follows that |PF_{1}|^{2} + |PF_{2}|^{2} = |F_{1}F_{2}|^{2} = 4c^{2}. Also, the area of triangle PF_{1}F_{2} is 9, so \\frac{1}{2}|PF_{1}| \\cdot |PF_{2}| = 9, which implies |PF_{1}| \\cdot |PF_{2}| = 18. Thus we obtain 4c^{2} + 2 \\times 18 = 64. Solving gives c^{2} = 7, and therefore b^{2} = a^{2} - c^{2} = 9. Hence, the standard equation of ellipse C is \\frac{x^{2}}{16} + \\frac{y^{2}}{9} = 1." }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and its directrix intersects the $x$-axis at point $A$. A line $l$ is drawn through $A$ intersecting the parabola at points $M$ and $N$. Then the range of values of $|FM|^{2}+|FN|^{2}$ is?", "fact_expressions": "l: Line;G: Parabola;F: Point;M: Point;N: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Intersection(Directrix(G), xAxis) = A;PointOnCurve(A, l);Intersection(l, G) = {M, N};A:Point", "query_expressions": "Range(Abs(LineSegmentOf(F, M))^2 + Abs(LineSegmentOf(F, N))^2)", "answer_expressions": "", "fact_spans": "[[[44, 49]], [[2, 16], [24, 25], [50, 53]], [[20, 23]], [[55, 58]], [[59, 62]], [[2, 16]], [[2, 23]], [[24, 38]], [[39, 49]], [[44, 64]], [[34, 38], [40, 43]]]", "query_spans": "[[[66, 94]]]", "process": "" }, { "text": "Given that $O$ is the coordinate origin, on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ there are two points $A$ and $B$ satisfying $O A \\perp O B$, and the distance from point $O$ to the line $A B$ is $c$. Then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;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);c:Number;IsPerpendicular(LineSegmentOf(O,A),LineSegmentOf(O,B));Distance(O, LineOf(A,B)) = c", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)+1)/2", "fact_spans": "[[[11, 67], [118, 121]], [[14, 67]], [[14, 67]], [[69, 72]], [[73, 76]], [[2, 5], [97, 101]], [[14, 67]], [[14, 67]], [[11, 67]], [[11, 78]], [[11, 78]], [[113, 116]], [[80, 95]], [[97, 116]]]", "query_spans": "[[[118, 127]]]", "process": "① When the slope of line AB does not exist, the distance from point O to line AB is c, so the equation of line AB is x = ±c. Since OA ⊥ OB, by the property of an isosceles right triangle and the symmetry of the hyperbola, we have c = \\frac{b^{2}}{a}, that is, b^{2} = ac. Also, in a hyperbola, b^{2} = c^{2} - a^{2}, so ac = c^{2} - a^{2}, which implies c^{2} - ac - a^{2} = 0. Dividing both sides by a^{2} gives e^{2} - e - 1 = 0. Solving yields e = \\frac{1 \\pm \\sqrt{5}}{2}. Since e > 1, we have e = \\frac{1 + \\sqrt{5}}{2}; \n② When the slope of line AB exists, let the equation of the line be y = kx + m. Combining equations: \n\\begin{cases} y = kx + m \\\\ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 \\end{cases}, \nsimplifying gives (b^{2} - a^{2}k^{2})x^{2} - 2a^{2}kmx - a^{2}m^{2} - a^{2}b^{2} = 0. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), then x_{1} + x_{2} = \\frac{2a^{2}km}{b^{2} - a^{2}k^{2}}, x_{1}x_{2} = \\frac{a^{2}m^{2} + a^{2}b^{2}}{a^{2}k^{2} - b^{2}}. Thus, y_{1}y_{2} = (kx_{1} + m)(kx_{2} + m) = k^{2}x_{1}x_{2} + km(x_{1} + x_{2}) + m^{2} = \\frac{a^{2}k^{2}b^{2} - b^{2}m^{2}}{a^{2}k^{2} - b^{2}}. Since the distance from point O to line AB is c, we have \\frac{|m|}{\\sqrt{k^{2} + 1}} = c, simplifying gives m^{2} = k^{2}c^{2} + c^{2}. Also, since OA ⊥ OB, we have x_{1}x_{2} + y_{1}y_{2} = \\frac{a^{2}m^{2} + a^{2}b^{2}}{a^{2}k^{2} - b^{2}} + \\frac{a^{2}k^{2}b^{2} - b^{2}m^{2}}{a^{2}k^{2} - b^{2}} = 0. Simplifying yields (a^{2} - b^{2})(k^{2}c^{2} + c^{2}) + a^{2}b^{2} + a^{2}k^{2}b^{2} = 0, that is, (a^{2}c^{2} - b^{4})(1 + k^{2}) = 0. Therefore, a^{2}c^{2} - b^{4} = 0. In a hyperbola, b^{2} = c^{2} - a^{2}, substituting gives a^{2}c^{2} - (c^{2} - a^{2})^{2} = 0. Rearranging yields e^{2} = \\frac{3 \\pm \\sqrt{5}}{2}, so e = \\frac{1 \\pm \\sqrt{5}}{2}. Since e > 1, we have e = \\frac{1 + \\sqrt{5}}{2}." }, { "text": "The coordinates of the foci of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$ are?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/4 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, pm*sqrt(2))", "fact_spans": "[[[0, 37]], [[0, 37]]]", "query_spans": "[[[0, 44]]]", "process": "The ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$, gives $a=2$, $b=\\sqrt{2}$, $c=\\sqrt{2}$, the coordinates of the foci of the ellipse are: $(0,\\pm\\sqrt{2})$" }, { "text": "Given that a moving point $P$ lies on the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{40}=1$. If point $A$ has coordinates $(3,0)$, point $M$ satisfies $|\\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;A: Point;M: Point;P: Point;Expression(G) = (x^2/49 + y^2/40 = 1);PointOnCurve(P,G);Coordinate(A) = (3, 0);Abs(VectorOf(A, M)) = 1;DotProduct(VectorOf(P, M), VectorOf(A, M)) = 0", "query_expressions": "Min(Abs(VectorOf(P, M)))", "answer_expressions": "sqrt(15)", "fact_spans": "[[[8, 47]], [[49, 53]], [[65, 69]], [[4, 7]], [[8, 47]], [[4, 48]], [[49, 64]], [[71, 97]], [[99, 150]]]", "query_spans": "[[[152, 182]]]", "process": "In the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{40}=1$, $a=7$, $b=\\sqrt{40}$, we obtain $c=3$. According to $\\overrightarrow{PM}\\cdot\\overrightarrow{AM}=0$, hence $\\overrightarrow{PM}\\bot\\overline{A}$, $|\\overrightarrow{PM}|^{2}=|\\overrightarrow{PA}|^{2}-|\\overrightarrow{AM}|^{2}=|\\overrightarrow{PA}|^{2}-1$ because $|\\overrightarrow{AM}|=1$. The trajectory of point $M$ is a circle with center at point $A$ and radius $1$. Combining with the graph, the answer can be obtained. In the ellipse $\\frac{x^{2}}{49}+\\frac{y^{2}}{40}=1$, $a=7$, $b=\\sqrt{40}$, $\\frac{1}{2}-b^{2}=3$ \n$$\n\\begin{matrix}\nPM & AM \\\\\n\\frac{PM}{AM} & =|\\overrightarrow{PA}|\n\\end{matrix}^{2}-|\\overrightarrow{AM}^{2}=|\\overrightarrow{PA}|^{2}-1\n$$\n$\\therefore$ The trajectory of point $M$ is a circle with center at point $A$ and radius $1$. \n$\\therefore$ The smaller $|\\overrightarrow{AP}|$, the smaller $|\\overrightarrow{PM}|$. Draw the graph as shown: \nCombining with the graph, when point $P$ is the right vertex of the ellipse, $|\\overrightarrow{AP}|$ reaches its minimum value $a-c=7-3=4$. \n$\\therefore$ The minimum value of $|\\overrightarrow{PM}|$ is $\\sqrt{4^{2}-1}=\\sqrt{15}$." }, { "text": "The ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has a focus at $(2,0)$, and the ellipse passes through the point $A(2, 3)$. Then the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (2, 3);Coordinate(OneOf(Focus(C))) = (2,0);PointOnCurve(A, C)", "query_expressions": "Eccentricity(C)", "answer_expressions": "1/2", "fact_spans": "[[[0, 57], [74, 76], [90, 95]], [[6, 57]], [[6, 57]], [[77, 88]], [[6, 57]], [[6, 57]], [[0, 57]], [[77, 88]], [[0, 72]], [[74, 88]]]", "query_spans": "[[[90, 101]]]", "process": "From the given conditions, we easily obtain: \n\\begin{cases}c=2\\\\\\frac{4}{a^{2}}+\\frac{9}{b^{2}}=1\\end{cases}, \nthus solving gives: \n\\begin{cases}a=4\\\\c=2\\end{cases} \n\\therefore the eccentricity \\( 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)$ with right focus $F$, a line $l$ passing through point $F$: $y=\\sqrt{3}(x-2 a)$ intersects the right branch of hyperbola $C$ at point $A$ and intersects the $y$-axis at point $B$. If the area of $\\triangle O B F$ is $8 \\sqrt{3}$, where $O$ is the origin, then $\\frac{|A F|}{|B F|}$=?", "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;l: Line;PointOnCurve(F, l) = True;Expression(l) = (y = sqrt(3)*(-2*a + x));A: Point;Intersection(l,RightPart(C)) = A;B: Point;Intersection(l,yAxis) = B;Area(TriangleOf(O, B, F)) = 8*sqrt(3);O: Origin", "query_expressions": "Abs(LineSegmentOf(A, F))/Abs(LineSegmentOf(B, F))", "answer_expressions": "3/8", "fact_spans": "[[[2, 63], [105, 111]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [73, 77]], [[2, 71]], [[78, 104]], [[72, 104]], [[78, 104]], [[116, 120]], [[78, 120]], [[129, 133]], [[78, 133]], [[136, 169]], [[173, 176]]]", "query_spans": "[[[183, 206]]]", "process": "According to the problem, we have c = 2a. Then from S_{\\triangleOBF} = \\frac{1}{2} \\cdot 2a \\cdot 2\\sqrt{3}a = 8\\sqrt{3}, solve for a. Using b^{2} = c^{2} - a^{2}, we can find the solution. Let y = 0, then x = 2a, so c = 2a. Let x = 0, then y = -2\\sqrt{3}a. Therefore, S_{\\triangleOBF} = \\frac{1}{2} \\cdot 2a \\cdot |-2\\sqrt{3}a| = 8\\sqrt{3}. Solving gives a = 2, so c = 4, and thus b^{2} = c^{2} - a^{2} = 12. The standard equation of the hyperbola is: \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1. The line l: y = \\sqrt{3}(x - 4). Solve the system \\begin{cases} \\frac{x^{2}}{4} - \\frac{y^{2}}{12} = 1 \\\\ y = \\sqrt{3}(x - 4) \\end{cases}, yielding x = \\frac{5}{2}, y = -\\frac{3\\sqrt{3}}{2}, so A(\\frac{5}{2}, -\\frac{3\\sqrt{3}}{2}), B(0, -4\\sqrt{3}). Draw a perpendicular from A to the y-axis, with foot D, then D(0, \\frac{-3\\sqrt{3}}{2}). Therefore, \\frac{|AF|}{|BF|} = \\frac{|DO|}{|BO|} = \\frac{3}{8}" }, { "text": "The focus of the parabola $y^{2}=4 x$ is $F$. A line passing through the point $P(2,0)$ intersects the parabola at points $A$ and $B$. The lines $A F$ and $B F$ intersect the parabola again at points $C$ and $D$, respectively. If the slopes of lines $A B$ and $C D$ are $k_{1}$ and $k_{2}$, respectively, then $\\frac{k_{1}}{k_{2}}$=?", "fact_expressions": "G: Parabola;H: Line;F: Point;A: Point;B: Point;C: Point;D: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (2, 0);Focus(G) = F;PointOnCurve(P, H);Intersection(H, G) = {A, B};Intersection(LineOf(A, F), G) = C;Intersection(LineOf(B, F), G) = D;Slope(LineOf(A, B)) = k1;Slope(LineOf(C, D)) = k2;k1: Number;k2: Number", "query_expressions": "k1/k2", "answer_expressions": "1/2", "fact_spans": "[[[0, 14], [37, 40], [70, 73]], [[33, 35]], [[18, 21]], [[43, 46]], [[47, 50]], [[74, 78]], [[79, 82]], [[23, 32]], [[0, 14]], [[23, 32]], [[0, 21]], [[22, 35]], [[33, 52]], [[53, 82]], [[53, 82]], [[85, 121]], [[85, 121]], [[105, 112]], [[114, 121]]]", "query_spans": "[[[123, 146]]]", "process": "" }, { "text": "$\\frac{x^{2}}{4-t}+\\frac{y^{2}}{t-1}=1$ represents a hyperbola; then the range of real number $t$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(4 - t) + y^2/(t - 1) = 1);t: Real", "query_expressions": "Range(t)", "answer_expressions": "(-oo, 1) + (4, +oo)", "fact_spans": "[[[41, 44]], [[0, 44]], [[46, 51]]]", "query_spans": "[[[46, 58]]]", "process": "When the focus is on the x-axis, \\begin{cases}4-t>0\\\\t-1<0\\end{cases}, solving yields t<1; when the focus is on the y-axis, \\begin{cases}4-t<0\\\\t-1>0\\end{cases}, solving yields t>4;" }, { "text": "Given that $P$ is a moving point on the parabola $x^{2}=4 y$, find the minimum value of the sum of the distances from point $P$ to the lines $l_{1}$: $4 x-3 y-7=0$ and $l_{2}$: $y=-1$.", "fact_expressions": "G: Parabola;Expression(G) = (x^2 = 4*y);P: Point;PointOnCurve(P, G);l1: Line;Expression(l1) = (4*x - 3*y - 7 = 0);l2: Line;Expression(l2) = (y = -1)", "query_expressions": "Min(Distance(P, l1) + Distance(P, l2))", "answer_expressions": "2", "fact_spans": "[[[6, 20]], [[6, 20]], [[2, 5], [27, 31]], [[2, 25]], [[32, 56]], [[32, 56]], [[57, 72]], [[57, 72]]]", "query_spans": "[[[27, 83]]]", "process": "Let $ P(x_{0},\\frac{x_{0}^{2}}{4}) $. Then the distance from $ P $ to $ l_{1} $ is $ \\underline{\\hspace{1cm}} $. The distance from $ P $ to $ l_{2} $ is $ \\frac{x_{0}^{2}}{4}+1 $. Therefore, the sum of distances from point $ P $ to the two lines is $ \\frac{-\\left(-\\frac{8}{3}\\right)^{2}-\\frac{5}{3}<0,}{5}+\\frac{x_{0}^{2}}{4}+1=\\frac{2}{5}(x_{0}-1)^{2}+2 $. Hence, when $ x_{0}=1 $, the sum of distances is minimized and equals 2." }, { "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ has an eccentricity of $2$, and its foci coincide with the foci of the ellipse $\\frac{x^{2}}{25} + \\frac{y^{2}}{9}=1$, what are the coordinates of the vertices of the hyperbola? What are the equations of its asymptotes?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/25 + y^2/9 = 1);Eccentricity(G) = 2;Focus(G) = Focus(H)", "query_expressions": "Coordinate(Vertex(G));Expression(Asymptote(G))", "answer_expressions": "(pm*2, 0)\ny = pm*sqrt(3)*x", "fact_spans": "[[[2, 48], [108, 111]], [[5, 48]], [[5, 48]], [[60, 100]], [[2, 48]], [[60, 100]], [[2, 56]], [[2, 105]]]", "query_spans": "[[[108, 118]], [[108, 125]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the ellipse $C$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{m}=1$. If there exists a point $P$ on $C$ such that $\\angle F_{1} P F_{2}=90^{\\circ}$, then what is the range of values for $m$?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/4 + y^2/m = 1);m: Number;F1: Point;F2: Point;Focus(C) = {F1, F2};P: Point;PointOnCurve(P, C);AngleOf(F1, P, F2) = ApplyUnit(90, degree)", "query_expressions": "Range(m)", "answer_expressions": "(0, 2]+[8, +oo)", "fact_spans": "[[[17, 59], [66, 69]], [[17, 59]], [[113, 116]], [[1, 8]], [[9, 16]], [[1, 64]], [[72, 76]], [[66, 76]], [[78, 111]]]", "query_spans": "[[[113, 123]]]", "process": "If the foci of the ellipse lie on the x-axis, then 04, and similarly it is required that \\frac{\\sqrt{m-4}}{2}\\geqslant1, solving which gives m\\geqslant8. In conclusion, m\\in(0,2]\\cup[8,+\\infty)." }, { "text": "Given that $M$ is a point on the parabola $y^{2}=4x$, $F$ is its focus, and point $A$ lies on the circle $(x-5)^{2}+(y+1)^{2}=1$, then the minimum value of $|MA|+|MF|$ is?", "fact_expressions": "M: Point;PointOnCurve(M, G) = True;G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;A: Point;PointOnCurve(A, H) = True;H: Circle;Expression(H) = ((x - 5)^2 + (y + 1)^2 = 1)", "query_expressions": "Min(Abs(LineSegmentOf(M, A)) + Abs(LineSegmentOf(M, F)))", "answer_expressions": "5", "fact_spans": "[[[2, 5]], [[2, 23]], [[6, 20], [28, 29]], [[6, 20]], [[24, 27]], [[24, 31]], [[32, 36]], [[32, 62]], [[37, 61]], [[37, 61]]]", "query_spans": "[[[64, 83]]]", "process": "As shown in the figure, draw a perpendicular line from M to the directrix x = -1, with foot E. Then |MA| + |MF| = |MA| + |ME|, and |MA| + |ME| \\geqslant AE \\geqslant 5 - 1 + 1 = 5. Therefore, the minimum value is 5." }, { "text": "The directrix of the parabola $y^{2}=8x$ intersects the $x$-axis at point $P$. A line with slope $k$ ($k>0$) passing through point $P$ intersects the parabola at points $A$ and $B$. $F$ is the focus of the parabola. If $|FA|=2|FB|$, then $k=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 8*x);P: Point;Intersection(Directrix(G), xAxis) = P;H: Line;PointOnCurve(P, H);k: Number;Slope(H) = k;k>0;A: Point;B: Point;Intersection(H, G) = {A, B};F: Point;Focus(G) = F;Abs(LineSegmentOf(F, A)) = 2*Abs(LineSegmentOf(F, B))", "query_expressions": "k", "answer_expressions": "2*sqrt(2)/3", "fact_spans": "[[[0, 14], [51, 54], [69, 72]], [[0, 14]], [[25, 29], [31, 35]], [[0, 29]], [[48, 50]], [[30, 50]], [[93, 96]], [[36, 50]], [[39, 47]], [[55, 58]], [[59, 62]], [[48, 64]], [[65, 68]], [[65, 75]], [[77, 91]]]", "query_spans": "[[[93, 98]]]", "process": "Problem Analysis: Let A(x_{1},y_{1}), B(x_{2},y_{2}). From the given FA = 2|FB, we get: x_{1} + 2 = 2(x_{2} + 2), that is, x_{1} = 2x_{2} + 2\\textcircled{1}. Since P(-2,0), the equation of AB is: y = kx + 2k. Solving simultaneously with y^{2} = 8x gives: k^{2}x^{2} + (4k^{2} - 8)x + 4k^{2} = 0. Then x_{1}x_{2} = 4\\textcircled{2}. From \\textcircled{1}\\textcircled{2} we get x_{2} = 1, then B(1,2\\sqrt{2}), so k = \\frac{2\\sqrt{2}}{3}" }, { "text": "Let $M$ and $N$ be the two endpoints of the transverse axis of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and let $Q$ be a point on the hyperbola (distinct from $M$ and $N$). If $\\angle Q M N=\\alpha$, $\\angle Q N M=\\beta$, then $\\tan \\alpha \\tan \\beta=$?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;M: Point;N: Point;Endpoint(RealAxis(G)) = {M, N};Q: Point;PointOnCurve(Q, G);Negation(Q=M);Negation(Q=N);alpha: Number;AngleOf(Q, M, N) = alpha;beta: Number;AngleOf(Q, N, M) = beta", "query_expressions": "Tan(alpha)*Tan(beta)", "answer_expressions": "-b^2/a^2", "fact_spans": "[[[9, 65], [77, 80]], [[9, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[1, 4], [87, 90]], [[5, 8], [91, 94]], [[1, 72]], [[73, 76]], [[73, 84]], [[73, 97]], [[73, 97]], [[98, 119]], [[98, 119]], [[121, 141]], [[121, 141]]]", "query_spans": "[[[144, 170]]]", "process": "Let Q(m,n) (m≠±a), then tanα = n/(m+a), tanβ = n/(a−m), so tanα tanβ = n²/(a²−m²). Since Q lies on the hyperbola, we have m²/a² − n²/b² = 1, thus m² = a²(1 + n²/b²), it follows that tanα tanβ = −b²/a²" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively, point $A$ is a point on the ellipse located in the first quadrant, $O$ is the origin, and satisfies $\\overrightarrow{O A} \\cdot \\overrightarrow{O F_{2}}=|\\overrightarrow{O F_{2}}|^{2}$. If the equation of line $O A$ is $y=\\frac{\\sqrt{2}}{2} x$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;O: Origin;A: Point;F2: Point;F1: Point;a: Number;b: Number;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Quadrant(A) = 1;PointOnCurve(A, G);DotProduct(VectorOf(O, A), VectorOf(O, F2)) = Abs(VectorOf(O, F2))^2;Expression(LineOf(O, A)) = (y = x*(sqrt(2)/2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[20, 70], [82, 84], [231, 233]], [[96, 99]], [[77, 81]], [[10, 17]], [[2, 9]], [[20, 70]], [[20, 70]], [[20, 70]], [[20, 70]], [[20, 70]], [[2, 76]], [[2, 76]], [[77, 95]], [[77, 95]], [[108, 192]], [[194, 229]]]", "query_spans": "[[[231, 240]]]", "process": "From $\\overrightarrow{OA}\\cdot\\overrightarrow{OF}_{2}=|\\overrightarrow{OF}_{2}||\\overrightarrow{OA}|\\cos\\angle AOF=|\\overrightarrow{OF_{2}}|^{2}$ it follows that $OF_{2}\\bot AF_{2}$, determine the coordinates of point $A$, then substitute into the line equation. $\\overrightarrow{OA}\\cdot\\overrightarrow{OF_{2}}=|\\overrightarrow{OF_{2}}||\\overrightarrow{OA}|\\cos\\angle AOF_{2}=|\\overrightarrow{OF_{2}}||$, thus $|\\overrightarrow{OA}|\\cos\\angle AOF_{2}=|\\overrightarrow{OF_{2}}|$, so $OF_{2}\\bot AF_{2}$, hence the abscissa of point $A$ is $c$, then $\\frac{c^{2}}{a^{2}}+\\frac{y_{A}^{2}}{b^{2}}=1$. Since $A$ lies in the first quadrant, $y_{A}=\\frac{b^{2}}{a}$, thus $A(c,\\frac{b^{2}}{a})$. Also, the equation of line $OA$ is $y=\\frac{\\sqrt{2}}{2}x$, giving $\\frac{b^{2}}{a}=\\frac{\\sqrt{2}}{2}c$, i.e., $a^{2}-c^{2}=\\frac{\\sqrt{2}}{2}ac$, so $1-e^{2}=\\frac{\\sqrt{2}}{2}e$, solving yields: $e=\\frac{\\sqrt{2}}{2}$." }, { "text": "Given circles $C_{1}$: $(x-2)^{2}+(y-2)^{2}=4$ and $C_{2}$: $(x+2)^{2}+(y+1)^{2}=2$, point $P$ is a moving point on circle $C_{1}$, and $AB$ is a moving chord of circle $C_{2}$ such that $|AB|=2$. Then the maximum value of $|\\overrightarrow{PA}+\\overrightarrow{PB}|$ is?", "fact_expressions": "C1: Circle;Expression(C1) = ((x - 2)^2 + (y - 2)^2 = 4);C2: Circle;Expression(C2) = ((x + 2)^2 + (y + 1)^2 = 2);A: Point;B: Point;P: Point;PointOnCurve(P, C1);IsChordOf(LineSegmentOf(A, B), C2);Abs(LineSegmentOf(A, B)) = 2", "query_expressions": "Max(Abs(VectorOf(P, A) + VectorOf(P, B)))", "answer_expressions": "16", "fact_spans": "[[[2, 34], [74, 82]], [[2, 34]], [[37, 68], [95, 103]], [[37, 68]], [[89, 94]], [[89, 94]], [[69, 73]], [[69, 88]], [[89, 108]], [[110, 119]]]", "query_spans": "[[[121, 172]]]", "process": "From the given conditions, the center of circle $ C_{1} $ is $ C_{1}(2,2) $ with radius 2, and the center of circle $ C_{2} $ is $ C_{2}(-2,-1) $ with radius $ \\sqrt{2} $. Draw $ C_{2}D \\perp AB $ intersecting $ AB $ at $ D $. Then $ D $ is the midpoint of $ AB $ and $ |C_{2}D| = \\sqrt{(\\sqrt{2})^{2} - 1^{2}} = 1 $. Therefore, the locus of point $ D $ is circle $ C_{3}: (x+2)^{2} + (y+1)^{2} = 1 $, with center $ C_{3}(-2,-1) $ and radius 1. By the vector parallelogram law, $ |\\overrightarrow{PA} + \\overrightarrow{PB}| = |2\\overrightarrow{PD}| $. Since $ (2+2)^{2} + (2+1)^{2} = 5 > 2 + 1 = 3 $, circles $ C_{1} $ and $ C_{3} $ are externally separated. Thus, the maximum value of $ |\\overrightarrow{PD}| $ is $ 5 + 2 + 1 = 8 $, and the maximum value of $ |\\overrightarrow{PA} + \\overrightarrow{PB}| $ is 16." }, { "text": "Given that point $O$ is the coordinate origin, point $M$ lies on the hyperbola $C$: $x^{2}-y^{2}=1$, and from point $M$ a perpendicular is drawn to one of the asymptotes of hyperbola $C$, with foot of the perpendicular at point $N$. What is the value of $|O N| \\cdot|M N|$?", "fact_expressions": "C: Hyperbola;O: Origin;N: Point;M: Point;Expression(C) = (x^2 - y^2 = 1);PointOnCurve(M, C);L:Line;PointOnCurve(M, L);IsPerpendicular(L, OneOf(Asymptote(C)));FootPoint(L, OneOf(Asymptote(C)))=N", "query_expressions": "Abs(LineSegmentOf(M, N))*Abs(LineSegmentOf(O, N))", "answer_expressions": "1/2", "fact_spans": "[[[17, 40], [48, 54]], [[2, 6]], [[68, 71]], [[12, 16], [43, 47]], [[17, 40]], [[12, 41]], [], [[42, 64]], [[42, 64]], [[42, 71]]]", "query_spans": "[[[73, 95]]]", "process": "Problem Analysis: Let point M(x₀, y₀), then x₀² - y₀² = 1, the line MN is y - y₀ = -(x - x₀); according to the problem, 0|ON|·|MN| = √2|(x₀ + y₀)/2| × |x₀ - y₀|/√2 = (x₀² - y₀²)/2 = 1/2" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{\\frac{75}{4}}=1$, and $P$ is a point on the ellipse such that $\\angle F_{1} PF_{2}=\\frac{\\pi}{3}$, find $S_{\\Delta F_{1} P F_{2}}=$?", "fact_expressions": "F1: Point;F2: Point;G: Ellipse;Expression(G) = (x^2/25 + y^2/((75/4)) = 1);LeftFocus(G) = F1;RightFocus(G) = F2;P: Point;PointOnCurve(P, G) ;AngleOf(F1, P, F2) = pi/3", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "25*sqrt(3)/4", "fact_spans": "[[[2, 9]], [[10, 17]], [[18, 67], [77, 79]], [[18, 67]], [[2, 72]], [[2, 72]], [[73, 76]], [[73, 82]], [[84, 119]]]", "query_spans": "[[[121, 149]]]", "process": "Test analysis: From the equation of the ellipse, it is known that a^{2}=25, b^{2}=\\frac{75}{4}, and the area of the focal triangle AF_{1}PF_{2} is S=b^{2}\\tan\\frac{\\angle F_{1}PF_{2}}{2}=\\frac{75}{4}\\times\\tan\\frac{\\pi}{6}=\\frac{25\\sqrt{3}}{4}" }, { "text": "Point $P$ moves along the curve $\\sqrt{\\frac{x^{2}}{25}}+\\sqrt{\\frac{y^{2}}{9}}=1$, and $E$ is a fixed point on the curve in the second quadrant with $y$-coordinate $\\frac{15}{8}$. Let $O(0,0)$, $F(4,0)$. If $\\overrightarrow{O P}=x \\overrightarrow{O F}+y \\overrightarrow{O E}$, then the maximum value of $x+y$ is?", "fact_expressions": "G: Curve;Expression(G) = (sqrt(x^2/25) + sqrt(y^2/9) = 1);P: Point;PointOnCurve(P, G);E: Point;Quadrant(E) = 2;PointOnCurve(E, G);YCoordinate(E) = 15/8;O: Origin;Coordinate(O) = (0, 0);F: Point;Coordinate(F) = (4,0);x1: Number;y1: Number;VectorOf(O, P) = x1*VectorOf(O, F) + y1*VectorOf(O, E)", "query_expressions": "Max(x1 + y1)", "answer_expressions": "47/20", "fact_spans": "[[[5, 57], [65, 67]], [[5, 57]], [[0, 4]], [[0, 60]], [[61, 64], [76, 79]], [[61, 75]], [[61, 75]], [[76, 98]], [[101, 109]], [[101, 109]], [[112, 120]], [[112, 120]], [[122, 190]], [[122, 190]], [[122, 190]]]", "query_spans": "[[[192, 203]]]", "process": "According to the problem, point P lies on the curve $\\sqrt{\\frac{x^{2}}{25}}+\\sqrt{\\frac{y^{2}}{9}}=1$, that is, moves along $\\frac{|x|}{5}+\\frac{|y|}{3}=1$. Also, E is a fixed point on the curve in the second quadrant, and the y-coordinate of E is $\\frac{15}{8}$. Substituting this value gives $x=-\\frac{15}{8}$, so $E(-\\frac{15}{8},\\frac{15}{8})$. Let point $P(5\\cos2\\theta,3\\sin^{2}\\theta)$. Since $\\overrightarrow{OP}=x\\overrightarrow{OF}+y\\overrightarrow{OE}$, that is, $(5\\cos2\\theta,3\\sin^{2}\\theta)=x(4,0)+y(-\\frac{15}{8},\\frac{15}{8})$, we have\n$$\n\\begin{cases}\n4x-\\frac{15}{8}y=5\\cos2\\theta\\\\\n3\\sin^{2}\\theta=\\frac{15}{8}y\n\\end{cases}\n$$\nSolving yields\n$$\n\\begin{cases}\nx=\\frac{1}{2}\\cos2\\theta+\\frac{3}{4}\\\\\ny=\\frac{8}{5}\\sin^{2}\\theta\n\\end{cases}\n$$\nThus,\n$$\nx+y=\\frac{3}{4}+\\frac{1}{2}\\cos2\\theta+\\frac{8}{5}\\sin^{2}\\theta=\\frac{5}{4}+\\frac{11}{10}\\sin^{2}\\theta.\n$$\nWhen $\\sin^{2}\\theta=1$, the maximum value is $(x+y)_{\\max}=\\frac{5}{4}+\\frac{11}{10}=\\frac{47}{20}$." }, { "text": "If the coordinates of point $A$ are $(1,1)$, $F_{1}$ is the lower focus of the ellipse $5 y^{2} + 9 x^{2} = 45$, and point $P$ is a moving point on this ellipse, then the maximum value of $|P A| + |P F_{1}|$ is $M$, the minimum value is $N$, and $M - N = $?", "fact_expressions": "G: Ellipse;P: Point;A: Point;F1: Point;M:Number;N:Number;Expression(G) = (9*x^2 + 5*y^2 = 45);Coordinate(A) = (1, 1);LowerFocus(G) = F1;PointOnCurve(P, G);Max(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1))) = M;Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, F1))) = N", "query_expressions": "M - N", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[26, 48], [59, 61]], [[53, 57]], [[1, 5]], [[17, 25]], [[89, 92]], [[97, 100]], [[26, 48]], [[1, 15]], [[18, 52]], [[53, 65]], [[67, 92]], [[67, 100]]]", "query_spans": "[[[102, 109]]]", "process": "The ellipse $5y^{2}+9x^{2}=45$ is equivalent to $\\frac{x^{2}}{5}+\\frac{y^{2}}{9}=1$, giving $a=3$, $b=\\sqrt{5}$, $c=2$. Let $F_{2}$ be the upper vertex of the ellipse. Since $|PF_{1}|+|PF_{2}|=2a=6$, then $|PF_{1}|=6-|PF_{2}|$, so $|PF_{1}|+|PA|=6-|PF_{2}|+|PA|=6+(|PA|-|PF_{2}|)$ ① According to the triangle inequality, when point $P$ is at $P_{1}$, the difference $|PA|-|PF_{2}|$ reaches its maximum value $|AF_{2}|$, at which time $F_{2}$, $A$, $P$ are collinear. It is easy to find $|AF_{2}|=\\sqrt{2}$. Therefore, the maximum value of $|PF_{1}|+|PA|$ is $6+\\sqrt{2}$ ② According to the triangle inequality, when point $P$ is at $P_{2}$, the difference $|PA|-|PF_{2}|$ reaches its minimum value $-|AF_{2}|$, at which time $F_{2}$, $A$, $P$ are collinear. Therefore, the minimum value of $|PF_{1}|+|PA|$ is $6-\\sqrt{2}$. Thus $M=6+\\sqrt{2}$, $n=6-\\sqrt{2}$, so $M-n=6+\\sqrt{2}-6+\\sqrt{2}=2\\sqrt{2}$." }, { "text": "Given that $P$ is a point on the hyperbola $\\frac{x^{2}}{36}-\\frac{y^{2}}{64}=1$, and $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola, respectively. If $|P F_{2}|=14$, then $|P F_{1}|$=?", "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;Expression(G) = (x^2/36 - y^2/64 = 1);PointOnCurve(P, G);LeftFocus(G) = F1;RightFocus(G) = F2;Abs(LineSegmentOf(P, F2)) = 14", "query_expressions": "Abs(LineSegmentOf(P, F1))", "answer_expressions": "26", "fact_spans": "[[[6, 46], [68, 71]], [[2, 5]], [[58, 65]], [[50, 57]], [[6, 46]], [[2, 49]], [[50, 78]], [[50, 78]], [[80, 94]]]", "query_spans": "[[[96, 109]]]", "process": "From the given, we have a^{2}=36, b^{2}=64, so a=6, b=8, c^{2}=100, that is, c=10. Since the minimum distance from a point on the left branch of the hyperbola to the right focus F_{2} is a+c=6+10=16. And |PF_{2}|=14<16, so point P must lie on the right branch of the hyperbola. According to the definition of the hyperbola, we obtain |PF_{1}|-|PF_{2}|=2a=12, so |PF_{1}|=26." }, { "text": "Given that $F$ is the left focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$. A line $l$ passing through point $F$ is tangent to the circle $x^{2}+y^{2}=a^{2}$ at point $A$, and intersects the right branch of the hyperbola at point $B$. If $\\overrightarrow{F A}=\\frac{1}{3} \\overrightarrow{F B}$, then the equations of the asymptotes of the hyperbola are?", "fact_expressions": "l: Line;G: Hyperbola;b: Number;a: Number;H: Circle;F: Point;A: Point;B: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2 + y^2 = a^2);LeftFocus(G) = F;PointOnCurve(F, l);TangentPoint(l, H) = A;Intersection(l, RightPart(G)) = B;VectorOf(F, A) = (1/3)*VectorOf(F, B)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*(3/2)*x", "fact_spans": "[[[73, 78]], [[6, 62], [109, 112], [181, 184]], [[9, 62]], [[9, 62]], [[79, 99]], [[2, 5], [68, 72]], [[102, 106]], [[118, 122]], [[9, 62]], [[9, 62]], [[6, 62]], [[79, 99]], [[2, 66]], [[67, 78]], [[73, 106]], [[73, 122]], [[124, 179]]]", "query_spans": "[[[181, 192]]]", "process": "As shown in the figure: let the right focus of the hyperbola be F. Since OA = a, AF ⊥ OA, OF = c, ∴ AF = √(c² - a²) = b, ∴ cos∠AFO = b/c. Since vector FA = (1/3) vector FB, and O is the midpoint of FF', ∴ BF = 3b, FF' = 2c. From the definition of hyperbola, we know: BF - BF' = 2a, ∴ BF' = 3b - 2a. In triangle BFF', by the law of cosines, we get: BF'² = 9b² + 4c² - 2·3b·2c·(b/c) = 4c² - 3b². ∴ (3b - 2a)² = 4c² - 3b², simplifying yields: b/a = 3/2. ∴ The asymptotes of the hyperbola are: y = ±(3/2)x. The correct answer is: y = ±(3/2)x" }, { "text": "Given that $F$ is a focus of the hyperbola $C$: $\\frac{x^{2}}{3}-y^{2}=1$, then the distance from point $F$ to an asymptote of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (x^2/3 - y^2 = 1);F: Point;OneOf(Focus(C)) = F", "query_expressions": "Distance(F, OneOf(Asymptote(C)))", "answer_expressions": "1", "fact_spans": "[[[6, 39], [51, 57]], [[6, 39]], [[2, 5], [46, 50]], [[2, 44]]]", "query_spans": "[[[46, 68]]]", "process": "Find the values of a, b, c for the hyperbola; assume F(2,0), then obtain an asymptote equation of the hyperbola, and apply the point-to-line distance formula to compute the required result. For hyperbola C: \\frac{x^2}{3}-y^{2}=1, we have a=\\sqrt{3}, b=1, c=\\sqrt{a^{2}+b^{2}}=2, thus let F(2,0); let an asymptote equation of the hyperbola be y=\\frac{\\sqrt{3}}{3}x, then the distance from F to the asymptote is d=\\frac{\\sqrt[2]{3}}{\\sqrt{1+\\frac{1}{3}}}=1" }, { "text": "Let $F$ be the focus of the parabola $C$: $x^{2}=16 y$, line $l$: $y=-1$, and point $A$ be any point on $C$. Draw $A P \\perp l$ from point $A$ meeting $l$ at $P$. Then $| A P|-| A F| $=?", "fact_expressions": "F: Point;Focus(C) = F;C: Parabola;Expression(C) = (x^2 = 16*y);l: Line;Expression(l) = (y = -1);A: Point;PointOnCurve(A, C) = True;PointOnCurve(A,LineSegmentOf(A, P)) = True;IsPerpendicular(LineSegmentOf(A,P),l) = True;FootPoint(LineSegmentOf(A,P),l) = P;P: Point", "query_expressions": "Abs(LineSegmentOf(A,P))-Abs(LineSegmentOf(A,F))", "answer_expressions": "3", "fact_spans": "[[[1, 4]], [[1, 28]], [[5, 25], [48, 51]], [[5, 25]], [[29, 42]], [[29, 42]], [[43, 47], [58, 62]], [[43, 56]], [[57, 76]], [[63, 76]], [[63, 80]], [[77, 80]]]", "query_spans": "[[[82, 101]]]", "process": "" }, { "text": "A point $P$ on the ellipse $C$ has the sum of distances to the two foci $F_{1}(-2,0)$, $F_{2}(2,0)$ equal to $6$. Then the standard equation of $C$ is?", "fact_expressions": "C: Ellipse;F1: Point;F2: Point;P: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);PointOnCurve(P, C);Distance(P, F1) + Distance(P, F2) = 6;Focus(C) = {F1, F2}", "query_expressions": "Expression(C)", "answer_expressions": "x^2/9 + y^2/5 = 1", "fact_spans": "[[[0, 5], [55, 58]], [[16, 29]], [[31, 43]], [[8, 11]], [[16, 29]], [[31, 43]], [[0, 11]], [[0, 53]], [[0, 43]]]", "query_spans": "[[[55, 65]]]", "process": "Since the sum of the distances from a point P on the ellipse C to the two foci F_{1}(-2,0) and F_{2}(2,0) is equal to 6, the semi-major axis length of the ellipse is a=3 and the semi-focal distance is c=2. Then the semi-minor axis length b satisfies b^{2}=a^{2}-c^{2}=5. Therefore, the standard equation of C is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1." }, { "text": "Let the left and right foci of the ellipse be $F_{1}$ and $F_{2}$, respectively, and the upper vertex be $B_{1}$. If $B_{1} F_{2}=F_{1} F_{2}=4$, then the standard equation of the ellipse is?", "fact_expressions": "E: Ellipse;F1: Point;F2: Point;B1: Point;LeftFocus(E) = F1;RightFocus(E) = F2;UpperVertex(E) = B1;LineSegmentOf(B1,F2) = LineSegmentOf(F1,F2);LineSegmentOf(F1,F2) = 4", "query_expressions": "Expression(E)", "answer_expressions": "x**2/16+y**2/12=1", "fact_spans": "[[[1, 3], [71, 73]], [[12, 19]], [[20, 27]], [[32, 39]], [[1, 27]], [[1, 27]], [[1, 39]], [[41, 68]], [[41, 68]]]", "query_spans": "[[[71, 80]]]", "process": "According to the given conditions and the relation c^{2}=a^{2}-b^{2}, we can obtain the answer. According to the problem, the left and right foci of the ellipse are F_{1} and F_{2}, respectively, and the upper vertex is B_{1}, then F_{1}F_{2}=2c, |OB_{1}|=b, so |B_{1}F_{2}|=\\sqrt{b^{2}+c^{2}}=a. If |B_{1}F_{2}|=|F_{1}F_{2}|=4, then a=2c=4, so a=4, c=2, then b=\\sqrt{a^{2}-c^{2}}=\\sqrt{12}. Then the equation of the ellipse is \\frac{x^{2}}{16}+\\frac{y^{2}}{12}=" }, { "text": "Given that point $A(-2,4)$ lies on the directrix of the parabola $C$: $ y^{2}=2 p x $, and the focus of the parabola is $F$, what is the slope of line $A F$?", "fact_expressions": "C: Parabola;p: Number;A: Point;F: Point;Expression(C) = (y^2 = 2*(p*x));Coordinate(A) = (-2, 4);PointOnCurve(A, Directrix(C));Focus(C) = F", "query_expressions": "Slope(LineOf(A,F))", "answer_expressions": "-1", "fact_spans": "[[[13, 34], [39, 42]], [[20, 34]], [[2, 12]], [[46, 49]], [[13, 34]], [[2, 12]], [[2, 38]], [[39, 49]]]", "query_spans": "[[[51, 63]]]", "process": "Problem analysis: From the given conditions, the directrix is x = -2, so \\frac{p}{2} = 2 \\therefore p = 4, and the focus is (2,0), therefore k_{AF} = \\frac{4-0}{-2-2} = -1" }, { "text": "The focus of a parabola is $F(1, -1)$, and the equation of the directrix is $x - y = 0$. Then, what are the coordinates of its vertex?", "fact_expressions": "G: Parabola;F: Point;Coordinate(F) = (1, -1);Focus(G) = F;Expression(Directrix(G)) = (x - y = 0)", "query_expressions": "Coordinate(Vertex(G))", "answer_expressions": "(1/2,-1/2)", "fact_spans": "[[[0, 3], [33, 34]], [[7, 17]], [[7, 17]], [[0, 17]], [[0, 30]]]", "query_spans": "[[[33, 41]]]", "process": "" }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\frac{3}{2}$, what are the equations of the asymptotes of this hyperbola?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;Eccentricity(G) = 3/2", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*sqrt(5)*x/2", "fact_spans": "[[[2, 58], [79, 82]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 76]]]", "query_spans": "[[[79, 90]]]", "process": "From the eccentricity $ e = \\frac{3}{2} $, we obtain $ \\frac{b}{a} = \\frac{\\sqrt{5}}{2} $; substituting into the asymptote equation gives the result. [Detailed solution] From the given information, the eccentricity is $ e = \\frac{c}{a} = \\frac{3}{2} $, so $ \\frac{c^{2}}{a^{2}} = \\frac{a^{2} + b^{2}}{a^{2}} = \\frac{9}{4} $, thus $ \\frac{b^{2}}{a^{2}} = \\frac{5}{4} $. Since the foci of the hyperbola lie on the x-axis, the equations of the asymptotes are $ y = \\pm \\frac{b}{a}x $, that is, $ y = \\pm \\frac{\\sqrt{5}}{2}x $." }, { "text": "If the equation $\\frac{x^{2}}{k-3}-\\frac{y^{2}}{k+3}=1(k \\in \\mathbb{R})$ represents a hyperbola, then what is the range of $k$?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/(k - 3) - y^2/(k + 3) = 1);k: Real", "query_expressions": "Range(k)", "answer_expressions": "(-3, 3)", "fact_spans": "[[[53, 56]], [[1, 56]], [[58, 61]]]", "query_spans": "[[[58, 66]]]", "process": "According to the problem: (k-3)(k+3)<0, solve to get -30)$, if the distance from point $P$ to the focus $F$ of the parabola is equal to $8$, then the distance from the focus $F$ to the directrix of the parabola is equal to?", "fact_expressions": "G: Parabola;p: Number;P: Point;F: Point;p>0;Expression(G)=(y^2 = 2*p*x);Coordinate(P)=(6, y1);PointOnCurve(P,G);Focus(G)=F;Distance(P,F) = 8;y1:Number", "query_expressions": "Distance(F,Directrix(G))", "answer_expressions": "4", "fact_spans": "[[[14, 35], [43, 46], [43, 46]], [[17, 35]], [[2, 13], [38, 42]], [[48, 51], [48, 51]], [[17, 35]], [[14, 35]], [[2, 13]], [[2, 36]], [[43, 51]], [[38, 59]], [[3, 13]]]", "query_spans": "[[[63, 78]]]", "process": "The directrix of the parabola \\( y^{2} = 2px \\) (\\( p > 0 \\)) is \\( x = -\\frac{p}{2} \\). Since point \\( P(6, y) \\) lies on the parabola, the distance from \\( P \\) to the focus \\( F \\) equals the distance from \\( P \\) to the directrix. Therefore, \\( 6 + \\frac{p}{2} = 8 \\), so \\( p = 4 \\). The distance from the focus \\( F \\) to the directrix of the parabola is 4." }, { "text": "The distance from point $A(0,1)$ to the asymptote of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$ is?", "fact_expressions": "G: Hyperbola;A: Point;Expression(G) = (x^2/4 - y^2 = 1);Coordinate(A) = (0, 1)", "query_expressions": "Distance(A, Asymptote(G))", "answer_expressions": "2*sqrt(5)/5", "fact_spans": "[[[10, 38]], [[0, 9]], [[10, 38]], [[0, 9]]]", "query_spans": "[[[0, 47]]]", "process": "" }, { "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, 79], [86, 89], [159, 162]], [[26, 79]], [[26, 79]], [[94, 97]], [[1, 8]], [[10, 17]], [[26, 79]], [[26, 79]], [[18, 79]], [[1, 84]], [[85, 97]], [[100, 121]], [[124, 156]]]", "query_spans": "[[[159, 168]]]", "process": "" }, { "text": "The line $y=k(x-1)$ intersects the parabola $y^{2}=4x$ at points $A$ and $B$. Given that the midpoint of $AB$ has coordinates $(\\frac{3}{2}, 1)$, then $|AB|$=?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = k*(x - 1));Coordinate(MidPoint(LineSegmentOf(A, B))) = (3/2, 1);Intersection(H, G) = {A, B};k:Number", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "5", "fact_spans": "[[[13, 27]], [[0, 12]], [[29, 32]], [[33, 36]], [[13, 27]], [[0, 12]], [[41, 70]], [[0, 38]], [[0, 12]]]", "query_spans": "[[[72, 81]]]", "process": "The parabola 2p=4, p=2, \\frac{p}{2}=1, the focus of the parabola is F(1,0), the directrix is x=-1, the line y=k(x-1) passes through (1,0), so k=\\frac{1-0}{3-1}=2. From \\begin{cases}y=2(x-1)\\\\y^2=4x\\end{cases}, eliminating y and simplifying yields x^{2}-3x+1=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then \\begin{cases}x_{1}+x_{2}=3\\\\x_{1}\\cdot x_{2}=1\\end{cases}, so |AB|=x_{1}+x_{2}+p=3+2=5" }, { "text": "Given that $F$ and $A$ are the left focus and the lower vertex of the ellipse $\\Gamma$: $\\frac{x^{2}}{2}+y^{2}=1$, respectively, and $M(x_{0}, y_{0})$ is a point on the ellipse $\\Gamma$ located in the first quadrant. The coordinates of point $N$ are $(0, \\frac{1}{y_{0}})$. If there exists a point $H$ on the segment $F M$ satisfying $|\\overrightarrow{F H}|=|\\overrightarrow{A H}|$ and $\\overrightarrow{F H} \\cdot \\overrightarrow{N H}=0$, then $x_{0} y_{0}$=?", "fact_expressions": "F: Point;A: Point;LeftFocus(Gamma) = F;LowerVertex(Gamma) = A;Gamma: Ellipse;Expression(Gamma) = (x^2/2 + y^2 = 1);M: Point;x0: Number;y0: Number;Coordinate(M) = (x0, y0);Quadrant(M) = 1;PointOnCurve(M, Gamma);N: Point;Coordinate(N) = (0, 1/y0);H: Point;PointOnCurve(H, LineSegmentOf(F, M));Abs(VectorOf(F, H)) = Abs(VectorOf(A, H));DotProduct(VectorOf(F, H), VectorOf(N, H)) = 0", "query_expressions": "x0*y0", "answer_expressions": "2/3", "fact_spans": "[[[2, 5]], [[6, 9]], [[2, 57]], [[2, 57]], [[12, 49], [76, 86]], [[12, 49]], [[58, 75]], [[58, 75]], [[58, 75]], [[58, 75]], [[58, 93]], [[58, 96]], [[97, 101]], [[97, 127]], [[139, 143]], [[129, 143]], [[147, 194]], [[195, 246]]]", "query_spans": "[[[248, 263]]]", "process": "Given F(-1,0), A(0,-1), since |\\overrightarrow{FH}| = |\\overrightarrow{AH}|, point H lies on the perpendicular bisector of segment FA, which is the line y = x. Therefore, we can set H(m,m) (m \\neq 0). Since H lies on segment FM, we have k_{FH} = k_{FM} \\Rightarrow \\frac{m}{m+1} = \\frac{y_{0}}{x_{0}+1} \\textcircled{1}. From \\overrightarrow{FH} \\cdot \\overrightarrow{NH} = 0, it follows that m(m+1) + m(m - \\frac{1}{y_{0}}) = 0 \\Rightarrow m = \\frac{1 - y_{0}}{2y_{0}}. Substituting into \\textcircled{1} gives \\frac{1 - y_{0}}{1 + y_{0}} = \\frac{y_{0}}{x_{0} + 1}, that is, 1 - y_{0}^{2} = (x_{0} + 2)y_{0} - x_{0}. Also, \\frac{x_{0}^{2}}{2} = 1 - y_{0}^{2}, so (x_{0} + 2)y_{0} - x_{0} = \\frac{x_{0}^{2}}{2} \\Rightarrow (x_{0} + 2)y_{0} = \\frac{x_{0}(x_{0} + 2)}{2}" }, { "text": "Through the point $P(1,1)$, draw a chord of the ellipse $\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1$ such that $P$ bisects the chord. Find the equation of the line containing this chord.", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/2 + y^2/4 = 1);P: Point;Coordinate(P) = (1, 1);H: LineSegment;PointOnCurve(P, H);IsChordOf(H, G);MidPoint(H) = P", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "2*x+y-3=0", "fact_spans": "[[[11, 48]], [[11, 48]], [[1, 10], [57, 61]], [[1, 10]], [], [[0, 52]], [[11, 52]], [[11, 63]]]", "query_spans": "[[[11, 76]]]", "process": "Let the line containing the chord intersect the ellipse at points A(x_{1},y_{1}) and B(x_{2},y_{2}). Since point P is the midpoint of the chord, then \\frac{x_{1}+x_{2}}{y_{1}+y_{2}}=1. From the given conditions, we have \\begin{cases}\\frac{x^{2}}{2}+\\frac{y^{2}}{4}=1\\\\\\frac{x^{2}}{2}+\\frac{y^{2}}{2}=1\\end{cases}. Subtracting these two equations gives \\frac{(x_{1}-x_{2})(x_{1}+x_{2})}{2}+\\frac{(y_{1}-y_{2})(y_{1}+y_{2})}{4}=0. Therefore, the slope of line AB is \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=-\\frac{4(x_{1}+x_{2})}{2(y_{1}+y_{2})}=-2. Hence, the equation of the line containing the chord is y-1=-2(x-1), that is, 2x+y-3=0." }, { "text": "The minimum value of the sum of distances from a moving point $P$ on the parabola $y=\\frac{1}{4} x^{2}$ to the lines $l_{1}: 3 x-4 y-6=0$ and $l_{2}: y=-1$ is?", "fact_expressions": "G: Parabola;l1: Line;l2: Line;Expression(G) = (y = x^2/4);Expression(l1) = (3*x - 4*y - 6 = 0);Expression(l2) = (y = -1);P: Point;PointOnCurve(P, G)", "query_expressions": "Min(Distance(P, l1) + Distance(P, l2))", "answer_expressions": "2", "fact_spans": "[[[0, 24]], [[32, 56]], [[57, 74]], [[0, 24]], [[32, 56]], [[57, 74]], [[28, 31]], [[0, 31]]]", "query_spans": "[[[28, 85]]]", "process": "Let the focus of the parabola be $ F $, then $ F(0,1) $; draw perpendiculars from $ P $ to $ l_{1} $ and $ l_{2} $, with feet of perpendiculars $ N $ and $ M $, respectively, as shown in the figure below: \n$ \\therefore $ The sum of the required distances is: $ |PM| + |PN| $. \nBy the definition of the parabola, we know: $ |PM| = |PF| $ \n$ \\therefore |PM| + |PN| = |PF| + |PN| \\geqslant |FN| $ (equality holds if and only if points $ F $, $ P $, $ N $ are collinear) \n$ \\therefore (|PM| + |PN|)_{\\min} = |FN| = \\frac{|-4 - 6|}{\\sqrt{3^{2} + 4^{2}}} = 2 $ \nThe correct answer to this problem: $ 2 $" }, { "text": "A straight line passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $P$ and $Q$. If the length of segment $PF$ is $3$, then what is the length of segment $FQ$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;L: Line;PointOnCurve(F, L);Intersection(L, G) = {P, Q};P: Point;Q: Point;Length(LineSegmentOf(P, F)) = 3", "query_expressions": "Length(LineSegmentOf(F, Q))", "answer_expressions": "3/2", "fact_spans": "[[[1, 15], [26, 29]], [[1, 15]], [[18, 21]], [[1, 21]], [[23, 25]], [[0, 25]], [[23, 39]], [[30, 33]], [[34, 37]], [[41, 53]]]", "query_spans": "[[[55, 65]]]", "process": "" }, { "text": "The line $y = kx - k$ intersects the parabola $y^2 = 4x$ at points $A$ and $B$. If $|AB| = 4$, then the distance from the midpoint of chord $AB$ to the directrix is?", "fact_expressions": "G: Parabola;H: Line;k: Number;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Expression(H) = (y = k*x - k);Intersection(H, G) = {A, B};Abs(LineSegmentOf(A, B)) = 4;IsChordOf(LineSegmentOf(A, B), G)", "query_expressions": "Distance(MidPoint(LineSegmentOf(A,B)),Directrix(G))", "answer_expressions": "2", "fact_spans": "[[[12, 26]], [[0, 11]], [[2, 11]], [[28, 31]], [[32, 35]], [[12, 26]], [[0, 11]], [[0, 37]], [[39, 48]], [[12, 56]]]", "query_spans": "[[[12, 67]]]", "process": "Problem Analysis: From the given conditions, the focus of the parabola y^{2}=4x is F(1,0), and its directrix is x=-1. The line y=kx-k=k(x-1) passes exactly through the point F(1,0). Let the x-coordinates of the intersection points of the line y=kx-k and the parabola y^{2}=4x be x_{1}, x_{2}. According to the definition of a parabola, the x-coordinate of the midpoint of AB is x_{0}=\\frac{x_{1}+x_{2}}{2}=\\frac{1}{2}|AB|-p=1. Therefore, the distance from the midpoint of chord AB to the directrix is 2." }, { "text": "$F_{1}$, $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$, and $A$ is a point on the ellipse such that $\\angle A F_{1} F_{2}=45^{\\circ}$. Then the area of $\\triangle A F_{1} F_{2}$ is?", "fact_expressions": "G: Ellipse;A: Point;F1: Point;F2: Point;Expression(G) = (x^2/9 + y^2/7 = 1);Focus(G) = {F1, F2};PointOnCurve(A, G);AngleOf(A, F1, F2) = ApplyUnit(45, degree)", "query_expressions": "Area(TriangleOf(A, F1, F2))", "answer_expressions": "7/2", "fact_spans": "[[[16, 53], [63, 65]], [[59, 62]], [[0, 7]], [[8, 15]], [[16, 53]], [[0, 58]], [[59, 68]], [[70, 103]]]", "query_spans": "[[[105, 135]]]", "process": "From $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$, we know $a^{2}=9$, $b^{2}=7$, $c^{2}=2$, so $a=3$, $b=\\sqrt{7}$, $c=\\sqrt{2}$, thus $|F_{1}F_{2}|=2\\sqrt{2}$. Let $|AF_{1}|=x$, then $|AF_{2}|=6-x$. Since $\\angle AF_{1}F_{2}=45^{\\circ}$, it follows that $(6-x)^{2}=x^{2}+8-4\\sqrt{2}x\\cdot\\frac{\\sqrt{2}}{2}$, solving gives $x=\\frac{7}{2}$, so $S_{\\triangle AF_{1}F_{2}}=\\frac{1}{2}\\times2\\sqrt{2}\\times\\frac{7}{2}\\times\\frac{\\sqrt{2}}{2}=\\frac{7}{2}$." }, { "text": "Given the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$, respectively. A line $l$ passing through $F_{1}$ with an inclination angle of $60^{\\circ}$ intersects the hyperbola at points $M$ and $N$. Find the inradius of $\\Delta M N F_{2}$.", "fact_expressions": "l: Line;G: Hyperbola;M: Point;N: Point;F2: Point;F1: Point;Expression(G) = (x^2/3 - y^2 = 1);RightFocus(G) = F2;LeftFocus(G)=F1;PointOnCurve(F1, l);Inclination(l) = ApplyUnit(60, degree);Intersection(l, G) = {M, N}", "query_expressions": "Radius(InscribedCircle(TriangleOf(M,N,F2)))", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[80, 85]], [[2, 30], [86, 89]], [[91, 94]], [[95, 98]], [[46, 53]], [[38, 45], [55, 62]], [[2, 30]], [[2, 53]], [[2, 53]], [[54, 85]], [[63, 85]], [[80, 100]]]", "query_spans": "[[[102, 128]]]", "process": "" }, { "text": "$A$ is an endpoint of the major axis of an ellipse, and $O$ is the center of the ellipse. If there exists a point $P$ on the ellipse such that $\\angle OPA = \\frac{\\pi}{2}$, then what is the range of the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;O: Origin;P: Point;A: Point;OneOf(Endpoint(MajorAxis(G)))=A;Center(G)=O;PointOnCurve(P, G);AngleOf(O, P, A) = pi/2", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(2)/2,1)", "fact_spans": "[[[4, 6], [4, 6], [4, 6], [4, 6]], [[14, 17]], [[32, 35]], [[0, 3]], [[0, 13]], [[14, 23]], [[25, 35]], [[37, 63]]]", "query_spans": "[[[65, 75]]]", "process": "" }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, point $N(4,0)$, and line $l$ passing through $F$ intersecting $C$ at points $A$ and $B$. If the circle with diameter $NF$ intersects $l$ at point $M$ (distinct from $F$), and $M$ is the midpoint of $AB$, then the length of segment $MF$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;N: Point;Coordinate(N) = (4, 0);l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};G: Circle;IsDiameter(LineSegmentOf(N, F), G);M: Point;Intersection(l, G) = M;Negation(M=F);MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Length(LineSegmentOf(M, F))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 21], [50, 53]], [[2, 21]], [[25, 28], [45, 48], [88, 91]], [[2, 28]], [[29, 38]], [[29, 38]], [[39, 44], [77, 80]], [[39, 48]], [[54, 57]], [[58, 61]], [[39, 63]], [[75, 76]], [[65, 76]], [[81, 85], [94, 97]], [[75, 85]], [[81, 92]], [[94, 105]]]", "query_spans": "[[[107, 118]]]", "process": "Given the focus of the parabola F(1,0), |NF|=3, so the midpoint of N and F is: (\\frac{5}{2},0), thus the equation of the circle with NF as diameter is (x-\\frac{5}{2})^{2}+y^{2}=\\frac{9}{4}. Let the equation of line AB be y=k(x-1). Let A(x_{1},y_{1}), B(x_{2},y_{2}). Solving the system \\begin{cases}y=k(x-1)\\\\y^{2}=4x\\end{cases}, we obtain: k^{2}x^{2}-2(2+k^{2})x+k^{2}=0, \\therefore x_{1}+x_{2}=2+\\frac{4}{k^{2}}, x_{M}=1+\\frac{2}{k^{2}}, y_{M}=k(x_{M}-1)=\\frac{2}{k}, so M(1+\\frac{2}{k^{2}},\\frac{2}{k}). Since point M lies on the circle with NF as diameter, MN\\bot MF. Since k_{MF}=k, k_{MN}=\\frac{\\frac{2}{k}}{1+\\frac{2}{12}-}, so k\\cdot\\frac{2k}{2-3k^{2}}k_{0}=-1, solving gives k=\\pm\\sqrt{2}, so M(2,\\pm\\sqrt{2}), thus |MF|=\\sqrt{(2-1)^{2}+(\\pm\\sqrt{2}-0)^{2}}=\\sqrt{3}" }, { "text": "Through the right focus $F_{2}$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, draw a chord $P Q$ perpendicular to the real axis, $F_{1}$ is the left focus. If $\\angle P F_{1} Q=90^{\\circ}$, 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;F2: Point;RightFocus(G) = F2;P: Point;Q: Point;IsChordOf(LineSegmentOf(P,Q),G) = True;PointOnCurve(F2,LineSegmentOf(P,Q)) = True;IsPerpendicular(LineSegmentOf(P,Q),RealAxis(G)) = True;F1: Point;LeftFocus(G) = F1;AngleOf(P, F1, Q) = ApplyUnit(90, degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1+sqrt(2)", "fact_spans": "[[[1, 47], [118, 121]], [[1, 47]], [[4, 47]], [[4, 47]], [[51, 58]], [[1, 58]], [[66, 71]], [[66, 71]], [[1, 71]], [[0, 71]], [[0, 71]], [[74, 81]], [[1, 85]], [[87, 116]]]", "query_spans": "[[[118, 127]]]", "process": "" }, { "text": "Given the ellipse $C_{1}$: $\\frac{x^{2}}{a_{1}^{2}}+\\frac{y^{2}}{b_{1}^{2}}=1(a_{1}>0, b_{1}>0)$ and the hyperbola $C_{2}$: $\\frac{x^{2}}{a_{2}^{2}}-\\frac{y^{2}}{b_{2}^{2}}=1(a_{2}>0, b_{2}>0)$ have the same foci $F_{1}$, $F_{2}$, point $P$ is a common point of curves $C_{1}$ and $C_{2}$, $e_{1}$, $e_{2}$ are the eccentricities of $C_{1}$ and $C_{2}$ respectively, if $P F_{1} \\perp P F_{2}$, then the minimum value of $4 e_{1}^{2}+e_{2}^{2}$ is?", "fact_expressions": "C1: Ellipse;Expression(C1) = (x^2/a1^2 + y^2/b1^2 = 1);a1: Number;b1: Number;a1 > 0;b1 > 0;C2: Hyperbola;Expression(C2) = (x^2/a2^2 - y^2/b2^2 = 1);a2: Number;b2: Number;a2 > 0;b2 > 0;F1: Point;F2: Point;Focus(C1) = {F1, F2};Focus(C2) = {F1, F2};OneOf(Intersection(C1, C2)) = P;P: Point;e1: Number;e2: Number;Eccentricity(C1) = e1;Eccentricity(C2) = e2;IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True", "query_expressions": "Min(4*e1^2 + e2^2)", "answer_expressions": "9/2", "fact_spans": "[[[2, 82], [191, 200], [234, 241]], [[2, 82]], [[13, 82]], [[13, 82]], [[13, 82]], [[13, 82]], [[83, 164], [201, 208], [242, 249]], [[83, 164]], [[95, 164]], [[95, 164]], [[95, 164]], [[95, 164]], [[170, 177]], [[178, 185]], [[2, 185]], [[2, 185]], [[186, 214]], [[186, 190]], [[215, 222]], [[224, 231]], [[215, 253]], [[215, 253]], [[255, 278]]]", "query_spans": "[[[280, 309]]]", "process": "According to the problem, let the focal distance be 2c, the major axis length of the ellipse be 2a_{1}, and the real axis length of the hyperbola be 2a_{2}. Let P lie on the right branch of the hyperbola. By the definition of the hyperbola, |PF_{1}| - |PF_{2}| = 2a_{2}; by the definition of the ellipse, |PF_{1}| + |PF_{2}| = 2a_{1}. Adding the squares of the two equations yields |PF_{1}|^{2} + |PF_{2}|^{2} = 2a_{1}^{2} + 2a_{2}^{2}. Since PF_{1} \\bot PF_{2}, it follows that |PF_{1}|^{2} + |PF_{2}|^{2} = 4c^{2}, so 2a_{1}^{2} + 2a_{2}^{2} = 4c^{2}. Therefore, 4e_{1}^{2} + e_{2}^{2} = \\frac{4c^{2} + c_{2}^{2}}{a_{1}^{2}} = \\frac{4(a_{1}^{2} + a_{2}^{2})}{2a_{1}^{2}} + \\frac{a_{1}^{2} + a_{2}^{2}}{a_{2}^{2}} = \\frac{5}{2} + \\frac{2a_{2}^{2}}{a_{1}^{2}} + \\frac{a_{1}^{2}}{a_{2}^{2}} \\geqslant \\frac{5}{2} + 2\\sqrt{\\frac{2a_{2}^{2}}{a_{1}^{2}} \\cdot \\frac{a_{1}^{2}}{a_{2}^{2}}} = \\frac{5}{2} + 2\\sqrt{2}, with equality if and only if a_{1} = 2a_{2}. Hence, the minimum value of 4e_{1}^{2} + e_{2}^{2} is \\frac{9}{2}." }, { "text": "Given that a moving point $P(x , y)$ lies on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$. If $F(3 , 0)$, $|PF|=2$, and $M$ is the midpoint of $P F$, then $|O M|$=?", "fact_expressions": "G: Ellipse;P: Point;F: Point;O: Origin;M: Point;x1:Number;y1:Number;Expression(G) = (x^2/25 + y^2/16 = 1);Coordinate(P) = (x1, y1);Coordinate(F) = (3, 0);PointOnCurve(P, G);Abs(LineSegmentOf(P, F)) = 2;MidPoint(LineSegmentOf(P, F)) = M", "query_expressions": "Abs(LineSegmentOf(O, M))", "answer_expressions": "4", "fact_spans": "[[[15, 54]], [[4, 14]], [[57, 67]], [[92, 99]], [[79, 82]], [[4, 14]], [[4, 14]], [[15, 54]], [[4, 14]], [[57, 67]], [[4, 55]], [[69, 77]], [[79, 90]]]", "query_spans": "[[[92, 101]]]", "process": "" }, { "text": "Given that $P$ is a moving point on the parabola $y=\\frac{1}{4} x^{2}$, find the minimum value of the sum of the distances from point $P$ to the line $l_{1}$: $4 x-3 y=7$ and the line $l_{2}$: $y=-1$.", "fact_expressions": "G: Parabola;l1: Line;l2: Line;P: Point;Expression(G) = (y = x^2/4);PointOnCurve(P, G);Expression(l1) = (4*x - 3*y = 7);Expression(l2) = (y = -1)", "query_expressions": "Min(Distance(P, l1) + Distance(P, l2))", "answer_expressions": "2", "fact_spans": "[[[6, 30]], [[42, 64]], [[65, 82]], [[2, 5], [37, 41]], [[6, 30]], [[2, 35]], [[42, 64]], [[65, 82]]]", "query_spans": "[[[37, 93]]]", "process": "Let P(x_{0},\\frac{x_{0}^{2}}{4}), then the distance from point P to line l_{1} is \\frac{4x_{0}-\\frac{3}{4}x_{0}^{2}-7}{5}, and the distance from point P to line l_{2} is \\frac{x_{0}^{2}}{4}+1. Since 4x_{0}-\\frac{3}{4}x_{0}-, the sum of distances from point P to the two lines is \\frac{3}{4}x_{0}-\\frac{8}{3}-\\frac{5}{5}x_{0}^{2}+7+\\frac{x_{0}^{2}}{4}+1=\\frac{2}{5}(x_{0}-1)^{2}+2. Therefore, when x_{0}=1, the minimum value of the sum of distances is 2." }, { "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}$ and $F_{2}$ respectively. A line passing through $F_{2}$ with slope $\\frac{b}{a}$ intersects the curve $C$ at point $P$. If $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);a: Number;b: Number;a>0;b>0;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F2, G);Slope(G) = b/a;P: Point;Intersection(G, C) = P;DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[2, 63], [117, 122], [191, 197]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[72, 79]], [[80, 87], [89, 96]], [[2, 87]], [[2, 87]], [[114, 116]], [[88, 116]], [[97, 116]], [[124, 128]], [[114, 128]], [[130, 189]]]", "query_spans": "[[[191, 203]]]", "process": "Analysis: The equation of $ PF_{2} $ is $ y = \\frac{b}{a}(x - c) $, the equation of $ PF_{1} $ is $ y = -\\frac{a}{b}(x + c) $. Solving these equations simultaneously gives the coordinates of point $ P $ as $ \\left( \\frac{b^{2} - a^{2}}{c}, \\frac{2ab}{c} \\right) $. Substituting into the hyperbola equation and simplifying yields the result. Consider the asymptotes of the hyperbola to be $ y = \\frac{b}{a}x $. Since $ F_{1}(-c, 0) $, $ F_{2}(c, 0) $, the line $ PF_{2} $ passing through $ F_{2} $ with slope $ \\frac{b}{a} $ has equation $ y = \\frac{b}{a}(x - c) $. Because $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = 0 $, it follows that $ PF_{1} \\perp PF_{2} $, so the equation of line $ PF_{1} $ is $ y = -\\frac{a}{b}(x + c) $. Solving the system of equations \n\\[\n\\begin{cases}\ny = \\frac{b}{a}(x - c) \\\\\ny = -\\frac{a}{b}(x + c)\n\\end{cases}\n\\]\ngives the coordinates of point $ P $ as $ \\left( \\frac{b^{2} - a^{2}}{c}, \\frac{2ab}{c} \\right) $. Since point $ P $ lies on the hyperbola, \n\\[\n\\frac{(b^{2} - a^{2})^{2}}{a^{2}c^{2}} - \\frac{\\left( \\frac{2ab}{c} \\right)^{2}}{b^{2}} = 1,\n\\]\nthat is, \n\\[\n\\frac{(b^{2} - a^{2})^{2}}{a^{2}c^{2}} - \\frac{4a^{2}}{c^{2}} = 1.\n\\]\nSince $ c^{2} = a^{2} + b^{2} $, it follows that \n\\[\n\\left( \\frac{c^{2} - 2a^{2}}{ac} \\right)^{2} - \\frac{4a^{2}}{c^{2}} = 1.\n\\]\nSimplifying yields $ c^{2} = 5a^{2} $. Since $ e = \\frac{c}{a} > 1 $, it follows that $ e = \\sqrt{5} $." }, { "text": "Parabola $C$: $y^{2}=2 p x(p>0)$, the intersection point of its directrix and the $x$-axis is $M$. From point $M$, draw two tangent lines to $C$, with points of tangency $P$ and $Q$. Then $\\angle P M Q$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;M: Point;Intersection(Directrix(C), xAxis) = M;P: Point;Q: Point;l1: Line;l2: Line;TangentOfPoint(M, C) = {l1, l2};TangentPoint(l1, C) = P;TangentPoint(l2, C) = Q", "query_expressions": "AngleOf(P, M, Q)", "answer_expressions": "pi/2", "fact_spans": "[[[0, 26], [48, 51]], [[0, 26]], [[8, 26]], [[8, 26]], [[38, 41], [43, 47]], [[0, 41]], [[62, 65]], [[67, 70]], [], [], [[42, 56]], [[42, 70]], [[42, 70]]]", "query_spans": "[[[73, 89]]]", "process": "The directrix of the parabola $ C: y^{2} = 2px $ ($ p > 0 $) intersects the x-axis at point $ M $, then $ M\\left(-\\frac{p}{2}, 0\\right) $. Let the equation of the line passing through point $ M $ be $ x = ky - \\frac{p}{2} $. From \n$$\n\\begin{cases}\nx = ky - \\frac{p}{2} \\\\\ny^{2} = 2px\n\\end{cases}\n$$\nwe obtain $ y^{2} - 2kpy + p^{2} = 0 $. Since $ \\Delta = 4k^{2}p^{2} - 4p^{2} = 0 $, we get $ k = \\pm 1 $, $ \\therefore PM \\perp QM $, therefore, $ \\angle PMQ = \\frac{\\pi}{2} $." }, { "text": "Given the line $ l $: $ x - \\sqrt{3} y = 0 $ intersects the hyperbola $ \\Gamma $: $ \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0 $, $ b > 0 $) at points $ A $ and $ B $. A perpendicular line $ AC $ to line $ l $ is drawn through point $ A $, intersecting the hyperbola $ \\Gamma $ at point $ C $. If $ \\angle ABC = 60^{\\circ} $, then the eccentricity of the hyperbola $ \\Gamma $ is?", "fact_expressions": "Gamma: Hyperbola;l: Line;a: Number;b: Number;a>0;b>0;A: Point;B: Point;Expression(l) = (x - sqrt(3)*y = 0);Expression(Gamma) = (-y^2/b^2 + x^2/a^2 = 1);Intersection(l, Gamma) = {A, B};C: Point;PointOnCurve(A, LineOf(A, C));IsPerpendicular(LineOf(A, C), l);Intersection(LineOf(A, C), Gamma) = C;AngleOf(A, B, C) = ApplyUnit(60, degree)", "query_expressions": "Eccentricity(Gamma)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[26, 92], [122, 133], [168, 179]], [[2, 25], [108, 113]], [[39, 92]], [[39, 92]], [[39, 92]], [[39, 92]], [[93, 96], [104, 107]], [[97, 100]], [[2, 25]], [[26, 92]], [[2, 102]], [[134, 138]], [[103, 121]], [[108, 121]], [[116, 138]], [[141, 166]]]", "query_spans": "[[[168, 185]]]", "process": "Solving the line $ x = \\sqrt{3}y $ and the hyperbola equation gives $ x^{2} = \\frac{3a^{2}b^{2}}{3b^{2} - a^{2}} $, $ y^{2} = \\frac{a^{2}b^{2}}{3b^{2} - a^{2}} $. In right triangle $ ABC $, $ \\angle ABC = 60^\\circ $, we get $ |AC| = \\sqrt{3}|AB| $. Let the equation of line $ AC $ be $ y = -\\sqrt{3}x + \\frac{4ab}{\\sqrt{3b^{2} - a^{2}}} $. Substituting into the hyperbola equation yields $ (b^{2} - 3a^{2})x^{2} + \\frac{8\\sqrt{3}a^{3}b}{\\sqrt{3b^{2} - a^{2}}}x - a^{2}b^{2} - \\frac{16a^{4}b^{2}}{3b^{2} - a^{2}} = 0 $. $ \\frac{b^{2} - a^{2}|b^{2} - 3}{3} $ implies $ AC = 2\\frac{2\\sqrt{3}ab}{\\sqrt{3b^{2} - a^{2}}} $, leading to $ a^{2} + b^{2} = |b^{2} - 3a^{2}| $, giving $ a = b $, $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{2} $." }, { "text": "Given that line $l$ is not perpendicular to the $x$-axis, and line $l$ passes through point $M(2,0)$ and intersects the parabola $y^{2}=4x$ at points $A$ and $B$, then $\\frac{1}{|AM|^{2}}+\\frac{1}{|BM|^{2}}$=?", "fact_expressions": "l: Line;G: Parabola;M: Point;A: Point;B: Point;Expression(G) = (y^2 = 4*x);Coordinate(M) = (2, 0);Negation(IsPerpendicular(l,xAxis));PointOnCurve(M, l);Intersection(l, G) = {A, B}", "query_expressions": "1/(Abs(LineSegmentOf(B, M))^2) + 1/(Abs(LineSegmentOf(A, M))^2)", "answer_expressions": "1/4", "fact_spans": "[[[2, 7], [2, 7]], [[33, 47]], [[23, 32]], [[49, 52]], [[53, 56]], [[33, 47]], [[23, 32]], [[2, 15]], [[17, 32]], [[17, 58]]]", "query_spans": "[[[60, 103]]]", "process": "Let $ l: \\begin{cases} x = 2 + t\\cos\\alpha \\\\ y = t\\sin\\alpha \\end{cases} $ ($ t $ is a parameter, $ \\alpha $ is the angle of inclination). Substituting into $ y^2 = 4x $ yields $ t^2\\sin^2\\alpha - 4t\\cos\\alpha - 8 = 0 $. Therefore, $ \\frac{1}{|AM|^2} + \\frac{1}{|BM|^2} = \\frac{1}{t_1} \\frac{+ \\frac{16}{\\sin^2\\alpha}}{4} = \\frac{1}{4}\\sin^4( $" }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{4}+y^{2}=1$, $F$ is the right focus, and the coordinates of point $A$ are $(0, \\sqrt{6})$, then the maximum perimeter of $\\triangle A F P$ is?", "fact_expressions": "G: Ellipse;A: Point;F: Point;P: Point;Expression(G) = (x^2/4 + y^2 = 1);PointOnCurve(P, G);RightFocus(G) = F;Coordinate(A) = (0, sqrt(6))", "query_expressions": "Max(Perimeter(TriangleOf(A, F, P)))", "answer_expressions": "10", "fact_spans": "[[[6, 33]], [[46, 50]], [[38, 41]], [[2, 5]], [[6, 33]], [[2, 37]], [[6, 45]], [[46, 69]]]", "query_spans": "[[[71, 96]]]", "process": "As shown in the figure, let the left focus of the ellipse be F. Using |AF| = |AF|, |PF| + |PF| = 2a = 4, and |PA| - |PF| ≤ |AF|, the result can be obtained. As shown in the figure, let the left focus of the ellipse be F. From the given conditions, a = 2, b = 1, c = \\sqrt{3}, then F(\\sqrt{3}, 0). Since the coordinates of A are (0, \\sqrt{6}), it follows that |AF| = |AF| = 3. By the definition of the ellipse, |PF| + |PF| = 2a = 4. Because |PA| - |PF| ≤ |AF|, the perimeter of \\triangle AFP is |AF| + |PA| + |PF| = |AF| + |PA| + 4 - |PF| ≤ 3 + 4 + 3 = 10. Equality holds if and only if points A, P, F are collinear. Therefore, the maximum perimeter of \\triangle AFP is 10." }, { "text": "Through the focus $F$ of the parabola $M$: $y^{2}=4x$, draw two mutually perpendicular chords $AB$ and $CD$, intersecting $M$ at points $A$, $B$, $C$, and $D$, respectively. Then the minimum value of $|AB|+|CD|$ is?", "fact_expressions": "M: Parabola;Expression(M) = (y^2 = 4*x);F: Point;Focus(M) = F;A: Point;B: Point;C: Point;D: Point;PointOnCurve(F, LineSegmentOf(A, B));PointOnCurve(F, LineSegmentOf(C, D));IsChordOf(LineSegmentOf(A, B), M);IsChordOf(LineSegmentOf(C, D), M);IsPerpendicular(LineSegmentOf(A, B), LineSegmentOf(C, D));Intersection(LineSegmentOf(A, B), M) = {A, B};Intersection(LineSegmentOf(C, D), M) = {C, D}", "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(C, D)))", "answer_expressions": "16", "fact_spans": "[[[1, 20], [50, 53]], [[1, 20]], [[23, 26]], [[1, 26]], [[54, 57]], [[58, 61]], [[62, 65]], [[66, 69]], [[0, 46]], [[0, 46]], [[1, 40]], [[1, 46]], [[27, 46]], [[35, 69]], [[35, 69]]]", "query_spans": "[[[71, 90]]]", "process": "It is clear that the slope of line AB exists and is non-zero, so let the equation of line AB be y = k(x - 1) (k ≠ 0), A(x₁, y₁), B(x₂, y₂). The equation of line AB and the parabola equation y² = 4x are combined, eliminating y, yielding: k²x² - (2k² + 4)x + k² = 0. ∴ x₁ + x₂ = (2k² + 4)/k², |AB| = x₁ + x₂ + p = 4/k² + 4. Similarly, |CD| = 4 + 4k². Then |AB| + |CD| = 8 + 4k² + 4/k² ≥ 16, with equality holding if and only if k = ±1, the minimum value being 16." }, { "text": "Given that $F_{1}$, $F_{2}$ are the left and right foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, if line $l$ passes through focus $F_{1}$ and intersects the ellipse at points $A$, $B$, then the perimeter of $\\Delta ABF_{2}$ is?", "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;F2: Point;F1: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftFocus(G)=F1;RightFocus(G) = F2;PointOnCurve(F1, l);Intersection(l,G) ={A,B}", "query_expressions": "Perimeter(TriangleOf(A,B,F2))", "answer_expressions": "8", "fact_spans": "[[[64, 69]], [[20, 57], [82, 84]], [[86, 89]], [[90, 93]], [[11, 19]], [[2, 9], [72, 79]], [[20, 57]], [[2, 62]], [[2, 62]], [[64, 79]], [[64, 93]]]", "query_spans": "[[[95, 116]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with left focus $F$, a line $l$ passing through the origin intersects the left and right branches of the hyperbola at points $P$ and $Q$, respectively, and satisfies that the value of $|Q F|-|P F|$ is equal to the length of the imaginary axis. Find the eccentricity of this hyperbola?", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;Q: Point;F: Point;P: Point;O:Origin;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F;PointOnCurve(O,l);Intersection(l,LeftPart(C))=P;Intersection(l,RightPart(C))=Q;Abs(LineSegmentOf(Q,F))-Abs(LineSegmentOf(P,F))=Length(ImageinaryAxis(C))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[76, 81]], [[2, 63], [82, 85], [132, 135]], [[10, 63]], [[10, 63]], [[100, 103]], [[68, 71]], [[94, 98]], [[73, 75]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 71]], [[72, 81]], [[76, 103]], [[76, 103]], [[82, 129]]]", "query_spans": "[[[132, 140]]]", "process": "Let the right focus be F'. Since line l passes through the origin and hyperbola C is symmetric about the origin, |PF| = |QF'|. Then |QF| - |PF| = |QF| - |QF'| = 2a = 2b, so the eccentricity is e = \\sqrt{1+\\frac{b^{2}}{a^{2}}} = \\sqrt{2}." }, { "text": "Given that $O$ is the coordinate origin, the parabola $C$: $y^{2}=2 p x(p>0)$ has focus $F$, $P$ is a point on $C$, $PF$ is perpendicular to the $x$-axis, $Q$ is a point on the $x$-axis, and $PQ \\perp OP$. If $|FQ|=6$, then $|PF|=$?", "fact_expressions": "C: Parabola;p: Number;P: Point;F: Point;Q: Point;O: Origin;p>0;Expression(C) = (y^2 = 2*(p*x));Focus(C)=F;PointOnCurve(P,C);IsPerpendicular(LineSegmentOf(P,F),xAxis);PointOnCurve(Q, xAxis);IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(O, P));Abs(LineSegmentOf(F, Q)) = 6", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "3", "fact_spans": "[[[11, 37], [49, 52]], [[19, 37]], [[45, 48]], [[41, 44]], [[69, 72]], [[2, 5]], [[19, 37]], [[11, 37]], [[11, 44]], [[45, 55]], [[56, 68]], [[69, 80]], [[82, 97]], [[99, 108]]]", "query_spans": "[[[110, 119]]]", "process": "The parabola $ C: y^2 = 2px $ ($ p > 0 $) has focus $ F\\left(\\frac{p}{2}, 0\\right) $. Since $ P $ is a point on $ C $ and $ PF $ is perpendicular to the x-axis, the x-coordinate of $ P $ is $ \\frac{p}{2} $. Substituting into the parabola equation gives the y-coordinate of $ P $ as $ \\pm p $. Without loss of generality, let $ P\\left(\\frac{p}{2}, p\\right) $. Since $ Q $ is a point on the x-axis and $ PQ \\perp OP $, $ Q $ lies to the right of $ F $. Given $ |FQ| = 6 $, we have $ Q\\left(6 + \\frac{p}{2}, 0\\right) $, so $ \\overrightarrow{PQ} = (6, -p) $. Because $ PQ \\perp OP $, it follows that $ \\overrightarrow{PQ} \\cdot \\overrightarrow{OP} = \\frac{p}{2} \\times 6 - p^2 = 0 $. Since $ p > 0 $, we get $ p = 3 $. Therefore, $ |PF| = 3 $." }, { "text": "The center of the ellipse is at the origin $O$, $A$ and $C$ are the top and bottom vertices of the ellipse, $B$ is the left vertex, $F$ is the left focus, and line $AF$ intersects line $BC$ at point $D$. If the eccentricity of the ellipse is $\\frac{1}{2}$, then what is the tangent of $\\angle BDF$?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;A: Point;C: Point;UpperVertex(G) = A;LowerVertex(G) = C;B: Point;LeftVertex(G) = B;F: Point;LeftFocus(G) = F;D: Point;Intersection(LineOf(A, F), LineOf(B, C)) = D;Eccentricity(G) = 1/2", "query_expressions": "Tan(AngleOf(B, D, F))", "answer_expressions": "3*sqrt(3)", "fact_spans": "[[[2, 4], [26, 28], [38, 40], [49, 51], [78, 80]], [[8, 15]], [[2, 15]], [[16, 19]], [[20, 23]], [[16, 33]], [[16, 33]], [[34, 37]], [[34, 44]], [[45, 48]], [[45, 55]], [[71, 75]], [[56, 75]], [[78, 98]]]", "query_spans": "[[[100, 117]]]", "process": "Problem Analysis: From the given conditions, A(0,b), B(-a,0), C(0,-b), F(-c,0), so \\overrightarrow{CB}=(-a,b), \\overrightarrow{FA}=(c,b), therefore \\cos\\angle BDF=\\frac{\\overrightarrow{CB}\\cdot\\overrightarrow{FA}}{|CB|\\cdot|FA|}=\\frac{\\sqrt{a^{2}+b^{2}}}{\\sqrt{c^{2}+b^{2}}}. Given the eccentricity is \\frac{1}{2}, so a=2c, b=\\sqrt{3}c, thus \\cos\\angle BDF=\\frac{1}{2\\sqrt{7}}, \\tan\\angle BDF=3\\sqrt{3}." }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4 x$, and $P$ is a point on the parabola $C$ in the first quadrant. If $M$ is the midpoint of $P F$, and $O$ is the vertex of the parabola $C$, then the maximum value of the slope of the line $O M$ is?", "fact_expressions": "C: Parabola;M: Point;O:Origin;P: Point;F: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(P, C);Quadrant(P)=1;MidPoint(LineSegmentOf(P, F)) = M;Vertex(C)=O", "query_expressions": "Max(Slope(LineOf(O,M)))", "answer_expressions": "1", "fact_spans": "[[[6, 25], [33, 39], [66, 72]], [[49, 52]], [[62, 65]], [[29, 32]], [[2, 5]], [[6, 25]], [[2, 28]], [[29, 47]], [[29, 47]], [[49, 61]], [[62, 75]]]", "query_spans": "[[[77, 92]]]", "process": "From the given conditions, we have F(1,0), let P(\\frac{y_{0}^{2}}{4}, y_{0}), (y_{0}>0), M(x,y). Since M is the midpoint of segment PF, then M(\\frac{1}{2}+\\frac{y_{0}^{2}}{8}, \\frac{y_{0}}{2}). Therefore, k_{OM} = \\frac{\\frac{y_{0}}{2}}{\\frac{1}{2}+\\frac{y_{0}^{2}}{8}} = \\frac{4y_{0}}{4+y_{0}^{2}} = \\frac{4}{\\frac{4}{y_{0}}+y_{0}} \\leqslant \\frac{4}{2\\sqrt{\\frac{4}{y_{0}}\\times y_{0}}} = 1, with equality if and only if y_{0}=2. Therefore, the maximum value of the slope of line OM is 1." }, { "text": "Given that $F$ is the focus of the parabola $y^{2}=-16x$, $O$ is the origin, point $P$ is a moving point on the directrix of the parabola, point $A$ lies on the parabola, and $|AF|=8$, then the minimum value of $|PA|+|PO|$ is?", "fact_expressions": "G: Parabola;A: Point;F: Point;P: Point;O: Origin;Expression(G) = (y^2 = -16*x);Focus(G) = F;PointOnCurve(P, Directrix(G));PointOnCurve(A, G);Abs(LineSegmentOf(A, F)) = 8", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, O)))", "answer_expressions": "4*sqrt(13)", "fact_spans": "[[[6, 22], [40, 43], [56, 59]], [[51, 55]], [[2, 5]], [[35, 39]], [[26, 29]], [[6, 22]], [[2, 25]], [[35, 50]], [[51, 60]], [[62, 71]]]", "query_spans": "[[[73, 92]]]", "process": "Since F is the focus of the parabola y^{2} = -16x, and point A lies on the parabola, |AF| = \\frac{p}{2} - x_{A} = 4 - x_{A} = 8, solving gives x_{A} = -4, so A(-4, \\pm8). O is the origin, and its symmetric point with respect to the directrix x = 4 is O(8,0), then |PO| = |PO|." }, { "text": "The minimum distance from a point on the parabola $y^{2}=2 p y(p>0)$ to its directrix is $1$, then $p=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2=2*p*y);p: Number;p>0;H: Point;PointOnCurve(H, G);Min(Distance(H, Directrix(G))) = 1", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[0, 21], [25, 26]], [[0, 21]], [[39, 42]], [[3, 21]], [[23, 24]], [[0, 24]], [[23, 37]]]", "query_spans": "[[[39, 44]]]", "process": "From the knowledge of parabolas, the minimum distance from the vertex of the parabola to the directrix is $\\frac{p}{2}$. Thus, $\\frac{p}{2}=1$, solving for $p$ gives $p=2$. Answer: 2" }, { "text": "On the hyperbola $4 x^{2}-y^{2}-4=0$, the distance from a point $P$ to one of its foci is equal to $3$. Then, what is the distance from point $P$ to the other focus?", "fact_expressions": "G: Hyperbola;P: Point;Expression(G) = (4*x^2 - y^2 - 4 = 0);PointOnCurve(P, G);F1:Point;F2:Point;OneOf(Focus(G))=F1;OneOf(Focus(G))=F2;Negation(F1=F2);Distance(P, F1) = 3", "query_expressions": "Distance(P,F2)", "answer_expressions": "5", "fact_spans": "[[[0, 22], [29, 30]], [[25, 28], [46, 50]], [[0, 22]], [[0, 28]], [], [], [[29, 35]], [[29, 56]], [[29, 56]], [[25, 43]]]", "query_spans": "[[[29, 62]]]", "process": "First simplify the hyperbola to the standard form $ x^{2}-\\frac{y^{2}}{4}=1 $, yielding $ a=1 $. Let the distance from point $ P $ to the other focus be $ PF_{2} $. Then by the definition of the hyperbola, $ ||PF_{2}|-3|=2a=2 $, $ |PF_{2}|>0 $, so solving gives the value of $ |PF_{2}| $. Since the hyperbola $ 4x^{2}-y^{2}-4=0 $ simplifies to the standard form $ x^{2}-\\frac{y^{2}}{4}=1 $, we obtain $ a^{2}=1 $, hence $ a=1 $. Let the distance from point $ P $ to the other focus be $ PF_{2} $, $ PF_{2}\\geqslant c-a=\\sqrt{5}-1 $. Then by the definition of the hyperbola, $ ||PF_{2}|-3|=2a=2 $, $ |PF_{2}|>0 $. Therefore, $ |PF_{2}|=1 $ (discarded) or $ |PF_{2}|=5 $." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$, respectively, and $A$ is a point on the hyperbola in the first quadrant. If $|A F_{2}|=2$ and $\\angle F_{1} A F_{2}=45^{\\circ}$, and the line $A F_{2}$ is extended to intersect the right branch of the hyperbola at point $B$, then the area of $\\triangle F_{1} A B$ equals?", "fact_expressions": "G: Hyperbola;b: Number;A: Point;F2: Point;F1: Point;B: Point;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;Quadrant(A) = 1;PointOnCurve(A, G);Abs(LineSegmentOf(A, F2)) = 2;AngleOf(F1, A, F2) = ApplyUnit(45, degree);Intersection(OverlappingLine(LineSegmentOf(A, F2)), RightPart(G)) = B", "query_expressions": "Area(TriangleOf(F1, A, B))", "answer_expressions": "4", "fact_spans": "[[[20, 59], [70, 73], [144, 147]], [[23, 59]], [[66, 69]], [[10, 17]], [[2, 9]], [[151, 155]], [[23, 59]], [[20, 59]], [[2, 65]], [[2, 65]], [[66, 82]], [[66, 82]], [[84, 97]], [[98, 131]], [[132, 155]]]", "query_spans": "[[[157, 184]]]", "process": "According to the definition of hyperbola, it can be transformed to find that $\\triangle BAF_{1}$ is an isosceles right triangle, then the area can be calculated. From the given conditions, $a=1$, by the definition of hyperbola, $|AF_{1}|-|AF_{2}|=2a=2$, $|BF_{1}|-|BF_{2}|=2a=2$, so $|AF_{1}|=2+|AF_{2}|=4$, $|BF_{1}|=2+|BF_{2}|$. From the given conditions, $|AB|=|AF_{2}|+|BF_{2}|=2+|BF_{2}|$, so $|BA|=|BF_{1}|$, therefore $\\triangle BAF_{1}$ is an isosceles triangle. Since $\\angle F_{1}AF_{2}=45^{\\circ}$, so $\\angle ABF_{1}=90^{\\circ}$, therefore $\\triangle BAF_{1}$ is an isosceles right triangle. So $|BA|=|BF_{1}|=\\frac{\\sqrt{2}}{2}|AF_{1}|=F_{1}|=\\frac{1}{2}\\times2\\sqrt{2}\\times2\\sqrt{2}=4$" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $E$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, respectively. A line $l$ passing through $F_{2}$ intersects $E$ at points $P$ and $Q$. If $|P F_{2}|=2|Q F_{2}|$ and $|Q F_{1}|=3|Q F_{2}|$, then what is the eccentricity of the ellipse $E$?", "fact_expressions": "l: Line;E: Ellipse;b: Number;a: Number;P: Point;F2: Point;Q: Point;F1: Point;a > b;b > 0;Expression(E) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(E) = F1;RightFocus(E) = F2;PointOnCurve(F2,l);Intersection(l, E) = {P, Q};Abs(LineSegmentOf(P, F2)) = 2*Abs(LineSegmentOf(Q,F2));Abs(LineSegmentOf(Q,F1))=3*Abs(LineSegmentOf(Q,F2))", "query_expressions": "Eccentricity(E)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[93, 98]], [[20, 77], [99, 102], [164, 169]], [[27, 77]], [[27, 77]], [[104, 107]], [[10, 17], [85, 92]], [[108, 111]], [[2, 9]], [[27, 77]], [[27, 77]], [[20, 77]], [[2, 83]], [[2, 83]], [[84, 98]], [[93, 113]], [[115, 138]], [[140, 162]]]", "query_spans": "[[[164, 175]]]", "process": "Let |QF₂| = m, then |QF₁| = 3m, therefore 4m = 2a, m = a/2, thus |PF₂| = a, and |QF₁| = 3a/2, |PF₁| = a, (a² + a² - 4c²)/(2a²) = (a² + (3a/2)² - (3a/2)²)/(2a(3a/2)) ∴ e² = 1/3, e = √3/3." }, { "text": "Given that $F$ is the focus of the parabola $y^{2}=4x$, a line with slope $k_{1}$ passing through $F$ intersects the parabola at points $A$ and $B$. Extending $AM$ and $BM$ to intersect the parabola again at points $C$ and $D$ respectively, the slope of line $CD$ is $k_{2}$. If $M(4,0)$, then $\\frac{k_{1}}{k_{2}}=$?", "fact_expressions": "G: Parabola;H: Line;D: Point;C: Point;A: Point;M: Point;B: Point;F: Point;k2: Number;k1: Number;Expression(G) = (y^2 = 4*x);Coordinate(M) = (4, 0);Focus(G) = F;PointOnCurve(F, H);Slope(H) = k1;Intersection(H, G) = {A, B};Intersection(OverlappingLine(LineSegmentOf(A, M)), G) = C;Intersection(OverlappingLine(LineSegmentOf(B, M)), G) = D;Slope(LineOf(C, D)) = k2", "query_expressions": "k1/k2", "answer_expressions": "4", "fact_spans": "[[[6, 20], [43, 46], [74, 77]], [[40, 42]], [[82, 85]], [[78, 81]], [[48, 51]], [[107, 115]], [[52, 55]], [[2, 5], [25, 28]], [[99, 106]], [[32, 39]], [[6, 20]], [[107, 115]], [[2, 23]], [[24, 42]], [[29, 42]], [[40, 57]], [[58, 87]], [[58, 87]], [[88, 106]]]", "query_spans": "[[[117, 140]]]", "process": "Let the line passing through point F with slope $k_{1}$ be given by: $y = k_{1}x - 1$, and consider the system of equations \n$$\n\\begin{cases}\ny^{2} = 4x \\\\\ny = k_{1}(x - 1)\n\\end{cases}\n$$ \nEliminating $x$ yields: $y^{2} - \\frac{4}{k_{1}}y - 4 = 0$. Let $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, $\\therefore y_{1}y_{2} = -4$. Let $C(x_{3}, y_{3})$, $D(x_{4}, y_{4})$. Then $k_{1} = \\frac{y_{1} - y_{2}}{\\frac{1}{4}(y_{1}^{2} - y_{2}^{2})} = \\frac{4}{y_{1} + y_{2}}$, similarly $k_{2} = \\frac{4}{y_{3} + y_{4}}$. Let the line AC be given by $y = m(x - 4)$, and consider the system \n$$\n\\begin{cases}\ny = m(x - 4) \\\\\ny^{2} = 4x\n\\end{cases}\n$$ \nEliminating $x$ gives: $my^{2} - 4y - 16m = 0$, $\\therefore y_{1}y_{3} = -16$, similarly $y_{2}y_{4} = -16$. Then \n$$\n\\frac{k_{1}}{k_{2}} = \\frac{y_{3} + y_{4}}{y_{1} + y_{2}} = \\frac{\\frac{-16}{y_{1}} + \\frac{-16}{y_{2}}}{y_{1} + y_{2}} = \\frac{-16}{y_{1}y_{2}} = 4\n$$ \n(1) The positional relationship between a line and a parabola is similar to that between a line and an ellipse or hyperbola; generally, one needs to use the relationship between roots and coefficients; (2) For problems concerning the chord length of a line and a parabola, pay attention to whether the line passes through the focus of the parabola; if it does, one can directly use the formula $|AB| = x_{1} + x_{2} + p$; if not, the general chord length formula must be used." }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, and let $H$ be the intersection point of the directrix and the $x$-axis. Point $P$ lies on $C$ such that $|PH|=\\frac{\\sqrt{5}}{2}|PF|$. What is the area of $\\Delta PFH$?", "fact_expressions": "C: Parabola;P: Point;F: Point;H: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;Intersection(Directrix(C), xAxis) = H;PointOnCurve(P, C);Abs(LineSegmentOf(P, H)) = (sqrt(5)/2)*Abs(LineSegmentOf(P, F))", "query_expressions": "Area(TriangleOf(P, F, H))", "answer_expressions": "4+pm*2*sqrt(3)", "fact_spans": "[[[2, 21], [49, 52]], [[44, 48]], [[25, 28]], [[40, 43]], [[2, 21]], [[2, 28]], [[2, 43]], [[44, 53]], [[55, 86]]]", "query_spans": "[[[88, 107]]]", "process": "Let P(\\frac{t^{2}}{4},t), (t>0), then |PF|=|PM|=\\frac{t^{2}}{4}+1, |PH|=\\sqrt{(\\frac{t^{2}}{4}+1)^{2}+t^{2}}. From |PH|=\\frac{\\sqrt{5}}{2}|PF| we obtain t^{2}-8t+4=0, solving gives t=4\\pm2\\sqrt{3}. Thus we can solve it. Given the parabola C: y^{2}=4x, we get the focus F(1,0), and the directrix equation x=-1. Draw PM perpendicular to the directrix at M, |PH|=\\sqrt{(\\frac{t^{2}}{4}+1)^{2}+t^{2}}. From |PH|=\\frac{\\sqrt{5}}{2}|PF| we obtain t^{2}-8t+4=0. Solving gives t=4\\pm2\\sqrt{3}. Then the area of APFH is \\frac{1}{2}\\times2\\timest=4\\pm2\\sqrt{3}" }, { "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}$ and $F_{2}$ respectively. Draw a line through $F_{2}$ perpendicular to the $x$-axis, intersecting the hyperbola at points $M$ and $N$, and $\\overrightarrow{N F_{1}} \\cdot \\overrightarrow{M F_{1}}<0$. Then the range of the eccentricity of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;N: 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;H: Line;PointOnCurve(F2,H);IsPerpendicular(H,xAxis);Intersection(H, C) = {M, N};DotProduct(VectorOf(N, F1), VectorOf(M, F1)) < 0", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "(sqrt(2) + 1, +oo)", "fact_spans": "[[[2, 63], [105, 108], [182, 188]], [[10, 63]], [[10, 63]], [[114, 117]], [[110, 113]], [[72, 79]], [[80, 87], [89, 96]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [], [[88, 104]], [[88, 104]], [[88, 119]], [[121, 180]]]", "query_spans": "[[[182, 199]]]", "process": "When x = c, \\frac{c^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1, solving gives: y = \\pm \\frac{b^{2}}{a}. Without loss of generality, let M(c, \\frac{b^{2}}{a}), N(c, -\\frac{b^{2}}{a}). Then \\overrightarrow{NF_{1}} \\cdot \\overrightarrow{MF_{1}} = (-2c, \\frac{b^{2}}{a}) \\cdot (-2c, -\\frac{b^{2}}{a}) = 4c^{2} - \\frac{b^{4}}{a^{2}} < 0, that is, 2ac < b^{2} = c^{2} - a^{2}. Dividing both sides of the inequality by a^{2} gives: e^{2} - 2e - 1 > 0, solving gives: e > \\sqrt{2} + 1. Hence the answer is: (\\sqrt{2}+1, +\\infty)" }, { "text": "The focus of the parabola $y^{2}=2 p x(p>0)$ is $F$, $O$ is the coordinate origin, $M$ is a point on the parabola, and $|M F|=3|O F|$, the area of $\\Delta M F O$ is $16 \\sqrt{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;O: Origin;M: Point;PointOnCurve(M, G) = True;Abs(LineSegmentOf(M, F)) = 3*Abs(LineSegmentOf(O, F));Area(TriangleOf(M, F, O)) = 16*sqrt(2)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=16*x", "fact_spans": "[[[0, 21], [42, 45], [99, 102]], [[0, 21]], [[3, 21]], [[3, 21]], [[25, 28]], [[0, 28]], [[29, 32]], [[38, 41]], [[38, 48]], [[50, 64]], [[66, 97]]]", "query_spans": "[[[99, 107]]]", "process": "Since the focus of the parabola y^{2}=2px (p>0) is F(\\frac{p}{2},0), let M(x_{0},y_{0}). From |MF|=3|OF| we get: (x_{0}-\\frac{p}{2})^{2}+y_{0}^{2}=(\\frac{3p}{2})^{2}. Since y_{0}^{2}=2px_{0}, it follows that (x_{0}-\\frac{p}{2})^{2}+2px_{0}=(\\frac{3p}{2})^{2}, which simplifies to x_{0}^{2}+px_{0}-2p^{2}=0. Solving gives x_{0}=p or x_{0}=-2p (discarded). Thus M(p,\\pm\\sqrt{2}p). Therefore S_{\\triangle MFO}=\\frac{1}{2}\\times\\frac{p}{2}\\times\\sqrt{2}p=16\\sqrt{2}, solving gives p=\\pm8. Since p>0, we have p=8. Hence the equation of the parabola is y^{2}=16x." }, { "text": "Let the hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0, b>0 $) have left and right foci $ F_{1} $, $ F_{2} $ respectively, and eccentricity $ \\sqrt{5} $. $ P $ is a point on $ C $ such that $ F_{1} P \\perp F_{2} P $. If the area of $ \\triangle P F_{1} F_{2} $ is $ 4 $, then $ a = $?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;F1: Point;P: 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);Eccentricity(C)=sqrt(5);IsPerpendicular(LineSegmentOf(F1, P), LineSegmentOf(F2, P));Area(TriangleOf(P, F1, F2)) = 4", "query_expressions": "a", "answer_expressions": "1", "fact_spans": "[[[1, 62], [107, 110]], [[174, 177]], [[9, 62]], [[71, 78]], [[103, 106]], [[79, 86]], [[9, 62]], [[9, 62]], [[1, 62]], [[1, 86]], [[1, 86]], [[103, 113]], [[1, 101]], [[115, 138]], [[140, 172]]]", "query_spans": "[[[174, 179]]]", "process": "" }, { "text": "The hyperbola $C_{1}$: $\\frac{y^{2}}{a^{2}}-\\frac{x^{2}}{b^{2}}=1$ $(a>0, b>0)$ and $C_{2}$: $\\frac{x^{2}}{4}-y^{2}=1$ have the same asymptotes. What is the eccentricity of the hyperbola $C_{1}$?", "fact_expressions": "C1: Hyperbola;a: Number;C2: Hyperbola;b: Number;a>0;b>0;Expression(C1) = (y^2/a^2 - x^2/b^2 = 1);Expression(C2) = (x^2/4 - y^2 = 1);Asymptote(C1) = Asymptote(C2)", "query_expressions": "Eccentricity(C1)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[0, 64], [107, 117]], [[11, 64]], [[65, 99]], [[11, 64]], [[11, 64]], [[11, 64]], [[0, 64]], [[65, 99]], [[0, 105]]]", "query_spans": "[[[107, 123]]]", "process": "The asymptotes of the hyperbola $ C_{2}: \\frac{x^{2}}{4} - y^{2} = 1 $ are $ y = \\pm\\frac{1}{2}x $, while the asymptotes of the hyperbola $ C_{1}: \\frac{y^{2}}{a^{2}} - \\frac{x^{2}}{b^{2}} = 1 $ $ (a > 0, b > 0) $ are $ y = \\pm\\frac{a}{b}x $. Therefore, $ \\frac{a}{b} = \\frac{1}{2} $. Thus, the eccentricity of the hyperbola $ C_{1} $ is $ e = \\frac{c}{a} = \\sqrt{1 + \\frac{b^{2}}{a^{2}}} = \\sqrt{1 + 4} = \\sqrt{5} $." }, { "text": "$AB$ is a chord passing through the right focus $F$ of $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$. From point $A$, draw a perpendicular $AA_{1}$ to the right directrix, with $A_{1}$ as the foot of the perpendicular. Connect $BA_{1}$ and let it intersect the $x$-axis at point $C$. Then the coordinates of point $C$ are?", "fact_expressions": "A: Point;B: Point;A1: Point;F: Point;G:Curve;C:Point;Expression(G)=(x^2/16-y^2/9=1);IsChordOf(LineSegmentOf(A,B),G);PointOnCurve(F,LineSegmentOf(A,B));RightFocus(G)=F;PointOnCurve(A,LineSegmentOf(A,A1));IsPerpendicular(RightDirectrix(G),LineSegmentOf(A,A1));FootPoint(RightDirectrix(G),LineSegmentOf(A,A1))=A1;Intersection(LineSegmentOf(B,A1),xAxis)=C", "query_expressions": "Coordinate(C)", "answer_expressions": "(11/10,0)", "fact_spans": "[[[52, 55]], [[0, 4]], [[71, 78]], [[45, 48]], [[6, 42]], [[105, 108], [99, 103]], [[6, 42]], [[0, 50]], [[0, 48]], [[6, 48]], [[51, 70]], [[6, 70]], [[6, 81]], [[82, 103]]]", "query_spans": "[[[105, 113]]]", "process": "" }, { "text": "The eccentricity of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{k}=1$ with foci on the $x$-axis is $\\frac{3}{5}$. Then $k$=?", "fact_expressions": "G: Ellipse;k: Number;Expression(G) = (x^2/25 + y^2/k = 1);PointOnCurve(Focus(G), xAxis);Eccentricity(G) = 3/5", "query_expressions": "k", "answer_expressions": "16", "fact_spans": "[[[9, 47]], [[67, 70]], [[9, 47]], [[0, 47]], [[9, 65]]]", "query_spans": "[[[67, 72]]]", "process": "From the problem, we know that a^{2}=25, b^{2}=k, and c^{2}=a^{2}-b^{2}, so c^{2}=25-b^{2}. Also, since e=\\frac{c}{a}=\\frac{3}{5}, it follows that e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{9}{25}, that is: \\frac{25-b^{2}}{25}=\\frac{9}{25}. Solving gives b^{2}=16, then k=16." }, { "text": "Given that the equation $\\frac{x^{2}}{m-1}+\\frac{y^{2}}{2-m}=1$ represents an ellipse with foci on the $x$-axis, what is the range of real values for $m$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/(m - 1) + y^2/(2 - m) = 1);m: Real;PointOnCurve(Focus(G), xAxis)", "query_expressions": "Range(m)", "answer_expressions": "(3/2, 2)", "fact_spans": "[[[54, 56]], [[2, 56]], [[58, 63]], [[45, 56]]]", "query_spans": "[[[58, 70]]]", "process": "According to the problem, the equation $\\frac{x2}{m-1}+\\frac{y^{2}}{2-m}=1$ represents an ellipse with foci on the x-axis, so it must satisfy $\\begin{cases}m-1>2^{n}-\\\\2-m>0\\end{cases}$. Solving this gives: $\\frac{3}{2}0 , b>0)$ is given by $y=\\sqrt{5} x$, 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;Expression(OneOf(Asymptote(G))) = (y = sqrt(5)*x)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(6)", "fact_spans": "[[[1, 58], [84, 87]], [[1, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[4, 58]], [[1, 81]]]", "query_spans": "[[[84, 93]]]", "process": "From the problem, we have $\\frac{b}{a}=\\sqrt{5}$, $\\therefore b^{2}=5a^{2}$, $\\therefore c^{2}-a^{2}=5a^{2}$, so $c^{2}=6a^{2}$, $\\therefore \\frac{c^{2}}{a^{2}}=e^{2}=6$, $\\therefore e=\\sqrt{6}$. Therefore, the eccentricity of the hyperbola is $\\sqrt{6}$." }, { "text": "Let $e$ be the eccentricity of the ellipse $\\frac{x^{2}}{a}+\\frac{y^{2}}{4}=1$, if $e \\in\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right]$, then the range of values for $a$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/4 + x^2/a = 1);a: Number;e: Number;Eccentricity(G) = e;In(e,(1/2, sqrt(3)/2])", "query_expressions": "Range(a)", "answer_expressions": "[1,3)+(16/3,16]", "fact_spans": "[[[5, 42]], [[5, 42]], [[90, 93]], [[1, 4]], [[1, 46]], [[48, 88]]]", "query_spans": "[[[90, 100]]]", "process": "When $ a > 4 $, $ e = \\frac{\\sqrt{a - 4}}{\\sqrt{a}} = \\sqrt{\\frac{a - 4}{a}} $, so $ \\frac{1}{2} < \\sqrt{\\frac{a - 4}{a}} \\leqslant \\frac{\\sqrt{3}}{2} $, hence $ \\frac{16}{3} < a \\leqslant 16 $. When $ 0 < a < 4 $, $ e = \\frac{\\sqrt{4 - a}}{2} $, so $ \\frac{1}{2} < \\frac{\\sqrt{4 - a}}{2} \\leqslant \\frac{\\sqrt{3}}{2} $, hence $ 1 \\leqslant a < 3 $. Therefore, the range of values for $ a $ is $ [1, 3) \\cup \\left( \\frac{10}{2} \\right) $." }, { "text": "The standard equation of a parabola with directrix $x=\\frac{1}{2}$ is?", "fact_expressions": "G: Parabola;Expression(Directrix(G)) = (x=1/2)", "query_expressions": "Expression(G)", "answer_expressions": "y^2=-2*x", "fact_spans": "[[[21, 24]], [[0, 24]]]", "query_spans": "[[[21, 31]]]", "process": "" }, { "text": "The standard equation of the hyperbola that has the same asymptotes as the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$ and passes through the point $(2,2)$ is?", "fact_expressions": "G: Hyperbola;H: Point;C:Hyperbola;Expression(G) = (x^2 - y^2/2 = 1);Coordinate(H) = (2, 2);Asymptote(G) = Asymptote(C);PointOnCurve(H, C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/2-y^2/4=1", "fact_spans": "[[[1, 29]], [[39, 47]], [[48, 51]], [[1, 29]], [[39, 47]], [[1, 51]], [[38, 51]]]", "query_spans": "[[[48, 58]]]", "process": "Test analysis: Let the hyperbola equation that shares the same asymptotes with the given hyperbola be $ x^{2} - \\frac{y^{2}}{2} = \\lambda $ ($ \\lambda > 0 $). Substituting the point $ (2,2) $ yields $ \\lambda = 2 $. Thus, the equation is $ \\frac{x^{2}}{2} - \\frac{y^{2}}{4} = 1 $." }, { "text": "Let a focus of the hyperbola $m x^{2}+n y^{2}=1$ coincide with the focus of the parabola $y=\\frac{1}{8} x^{2}$, and let the eccentricity be $2$. Then the distance from the focus of the parabola to an asymptote of the hyperbola is?", "fact_expressions": "G: Hyperbola;m: Number;n: Number;H: Parabola;Expression(G) = (m*x^2 + n*y^2 = 1);Expression(H) = (y = x^2/8);OneOf(Focus(G)) = Focus(H);Eccentricity(G)=2", "query_expressions": "Distance(Focus(H),OneOf(Asymptote(G)))", "answer_expressions": "", "fact_spans": "[[[1, 23], [75, 78]], [[4, 23]], [[4, 23]], [[29, 53], [68, 71]], [[1, 23]], [[29, 53]], [[1, 58]], [[1, 66]]]", "query_spans": "[[[68, 89]]]", "process": "" }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4x$, point $A$ lies on the parabola, point $P$ is a moving point on the directrix of the parabola, $O$ is the origin, and $|A F|=5$, then the minimum value of $|P A|+|P O|$ is?", "fact_expressions": "C: Parabola;A: Point;F: Point;P: Point;O: Origin;Expression(C) = (y^2 = 4*x);Focus(C) = F;PointOnCurve(A, C);PointOnCurve(P, Directrix(C));Abs(LineSegmentOf(A, F)) = 5", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, O)))", "answer_expressions": "2*sqrt(13)", "fact_spans": "[[[6, 25], [34, 37], [45, 48]], [[29, 33]], [[2, 5]], [[40, 44]], [[55, 58]], [[6, 25]], [[2, 28]], [[29, 38]], [[40, 54]], [[65, 74]]]", "query_spans": "[[[76, 95]]]", "process": "\\because|AF|=5,\\therefore the distance from point A to the directrix is 5, i.e., the x-coordinate of point A is 4; since point A lies on the parabola, \\therefore the coordinates of point A are (4,\\pm4); \\because the coordinates of the point symmetric to the origin with respect to the directrix are B(-2,0), then |PA|+|PO|=|PA|+|PB|>|AB|=2\\sqrt{13}. [Analysis] This problem examines the definition of a parabola and using symmetry to find minimum values, testing basic analytical and solving abilities, classified as a fundamental problem." }, { "text": "Let the ellipse $ C $: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 $ ($ a>b>0 $) have right focus $ F $, let $ P $ be a point on the ellipse $ C $ in the first quadrant, and let $ O $ be the origin. If $\\overrightarrow{F O} \\cdot \\overrightarrow{F P}=|P F|^{2}$, then what is the range of values for the eccentricity of ellipse $ C $?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F: Point;O: Origin;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);RightFocus(C)=F;PointOnCurve(P,C);Quadrant(P)=1;DotProduct(VectorOf(F, O), VectorOf(F, P)) = Abs(LineSegmentOf(P, F))^2", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[sqrt(3)/2,1)", "fact_spans": "[[[1, 58], [71, 76], [155, 160]], [[8, 58]], [[8, 58]], [[67, 70]], [[63, 66]], [[85, 88]], [[8, 58]], [[8, 58]], [[1, 58]], [[1, 66]], [[67, 84]], [[67, 84]], [[94, 153]]]", "query_spans": "[[[155, 170]]]", "process": "[Points and T] Let the coordinates of point P be (x_{0},y_{0}). According to \\overrightarrow{FO}\\cdot\\overrightarrow{FP}=|PF|, obtain a quadratic equation in x_{0}. By requiring the equation to have solutions, i.e., \\Delta\\geqslant0, derive an inequality relation between a and b. Combining with a^{2}=b^{2}+c^{2}, determine the range of possible values for the eccentricity. Let P(x_{0},y_{0}), F(c,0), so \\overrightarrow{FO}=(-c,0), \\overrightarrow{FP}=(x_{0}-c,y_{0}), thus \\overrightarrow{FO}\\cdot\\overrightarrow{FP}=c^{2}-cx_{0}. Also |PF|^{2}=(x_{0}-c)^{2}+y_{0}^{2}, so (x_{0}-c)^{2}+y_{0}^{2}=c^{2}-cx_{0}, hence x_{0}^{2}-cx_{0}+y_{0}^{2}=0. Since \\frac{x_{0}^{2}}{a^{2}}+\\frac{y_{0}^{2}}{b^{2}}=1, it follows that x_{0}^{2}-cx_{0}+b^{2}-\\frac{b^{2}x_{0}^{2}}{a^{2}}=0, so \\frac{c^{2}x_{0}^{2}}{a^{2}}-cx_{0}+b^{2}=0. Since the above equation has positive solutions, it suffices that \\Delta=c^{2}-4b^{2}\\cdot\\frac{c^{2}}{a^{2}}\\geqslant0, so a^{2}\\geqslant4b^{2}, hence 4c^{2}\\geqslant3a^{2}, so e^{2}\\geqslant\\frac{3}{4} and 00)$ is a tangent line to the circle $x^{2}+y^{2}-4 x=0$, then the standard equation of the parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = -2*p*x);p: Number;p>0;H: Circle;Expression(H) = (-4*x + x^2 + y^2 = 0);IsTangent(H, Directrix(G)) = True", "query_expressions": "Expression(G)", "answer_expressions": "y^2 = -16*x", "fact_spans": "[[[1, 23], [54, 57]], [[1, 23]], [[4, 23]], [[4, 23]], [[27, 47]], [[27, 47]], [[1, 52]]]", "query_spans": "[[[54, 64]]]", "process": "Using the fact that the distance from the center of the circle to the tangent line equals the radius, set up an equation to obtain the solution. Since the standard equation of the parabola is y^{2}=-2px (p>0), its directrix is x=\\frac{p}{2}. The standard equation of the circle is (x-2)^{2}+y^{2}=4, with center (2,0) and radius r=2. Since x=\\frac{p}{2} is a tangent to the circle, we have |2-\\frac{p}{2}|=2, which gives p=0 (discarded) or p=8. The standard equation of the parabola is y^{2}=-16x." }, { "text": "The directrix of the parabola $C$: $y^{2}=4 x$ intersects the $x$-axis at point $M$. A line passing through the focus $F$ with an inclination angle of $60^{\\circ}$ intersects $C$ at points $A$ and $B$. Then $\\tan \\angle A M B$ = ?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);Intersection(Directrix(C), xAxis) = M;M: Point;F: Point;Focus(C) = F;L: Line;Inclination(L) = ApplyUnit(60,degree);PointOnCurve(F,L) = True;Intersection(L,C) = {A,B};A: Point;B: Point", "query_expressions": "Tan(AngleOf(A, M, B))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[0, 19], [61, 64]], [[0, 19]], [[0, 33]], [[29, 33]], [[37, 40]], [[0, 40]], [[58, 60]], [[41, 60]], [[34, 60]], [[58, 75]], [[66, 69]], [[70, 73]]]", "query_spans": "[[[77, 98]]]", "process": "The focus coordinates of the parabola $ C: y^{2} = 4x $ are $ F(1,0) $, $ M(-1,0) $. Then 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 $ x = \\frac{1}{3} $ or $ 3 $, and thus $ A\\left(\\frac{1}{3}, -\\frac{2\\sqrt{3}}{3}\\right) $, $ B(3, 2\\sqrt{3}) $. Then $ k_{AM} = -\\frac{\\sqrt{3}}{2} $, $ k_{BM} = \\frac{\\sqrt{3}}{2} $, so $ \\tan\\angle AMB = \\frac{\\frac{\\sqrt{3}}{2} + \\frac{\\sqrt{3}}{2}}{1 + \\frac{\\sqrt{3}}{2} \\cdot \\left(-\\frac{\\sqrt{3}}{2}\\right)} = 4\\sqrt{3} $." }, { "text": "Let the trajectory of moving point $A$ be the parabola $y^{2}=4 x$, and let point $B(2,0)$ be a fixed point. If the midpoint of segment $A B$ is point $P$, then the equation of the trajectory of point $P$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(B) = (2, 0);MidPoint(LineSegmentOf(A, B)) = P;PointOnCurve(A, G)", "query_expressions": "LocusEquation(P)", "answer_expressions": "y^2 = 2*x - 2", "fact_spans": "[[[10, 24]], [[3, 6]], [[25, 34]], [[50, 54], [56, 60]], [[10, 24]], [[25, 34]], [[39, 54]], [[1, 24]]]", "query_spans": "[[[56, 67]]]", "process": "Let point P(x, y), then we can obtain point A(2x-4, 2y). Substituting the coordinates of point A into the equation of the parabola and simplifying gives the trajectory equation of point P. Let point A(x_{0}, y_{0}) and point P(x, y). By the midpoint formula, we have\n\\begin{cases}x=\\frac{x_{0}+2}{2}\\\\y=\\frac{y_{0}}{2}\\end{cases},\nwhich yields\n\\begin{cases}x_{0}=2x-2\\\\y_{0}=2y\\end{cases}.\nSince point A lies on the parabola y^{2}=4x, that is, y_{0}^{2}=4x_{0}, it follows that 4y^{2}=4(2x-2). Simplifying gives y^{2}=2x-2. Therefore, the trajectory equation of point P is y^{2}=2x-2." }, { "text": "The distance from the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{8}=1$ to the line $x+2 y+7=0$ is?", "fact_expressions": "G: Hyperbola;H: Line;Expression(G) = (x^2 - y^2/8 = 1);Expression(H) = (x + 2*y + 7 = 0)", "query_expressions": "Distance(RightFocus(G),H)", "answer_expressions": "2*sqrt(5)", "fact_spans": "[[[0, 28]], [[33, 46]], [[0, 28]], [[33, 46]]]", "query_spans": "[[[0, 51]]]", "process": "The right focus of the hyperbola \\( x^{2} - \\frac{y^{2}}{8} = 1 \\) is \\( (3,0) \\), so the distance from the right focus to the line \\( x + 2y + 7 = 0 \\) is \\( d = \\frac{|3 + 0 + 7|}{\\sqrt{1^{2} + 2^{2}}} = 2\\sqrt{5} \\)." }, { "text": "Given that the circumradius of $\\triangle ABC$ is $R=\\frac{14 \\sqrt{3}}{3}$, $\\angle ABC=120^{\\circ}$, $BC=10$, and side $BC$ lies on the $x$-axis with the $y$-axis perpendicularly bisecting $BC$, what is the equation of the hyperbola passing through point $A$ with $B$ and $C$ as foci?", "fact_expressions": "A: Point;B: Point;C: Point;R: Number;Radius(CircumCircle(TriangleOf(A, B, C))) = R;R = 14*sqrt(3)/3;AngleOf(A, B, C) = ApplyUnit(120, degree);LineSegmentOf(B, C) = 10;OverlappingLine(LineSegmentOf(B, C), xAxis);PerpendicularBisector(LineSegmentOf(B, C)) = yAxis;G: Hyperbola;PointOnCurve(A, G);Focus(G) = {B, C}", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16-y^2/9=1", "fact_spans": "[[[111, 115]], [[117, 120]], [[121, 124]], [[22, 47]], [[2, 47]], [[22, 47]], [[49, 73]], [[75, 82]], [[84, 94]], [[95, 107]], [[128, 131]], [[110, 131]], [[116, 131]]]", "query_spans": "[[[128, 135]]]", "process": "" }, { "text": "The coordinates of the foci of the hyperbola $y^{2}-x^{2}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2 + y^2 = 1)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, pm*sqrt(2))", "fact_spans": "[[[0, 18]], [[0, 18]]]", "query_spans": "[[[0, 25]]]", "process": "" }, { "text": "Given that $O$ is the origin, and for a point $A$ on the parabola $C$: $y^{2}=2 p x$, the distance from $A$ to the focus $F$ is $4$. If point $M$ is a moving point on the directrix of parabola $C$, and $\\overrightarrow{M F}=3 \\overrightarrow{F A}$, then $p=?$", "fact_expressions": "C: Parabola;p: Number;M: Point;F: Point;A: Point;O: Origin;Expression(C) = (y^2 = 2*p*x);PointOnCurve(A, C);Focus(C) = F;Distance(A, F) = 4;PointOnCurve(M, Directrix(C));VectorOf(M, F) = 3*VectorOf(F, A)", "query_expressions": "p", "answer_expressions": "pm*3", "fact_spans": "[[[11, 32], [58, 64]], [[119, 122]], [[53, 57]], [[41, 44]], [[35, 38]], [[2, 5]], [[11, 32]], [[11, 38]], [[11, 44]], [[35, 51]], [[53, 70]], [[72, 117]]]", "query_spans": "[[[119, 124]]]", "process": "Draw AE perpendicular to the directrix at E. By the definition of the parabola, |AE| = |AF| = 4. The solution can then be obtained using knowledge of similarity. If p > 0, draw AE perpendicular to the directrix at E, and let N be the point of intersection between the directrix and the x-axis. As shown in the figure, since \\overrightarrow{MF} = 3\\overrightarrow{FA}, it follows that |MF| = 3|FA| = 12 and |MA| = 16. By the property of similarity, \\frac{|MF|}{|MA|} = \\frac{|NF|}{|AE|}, that is, \\frac{12}{16} = \\frac{|NF|}{4}, so |NF| = 3, hence D = 3. When D < 0, similarly we obtain p = -3." }, { "text": "Given $x_{0}<0$, $M(x_{0}, 0)$, $O$ is the origin. If on the parabola $C$: $y^{2}=4 x$ there exists a point $N$ such that $\\angle O M N=45^{\\circ}$, then the range of values for $x_{0}$ is?", "fact_expressions": "x0: Number;x0 < 0;M: Point;Coordinate(M) = (x0, 0);O: Origin;C: Parabola;Expression(C) = (y^2 = 4*x);N: Point;PointOnCurve(N, C);AngleOf(O, M, N) = ApplyUnit(45, degree)", "query_expressions": "Range(x0)", "answer_expressions": "[-1,0)", "fact_spans": "[[[93, 100]], [[2, 11]], [[12, 25]], [[12, 25]], [[26, 29]], [[37, 56]], [[37, 56]], [[59, 63]], [[36, 63]], [[66, 91]]]", "query_spans": "[[[93, 107]]]", "process": "Draw a tangent line from point M to curve C, with Q as the point of tangency, as shown in the figure: Let ∠OMQ = θ. Since there exists a point N on the parabola C: y² = 4x such that ∠OMN = 45°, it follows that θ ≥ 45°. When θ = 45°, the equation of line MQ is y = x - x₀. Substituting y = x - x₀ into y² = 4x yields y² - 4y - 4x₀ = 0. From Δ = 16 + 16x₀ = 0, solving gives x₀ = -1. Therefore, the range of values for x₀ is [-1, 0)." }, { "text": "The foci of the ellipse $\\frac{x^{2}}{4}+y^{2}=1$ are $F_{1}$ and $F_{2}$. Point $P$ lies on the ellipse, and the midpoint of segment $P F_{1}$ lies exactly on the $y$-axis. If $|P F_{1}|=\\lambda|P F_{2}|$, then $\\lambda=$?", "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);PointOnCurve(MidPoint(LineSegmentOf(P,F1)), yAxis);Abs(LineSegmentOf(P, F1)) = lambda*Abs(LineSegmentOf(P, F2));lambda:Number", "query_expressions": "lambda", "answer_expressions": "7", "fact_spans": "[[[0, 27], [52, 54]], [[31, 38]], [[47, 51]], [[39, 46]], [[0, 27]], [[0, 46]], [[47, 55]], [[57, 79]], [[80, 108]], [[110, 119]]]", "query_spans": "[[[110, 121]]]", "process": "" }, { "text": "Given point $A(-\\frac{1}{2}, 0)$, the focus of the parabola $y^{2}=2 x$ is $F$, point $P$ lies on the parabola, and $|A P|=\\sqrt{2}|P F|$, then $|O P|$=?", "fact_expressions": "G: Parabola;A: Point;P: Point;F: Point;O: Origin;Expression(G) = (y^2 = 2*x);Coordinate(A) = (-1/2, 0);Focus(G) = F;PointOnCurve(P, G);Abs(LineSegmentOf(A, P)) = sqrt(2)*Abs(LineSegmentOf(P, F))", "query_expressions": "Abs(LineSegmentOf(O, P))", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[24, 38], [51, 54]], [[2, 23]], [[46, 50]], [[42, 45]], [[80, 87]], [[24, 38]], [[2, 23]], [[24, 45]], [[46, 55]], [[57, 78]]]", "query_spans": "[[[80, 89]]]", "process": "Let P(\\frac{1}{2}m^{2},m), use the given condition and distance formula to find m^{2}=1, then |OP| can be determined. From the given, F(\\frac{1}{2},0), let P(\\frac{1}{2}m^{2},m). Since |AP|=\\sqrt{2}|PF|, we have |AP|^{2}=2|PF|^{2}. Then (\\frac{m^{2}}{2}+\\frac{1}{2})^{2}+m^{2}=2[(\\frac{m^{2}}{2}-\\frac{1}{2})^{2}+m^{2}], solving gives m^{2}=1. Therefore, |OP|=\\sqrt{\\frac{1}{4}m^{4}+m^{2}}=\\sqrt{\\frac{1}{4}+1}=\\frac{\\sqrt{5}}{2}." }, { "text": "Let the hyperbola $C$: $\\frac{x^{2}}{5}-\\frac{y^{2}}{b^{2}}=1(b>0)$ have left and right vertices $A_{1}$, $A_{2}$, and left and right foci $F_{1}$, $F_{2}$, respectively. A circle with diameter $F_{1} F_{2}$ intersects the left branch of the hyperbola at a point $P$. If the circle with diameter $A_{1} A_{2}$ is tangent to the line $P F_{2}$, then the area of $\\Delta F_{1} P F_{2}$ is?", "fact_expressions": "C: Hyperbola;b: Number;G: Circle;H: Circle;F2: Point;P: Point;F1: Point;A1: Point;A2: Point;b>0;Expression(C) = (x^2/5 - y^2/b^2 = 1);LeftVertex(C) = A1;RightVertex(C) = A2;LeftFocus(C)=F1;RightFocus(C)=F2;IsDiameter(LineSegmentOf(F1,F2),G);OneOf(Intersection(G,LeftPart(C)))=P;IsDiameter(LineSegmentOf(A1,A2),H);IsTangent(H,LineOf(P,F2))", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "20", "fact_spans": "[[[1, 53], [122, 125]], [[9, 53]], [[120, 121]], [[157, 158]], [[94, 101]], [[133, 136]], [[86, 93]], [[62, 69]], [[70, 77]], [[9, 53]], [[1, 53]], [[1, 77]], [[1, 77]], [[1, 101]], [[1, 101]], [[102, 121]], [[120, 136]], [[139, 158]], [[157, 172]]]", "query_spans": "[[[174, 201]]]", "process": "Let the circle with diameter A₁A₂ be tangent to the line PF₂ at point M. From the hyperbola equation, we have: |OA₁| = |OA₂| = √5; since PF₁ ⊥ PF₂, OM ⊥ PF₂, therefore OM ∥ PF₁. Also, O is the midpoint of F₁F₂, so |PF₁| = 2|OM| = 2√5. By the definition of the hyperbola: |PF₂| − |PF₁| = 2√5, thus |PF₂| = 4√5, hence SΔF₁PF₂ = (1/2)|PF₁|·|PF₂| = (1/2) × 2√5 × 4√5 = 20" }, { "text": "If the graph represented by the equation $\\frac{x^{2}}{2-k}+\\frac{y^{2}}{2k-3}=1$ is a hyperbola, then what is the range of values for $k$?", "fact_expressions": "G: Hyperbola;Expression(G)=(x^2/(2-k)+y^2/(2*k-3)=1);k:Number", "query_expressions": "Range(k)", "answer_expressions": "{(-oo,3/2),(2,+oo)}", "fact_spans": "[[[50, 53]], [[1, 53]], [[55, 58]]]", "query_spans": "[[[55, 65]]]", "process": "" }, { "text": "Given the ellipse $M$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$ has left and right foci $F_{1}$, $F_{2}$ respectively, and the parabola $N$: $y^{2}=2 p x$ has focus $F_{2}$. If $P$ is a common point of $M$ and $N$, and $|P F_{1}|=\\sqrt{2}|P F_{2}|$, then the eccentricity of $M$ is?", "fact_expressions": "N: Parabola;p: Number;M: Ellipse;a: Number;b: Number;P: Point;F1: Point;F2: Point;Expression(N) = (y^2 = 2*(p*x));a > b;b > 0;Expression(M) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(M) = F1;RightFocus(M) = F2;Focus(N) = F2;OneOf(Intersection(M,N))=P;Abs(LineSegmentOf(P, F1)) = sqrt(2)*Abs(LineSegmentOf(P, F2))", "query_expressions": "Eccentricity(M)", "answer_expressions": "sqrt(2)-1", "fact_spans": "[[[84, 105], [127, 130]], [[91, 105]], [[2, 59], [123, 126], [169, 172]], [[9, 59]], [[9, 59]], [[119, 122]], [[68, 75]], [[76, 83], [109, 116]], [[84, 105]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[84, 116]], [[119, 136]], [[138, 167]]]", "query_spans": "[[[169, 178]]]", "process": "Solve by transforming the edge-angle relationship according to the definitions of the parabola and ellipse. By the definition of the parabola, the directrix is the line passing through the left focus and perpendicular to the x-axis. Draw PQ\\botF_{1}Q, then PF_{2}=PQ. Also |PF_{1}|=\\sqrt{2}|PF_{2}|, hence QF_{1}=\\sqrt{PF_{1}^{2}-PQ^{2}}=\\sqrt{PF_{1}^{2}-PF_{2}^{2}}=PF_{2}=PQ. Therefore, \\trianglePF_{1}Q is an isosceles right triangle. Thus \\anglePF_{1}Q=\\frac{\\pi}{4}. Since \\anglePF_{1}Q+\\anglePF_{1}F_{2}=\\frac{\\pi}{2}, it follows that \\anglePF_{1}F_{2}=\\frac{\\pi}{4}. Also |PF_{1}|=\\sqrt{2}|PF_{2}|, similarly we get F_{1}F_{2}=PF_{2}. Hence \\trianglePF_{1}F_{2} is also an isosceles right triangle." }, { "text": "Given that $A$ and $B$ are the left and right vertices of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, respectively, and points $M$ and $N$ are two moving points on the ellipse such that the segment $MN$ is perpendicular to the $x$-axis, find the equation of the locus of the intersection point of lines $MA$ and $NB$.", "fact_expressions": "G: Ellipse;A: Point;M: Point;N: Point;B: Point;Expression(G) = (x^2/4 + y^2/3 = 1);LeftVertex(G) = A;RightVertex(G) = B;PointOnCurve(M,G);PointOnCurve(N,G);IsPerpendicular(LineSegmentOf(M,N),xAxis)", "query_expressions": "LocusEquation(Intersection(LineOf(M,A),LineOf(N,B)))", "answer_expressions": "x^2/4-y^2/3=1", "fact_spans": "[[[12, 49], [65, 67]], [[2, 5]], [[57, 60]], [[61, 64]], [[6, 9]], [[12, 49]], [[2, 55]], [[2, 55]], [[57, 73]], [[57, 73]], [[76, 90]]]", "query_spans": "[[[92, 114]]]", "process": "First find the coordinates of A and B. Since MN is perpendicular to the x-axis, let the coordinates of M and N be set as follows. Let P be the intersection point of MA and NB, and use the collinearity of three points to obtain k_{PA} = k_{MA}, k_{PB} = k_{NB}. Express these in terms of coordinates, multiply the two equations, and eliminate x_{1}, y_{1} to obtain the trajectory equation of the intersection point of lines MA and NB. \nSolution: Since A and B are the left and right vertices of the ellipse \\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1, respectively, we have A(-2,0), B(2,0). Let P be the intersection point of MA and NB, with P(x,y), M(x_{1},y_{1}), N(x_{1},-y_{1}). From k_{PA} = k_{MA}, k_{PB} = k_{NB}, we get \\frac{y}{x+2} = \\frac{y_{1}}{x_{1}+2}, \\frac{y}{x-2} = \\frac{-y_{1}}{x_{1}-2}. Multiplying the two equations gives \\frac{y^{2}}{x^{2}-4} = \\frac{-y_{1}^{2}}{x_{1}^{2}-4} = \\frac{-3(1-\\frac{x_{1}^{2}}{4})}{x_{1}^{2}-4} = \\frac{3}{4}, simplifying yields \\frac{x^{2}}{4} - \\frac{y^{2}}{3} = 1." }, { "text": "Given that the line $y=2x-2$ intersects the parabola $y^2=8x$ at points $A$ and $B$, and the focus of the parabola is $F$, then the value of $\\overrightarrow{FA} \\cdot \\overrightarrow{FB}$ is?", "fact_expressions": "G: Parabola;H: Line;F: Point;A: Point;B: Point;Expression(G) = (y^2 = 8*x);Expression(H) = (y = 2*x - 2);Intersection(H, G) = {A, B};Focus(G) = F", "query_expressions": "DotProduct(VectorOf(F, A), VectorOf(F, B))", "answer_expressions": "-11", "fact_spans": "[[[14, 28], [40, 43]], [[2, 13]], [[47, 50]], [[30, 33]], [[34, 37]], [[14, 28]], [[2, 13]], [[2, 39]], [[40, 49]]]", "query_spans": "[[[52, 105]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), F(2,0). Substituting y=2x-2 into y^{2}=8x, we obtain 4x^{2}-8x+4=8x, that is, x^{2}-4x+1=0. Therefore, x_{1}+x_{2}=4, x_{1}x_{2}=1, then (x_{1}-2)(x_{2}-2)=x_{1}x_{2}-2(x_{1}+x_{2})+4=1-8+4=-3. Also y_{1}=2x_{1}-2, y_{2}=2x_{2}-2, hence y_{1}y_{2}=4(x_{1}-1)(x_{2}-1)=4(x_{1}x_{2}-x_{1}-x_{2}+1)=4(1-4+1)=-8. Since \\overrightarrow{FA}=(x_{1}-2,y_{1}), \\overrightarrow{FB}=(x_{2}-2,y_{2}), then \\overrightarrow{FA}\\cdot\\overrightarrow{FB}=(x_{1}-2)(x_{2}-2)+y_{1}y_{2}=-3-8=" }, { "text": "Given that line $l$ passes through the left focus $F$ of the ellipse $C$: $\\frac{x^{2}}{2}+y^{2}=1$ and intersects the ellipse $C$ at points $A$ and $B$, and $O$ is the coordinate origin. If $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$, a perpendicular is drawn from point $O$ to line $AB$, with foot of perpendicular at $H$, then point $H$ is?", "fact_expressions": "C: Ellipse;Expression(C) = (x^2/2 + y^2 = 1);F: Point;LeftFocus(C) = F;l: Line;PointOnCurve(F, l);A: Point;B: Point;Intersection(l, C) = {A, B};O: Origin;DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0;Z: Line;PointOnCurve(O, Z);IsPerpendicular(Z, LineOf(A, B));H: Point;FootPoint(Z, LineOf(A, B)) = H", "query_expressions": "Coordinate(H)", "answer_expressions": "{(-2/3, sqrt(2)/3), (-2/3, -sqrt(2)/3)}", "fact_spans": "[[[8, 40], [49, 54]], [[8, 40]], [[44, 47]], [[8, 47]], [[2, 7]], [[2, 47]], [[55, 58]], [[59, 62]], [[2, 64]], [[65, 68], [129, 133]], [[76, 127]], [], [[128, 144]], [[128, 144]], [[148, 151], [153, 157]], [[128, 151]]]", "query_spans": "[[[153, 159]]]", "process": "From the ellipse $ C: \\frac{x^{2}}{2} + y^{2} = 1 $, we get $ F(-1, 0) $. \n① If line $ l $ has no slope, the line equation is $ x = -1 $, then $ A\\left(-1, \\frac{\\sqrt{2}}{2}\\right), B\\left(-1, -\\frac{\\sqrt{2}}{2}\\right) $, \n$ \\therefore \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 1 - \\frac{1}{2} = \\frac{1}{2} $, which does not satisfy the condition. \n② If line $ l $ has a slope, let the equation of line $ l $ be $ y = k(x + 1) $, solve the system of equations \n$$\n\\begin{cases}\ny = k(x + 1) \\\\\nx^{2} + 2y^{2} = 2\n\\end{cases}\n$$\neliminating variables gives: $ (1 + 2k^{2})x^{2} + 4k^{2}x + 2k^{2} - 2 = 0 $, \nlet $ A(x_{1}, y_{1}), B(x_{2}, y_{2}) $, then \n$ x_{1}x_{2} = \\frac{2k^{2} - 2}{1 + 2k^{2}}, x_{1} + x_{2} = -\\frac{4k^{2}}{1 + 2k^{2}} $, \n$ \\therefore y_{1}y_{2} = k^{2}(x_{1} + 1)(x_{2} + 1) = k^{2}[x_{1}x_{2} + (x_{1} + x_{2}) + 1] $, \n$ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = 0 \\Rightarrow x_{1}x_{2} + y_{1}y_{2} = 0 $, \n$ \\therefore (k^{2} + 1)x_{1}x_{2} + k^{2}(x_{1} + x_{2}) + k^{2} = 0 $, \n$ (k^{2} + 1)\\frac{2k^{2} - 2}{1 + 2k^{2}} + k^{2}\\left(-\\frac{4k^{2}}{1 + 2k^{2}}\\right) + k^{2} = 0 $, \nsimplifies to: $ k^{2} = 2 $. Solving gives $ k = \\pm\\sqrt{2} $. \n$ \\therefore $ The equation of line $ l $ is $ \\sqrt{2}x - y + \\sqrt{2} = 0 $, or $ \\sqrt{2}x + y + \\sqrt{2} = 0 $. \nThe line passing through $ O $ and perpendicular to line $ l $ has equation: $ y = \\pm\\frac{1}{\\sqrt{2}}x $. \nSolving $ \\sqrt{2}x + y + \\sqrt{2} = 0 $ and $ y = \\frac{1}{\\sqrt{2}}x $, we get $ H\\left(-\\frac{2}{3}, -\\frac{\\sqrt{2}}{3}\\right) $, or $ \\left(\\frac{2}{3}, -\\frac{\\sqrt{2}}{3}\\right) $." }, { "text": "Given that point $P$ lies on the curve $C$: $y=\\frac{1}{2} x^{2}$, the tangent line to curve $C$ at point $P$ is $l$, and the line passing through point $P$ and perpendicular to line $l$ intersects curve $C$ again at point $Q$. Let $O$ be the origin. If $OP \\perp OQ$, then the $y$-coordinate of point $P$ is?", "fact_expressions": "l: Line;C: Curve;O: Origin;P: Point;Q: Point;Expression(C) = (y = x^2/2);PointOnCurve(P, C);TangentOnPoint(P, C) = l;Z: Line;PointOnCurve(P, Z);IsPerpendicular(Z, l);Intersection(Z, C) = Q;IsPerpendicular(LineSegmentOf(O, P), LineSegmentOf(O, Q))", "query_expressions": "YCoordinate(P)", "answer_expressions": "1", "fact_spans": "[[[52, 55], [63, 68]], [[7, 35], [74, 79], [37, 42]], [[89, 92]], [[2, 6], [43, 47], [57, 61], [114, 118]], [[85, 88]], [[7, 35]], [[2, 36]], [[37, 55]], [[71, 73]], [[56, 73]], [[62, 73]], [[71, 88]], [[99, 112]]]", "query_spans": "[[[114, 124]]]", "process": "" }, { "text": "The distance $|P F|$ from the point $P(1,4)$ on the parabola $y^{2}=a x$ $(a \\neq 0)$ to its focus $F$ is $?$.", "fact_expressions": "G: Parabola;a: Number;P: Point;F: Point;Expression(G) = (y^2 = a*x);Coordinate(P) = (1, 4);PointOnCurve(P, G);Focus(G) = F;Distance(P, F) = Abs(LineSegmentOf(P, F));Negation(a = 0)", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "5", "fact_spans": "[[[0, 24], [37, 38]], [[3, 24]], [[28, 36]], [[40, 43]], [[0, 24]], [[28, 36]], [[0, 36]], [[37, 43]], [[28, 53]], [[3, 24]]]", "query_spans": "[[[46, 55]]]", "process": "Substitute the coordinates of point P into the parabola equation to find the value of a, thereby obtaining the standard equation of the parabola; use the definition of the parabola to find |PF|. Substituting the coordinates of point P into the parabola equation gives a = 4^{2} = 16, so the standard equation of the parabola is y^{2} = 16x. The directrix of the parabola is x = -4, therefore, |PF| = 1 + 4 = 5. The answer is: 5" }, { "text": "Given that point $A(3,4)$ lies on the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $F_{1}$, $F_{2}$ are the left and right foci of the hyperbola $C$, respectively. If the circle with diameter $F_{1} F_{2}$ passes through point $A$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;a: Number;b: Number;G: Circle;F1: Point;F2: Point;A: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (3, 4);PointOnCurve(A, C);LeftFocus(C) = F1;RightFocus(C) = F2;IsDiameter(LineSegmentOf(F1, F2), G);PointOnCurve(A, G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[12, 73], [95, 101], [136, 142]], [[20, 73]], [[20, 73]], [[127, 128]], [[77, 84]], [[85, 92]], [[2, 11], [130, 134]], [[20, 73]], [[20, 73]], [[12, 73]], [[2, 11]], [[2, 76]], [[77, 107]], [[77, 107]], [[109, 128]], [[127, 134]]]", "query_spans": "[[[136, 148]]]", "process": "From the given, we have AF₁ ⊥ AF₂, so |F₁F₂| = 2|AO| = 10, hence c = 5. Also, √((3+5)² + 4²) − √((3−5)² + 4²) = 2a, so a = √5. Therefore, the eccentricity of hyperbola C is e = √5." }, { "text": "If the line $ m x + n y = 4 $ and the circle $ O $: $ x^{2} + y^{2} = 4 $ have no intersection points, then the number of intersection points between the line passing through the point $ (m, n) $ and the ellipse $ \\frac{x^{2}}{9} + \\frac{y^{2}}{4} = 1 $ is ?", "fact_expressions": "G: Ellipse;O: Circle;H: Line;m: Number;n: Number;I: Point;L:Line;Expression(G) = (x^2/9 + y^2/4 = 1);Expression(O) = (x^2 + y^2 = 4);Expression(H) = (m*x + n*y = 4);Coordinate(I) = (m, n);NumIntersection(H, O)=0;PointOnCurve(I,L)", "query_expressions": "NumIntersection(L,G)", "answer_expressions": "2", "fact_spans": "[[[56, 93]], [[15, 36]], [[1, 14]], [[44, 52]], [[44, 52]], [[43, 52]], [[53, 55]], [[56, 93]], [[15, 36]], [[1, 14]], [[43, 52]], [[1, 40]], [[42, 55]]]", "query_spans": "[[[53, 102]]]", "process": "" }, { "text": "The foci of an ellipse are $F_{1}(-4,0)$, $F_{2}(4,0)$, and point $P(5,3)$ lies on the ellipse. What is the equation of the ellipse?", "fact_expressions": "G: Ellipse;F1: Point;F2: Point;Coordinate(F1) = (-4, 0);Coordinate(F2) = (4, 0);Focus(G) = {F1, F2};P: Point;Coordinate(P) = (5, 3);PointOnCurve(P, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/40+y^2/24=1", "fact_spans": "[[[0, 2], [43, 45], [48, 50]], [[5, 18]], [[20, 32]], [[5, 18]], [[20, 32]], [[0, 32]], [[34, 42]], [[34, 42]], [[34, 46]]]", "query_spans": "[[[48, 54]]]", "process": "The two foci of the ellipse are F_{1}(-4,0) and F_{2}(4,0), so c=4. The point P(5,3) lies on the ellipse, then |PF_{1}|+|PF_{2}|=\\sqrt{(5+4)^{2}+(3-0)^{2}}+\\sqrt{(5-4)^{2}+(3-0)^{2}}=4\\sqrt{10}=2a \\therefore a=2\\sqrt{10} \\therefore b=\\sqrt{a^{2}-c^{2}}=\\sqrt{40-16}=\\sqrt{24}. The standard equation of the ellipse is \\frac{x^{2}}{40}+\\frac{y^{2}}{24}=1." }, { "text": "It is known that the asymptotes of a hyperbola with foci on the $x$-axis are given by $y = \\pm 2x$. What is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;PointOnCurve(Focus(G), xAxis);Expression(Asymptote(G)) = (y = pm*(2*x))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[11, 14], [35, 38]], [[2, 14]], [[11, 32]]]", "query_spans": "[[[35, 44]]]", "process": "" }, { "text": "Given that the equation $\\frac{x^{2}}{5+k}+\\frac{y^{2}}{4-2 k}=1$ represents an ellipse, find the range of values for $k$.", "fact_expressions": "G: Ellipse;Expression(G)=(x^2/(k + 5) + y^2/(4 - 2*k) = 1) ;k:Number", "query_expressions": "Range(k)", "answer_expressions": "(-5,2)&Negation(k=-1/3)", "fact_spans": "[[[47, 49]], [[2, 49]], [[51, 54]]]", "query_spans": "[[[51, 61]]]", "process": "According to the problem, the equation $\\frac{x^{2}}{5+k} + \\frac{y^{2}}{4-2k} = 1$ represents an ellipse, so it satisfies\n\\[\n\\begin{cases}\n5+k>0 \\\\\n4-2k>0 \\\\\n5+k \\neq 4-2k\n\\end{cases}\n\\]\nSolving gives $-5 < k < 2$ and $k \\neq -\\frac{1}{3}$, that is, the range of real values of $k$ is $-5 < k < 2$ and $k \\neq -\\frac{1}{3}$." }, { "text": "Let $F_{1}$ and $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)$, respectively. A line passing through $F_{2}$ intersects the ellipse at two points $P$ and $Q$. If $\\angle F_{1} P Q=60^{\\circ}$ and $|P F_{1}|=|P Q|$, 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;H: Line;P: Point;F1: Point;Q: Point;F2: Point;a > b;b > 0;LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);Intersection(H, G) = {P, Q};AngleOf(F1, P, Q) = ApplyUnit(60, degree);Abs(LineSegmentOf(P,F1))=Abs(LineSegmentOf(P,Q))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[19, 71], [89, 91], [154, 156]], [[19, 71]], [[21, 71]], [[21, 71]], [[86, 88]], [[94, 97]], [[1, 8]], [[98, 101]], [[9, 16], [78, 85]], [[21, 71]], [[21, 71]], [[1, 76]], [[1, 76]], [[77, 88]], [[86, 101]], [[103, 133]], [[134, 152]]]", "query_spans": "[[[154, 162]]]", "process": "From geometric relations, it follows that $\\triangle PF_{1}Q$ is an equilateral triangle. Combining this with the definition of the ellipse, we obtain $PQ \\perp x$-axis. Using the equation of the ellipse, we find $|PF_{2}| = \\frac{b^{2}}{a}$. Then, using the side-angle relationships in the right triangle, we derive $\\sqrt{3}a^{2} - \\sqrt{3}c^{2} = 2ac$. Solving the equation $\\sqrt{3}e^{2} + 2e - \\sqrt{3} = 0$ gives the answer. \n**Detailed Solution:** \n$\\because \\angle F_{1}PQ = 60^{\\circ}, |PF_{1}| = |PQ|$, and $\\triangle PF_{1}Q$ is an equilateral triangle, then $|PF_{1}| = |PQ| = |F_{1}Q|$. \nFrom the definition of the ellipse, $|PF_{1}| + |PF_{2}| = 2a$, $|QF_{1}| + |QF_{2}| = 2a$. \nThen $|PF_{1}| + |PF_{2}| = |PF_{1}| + |QF_{2}|$, so $|PF_{2}| = |QF_{2}|$. \n$\\therefore PQ \\perp x$-axis. \nLet point $P(c, y_{0})$, $y_{0} > 0$, from \n\\[\n\\begin{cases}\n\\frac{c^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1 \\\\\na^{2} = b^{2} + c^{2}\n\\end{cases}\n\\]\nsolving gives $y_{0} = \\frac{b^{2}}{a}$, so $|PF_{2}| = \\frac{b^{2}}{a}$. \nIn $\\triangle F_{1}PF_{2}$, $\\tan \\angle F_{1}PF_{2} = \\frac{|PF_{2}|}{|F_{1}F_{2}|} = \\frac{\\frac{b^{2}}{a}}{2c} = \\frac{b^{2}}{2ac} = \\frac{1}{\\sqrt{3}}$, but since $\\angle F_{1}PF_{2} = 60^\\circ$, we have $\\frac{b^{2}}{2ac} = \\frac{1}{\\sqrt{3}}$, thus $\\sqrt{3}b^{2} = 2ac$, i.e., $\\sqrt{3}a^{2} - \\sqrt{3}c^{2} = 2ac$. \n$\\therefore \\sqrt{3}e^{2} + 2e - \\sqrt{3} = 0$, solving gives $e = \\frac{\\sqrt{3}}{3}$." }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1$, and let point $P$ lie on the ellipse. If the midpoint of segment $P F_{1}$ lies on the $y$-axis, then the value of $\\frac{| P F_{2} |}{|P F_{1}|}$ is?", "fact_expressions": "F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;G: Ellipse;Expression(G) = (x^2/9 + y^2/5 = 1);P: Point;PointOnCurve(P, G) = True;PointOnCurve(MidPoint(LineSegmentOf(P,F1)),yAxis) = True", "query_expressions": "Abs(LineSegmentOf(P, F2))/Abs(LineSegmentOf(P, F1))", "answer_expressions": "5/13", "fact_spans": "[[[1, 8]], [[9, 16]], [[1, 61]], [[1, 61]], [[17, 54], [67, 69]], [[17, 54]], [[62, 66]], [[62, 70]], [[72, 92]]]", "query_spans": "[[[94, 129]]]", "process": "According to the problem, |PF₁| + |PF₂| = 6, the right focus F₂ is (2,0). As shown in the figure, since the midpoint of segment PF lies on the y-axis and O is the midpoint of segment F₁F₂, it follows that PF₂ // y-axis, that is, PF₂ ⊥ x-axis. From \\frac{2^{2}}{9} + \\frac{y^{2}}{5} = 1 we obtain |y| = \\frac{5}{3}, then |PF₂| = \\frac{5}{3}, thus |PF₁| = 6 - |PF₂| = \\frac{13}{3}, \\frac{|PF_{2}|}{|PF|} = \\frac{5}{13}. Therefore, the value of \\frac{|PF_{2}|}{|PF|} is \\frac{5}{13}." }, { "text": "If the eccentricity of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\frac{2 \\sqrt{3}}{3}$, then what is the inclination angle of the line $y=\\frac{b}{a} x$?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Eccentricity(G) = 2*sqrt(3)/3;H: Line;Expression(H) = (y = x*(b/a))", "query_expressions": "Inclination(H)", "answer_expressions": "pi/6", "fact_spans": "[[[1, 57]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 84]], [[86, 105]], [[86, 105]]]", "query_spans": "[[[86, 111]]]", "process": "\\because the eccentricity of the hyperbola is \\frac{2\\sqrt{3}}{3}, \\therefore 1+\\frac{b^{2}}{a^{2}}=\\frac{4}{3}, \\therefore \\frac{b}{a}=\\frac{\\sqrt{3}}{3}, \\therefore the inclination angle of the line y=\\frac{b}{a}x is \\frac{\\pi}{6}" }, { "text": "Given that the ellipse $\\frac{x^{2}}{3 m^{2}}+\\frac{y^{2}}{5 n^{2}}=1$ and the hyperbola $\\frac{x^{2}}{2 m^{2}}-\\frac{y^{2}}{3 n^{2}}=1$ have common foci, what is the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;n: Number;m: Number;H: Ellipse;Expression(G) = (-y^2/(3*n^2) + x^2/(2*m^2) = 1);Expression(H) = (y^2/(5*n^2) + x^2/(3*m^2) = 1);Focus(H) = Focus(G)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*(sqrt(3)/4)*x", "fact_spans": "[[[52, 102], [111, 114]], [[55, 102]], [[55, 102]], [[2, 51]], [[52, 102]], [[2, 51]], [[2, 108]]]", "query_spans": "[[[111, 122]]]", "process": "" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, the left focus of $C$ is $F$. A line passing through the origin intersects $C$ at points $A$ and $B$. Connect $AF$ and $BF$. If $|AB|=10$, $|AF|=6$, $\\cos \\angle ABF=\\frac{4}{5}$, then the eccentricity $e$ 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;F: Point;LeftFocus(C) = F;G: Line;O: Origin;PointOnCurve(O, G);A: Point;B: Point;Intersection(C, G) = {A,B};Abs(LineSegmentOf(A, B)) = 10;Abs(LineSegmentOf(A, F)) = 6;Cos(AngleOf(A, B, F)) = 4/5;e: Number;Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "5/7", "fact_spans": "[[[2, 59], [68, 71], [160, 163]], [[2, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[9, 59]], [[64, 67]], [[2, 67]], [[76, 78]], [[73, 75]], [[72, 78]], [[81, 84]], [[87, 90]], [[68, 92]], [[106, 116]], [[117, 126]], [[127, 158]], [[167, 170]], [[160, 170]]]", "query_spans": "[[[167, 172]]]", "process": "" }, { "text": "The hyperbola $C_{1}$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has eccentricity $e_{1}$, the hyperbola $C_{2}$: $\\frac{y^{2}}{m^{2}}-\\frac{x^{2}}{n^{2}}=1(m>0, n>0)$ has eccentricity $e_{2}$. If hyperbolas $C_{1}$ and $C_{2}$ have the same asymptotes, then the range of $\\frac{1}{e_{1}}+\\frac{1}{e_{2}}$ is?", "fact_expressions": "C1:Hyperbola;C2:Hyperbola;a:Number;b:Number;a>0;b>0;m:Number;n:Number;m>0;n>0;e1:Number;e2:Number;Expression(C1)=(x^2/a^2-y^2/b^2=1);Expression(C2)=(y^2/m^2-x^2/n^2=1);Eccentricity(C1)=e1;Eccentricity(C2)=e2;Asymptote(C1)=Asymptote(C2)", "query_expressions": "Range(1/e2 + 1/e1)", "answer_expressions": "(1, sqrt(2)]", "fact_spans": "[[[0, 65], [156, 166]], [[78, 143], [167, 174]], [[12, 65]], [[12, 65]], [[12, 65]], [[12, 65]], [[90, 143]], [[90, 143]], [[90, 143]], [[90, 143]], [[70, 77]], [[148, 155]], [[0, 65]], [[78, 143]], [[0, 77]], [[78, 155]], [[156, 181]]]", "query_spans": "[[[183, 223]]]", "process": "The hyperbola $ C_{1}: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1 $ ($ a > 0, b > 0 $) has eccentricity $ e_{1} = \\frac{\\sqrt{a^{2} + b^{2}}}{a} $, the hyperbola $ C_{2}: \\frac{y^{2}}{m^{2}} - \\frac{x^{2}}{n^{2}} = 1 $ ($ m > 0, n > 0 $) has eccentricity $ e_{2} = \\frac{\\sqrt{m^{2} + n^{2}}}{m} $. Since the hyperbolas $ C_{1} $ and $ C_{2} $ have the same asymptotes, it follows that $ \\frac{b}{a} = \\frac{m}{n} $, so $ m = \\frac{b}{a}n $. Then $ 1 < \\sqrt{1 + \\frac{2ab}{a^{2} + b^{2}}} $, with equality if and only if $ a = b $, thus $ 0 < \\frac{2ab}{a^{2} + b^{2}} \\leqslant 1 $, hence the range of $ \\frac{1}{e_{1}} + \\frac{1}{e_{2}} $ is $ (1, \\sqrt{2}] $." }, { "text": "$F$ is the focus of the parabola $C$: $y^{2}=4x$. The line $l$ passing through $F$ with slope $k$ intersects the parabola at points $P$ and $Q$. The perpendicular bisector of segment $PQ$ intersects the $x$-axis at point $M$, and $|PQ|=6$. Then $|MF|=$?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l: Line;PointOnCurve(F, l);k: Number;Slope(l) = k;P: Point;Q: Point;Intersection(l, C) = {P, Q};M: Point;Intersection(PerpendicularBisector(LineSegmentOf(P, Q)), xAxis) = M;Abs(LineSegmentOf(P, Q)) = 6", "query_expressions": "Abs(LineSegmentOf(M, F))", "answer_expressions": "3", "fact_spans": "[[[4, 23], [45, 48]], [[4, 23]], [[0, 3], [28, 31]], [[0, 26]], [[39, 44]], [[27, 44]], [[35, 38]], [[32, 44]], [[50, 53]], [[54, 57]], [[39, 59]], [[79, 83]], [[60, 83]], [[85, 94]]]", "query_spans": "[[[96, 105]]]", "process": "First, find the value of p from the parabola equation, then use the properties of the parabola to find the equation of the perpendicular bisector of PQ, and thus obtain the answer. \n∵ the parabola C: y^{2} = 4x, ∴ p = 2, focus F(1,0). \nLet line l: y = k(x - 1), P(x_{1}, y_{1}), Q(x_{2}, y_{2}). \nSubstitute y = k(x - 1) into the parabola C: y^{2} = 4x, we get: k^{2}x^{2} - (2k^{2} + 4)x + k^{2} = 0. \n∴ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}}. \n∵ |PQ| = x_{1} + x_{2} + p = 2 + \\frac{4}{k^{2}} + 2 = 6, ∴ k = \\pm\\sqrt{2}. \nLet N(x_{0}, y_{0}) be the midpoint of PQ, then \nx_{0} = \\frac{x_{1} + x_{2}}{2} = \\frac{2 + \\frac{4}{k^{2}}}{2} = \\frac{4}{2} = 2, \ny_{0} = k(x_{0} - 1) = k. \nSo the equation of the perpendicular bisector of segment PQ is: y - k = -\\frac{1}{k}(x - 2). \nLet y = 0, we get x = 4, so |MF| = 4 - 1 = 3." }, { "text": "It is known that point $P$ is a moving point on the parabola $y^{2}=4x$, the projection of point $P$ on the $y$-axis is $M$, and point $A(4, 6)$. Then what is the minimum value of $|PA|+|PM|$?", "fact_expressions": "G: Parabola;A: Point;P: Point;M: Point;Expression(G) = (y^2 = 4*x);Coordinate(A) = (4, 6);PointOnCurve(P, G);Projection(P, yAxis) = M", "query_expressions": "Min(Abs(LineSegmentOf(P, A)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "", "fact_spans": "[[[7, 21]], [[43, 54]], [[26, 30], [2, 6]], [[39, 42]], [[7, 21]], [[43, 54]], [[2, 25]], [[26, 42]]]", "query_spans": "[[[56, 72]]]", "process": "" }, { "text": "Given a point $A$ on the parabola $x^{2}=4 y$, the distance from $A$ to the $x$-axis is $m$, and the distance from $A$ to the line $x+2 y+8=0$ is $n$. Then the minimum value of $m+n$ is?", "fact_expressions": "G: Parabola;H: Line;A: Point;m:Number;n:Number;Expression(G) = (x^2 = 4*y);Expression(H) = (x + 2*y + 8 = 0);PointOnCurve(A, G);Distance(A, xAxis) = m;Distance(A,H)=n", "query_expressions": "Min(m + n)", "answer_expressions": "2*sqrt(5) - 1", "fact_spans": "[[[2, 16]], [[36, 49]], [[19, 22]], [[31, 34]], [[53, 56]], [[2, 16]], [[36, 49]], [[2, 22]], [[19, 34]], [[19, 56]]]", "query_spans": "[[[58, 69]]]", "process": "According to the definition of a parabola, the distance from point A to the directrix is equal to the distance from A to the focus, so m = |AF| - 1. As shown in the figure, |AP| = n, then m + n = |AF| + |AP| - 1. The minimum value of |AF| + |AP| is the distance d from point F(0,1) to the line x + 2y + 8 = 0, which is d = \\frac{|2 + 8|}{\\sqrt{1^{2} + 2^{2}}} = 2\\sqrt{5}. Therefore, the minimum value of m + n is 2\\sqrt{5} - 1." }, { "text": "Given that a line with slope $\\sqrt{2}$ intersects the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ at points $A$ and $B$, $O$ is the origin, and the midpoint of $AB$ is $P$. If the slope of the line $OP$ is $2 \\sqrt{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;G: Line;Slope(G) = sqrt(2);A: Point;B: Point;Intersection(G, C) = {A, B};O: Origin;P: Point;MidPoint(LineSegmentOf(A, B)) = P;Slope(LineOf(O, P)) = 2*sqrt(2)", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)", "fact_spans": "[[[19, 80], [141, 147]], [[19, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[16, 18]], [[2, 18]], [[83, 86]], [[87, 90]], [[16, 92]], [[93, 96]], [[111, 114]], [[102, 114]], [[116, 139]]]", "query_spans": "[[[141, 153]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), P(x_{0},y_{0}), then \\begin{cases}\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\\\\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\end{cases}. Subtracting the two equations gives \\frac{x_{1}^{2}-x_{2}^{2}}{a^{2}}=\\frac{y_{1}^{2}-y_{2}^{2}}{b^{2}}, so \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{b^{2}}{a^{2}}\\cdot\\frac{x_{1}+x_{2}}{y_{1}+y_{2}}. Since x_{1}+x_{2}=2x_{0}, y_{1}+y_{2}=2y_{0}, we have \\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\frac{b^{2}}{a^{2}}\\cdot\\frac{x_{0}}{y_{0}}. Because k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\\sqrt{2}, k_{OP}=\\frac{y_{0}}{x_{0}}=2\\sqrt{2}, it follows that \\sqrt{2}=\\frac{b^{2}}{a^{2}}\\cdot\\frac{x_{0}}{y_{0}}=\\frac{b^{2}}{a^{2}}\\cdot\\frac{1}{2\\sqrt{2}}, so \\frac{b^{2}}{a^{2}}=4. Hence e=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{5}." }, { "text": "Given that the foci of the ellipse are $F_{1}(-2,0)$, $F_{2}(2,0)$, and it passes through the point $P\\left(\\frac{5}{2},-\\frac{3}{2}\\right)$, then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;F1: Point;Coordinate(F1) = (-2, 0);F2: Point;Coordinate(F2) = (2, 0);Focus(G) = {F1, F2};P: Point;Coordinate(P) = (5/2, -3/2);PointOnCurve(P, G) = True", "query_expressions": "Expression(G)", "answer_expressions": "x^2/10+y^2/6=1", "fact_spans": "[[[2, 4], [71, 73]], [[8, 21]], [[8, 21]], [[24, 36]], [[24, 36]], [[2, 36]], [[40, 69]], [[40, 69]], [[2, 69]]]", "query_spans": "[[[71, 78]]]", "process": "Let the standard equation of ellipse F be: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). According to the given conditions, c=2, \\therefore a=\\sqrt{10}, then b^{2}=a^{2}-c^{2}=6. Hence, the standard equation of ellipse F is \\frac{x^{2}}{10}+\\frac{y^{2}}{6}=1;" }, { "text": "A line $l$ passes through the focus of the parabola $y^{2}=6x$, intersecting the parabola at points $A$ and $B$. If the x-coordinate of the midpoint $M$ of segment $AB$ is $2$, then $|AB|$ equals?", "fact_expressions": "l: Line;G: Parabola;B: Point;A: Point;M:Point;Expression(G) = (y^2 = 6*x);PointOnCurve(Focus(G),l);Intersection(l,G)={A,B};MidPoint(LineSegmentOf(A,B))=M;XCoordinate(M)=2", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "7", "fact_spans": "[[[18, 23]], [[1, 15], [25, 28]], [[33, 36]], [[29, 32]], [[48, 51]], [[1, 15]], [[0, 23]], [[18, 38]], [[39, 51]], [[48, 59]]]", "query_spans": "[[[61, 71]]]", "process": "According to the equation of the parabola, we can find p=3, then using the midpoint coordinate formula we can find x_{1}+x_{2}, and finally using the focal chord formula of the parabola we can find |AB|. \\because y2=6x, then p=3, let A(x_{1},y_{1}), B(x_{2},y_{2}) \\because the horizontal coordinate of the midpoint M of segment AB is 2, \\therefore x_{1}+x_{2}=2\\times2=4, \\therefore |AB|=x_{1}+x_{2}+p=4+3=7" }, { "text": "The minimum distance from a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ to the line $4 x-5 y+40=0$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);P: Point;PointOnCurve(P, G);H: Line;Expression(H) = (4*x - 5*y + 40 = 0)", "query_expressions": "Min(Distance(P, H))", "answer_expressions": "(15*sqrt(41))/41", "fact_spans": "[[[0, 38]], [[0, 38]], [[40, 41]], [[0, 41]], [[42, 58]], [[42, 58]]]", "query_spans": "[[[40, 65]]]", "process": "Problem Analysis: Let a point on the ellipse be (5\\cos\\alpha,3\\sin\\alpha). The distance from the point to the line is d=\\frac{|20\\cos\\alpha-15\\sin\\alpha+40|}{\\sqrt{4^{2}+5^{2}}}=\\frac{|25\\cos(\\alpha+\\beta)+40|}{\\sqrt{41}},\\tan\\beta=\\frac{3}{4}. When \\cos(\\alpha+\\beta)=-1, d_{\\min}=\\frac{15}{\\sqrt{41}}=\\frac{15}{41}\\sqrt{41}. Topic: Line and ellipse" }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1$, the left and right foci are denoted as $F_{1}$ and $F_{2}$ respectively. A circle $F_{2}$ with radius $1$ is drawn centered at $F_{2}$. Let $P$ be a point on the ellipse $C$ and $Q$ be a point on the circle $F_{2}$. Then the range of values of $|P F_{1}|+|P Q|$ is?", "fact_expressions": "C: Ellipse;F2: Point;P: Point;F1: Point;Q: Point;F: Circle;Expression(C) = (x^2/9 + y^2/6 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;Center(F) = F2;Radius(F) = 1;PointOnCurve(P, C);PointOnCurve(Q, F)", "query_expressions": "Range(Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, Q)))", "answer_expressions": "[5, 7]", "fact_spans": "[[[2, 44], [104, 109]], [[63, 70], [73, 80]], [[100, 103]], [[53, 61]], [[113, 116]], [[91, 99], [117, 125]], [[2, 44]], [[2, 70]], [[2, 70]], [[72, 99]], [[84, 99]], [[100, 112]], [[113, 128]]]", "query_spans": "[[[130, 154]]]", "process": "From the ellipse equation, we know: a=3. By the definition of an ellipse: |PF₁| + |PF₂| = 2a = 6 ⇒ |PF₁| = 6 - |PF₂|. Therefore, |PF₁| + |PQ| = 6 + |PQ| - |PF₂|. Also, -|QF₂| ≤ |PQ| - |PF₂| ≤ |QF₂| and |QF₂| = 1. Therefore, |PF₁| + |PQ| = 6 + |PQ| - |PF₂| ∈ [5,7]. The correct answer to this problem: [5,7]" }, { "text": "Let $A(-2,0)$, $B(2,0)$. If there exists a point $P$ on the line $y = ax$ ($a > 0$) such that $|PA| + |PB| = 6$, and the distance from the incenter of $\\Delta PAB$ to the $x$-axis is $\\frac{3\\sqrt{30}}{20}$, then $a = $?", "fact_expressions": "G: Line;a: Number;A: Point;B: Point;P: Point;a>0;Expression(G) = (y = a*x);Coordinate(A) = (-2, 0);Coordinate(B) = (2, 0);PointOnCurve(P,G);Abs(LineSegmentOf(P,A))+Abs(LineSegmentOf(P,B))=6;Distance(Incenter(TriangleOf(P,A,B)), xAxis) = (3*sqrt(30))/20", "query_expressions": "a", "answer_expressions": "sqrt(3)", "fact_spans": "[[[22, 36]], [[115, 118]], [[1, 10]], [[11, 20]], [[41, 44]], [[24, 36]], [[22, 36]], [[1, 10]], [[11, 20]], [[22, 44]], [[46, 61]], [[63, 113]]]", "query_spans": "[[[115, 120]]]", "process": "According to the problem, point P is the intersection point of the line y = ax (a > 0) and the ellipse \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1. Solving the system of equations formed by the line and the ellipse yields y^{2} = \\frac{45a^{2}}{9a^{2} + 5}. Given that the distance from the incenter of \\triangle PAB to the x-axis is \\frac{3\\sqrt{30}}{20}, i.e., the inradius r of \\triangle PAB is \\frac{3\\sqrt{30}}{20}, the value of parameter a can be found using the equal area method. Point P satisfies |PA| + |PB| = 6, so point P lies on the ellipse \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1. According to the problem, point P is the intersection of the line y = ax (a > 0) and the ellipse \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1. Substituting y = ax into \\frac{x^{2}}{9} + \\frac{y^{2}}{5} = 1 and eliminating y gives x^{2} = \\frac{45}{9a^{2} + 5}, then y^{2} = \\frac{45a^{2}}{9a^{2} + 5}. Since the distance from the incenter of \\triangle APB to the x-axis is \\frac{3\\sqrt{30}}{20}, the inradius r of \\triangle PAB is \\frac{3\\sqrt{30}}{20}. Therefore, the area of \\triangle APB is \\frac{1}{2} \\times |AB| \\times |y| = \\frac{1}{2} \\times r \\times (|AB| + |PA| + |PB|), that is, |y| = \\frac{5}{2}r, y^{2} = \\frac{45a^{2}}{9a^{2} + 5} = \\frac{5}{4}r^{2} = \\frac{25}{4} \\times \\frac{27}{40}, solving gives a^{2} = 3, and since a > 0, then a = \\sqrt{3}. [Note] This problem examines the relationship between a line and an ellipse, and finding parameters based on properties of focal triangles of ellipses, and is of medium difficulty." }, { "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. The line passing through $F_{1}$ and perpendicular to the $x$-axis intersects the ellipse at points $A$ and $B$. The line $A F_{2}$ intersects the ellipse again at point $C$. If $\\overrightarrow{A F_{2}}=2 \\overrightarrow{F_{2} C}$, 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);IsPerpendicular(H, xAxis);A: Point;B: Point;Intersection(H, G) = {A, B};C: Point;Intersection(LineOf(A, F2), G) = {A, C};VectorOf(A, F2) = 2*VectorOf(F2, C)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(5)/5", "fact_spans": "[[[2, 54], [101, 103], [126, 128], [195, 197]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[63, 70], [82, 89]], [[72, 79]], [[2, 79]], [[2, 79]], [[98, 100]], [[81, 100]], [[90, 100]], [[104, 107]], [[108, 111]], [[98, 113]], [[135, 138]], [[104, 138]], [[140, 193]]]", "query_spans": "[[[195, 203]]]", "process": "F_{1}(-c,0),F_{2}(c,0) Let A(-c,\\frac{b^{2}}{a}). Draw CD\\bot x-axis from point C, with D as the foot of the perpendicular. As shown in the figure, since Rt\\Delta AF_{1}F_{2} \\sim Rt\\Delta CDF_{2} and \\overrightarrow{AF}_{2}=2\\overrightarrow{F_{2}C}, \\therefore \\frac{CD}{AF}=\\frac{DF_{2}}{F_{1}F_{2}}=\\frac{F_{2}C}{AF}=\\frac{1}{2}, \\therefore C(2c,-\\frac{b^{2}}{\\frac{b^{2}}{a}}). Substituting into the ellipse equation gives \\frac{4c^{2}}{a^{2}}+\\frac{b^{2}}{4a^{2}}=1 \\Rightarrow \\frac{4c^{2}}{a^{2}}+\\frac{a^{2}-c^{2}}{4a^{2}}=1. Solving yields: e=\\frac{c}{a}=\\frac{\\sqrt{5}}{5}" }, { "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;Expression(G) = (x^2/2 + y^2 = 1);H: LineSegment;IsChordOf(H,G) = True;Coordinate(MidPoint(H)) = (1/2, 1/2)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "", "fact_spans": "[[[0, 27]], [[0, 27]], [], [[0, 29]], [[0, 62]]]", "query_spans": "[[[0, 76]]]", "process": "" }, { "text": "The line passes through the point $P(-1,0)$ and intersects the parabola $y^{2}=4x$ at points $A$ and $B$. If $A$ is the midpoint of segment $PB$, then the slope of line $AB$ is?", "fact_expressions": "G: Parabola;H: Line;A: Point;B: Point;P: Point;Expression(G) = (y^2 = 4*x);Coordinate(P) = (-1, 0);PointOnCurve(P, H);Intersection(H, G) = {A, B};MidPoint(LineSegmentOf(P, B)) = A", "query_expressions": "Slope(LineOf(A, B))", "answer_expressions": "pm*2*sqrt(2)/3", "fact_spans": "[[[15, 29]], [[0, 2]], [[41, 44], [30, 33]], [[34, 37]], [[3, 13]], [[15, 29]], [[3, 13]], [[0, 13]], [[0, 39]], [[41, 55]]]", "query_spans": "[[[57, 69]]]", "process": "Let B(x_{0},y_{0}), then A(\\frac{x_{0}-1}{2},\\frac{y_{0}}{2}). Substituting the coordinates of points A and B into the parabola equation gives \\begin{cases}y_{0}^{2}=4x_{0}\\\\(\\frac{y_{0}}{2})^{2}=4\\times\\frac{x_{0}-1}{2}\\end{cases}. Then A(\\frac{1}{2},\\sqrt{2}), B(2,2\\sqrt{2}) or A(\\frac{1}{2},-\\sqrt{2}), B(2,-2\\sqrt{2}), yielding a slope of \\underline{2\\sqrt{2}}" }, { "text": "The hyperbola $C$: $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1$ has a focal distance of?", "fact_expressions": "C: Hyperbola;Expression(C) = (-x^2/16 + y^2/9 = 1)", "query_expressions": "FocalLength(C)", "answer_expressions": "", "fact_spans": "[[[0, 44]], [[0, 44]]]", "query_spans": "[[[0, 49]]]", "process": "" }, { "text": "Given that the curve $C$: $y=\\frac{\\sqrt{20-x^{2}}}{2}$ and the line $l$: $y=-x+m$ have exactly one common point, what is the range of values for $m$?", "fact_expressions": "l: Line;C: Curve;m:Number;Expression(C) = (y = sqrt(20 - x^2)/2);Expression(l)=(y=m-x);NumIntersection(l,C)=1", "query_expressions": "Range(m)", "answer_expressions": "{[-2*sqrt(5),2*sqrt(5)),(m=5)}", "fact_spans": "[[[39, 54]], [[2, 38]], [[63, 66]], [[2, 38]], [[39, 54]], [[2, 61]]]", "query_spans": "[[[63, 73]]]", "process": "As shown in the figure, the curve \\( C: y = \\frac{\\sqrt{20 - x^{2}}}{2} \\) is graphed. To make the line \\( l: y = -x + m \\) intersect it at exactly one point, there are two cases: \n① The curve and the line are tangent. Solving the system of equations: \n\\[\n\\begin{cases}\ny = -x + m \\\\\ny = \\frac{\\sqrt{20 - x^{2}}}{2}\n\\end{cases}\n\\] \nEliminating \\( y \\) gives: \\( 5x^{2} - 8mx + 4m^{2} - 20 = 0 \\). \nSince the equation has only one solution, setting discriminant \\( \\Delta = 0 \\), we get \\( m = 5 \\) or \\( m = -5 \\) (discarded). \n② The line intersects the curve. From the graph, the critical condition occurs when the line passes through \\( (-2\\sqrt{5}, 0) \\) or \\( (2\\sqrt{5}, 0) \\), so \\( -2\\sqrt{5} \\leqslant m < 2\\sqrt{5} \\). \nIn summary, \\( -2\\sqrt{5} \\leqslant m < 2\\sqrt{5} \\) or \\( m = 5 \\)." }, { "text": "Let $F_{1}$, $F_{2}$ be the two foci of the hyperbola $x^{2}-4 y^{2}=4$, and let $P$ lie on the hyperbola such that $\\overrightarrow{P F}_{1} \\cdot \\overrightarrow{P F_{2}}=0$. Then $|\\overrightarrow{P F_{1}}| \\cdot|\\overrightarrow{P F_{2}}|$=?", "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;Expression(G) = (x^2 - 4*y^2 = 4);Focus(G) = {F1, F2};PointOnCurve(P, G);DotProduct(VectorOf(P,F1),VectorOf(P,F2))=0", "query_expressions": "Abs(VectorOf(P, F1))*Abs(VectorOf(P, F2))", "answer_expressions": "2", "fact_spans": "[[[17, 37], [47, 50]], [[43, 46]], [[9, 16]], [[1, 8]], [[17, 37]], [[1, 42]], [[43, 51]], [[53, 112]]]", "query_spans": "[[[114, 176]]]", "process": "Hyperbola $ x^{2} - 4y^{2} = 4 \\therefore \\frac{x^{2}}{4} - y^{2} = 1 $, without loss of generality, assume $ |\\overrightarrow{PF_{1}}| > |\\overrightarrow{PF_{2}}| $, then $ |\\overrightarrow{PF_{1}}| - |\\overrightarrow{PF_{2}}| = 4 $, hence $ |\\overrightarrow{PF}|^{2} + |\\overrightarrow{PF_{2}}|^{2} - 2|\\overrightarrow{PF}| \\cdot |\\overrightarrow{PF_{2}}| = 16 $, $ \\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}} = 0 \\therefore \\overrightarrow{PF_{1}} \\bot \\overrightarrow{PF_{2}} $, hence $ |\\overrightarrow{PF}|^{2} + |\\overrightarrow{PF_{2}}|^{2} = (2\\sqrt{5})^{2} = 20 $, hence $ |\\overrightarrow{PF}| \\cdot |\\overrightarrow{PF}| = 2 $" }, { "text": "The parabola $x^{2}=m y$ passes through the point $(-1,1)$. Then the coordinates of the focus of the parabola are?", "fact_expressions": "G: Parabola;m: Number;H: Point;Expression(G) = (x^2 = m*y);Coordinate(H) = (-1, 1);PointOnCurve(H, G)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, 1/4)", "fact_spans": "[[[0, 14], [26, 29]], [[3, 14]], [[15, 24]], [[0, 14]], [[15, 24]], [[0, 24]]]", "query_spans": "[[[26, 36]]]", "process": "Since the parabola \\( x^{2} = my \\) passes through the point \\( (-1, 1) \\), we have \\( m = 1 \\), so \\( x^{2} = y \\), therefore the coordinates of the focus of the parabola are \\( \\left(0, \\frac{1}{4}\\right) \\)." }, { "text": "Given the line $ l $: $ a x+(a+1) y+2=0 $, and point $ P(-2,0) $. Draw $ P M \\perp l $ at point $ M $. Then the equation of the trajectory of point $ M $ is?", "fact_expressions": "l: Line;P: Point;M: Point;Coordinate(P) = (-2, 0);Expression(l) = (a*x + y*(a + 1) + 2 = 0);IsPerpendicular(LineSegmentOf(P,M),l);FootPoint(LineSegmentOf(P,M),l)=M", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2+(y+1)^2=5", "fact_spans": "[[[2, 26]], [[27, 37]], [[53, 57], [59, 63]], [[27, 37]], [[2, 26]], [[39, 52]], [[39, 57]]]", "query_spans": "[[[59, 70]]]", "process": "The equation of line l can be rewritten as a(x+y)+y+2=0. Solving the system of equations \\begin{cases}x+y=0\\\\y+2=0\\end{cases}, we obtain \\begin{cases}x=2\\\\y=-2\\end{cases}, which means line l passes through the fixed point Q(2,-2). Since PM\\bot l, triangle PMQ is a right triangle. Based on the fact that in a circle the angle subtended by the diameter at the circumference is a right angle, it follows that point M lies on the circle with PQ as diameter. Given point P(-2,0), the midpoint of segment PQ has coordinates (0,-1). Therefore, the trajectory equation of point M is x^{2}+(y+1)^{2}=5." }, { "text": "Given fixed circle $C_{1}$: $x^{2}+y^{2}+4 x=0$, fixed circle $C_{2}$: $x^{2}+y^{2}-4 x-60=0$, a moving circle $M$ is externally tangent to fixed circle $C_{1}$ and internally tangent to circle $C_{2}$. Find the equation of the locus of the center $M$ of the moving circle?", "fact_expressions": "C1: Circle;Expression(C1) = (x^2 +y^2 + 4*x = 0);C2: Circle;Expression(C2) = (x^2 +y^2 - 4*x -60 = 0);M: Circle;M1: Point;Center(M) = M1;IsOutTangent(M, C1);IsInTangent(M, C2)", "query_expressions": "LocusEquation(M1)", "answer_expressions": "x^2/25+y^2/21=1", "fact_spans": "[[[4, 32], [75, 82]], [[4, 32]], [[35, 66], [86, 94]], [[35, 66]], [[69, 72]], [[103, 106]], [[98, 106]], [[69, 84]], [[69, 96]]]", "query_spans": "[[[103, 112]]]", "process": "C₁ has center (-2,0) and radius 2, C₂ has center (2,0) and radius 8. Let the radius of the moving circle be r. According to the concepts of internal and external tangency of two circles, |MC₁| = r + 2, |MC₂| = 8 - r, |MC₁| + |MC₂| = 10 satisfies the definition of an ellipse, with 2a = 10, 2c = 4, b² = a² - c² = 21. Hence, the equation of the ellipse is \\frac{x^{2}}{25} + \\frac{y^{2}}{21} = 1" }, { "text": "If the distance from point $P$ on the parabola $x^{2}=8 y$ to the focus is $8$, then what is the distance from $P$ to the $x$-axis?", "fact_expressions": "G: Parabola;P: Point;Expression(G) = (x^2 = 8*y);PointOnCurve(P, G);Distance(P, Focus(G)) = 8", "query_expressions": "Distance(P, xAxis)", "answer_expressions": "6", "fact_spans": "[[[1, 15]], [[17, 21], [33, 36]], [[1, 15]], [[1, 21]], [[1, 31]]]", "query_spans": "[[[33, 46]]]", "process": "Let point P(x_{0},y_{0}), then y_{0}+2=8, that is, y_{0}=6, that is, the distance from P to the x-axis is 6." }, { "text": "On a plane, two fixed points $A$ and $B$ are at a distance of $10$. A moving point $P$ satisfies $PB - PA = 6$. What is the minimum value of the distance from point $P$ to the midpoint of $AB$?", "fact_expressions": "A: Point;B: Point;P: Point;Distance(A, B) = 10;-LineSegmentOf(P, A) + LineSegmentOf(P, B) = 6", "query_expressions": "Min(Distance(P, MidPoint(LineSegmentOf(A, B))))", "answer_expressions": "3", "fact_spans": "[[[6, 9]], [[12, 15]], [[28, 31], [44, 48]], [[6, 25]], [[33, 42]]]", "query_spans": "[[[44, 63]]]", "process": "" }, { "text": "Given that a line with slope $k$ ($k>0$) passes through the focus $F$ of the parabola $C$: $y^{2}=4x$ and intersects the parabola $C$ at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the directrix of the parabola, with feet of perpendiculars denoted as $A_{1}$ and $B_{1}$, respectively. If $\\frac{S_{\\Delta A B B_{1}}}{S_{\\triangle AB A_{1}}}=4$, then $k=$?", "fact_expressions": "C: Parabola;k: Number;k > 0;F: Point;A: Point;B: Point;A1: Point;B1: Point;Expression(C) = (y^2 = 4*x);Focus(C) = F;H: Line;PointOnCurve(F, H);Slope(H) = k;Intersection(H, C) = {A, B};l1: Line;l2: Line;PointOnCurve(A, l1);PointOnCurve(B, l2);IsPerpendicular(l1, Directrix(C));IsPerpendicular(l2, Directrix(C));FootPoint(l1, Directrix(C)) = A1;FootPoint(l2, Directrix(C)) = B1;Area(TriangleOf(A, B, B1))/Area(TriangleOf(A, B, A1)) = 4", "query_expressions": "k", "answer_expressions": "4/3", "fact_spans": "[[[17, 36], [44, 50], [75, 78]], [[163, 166]], [[4, 12]], [[39, 42]], [[53, 56], [64, 67]], [[57, 60], [68, 71]], [[89, 96]], [[97, 104]], [[17, 36]], [[17, 42]], [[14, 16]], [[14, 42]], [[2, 16]], [[14, 62]], [], [], [[63, 83]], [[63, 83]], [[63, 83]], [[63, 83]], [[63, 104]], [[63, 104]], [[106, 161]]]", "query_spans": "[[[163, 168]]]", "process": "As shown in the figure: Parabola C: y^{2}=4x, F(1,0). Let line AB: y=k(x-1), A(x_{1},y_{1}), B(x_{2},y_{2}). From the problem: \\begin{cases}y=k(x-1)\\end{cases}\\begin{cases}y^{2}=4x\\\\y=1\\end{cases}x_{1}x_{2}=1. Since \\frac{S_{\\DeltaABB_{1}}}{S_{\\DeltaABA_{1}}}=\\frac{\\frac{1}{2}|BB_{1}|\\cdot|A_{1}B_{1}|}{\\frac{1}{2}|AA_{1}|\\cdot|A_{1}B_{1}|}=\\frac{|BB_{1}|}{|AA_{1}|}=\\frac{x_{2}+1}{x_{1}+1}=4, therefore \\begin{cases}\\frac{x_{2}+1}{x_{1}+1}=4\\\\x_{1}x_{2}=1\\end{cases}\\Rightarrow\\begin{cases}x_{1}=\\frac{1}{4}\\\\x_{2}=4\\end{cases}, solving gives A(\\frac{1}{4},-1), B(4,4). Thus k=\\frac{4-(-1)}{4-\\frac{1}{4}}=\\frac{4}{3}" }, { "text": "The line passing through the right focus of the hyperbola $\\frac{x^{2}}{2}-y^{2}=1$ with an inclination angle of $45^{\\circ}$ intersects the hyperbola at points $A$ and $B$. Then $|A B|$=?", "fact_expressions": "G: Hyperbola;H: Line;A: Point;B: Point;Expression(G) = (x^2/2 - y^2 = 1);PointOnCurve(RightFocus(G), H);Inclination(H)=ApplyUnit(45, degree) ;Intersection(H, G) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "", "fact_spans": "[[[1, 29], [55, 58]], [[52, 54]], [[59, 63]], [[64, 67]], [[1, 29]], [[0, 54]], [[35, 54]], [[52, 67]]]", "query_spans": "[[[69, 78]]]", "process": "" }, { "text": "The equation of the circle with center at the right focus of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$ and tangent to the asymptotes of the hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/5 = 1);H: Circle;Center(H) = RightFocus(G);IsTangent(H, Asymptote(G))", "query_expressions": "Expression(H)", "answer_expressions": "(x-3)^2+y^2=5", "fact_spans": "[[[1, 39], [49, 52]], [[1, 39]], [[59, 60]], [[0, 60]], [[48, 60]]]", "query_spans": "[[[59, 65]]]", "process": "Given that for the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$, $c^{2}=5+4$, $c=3$, the foci lie on the x-axis, so the center of the circle is $(3,0)$, and the asymptotes are given by: $y=\\frac{+\\sqrt{5}}{2}x$. Since the circle is tangent to the asymptotes, the distance from the center of the circle to an asymptote equals the radius, $r=\\frac{3\\times\\frac{\\sqrt{5}}{2}}{\\sqrt{\\left(\\frac{\\sqrt{5}}{2}\\right)^{2}+1^{2}}}=\\sqrt{5}$. Therefore, the equation of the required circle is $(x-3)^{2}+y^{2}=5$." }, { "text": "Given that the center of ellipse $G$ is at the origin of the coordinate system, the focal distance is $4$, and the minimum distance from a point on the ellipse to one of its foci is $6$, what is the eccentricity of the ellipse?", "fact_expressions": "G:Ellipse;O:Origin;P:Point;Center(G)=O;FocalLength(G) = 4;PointOnCurve(P,G);Min(Distance(P,Focus(G)))=6", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/4", "fact_spans": "[[[2, 7], [24, 26], [24, 26], [24, 26]], [[11, 15]], [], [[2, 15]], [[2, 22]], [[24, 29]], [[24, 43]]]", "query_spans": "[[[45, 53]]]", "process": "Using the given conditions, set up a system of equations to solve for a and c, and find the eccentricity of the ellipse. The ellipse G has its center at the coordinate origin, a focal distance of 4, and the minimum distance from a point on the ellipse to one of its foci is 6. \n\\begin{cases}2c=4\\\\a-c=6\\end{cases}, solving gives a=8, c=2, so the eccentricity of the ellipse is: e=\\frac{c}{a}=\\frac{1}{4}" }, { "text": "Let point $P(x, y)$ be an arbitrary point on the ellipse $\\frac{x^{2}}{9}+y^{2}=1$. Then the maximum value of $x+3 y$ is?", "fact_expressions": "P: Point;Coordinate(P) = (x1, y1);x1: Number;y1: Number;G: Ellipse;Expression(G) = (x^2/9 + y^2 = 1);PointOnCurve(P, G)", "query_expressions": "Max(x1 + 3*y1)", "answer_expressions": "3*sqrt(2)", "fact_spans": "[[[0, 10]], [[0, 10]], [[1, 10]], [[1, 10]], [[11, 38]], [[11, 38]], [[0, 44]]]", "query_spans": "[[[46, 59]]]", "process": "Test analysis: From the ellipse equation we get x^{2}+9y^{2}=9. Since 2(x^{2}+9y^{2})\\geqslant(x+3y)^{2}, it follows that \\sqrt{18}\\geqslant\\sqrt{(x+3y)^{2}}=x+3y, with equality holding if and only if x=3y, that is, when x=\\frac{3\\sqrt{2}}{2}, y=\\frac{\\sqrt{2}}{2}, x+3y attains the maximum value 3\\sqrt{2}. Therefore, the answer should be filled in as: 3\\sqrt{2}" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, the left and right foci are $F_{1}$ and $F_{2}$ respectively. A line passing through point $F_{2}$ and perpendicular to the $x$-axis intersects the hyperbola at a point $P$. Given that $\\angle P F_{1} F_{2}=\\frac{\\pi}{6}$, find the equation of the asymptotes of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;F1: Point;F2: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(F2, H);IsPerpendicular(H, xAxis);OneOf(Intersection(H, G)) = P;AngleOf(P, F1, F2) = pi/6", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*sqrt(2)*x", "fact_spans": "[[[2, 48], [102, 105], [155, 158]], [[5, 48]], [[5, 48]], [[93, 95], [99, 101]], [[111, 114]], [[57, 64]], [[67, 74], [76, 84]], [[2, 48]], [[2, 74]], [[2, 74]], [[75, 95]], [[85, 95]], [[99, 114]], [[117, 153]]]", "query_spans": "[[[155, 166]]]", "process": "" }, { "text": "Let $P$ be a point on the ellipse $C$, and let $F_1$, $F_2$ be the two foci. Given $|P F_1| = 13$, $|P F_2| = 15$, and $\\tan \\angle P F_1 F_2 = \\frac{12}{5}$, find the eccentricity $e$ of the ellipse $C$.", "fact_expressions": "C: Ellipse;P: Point;F1: Point;F2: Point;e: Number;PointOnCurve(P, C);Focus(C) = {F1, F2};Abs(LineSegmentOf(P, F1)) = 13;Abs(LineSegmentOf(P, F2)) = 15;Tan(AngleOf(P, F1, F2)) = 12/5;Eccentricity(C) = e", "query_expressions": "e", "answer_expressions": "1/2", "fact_spans": "[[[4, 9], [106, 111]], [[0, 3]], [[13, 20]], [[21, 28]], [[115, 118]], [[0, 12]], [[4, 32]], [[33, 47]], [[48, 62]], [[64, 104]], [[106, 118]]]", "query_spans": "[[[115, 120]]]", "process": "Given that $\\tan\\angle PF_{1}F_{2}=\\frac{12}{5}$, $\\cos\\angle PF_{1}F_{2}=\\frac{5}{13}$, by the law of cosines we have $|PF_{2}|^{2}=|PF_{1}|^{2}+|F_{1}F_{2}|^{2}-2|PF_{1}|\\times|F_{1}F_{2}|\\times\\cos\\angle PF_{1}F_{2}$, so $15^{2}=13^{2}+|F_{1}F_{2}|^{2}-2\\times13\\times|F_{1}F_{2}|\\times\\frac{5}{13}$, therefore $|F_{1}F_{2}|=14$, and $2a=|PF_{1}|+|PF_{2}|=28$, thus the eccentricity of ellipse $C$ is $e=\\frac{1}{2}$." }, { "text": "Given the parabola $y^{2}=4x$, $F$ is the focus of the parabola, and line $l$ intersects the parabola at points $A$ and $B$. If $|AF|+|BF|=6$, then what is the horizontal coordinate of the midpoint of segment $AB$?", "fact_expressions": "l: Line;G: Parabola;B: Point;A: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;Intersection(l, G) = {A, B};Abs(LineSegmentOf(A, F)) + Abs(LineSegmentOf(B, F)) = 6", "query_expressions": "XCoordinate(MidPoint(LineSegmentOf(A,B)))", "answer_expressions": "2", "fact_spans": "[[[30, 35]], [[2, 16], [23, 26], [36, 39]], [[45, 48]], [[41, 44]], [[19, 22]], [[2, 16]], [[19, 29]], [[30, 50]], [[52, 67]]]", "query_spans": "[[[69, 84]]]", "process": "\\because the parabola y^{2}=4x, \\therefore p=2, the line l intersects the parabola at points A and B, with their x-coordinates denoted as x_{1}, x_{2}. Since |AF|+|BF|=6, by using the definition of the parabola, we have x_{1}+\\frac{p}{2}+x_{2}+\\frac{p}{2}=x_{1}+x_{2}+p=6, so x_{1}+x_{2}=4. Therefore, the x-coordinate of the midpoint of AB is x_{0}=\\frac{1}{2}(x_{1}+x_{2})=2" }, { "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=2x$, and one of its foci coincides with the focus of the parabola $y^{2}=20x$, 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;Expression(OneOf(Asymptote(G))) = (y = 2*x);H: Parabola;Expression(H) = (y^2 = 20*x);OneOf(Focus(G)) = Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/5-y^2/20=1", "fact_spans": "[[[2, 58], [75, 76], [104, 107]], [[2, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[5, 58]], [[2, 74]], [[82, 97]], [[82, 97]], [[75, 102]]]", "query_spans": "[[[104, 112]]]", "process": "Analysis: Using the equation of the asymptotes of the hyperbola, we obtain $\\frac{b}{a}=2$. Since the focus coincides with the focus of the parabola $y^{2}=20x$, we can determine $c$, and thus obtain the conclusion. Detailed solution: From the problem we have $\\frac{b}{a}=2$, $c=5$, and from $c^{2}=a^{2}+b^{2}$ we get $a^{2}=5$, $b^{2}=20$. Hence, the equation of the hyperbola is $\\frac{x^{2}}{5}-\\frac{y^{2}}{20}=1$." }, { "text": "If real numbers $x$, $y$ satisfy $\\frac{x^{2}}{2}+y^{2}=1$ and $x+y 0$), draw a straight line intersecting the parabola at points $P$ and $Q$. If the lengths of segments $PF$ and $FQ$ are $p$ and $q$ respectively, then $\\frac{1}{p} + \\frac{1}{q}$ equals?", "fact_expressions": "G: Parabola;a: Number;H: Line;F: Point;P: Point;Q: Point;a>0;Expression(G) = (y = a*x^2);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {P, Q};p:Number;q:Number;Length(LineSegmentOf(P,F))=p;Length(LineSegmentOf(F,Q))=q", "query_expressions": "1/q + 1/p", "answer_expressions": "4*a", "fact_spans": "[[[1, 20], [31, 34]], [[4, 20]], [[28, 30]], [[23, 26]], [[35, 38]], [[39, 42]], [[4, 20]], [[1, 20]], [[1, 26]], [[0, 30]], [[28, 44]], [[63, 66]], [[67, 71]], [[46, 72]], [[46, 72]]]", "query_spans": "[[[74, 102]]]", "process": "The standard equation of the parabola is $x^{2}=\\frac{1}{a}y$, with focus $F(0,\\frac{1}{4a})$. Let points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$. Suppose the equation of line $PQ$ is $y=kx+\\frac{1}{4a}$. Solving the system\n\\[\n\\begin{cases}\ny=kx+\\frac{1}{4a} \\\\\ny=ax^{2}\n\\end{cases}\n\\]\ngives $ax^{2}-kx-\\frac{1}{4a}=0$. By Vieta's formulas, we obtain $x_{1}+x_{2}=\\frac{k}{a}$, $x_{1}x_{2}=-\\frac{1}{4a^{2}}$. From the definition of the parabola, we have" }, { "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}$ and $F_{2}$ respectively. Point $M$ lies on an asymptote of $C$, with $M F_{1} \\perp M F_{2}$ and $|M F_{1}|=2 b+|M F_{2}|$. Find $\\frac{b^{2}}{a^{2}}$=?", "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;PointOnCurve(M, Asymptote(C));IsPerpendicular(LineSegmentOf(M, F1), LineSegmentOf(M, F2));Abs(LineSegmentOf(M, F1)) = 2*b + Abs(LineSegmentOf(M, F2))", "query_expressions": "b^2/a^2", "answer_expressions": "(sqrt(5)-1)/2", "fact_spans": "[[[2, 63], [93, 96]], [[10, 63]], [[10, 63]], [[88, 92]], [[72, 79]], [[80, 87]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 87]], [[2, 87]], [[88, 101]], [[103, 126]], [[127, 152]]]", "query_spans": "[[[154, 177]]]", "process": "Without loss of generality, assume point M is in the first quadrant. Let |MF₁| = m, |MF₂| = n, then m = 2b + n, and MF₁ ⊥ MF₂, hence m² + n² = 4c². Solving these two equations together yields mn = 2c² - 2b². Also, point C lies on a circle centered at the origin with radius c. Solving the system \n\\begin{cases} y = \\frac{b}{a}x \\\\ x^{2} + y^{2} = c^{2} \\end{cases} \ngives M(a, b). By the triangle area formula, we obtain \\frac{1}{2}mn = \\frac{1}{2} \\cdot 2cb, that is, c² - b² = cb, hence a² = bc, so a⁴ = b²c², thus a⁴ = b²(a² + b²), therefore b⁴ + a²b² - a⁴ = 0, then (\\frac{b}{a})⁴ + (\\frac{b}{a})² - 1 = 0, solving gives \\frac{b^{2}}{a^{2}} = \\frac{\\sqrt{5}-1}{2}. This problem examines geometric properties of hyperbolas and is of medium difficulty." }, { "text": "Given the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ has left and right foci $F_{1}$ and $F_{2}$ respectively, and point $P$ lies on the line $l$: $x-\\sqrt{3} y+8+2 \\sqrt{3}=0$. When $\\angle F_{1} P F_{2}$ takes its maximum value, $\\frac{|P F_{1}|}{|P F_{2}|}$=?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/16 + y^2/4 = 1);F1: Point;F2: Point;LeftFocus(G) = F1;RightFocus(G) = F2;l: Line;Expression(l) = (x - sqrt(3)*y + 8 + 2*sqrt(3) = 0);P: Point;PointOnCurve(P, l);WhenMax(AngleOf(F1, P, F2))", "query_expressions": "Abs(LineSegmentOf(P, F1))/Abs(LineSegmentOf(P, F2))", "answer_expressions": "sqrt(3)-1", "fact_spans": "[[[2, 40]], [[2, 40]], [[49, 56]], [[57, 64]], [[2, 64]], [[2, 64]], [[70, 106]], [[70, 106]], [[65, 69]], [[65, 107]], [[108, 136]]]", "query_spans": "[[[137, 168]]]", "process": "First, from the equation of the ellipse \\frac{x^2}{16} + \\frac{y^2}{4} = 1, we obtain its left and right foci as F_{1}(-2\\sqrt{3},0) and F_{2}(2\\sqrt{3},0). As shown in the figure, according to plane geometry knowledge, when \\angle F_{1}PF_{2} takes its maximum value, the circle passing through F_{1} and F_{2} is tangent to the line l. Find the coordinates of the center of the circle, then use the properties of similar triangles to derive \\frac{|PF_{1}|}{|PF_{2}|} = \\frac{PB}{BF_{2}}, and finally use the similarity ratio to find the answer. [Solution] The left and right foci of the ellipse \\frac{x^2}{16} + \\frac{y^2}{4} = 1 are F_{1}(-2\\sqrt{3},0) and F_{2}(2\\sqrt{3},0). As shown in the figure, according to plane geometry knowledge, when \\angle F_{1}PF_{2} takes its maximum value, the circle passing through F_{1} and F_{2} is tangent to the line l, and at this time the center lies on the y-axis with coordinates A(0,2). In the line l: x - \\sqrt{3}y + 8 + 2\\sqrt{3} = 0, setting y = 0 gives the coordinates of B: B(-8 - 2\\sqrt{3}, 0). In triangles BPF_{1} and BF_{2}P, \\angle BPF_{1} = \\angle BF_{2}P, \\therefore \\triangle BPF_{1} \\sim \\triangle BF_{2}P, \\therefore \\frac{|PF_{1}|}{|PF_{2}|} = \\frac{PB}{BF_{2}} = \\frac{\\sqrt{AB^{2} - PA^{2}}}{BO + OF_{2}} = \\sqrt{3} - 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. Let $P$ be a point on the right branch of the hyperbola. The circle $E$: $x^{2}+(y-\\frac{1}{2})^{2}=\\frac{49}{4}$, having $P F_{1}$ as its diameter, passes through the point $F_{2}$. Then the equation of the hyperbola is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;E: Circle;P: Point;F1: Point;F2: Point;a>0;b>0;Expression(C)=(-y^2/b^2 + x^2/a^2=1);Expression(E)=(x^2 + (y-1/2)^2=49/4);LeftFocus(C)=F1;RightFocus(C)=F2;PointOnCurve(P,RightPart(C));IsDiameter(LineSegmentOf(P,F1),E);PointOnCurve(F2,E)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/9-y^2/3=1", "fact_spans": "[[[2, 63], [90, 93], [168, 171]], [[9, 63]], [[9, 63]], [[111, 157]], [[86, 89]], [[70, 77]], [[78, 85], [158, 166]], [[9, 63]], [[9, 63]], [[2, 63]], [[111, 157]], [[2, 85]], [[2, 85]], [[86, 96]], [[97, 157]], [[111, 166]]]", "query_spans": "[[[168, 175]]]", "process": "According to the problem, E(0,\\frac{1}{2}), |EF_{1}|=\\frac{7}{2}, PF_{2}\\bot F_{1}F_{2}, so c=|OF_{1}|=\\sqrt{|EF_{1}|^{2}-|OE|^{2}}=2\\sqrt{3}, |PF_{1}|=2|EF_{1}|=7, |PF_{2}|=2|OE|=1. Since |PF_{1}|-|PF_{2}|=2a=6, it follows that a=3. Also, c^{2}=a^{2}+b^{2}, so b^{2}=3. Therefore, the equation of the hyperbola is \\frac{x^{2}}{9}-\\frac{y^{2}}{3}=1" }, { "text": "The coordinates of the focus of the parabola $y^{2}-x=0$ are? The equation of the directrix is?", "fact_expressions": "G: Parabola;Expression(G) = (-x + y^2 = 0)", "query_expressions": "Coordinate(Focus(G));Expression(Directrix(G))", "answer_expressions": "(1/4,0)\nx=-1/4", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]], [[0, 27]]]", "process": "" }, { "text": "Let $P$ be a moving point on the hyperbola $C$: $\\frac{x^{2}}{2}-y^{2}=1$. The distances from point $P$ to the two asymptotes of $C$ are $d_{1}$ and $d_{2}$, respectively. Then the minimum value of $3 d_{1}+d_{2}$ is?", "fact_expressions": "C: Hyperbola;P:Point;Expression(C) = (x^2/2 - y^2 = 1);PointOnCurve(P, C);l1:Line;l2:Line;Asymptote(C)={l1,l2};Distance(P,l1)=d1;Distance(P,l2)=d2;d1:Number;d2:Number", "query_expressions": "Min(3*d1 + d2)", "answer_expressions": "2*sqrt(2)", "fact_spans": "[[[5, 38], [50, 53]], [[1, 4], [45, 49]], [[5, 38]], [[1, 44]], [], [], [[50, 59]], [[45, 80]], [[45, 80]], [[65, 72]], [[73, 80]]]", "query_spans": "[[[82, 103]]]", "process": "P(x,y) is any point on the hyperbola. Find d_{1}, d_{2}, and then find 3d_{1}+d_{2}. Using the fact that the point lies on the hyperbola, \\frac{x^{2}}{2}-y^{2}=1, i.e., (x+\\sqrt{2}y)(x-\\sqrt{2}y)=2, apply the basic inequality to obtain the minimum value. The asymptotes of the hyperbola are x\\pm\\sqrt{2}y=0. Let P(x,y) be any point on the hyperbola. Without loss of generality, let d_{1}=\\frac{|x-\\sqrt{2}y|}{\\sqrt{3}}, d_{2}=\\frac{|x+\\sqrt{2}y|}{\\sqrt{3}}, 3d. Since P(x,y) lies on the hyperbola, \\therefore \\frac{x^{2}}{2}-y^{2}=1, so 3d_{1}+d_{2}=\\frac{3|x-\\sqrt{2}y|+|x+\\sqrt{2}}{\\sqrt{3}}\\frac{2y)}{2}=2\\sqrt{2}, when 3|x-\\sqrt{2}y|=|x+\\sqrt{2}y|, i.e., \\begin{cases}x=\\frac{2\\sqrt{6}}{3}\\\\y=\\frac{\\sqrt{3}}{3}\\end{cases} or \\begin{cases}x=-\\frac{2\\sqrt{6}}{3}\\\\y=-\\frac{\\sqrt{3}}{3}\\end{cases}, equality holds. \\cdot The minimum value of 3d_{1}+d_{2} is 2\\sqrt{2}." }, { "text": "If the foci of an ellipse lie on the $x$-axis, and the ellipse passes through the point $(1, \\frac{3}{2})$ with a focal distance of $2$, then what is the standard equation of the ellipse?", "fact_expressions": "PointOnCurve(Focus(G), xAxis) = True;H: Point;Coordinate(H) = (1, 3/2);PointOnCurve(H, G) = True;G: Ellipse;FocalLength(G) = 2", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4+y^2/3=1", "fact_spans": "[[[1, 33]], [[11, 30]], [[11, 30]], [[10, 33]], [[31, 33], [41, 43]], [[31, 39]]]", "query_spans": "[[[41, 50]]]", "process": "Problem Analysis: According to the given conditions, the equation of the ellipse can be written as \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1, (a>b>0), where c=1 and the coordinates of the foci are (-1,0), (1,0). \\therefore b^{2}=a^{2}-c^{2}=3, so the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1" }, { "text": "Through the focus of the parabola $y^{2}=2x$, two mutually perpendicular chords $AB$ and $CD$ are drawn, and $|AB|+|CD|=\\lambda|AB| \\cdot |CD|$. Then the value of $\\lambda$ is?", "fact_expressions": "G: Parabola;A: Point;B: Point;C: Point;D: Point;lambda:Number;Expression(G) = (y^2 = 2*x);PointOnCurve(Focus(G),LineSegmentOf(A,B));PointOnCurve(Focus(G),LineSegmentOf(C,D));IsChordOf(LineSegmentOf(A,B),G);IsChordOf(LineSegmentOf(C,D),G);IsPerpendicular(LineSegmentOf(A,B),LineSegmentOf(C,D));Abs(LineSegmentOf(A, B)) + Abs(LineSegmentOf(C, D)) = (lambda*Abs(LineSegmentOf(A, B)))*Abs(LineSegmentOf(C, D))", "query_expressions": "lambda", "answer_expressions": "1/2", "fact_spans": "[[[1, 15]], [[27, 32]], [[27, 32]], [[34, 39]], [[34, 39]], [[80, 89]], [[1, 15]], [[0, 39]], [[0, 39]], [[1, 39]], [[1, 39]], [[19, 39]], [[41, 78]]]", "query_spans": "[[[80, 93]]]", "process": "The focus is at $ F\\left(\\frac{1}{2},0\\right) $, $ p=1 $. Let the inclination angle of line $ AB $ be $ \\theta $, and the inclination angle of line $ CD $ be $ \\theta+\\frac{\\pi}{2} $ or $ \\theta-\\frac{\\pi}{2} $. Clearly, when $ \\theta \\neq \\frac{\\pi}{2} $, let $ k = \\tan\\theta $, then the equation of line $ AB $ is $ y = k\\left(x - \\frac{1}{2}\\right) $. Let $ A(x_{1},y_{1}) $, $ B(x_{2},y_{2}) $. From\n$$\n\\begin{cases}\ny = k\\left(x - \\frac{1}{2}\\right) \\\\\ny^2 = 2x\n\\end{cases}\n$$\nwe obtain $ k^{2}x^{2} - (k^{2}+2)x + \\frac{k^{2}}{4} = 0 $, then $ x_{1} + x_{2} = \\frac{k^{2}+2}{k^{2}} $, $ x_{1}x_{2} = \\frac{1}{4} $. Therefore,\n$$\n|AB| = |AF| + |BF| = x_{1} + \\frac{1}{2} + x_{2} + \\frac{1}{2} = \\frac{k^{2}+2}{k^{2}} + 1 = 2 + \\frac{2}{k^{2}} = 2 + \\frac{2\\cos^{2}\\theta}{\\sin^{2}\\theta} = \\frac{2}{\\sin^{2}\\theta}\n$$\n$$\n|AB| = \\frac{2p}{\\sin^{2}\\theta},\\quad |CD| = \\frac{2p}{\\sin^{2}\\left(\\theta \\pm \\frac{\\pi}{2}\\right)} = \\frac{2p}{\\cos^{2}\\theta}.\n$$\n$$\n\\therefore \\lambda = \\frac{1}{|AB|} + \\frac{1}{|CD|} = \\frac{\\sin^{2}\\theta}{2p} + \\frac{\\cos^{2}\\theta}{2p} = \\frac{1}{2p} = \\frac{1}{2}\n$$" }, { "text": "Given that the focus of the parabola $x^{2}=m y$ has coordinates $(0,2)$, what is the value of $m$?", "fact_expressions": "G: Parabola;m:Number;Coordinate(Focus(G)) = (0, 2);Expression(G)=(x**2=m*y)", "query_expressions": "m", "answer_expressions": "8", "fact_spans": "[[[2, 5]], [[33, 36]], [[2, 31]], [[2, 18]]]", "query_spans": "[[[33, 40]]]", "process": "Since the focus of the parabola $ x^{2} = my $ is at $ (0, 2) $, we have $ 2 = \\frac{m}{4} $, solving gives $ m = 8 $." }, { "text": "The foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$ and $F_{2}$, and point $P$ lies on the ellipse. If $|PF_{1}|=4$, then what is $|PF_{2}|$, and what is the measure of $\\angle F_{1} PF_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4", "query_expressions": "Abs(LineSegmentOf(P, F2));AngleOf(F1, P, F2)", "answer_expressions": "2\nApplyUnit(120, degree)", "fact_spans": "[[[0, 37], [64, 66]], [[0, 37]], [[41, 49]], [[50, 58]], [[0, 58]], [[59, 63]], [[59, 67]], [[70, 82]]]", "query_spans": "[[[85, 97]], [[99, 125]]]", "process": "" }, { "text": "Given that the equation $\\frac{x^{2}}{a}+\\frac{y^{2}}{2-a}=1$ represents a hyperbola with foci on the $y$-axis, what is the range of values for $a$?", "fact_expressions": "G: Hyperbola;Expression(G) = (y^2/(2 - a) + x^2/a = 1);PointOnCurve(Focus(G), yAxis);a:Number", "query_expressions": "Range(a)", "answer_expressions": "(-oo,0)", "fact_spans": "[[[52, 55]], [[2, 55]], [[43, 55]], [[57, 60]]]", "query_spans": "[[[57, 67]]]", "process": "Since the foci of the hyperbola lie on the y-axis, we have \\begin{cases}a<0\\\\2-a>0\\end{cases}, thus a<0, so the range of values for a is (-\\infty,0)." }, { "text": "The focal distance of the ellipse $\\frac{x^{2}}{10}+\\frac{y^{2}}{6}=1$ is equal to?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/10 + y^2/6 = 1)", "query_expressions": "FocalLength(G)", "answer_expressions": "4", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 44]]]", "process": "According to the problem, for the ellipse $\\frac{x^{2}}{10}+\\frac{y^{2}}{6}=1$, we have $a^{2}=10$, $b^{2}=6$, then $c=\\sqrt{a^{2}-b^{2}}=2$, so the focal distance of the ellipse is $2c=4$." }, { "text": "Given that point $M$ lies on the circle $(x-6)^{2}+(y-4)^{2}=1$, point $P$ lies on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $F(-3,0)$, then the minimum value of $|P M|-|P F|$ is?", "fact_expressions": "M: Point;H: Circle;Expression(H) = ((x - 6)^2 + (y - 4)^2 = 1);PointOnCurve(M, H);P: Point;G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);PointOnCurve(P, G);F: Point;Coordinate(F) = (-3, 0)", "query_expressions": "Min(-Abs(LineSegmentOf(P, F)) + Abs(LineSegmentOf(P, M)))", "answer_expressions": "-6", "fact_spans": "[[[2, 6]], [[7, 31]], [[7, 31]], [[2, 32]], [[33, 37]], [[38, 77]], [[38, 77]], [[33, 78]], [[79, 88]], [[79, 88]]]", "query_spans": "[[[90, 109]]]", "process": "Analysis: According to the problem, when points P, C, F are collinear, |PM| - |PF||PM| - |PF| = |PC| - 1 - (2a - |PF|) = |PF| + |PF| - 11 \\geqslant |CF| - 11 = 5 - 11 = -6" }, { "text": "Let $F_{1}$ and $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)$, respectively. A line passing through $F_{1}$ intersects the left branch of hyperbola $C$ at points $A$ and $B$, with $|A F_{2}|=3$, $|B F_{2}|=5$, and $|A B|=4$. Then the area of $\\Delta B F_{1} 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;F1: Point;F2: Point;LeftFocus(C) = F1;RightFocus(C) = F2;G: Line;PointOnCurve(F1, G);A: Point;B: Point;Intersection(G, LeftPart(C)) = {A, B};Abs(LineSegmentOf(A, F2)) = 3;Abs(LineSegmentOf(B, F2)) = 5;Abs(LineSegmentOf(A, B)) = 4", "query_expressions": "Area(TriangleOf(B, F1, F2))", "answer_expressions": "9/2", "fact_spans": "[[[19, 80], [99, 105]], [[19, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[27, 80]], [[1, 8], [88, 95]], [[9, 16]], [[1, 86]], [[1, 86]], [[96, 98]], [[87, 98]], [[109, 112]], [[113, 116]], [[96, 118]], [[120, 133]], [[134, 147]], [[148, 157]]]", "query_spans": "[[[159, 186]]]", "process": "\\because|AF_{2}|=3,BF_{2}|=5, also |AF_{2}|\\cdot|AF_{1}|=2a,|BF_{2}|\\cdot|BF_{1}|=2a,\\therefore|AF_{2}|+|BF_{2}|-|AB|=4a=3+5-4=4,\\thereforea=1,\\therefore|BF_{1}|=3 Also |AF_{2}|^{2}+|AB|^{2}=|BF_{2}|^{2}, then \\angleF_{2}AB=90^{\\circ},\\therefore\\cosB=\\frac{4}{5},\\therefore\\sinB=\\frac{3}{5}\\thereforeS_{\\DeltaBF_{1}F_{2}}=\\frac{1}{2}\\times5\\times3\\times\\sinB==\\frac{1}{2}\\times5\\times3\\times\\frac{3}{5}=\\frac{9}{2} Therefore the answer. This problem mainly examines the definition of hyperbola and the triangle area formula, it is a medium-level problem." }, { "text": "A focus of the hyperbola $8 k x^{2}-k y^{2}=8$ is $(0,3)$, then what is 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, 24]], [[40, 43]], [[0, 24]], [[0, 37]]]", "query_spans": "[[[40, 47]]]", "process": "" }, { "text": "If one asymptote of the hyperbola $x^{2}-\\frac{y^{2}}{b^{2}}=1(b>0)$ has at most one common point with the circle $x^{2}+(y-2)^{2}=1$, then the range of the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;H: Circle;b>0;Expression(G) = (x^2 - y^2/b^2 = 1);Expression(H) = (x^2 + (y - 2)^2 = 1);NumIntersection(OneOf(Asymptote(G)), H)<=1", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(1,2]", "fact_spans": "[[[1, 38], [75, 78]], [[4, 38]], [[45, 65]], [[4, 38]], [[1, 38]], [[45, 65]], [[1, 73]]]", "query_spans": "[[[75, 88]]]", "process": "The asymptotes of the hyperbola are given by $ y = \\pm bx $, then $ \\frac{|0 - 2|}{\\sqrt{1 + b^{2}}} \\geqslant $ solving yields $ b^{2} \\leqslant 3 $, then $ e^{2} = 1 + b^{2} \\leqslant 4 $, $ \\because e > 1 $, $ < e \\leqslant 2 $" }, { "text": "If the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3^{2}}=1$ $(a>0)$ has eccentricity $2$, then what is the equation of its asymptotes?", "fact_expressions": "G: Hyperbola;a: Number;a>0;Expression(G) = (-y^2/3^2 + x^2/a^2 = 1);Eccentricity(G) = 2", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*sqrt(3)*x", "fact_spans": "[[[1, 52], [62, 63]], [[4, 52]], [[4, 52]], [[1, 52]], [[1, 60]]]", "query_spans": "[[[62, 71]]]", "process": "" }, { "text": "The line $ l $ passing through $ M(-2,0) $ with slope $ \\frac{2}{3} $ intersects the parabola $ C: y^{2}=2 p x(p>0) $ at points $ A $ and $ B $, $ F $ is the focus of $ C $. If the area of $ \\Delta M F B $ is twice the area of $ \\Delta M F A $, then the value of $ p $ is?", "fact_expressions": "l: Line;C: Parabola;p: Number;M: Point;F: Point;B: Point;A: Point;p>0;Expression(C) = (y^2 = 2*(p*x));Coordinate(M) = (-2, 0);PointOnCurve(M, l);Slope(l) = 2/3;Intersection(l, C) = {A, B};Focus(C)=F;Area(TriangleOf(M,F,B))=2*Area(TriangleOf(M,F,A))", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[28, 33]], [[34, 60], [75, 78]], [[126, 129]], [[1, 10]], [[71, 74]], [[65, 68]], [[61, 64]], [[42, 60]], [[34, 60]], [[1, 10]], [[0, 33]], [[11, 33]], [[28, 70]], [[71, 81]], [[83, 124]]]", "query_spans": "[[[126, 133]]]", "process": "" }, { "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)$, and $P$ is a point on the left branch of the hyperbola, if $\\frac{|P F_{2}|^{2}}{|P F_{1}|}=8 a$, then the range of values for 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, 76], [86, 89], [86, 89]], [[23, 76]], [[23, 76]], [[82, 85]], [[10, 17]], [[2, 9]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 81]], [[2, 81]], [[82, 95]], [[97, 134]]]", "query_spans": "[[[136, 150]]]", "process": "" }, { "text": "Given that one focus of the ellipse $5 x^{2} + k y^{2} = 5$ is $(0, 2)$, find the value of the real number $k$.", "fact_expressions": "G: Ellipse;Expression(G) = (k*y^2 + 5*x^2 = 5);k: Real;F: Point;OneOf(Focus(G)) = F;Coordinate(F) = (0, 2)", "query_expressions": "k", "answer_expressions": "1", "fact_spans": "[[[2, 23]], [[2, 23]], [[40, 45]], [[29, 38]], [[2, 38]], [[29, 38]]]", "query_spans": "[[[40, 49]]]", "process": "" }, { "text": "Given that the distance from the point $M(1, m)$ on the parabola $E$: $y^{2}=2 p x$ to its focus is $2$, then the standard equation of the parabola is?", "fact_expressions": "E: Parabola;Expression(E) = (y^2 = 2*p*x);p: Number;M: Point;Coordinate(M) = (1, m);m: Number;PointOnCurve(M, E) = True;Distance(M, Focus(E)) = 2", "query_expressions": "Expression(E)", "answer_expressions": "y^2 = 4*x", "fact_spans": "[[[2, 22], [35, 36], [48, 51]], [[2, 22]], [[9, 22]], [[24, 34]], [[24, 34]], [[25, 34]], [[2, 34]], [[24, 45]]]", "query_spans": "[[[48, 58]]]", "process": "Since the distance from point M(1, m) on the parabola E: y^{2}=2px to its focus is 2, we have 1+\\frac{p}{2}=2, \\therefore p=2, so the standard equation of the parabola is y^{2}=4x" }, { "text": "Given the ellipse $ C $: $ \\frac{x^{2}}{2} + y^{2} = 1 $ with foci $ F_{1} $, $ F_{2} $. Point $ P(x_{0}, y_{0}) $ satisfies $ \\frac{x_{0}^{2}}{2} + y_{0}^{2} \\leq 1 $. Then the range of values for $ PF_{1} + PF_{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);x0^2/2 + y0^2 <= 1", "query_expressions": "Range(LineSegmentOf(P, F1) + LineSegmentOf(P, F2))", "answer_expressions": "[2, 2*sqrt(2)]", "fact_spans": "[[[2, 34]], [[2, 34]], [[39, 46]], [[48, 55]], [[2, 55]], [[56, 75]], [[57, 75]], [[57, 75]], [[56, 75]], [[77, 115]]]", "query_spans": "[[[117, 139]]]", "process": "" }, { "text": "Let the semi-focal length of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be $c$. The distance from the origin to the line $l$: $a x+b y=a b$ is equal to $\\frac{1}{3} c+1$. Then the minimum value of $c$ is?", "fact_expressions": "l: Line;G: Hyperbola;O: Origin;c: Number;a: Number;b: Number;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);HalfFocalLength(G) = c;Expression(l) = (a*x + b*y = a*b);Distance(O, l) = (1/3)*c + 1", "query_expressions": "Min(c)", "answer_expressions": "6", "fact_spans": "[[[69, 89]], [[1, 57]], [[66, 68]], [[62, 65], [113, 116]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 65]], [[69, 89]], [[66, 111]]]", "query_spans": "[[[113, 122]]]", "process": "Test Analysis: According to the given condition, the distance from the origin O to the line l: ax + by = ab is d = \\frac{ab}{\\sqrt{a^{2}+b^{2}}} = \\frac{ab}{c} = 1 + \\frac{1}{3}c, that is, c^{2} + 3c = 3ab. Since ab \\leqslant \\frac{a^{2}+b^{2}}{2} (equality holds if and only if a = b), it follows that c^{2} + 3c = 3ab \\leqslant \\frac{3}{2}(a^{2}+b^{2}), i.e., c^{2} + 3c \\leqslant \\frac{3}{2}c^{2}. Solving this inequality yields c \\geqslant 6, so the minimum value is 6." }, { "text": "The ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, through the right focus $F$, draw a line $l$ intersecting the ellipse at points $P$ and $Q$, with $P$ in the second quadrant. Given $Q(x_{Q}, y_{Q})$, $Q^{\\prime}(x_{Q'}, y_{Q'})$ are both on the ellipse, and $y_{Q}+y_{Q'}=0$, $F Q^{\\prime} \\perp P Q$, then the equation of line $l$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F: Point;RightFocus(G) = F;l: Line;PointOnCurve(F, l) = True;Intersection(l, G) = {P, Q};Q: Point;P: Point;Quadrant(P) = 2;Q1: Point;PointOnCurve(Q, G) = True;PointOnCurve(Q1, G) = True;Coordinate(Q) = (xQ, yQ);xQ: Number;yQ: Number;Coordinate(Q1) = (xQ1, yQ1);xQ1: Number;yQ1: Number;yQ + yQ1 = 0;IsPerpendicular(LineSegmentOf(F, Q1), LineSegmentOf(P, Q)) = True", "query_expressions": "Expression(l)", "answer_expressions": "y=-x+1", "fact_spans": "[[[0, 37], [52, 54], [124, 126]], [[0, 37]], [[42, 45]], [[0, 45]], [[46, 51], [174, 179]], [[38, 51]], [[46, 64]], [[59, 62], [75, 92]], [[55, 58], [65, 68]], [[65, 73]], [[94, 122]], [[75, 127]], [[75, 127]], [[75, 92]], [[75, 92]], [[75, 92]], [[94, 122]], [[94, 122]], [[94, 122]], [[129, 145]], [[148, 172]]]", "query_spans": "[[[174, 184]]]", "process": "Given that $ Q(x_{Q}, y_{Q}) $ and $ Q'(x'_{Q}, y'_{Q}) $ both lie on the ellipse, and $ y_{Q} + y'_{Q} = 0 $, therefore $ Q $ and $ Q' $ are symmetric with respect to the x-axis. Furthermore, since $ FQ' \\perp PQ $, it follows that $ \\angle QFx = 45^{\\circ} $. Draw a line $ l $ through the right focus $ F $ intersecting the ellipse at points $ P $ and $ Q $, with $ P $ in the second quadrant. Therefore, the inclination angle of line $ l $ is $ 135^{\\circ} $, $ \\tan 135^{\\circ} = -1 $, so the slope of line $ l $ is $ -1 $, and it passes through point $ F(1, 0) $. Thus, the equation of line $ l $ is $ y = -(x - 1) $, or $ y = -x + 1 $." }, { "text": "Given that the directrix of the parabola $y^{2}=2 p x$ coincides with the left directrix of the hyperbola $x^{2}-y^{2}=2$, find the value of $p$.", "fact_expressions": "G: Hyperbola;H: Parabola;p: Number;Expression(G) = (x^2 - y^2 = 2);Expression(H) = (y^2 = 2*(p*x));Directrix(H) = LeftDirectrix(G)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[22, 40]], [[2, 18]], [[48, 51]], [[22, 40]], [[2, 18]], [[2, 46]]]", "query_spans": "[[[48, 55]]]", "process": "The directrix of the parabola \\( y^{2} = 2px \\) is: \\( x = -\\frac{p}{2} \\), the left directrix of the hyperbola \\( x^{2} - y^{2} = 2 \\) is: \\( x = -\\frac{a^{2}}{c} = -\\frac{(\\sqrt{2})^{2}}{2} \\). From the given condition, \\( -\\frac{p}{2} = -\\frac{(\\sqrt{2})^{2}}{2} \\), we obtain \\( p = 2 \\)." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $E$: $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, and $P$ is any point on the ellipse $E$, then the range of values of $\\overrightarrow{F_{1} P} \\cdot \\overrightarrow{F_{2} P}$ is?", "fact_expressions": "E: Ellipse;F1: Point;P: Point;F2: Point;Expression(E) = (x^2/4 + y^2/3 = 1);Focus(E) = {F1, F2};PointOnCurve(P, E)", "query_expressions": "Range(DotProduct(VectorOf(F1, P), VectorOf(F2, P)))", "answer_expressions": "[2, 3]", "fact_spans": "[[[18, 60], [70, 75]], [[2, 9]], [[66, 69]], [[10, 17]], [[18, 60]], [[2, 65]], [[66, 79]]]", "query_spans": "[[[81, 145]]]", "process": "From a^{2}=4, b^{2}=3, we obtain: c^{2}=a^{2}-b^{2}=1, so c=1. Without loss of generality, let F_{1}(-1,0), F_{2}(1,0). Since P is an arbitrary point on the ellipse E, let P(m,n) (-\\sqrt{3}\\leqslant n\\leqslant \\sqrt{3}), then \\frac{m^{2}}{4}+\\frac{n^{2}}{3}=1. \\overrightarrow{F_{1}P}\\cdot\\overrightarrow{F_{2}P}=(m+1,n)\\cdot(m-1,n)=m^{2}-1+n^{2}=3-\\frac{1}{3}n^{2}. Since -\\sqrt{3}\\leqslant n\\leqslant \\sqrt{3}, it follows that 0\\leqslant n^{2}\\leqslant 3, and 2\\leqslant 3-\\frac{1}{3}n^{2}\\leqslant 3. Therefore, the range of \\overrightarrow{F_{1}P}\\cdot\\overrightarrow{F_{2}P} is [2,3]." }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $A(2 , 0)$ is an endpoint of the major axis, chord $BC$ passes through the center $O$ of the ellipse, and $\\overrightarrow{A C} \\cdot \\overrightarrow{B C}=0$, $|\\overrightarrow{O C}-\\overrightarrow{O B}|=2|\\overrightarrow{B C}-\\overrightarrow{B A}|$. Then the equation of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;B: Point;C: Point;A: Point;O: Point;a > b;b > 0;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (2, 0);Center(G) = O;OneOf(Endpoint(MajorAxis(G)))=A;PointOnCurve(O, LineSegmentOf(B, C));IsChordOf(LineSegmentOf(B, C),G);DotProduct(VectorOf(A, C), VectorOf(B, C)) = 0;Abs(-VectorOf(O, B) + VectorOf(O, C)) = 2*Abs(-VectorOf(B, A) + VectorOf(B, C))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4+3*y^2/4=1", "fact_spans": "[[[2, 55], [83, 85], [237, 239]], [[4, 54]], [[4, 54]], [[77, 82]], [[77, 82]], [[57, 67]], [[88, 91]], [[4, 54]], [[4, 54]], [[2, 54]], [[57, 67]], [[83, 91]], [[2, 75]], [[77, 91]], [[76, 85]], [[93, 144]], [[145, 235]]]", "query_spans": "[[[237, 244]]]", "process": "\\because\\overrightarrow{AC}\\cdot\\overrightarrow{BC}=0,|\\overrightarrow{OB}-\\overrightarrow{OC}|=2|\\overrightarrow{BC}-\\overrightarrow{BA}|,\\therefore|BC|=2|AC|,AC\\bot BC.\\ By the structural characteristics of the ellipse, |OC|=|AC|.\\ \\because A(2,0) is an endpoint of the major axis, i.e., a=2,\\ \\therefore the x-coordinate of point C is 1, that is, C(1,\\pm1).\\ \\because point C lies on the ellipse \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1,\\ \\therefore b^{2}=\\frac{4}{3},\\ \\therefore the equation of the ellipse is \\frac{x^{2}}{4}+\\frac{3y^{2}}{4}=1." }, { "text": "Given the parabola $C$: $y^{2}=4x$ with focus $F$, a line $l$ passing through point $F$ intersects the parabola $C$ at points $A$ and $B$, and $|AF|=5$, then $|AB|$=?", "fact_expressions": "l: Line;C: Parabola;A: Point;F: Point;B: Point;Expression(C)=(y^2 = 4*x);Focus(C)=F;PointOnCurve(F, l);Intersection(l, C) = {A, B};Abs(LineSegmentOf(A, F)) = 5", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "25/4", "fact_spans": "[[[35, 40]], [[2, 21], [41, 47]], [[49, 52]], [[30, 34], [25, 28]], [[53, 56]], [[2, 21]], [[2, 28]], [[29, 40]], [[35, 58]], [[60, 69]]]", "query_spans": "[[[71, 80]]]", "process": "From the given conditions, we have F(1,0). Let A(m,n), then \n\\begin{cases}m+1=5\\\\n^{2}=4m\\end{cases} \nSolving gives \n\\begin{cases}m=4\\\\n=\\pm4\\end{cases} \nBy the symmetry of the parabola, without loss of generality, assume point A lies in the first quadrant, so A(4,4). Then the equation of line l is 4x-3y-4=0. Solving simultaneously \n\\begin{cases}4x-3y-\\\\y2=4x\\end{cases}-4=0 \nRearranging yields y^{2}-3y-4=0. Solving gives y=-1 or y=4, so B(\\frac{1}{4},-1). Hence |AB|=x_{A}+x_{B}+p=\\frac{1}{4}+4+2=\\frac{25}{4}" }, { "text": "Given that one focus of the ellipse $\\frac{x^{2}}{m}+\\frac{y^{2}}{3}=1$ $(m>0)$ is $(0,1)$, then $m=?$. If a point $P$ on the ellipse forms a triangle $P F_{1} F_{2}$ with the two foci $F_{1}$, $F_{2}$ of the ellipse and the area of the triangle is $\\sqrt{2}$, then what are the coordinates of point $P$?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/3 + x^2/m = 1);m: Number;m>0;T:Point;OneOf(Focus(G))=T;Coordinate(T) = (0, 1);PointOnCurve(P, G) ;P: Point;F1: Point;F2: Point;Focus(G) = {F1, F2};Area(TriangleOf(P, F1, F2)) = sqrt(2)", "query_expressions": "m;Coordinate(P)", "answer_expressions": "sqrt(2)\n(pm*sqrt(2), 0)", "fact_spans": "[[[2, 44], [65, 67], [74, 76]], [[2, 44]], [[59, 62]], [[4, 44]], [[50, 57]], [[2, 57]], [[50, 57]], [[65, 73]], [[70, 73], [133, 137]], [[81, 88]], [[89, 96]], [[74, 96]], [[99, 131]]]", "query_spans": "[[[59, 64]], [[133, 142]]]", "process": "Problem Analysis: From the given conditions, the foci lie on the y-axis, so a^{2}=3, b^{2}=m. From b^{2}=a^{2}-c^{2}=2, we get m=\\sqrt{2}; from S=\\frac{1}{2}|F_{1}F_{2}|\\times|x_{P}|=\\sqrt{2}, we get x_{P}=\\pm\\sqrt{2}. Substituting into the ellipse equation gives y_{P}=0. Therefore, the coordinates of point P are (\\pm\\sqrt{2},0)." }, { "text": "Given the parabola $y = a x^{2}$ ($a > 0$) with two moving points $A$ and $B$ (not at the origin) satisfying $\\overrightarrow{O A} \\perp \\overrightarrow{O B}$. If there exists a point $M$ such that $\\overrightarrow{O M} = \\lambda \\overrightarrow{O A} + \\mu \\overrightarrow{O B}$ and $\\lambda + \\mu = 1$, then what are the coordinates of $M$?", "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number;a>0;A: Point;B: Point;PointOnCurve(A, G);PointOnCurve(B, G);O: Origin;Negation(A=O);Negation(B=O);IsPerpendicular(VectorOf(O, A), VectorOf(O, B));M: Point;lambda: Number;mu: Number;VectorOf(O, M) = lambda*VectorOf(O, A) + mu*VectorOf(O, B);lambda + mu = 1", "query_expressions": "Coordinate(M)", "answer_expressions": "(0, 1/a)", "fact_spans": "[[[2, 21]], [[2, 21]], [[5, 21]], [[5, 21]], [[25, 28]], [[29, 32]], [[2, 32]], [[2, 32]], [[35, 37]], [[25, 38]], [[29, 38]], [[41, 90]], [[94, 98], [196, 199]], [[179, 194]], [[179, 194]], [[101, 177]], [[179, 194]]]", "query_spans": "[[[196, 203]]]", "process": "Test analysis: From $\\overrightarrow{OM}=\\lambda\\overrightarrow{OA}+\\mu\\overrightarrow{OB}$, and $\\lambda+\\mu=1$, we get $\\overrightarrow{OM}=\\lambda\\overrightarrow{OA}+(1-\\lambda)\\overrightarrow{OB}$, that is, $\\overrightarrow{BM}=\\lambda\\overrightarrow{BA}$, $\\therefore$ points $A$, $B$, $M$ are collinear. Thus, the problem is transformed into the moving line $AB$ passing through a fixed point. Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, then $y_{1}=ax_{1}^{2}$, $y_{2}=ax_{2}^{2}$. Subtracting these two equations gives $y_{1}-y_{2}=a(x_{1}+x_{2})\\cdot(x_{1}-x_{2})$, $\\therefore k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=a(x_{1}+x_{2})$. The equation of line $AB$ is $y-y_{1}=a(x_{1}+x_{2})(x-x_{1})$, that is, $y=a(x_{1}+x_{2})(x-x_{1})+y_{1}=a(x_{1}+x_{2})x-ax_{1}-ax_{1}x_{2}+ax_{1}^{2}=a(x_{1}+x_{2})x-ax_{1}x_{2}\\cdots(1)$. $\\because \\overrightarrow{OA}\\bot\\overrightarrow{OB}$, $\\therefore x_{1}x_{2}+y_{1}y_{2}=0$, that is, $x_{1}x_{2}+a^{2}x_{1}^{2}x_{2}^{2}=0$, or $a^{2}x_{1}x_{2}=-1\\cdots(2)$. Substituting (2) into (1) yields $y=a(x_{1}+x_{2})x+\\frac{1}{a}$. $\\therefore$ line $AB$ passes through the fixed point $(0,\\frac{1}{a})$, $\\therefore$ the coordinates of $M$ are $(0,\\frac{1}{a})$." }, { "text": "Let $P$ be 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 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": "sqrt(2)", "fact_spans": "[[[5, 19]], [[50, 58]], [[32, 41]], [[27, 31], [1, 4], [45, 49]], [[5, 19]], [[50, 58]], [[32, 41]], [[1, 25]]]", "query_spans": "[[[27, 69]]]", "process": "The directrix of $ y^{2} = 4x $ is $ x = -1 $. Therefore, the distance from $ P $ to $ x = -1 $ equals the distance from $ P $ to the focus $ F(1,0) $. Hence, the sum of the distance from point $ P $ to point $ A(0,1) $ and the distance from $ P $ to the line $ x = -1 $ equals $ |PA| + |PF| \\geqslant |AF| $, that is, 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 $ |FA| = \\sqrt{2} $." }, { "text": "If $P$ is the right vertex of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$, and a line passing through the origin $O$ intersects the ellipse at points $A$ and $B$, while lines $PA$ and $PB$ intersect the line $x=3$ at points $E$ and $F$ respectively, then the minimum value of $|EF|$ is?", "fact_expressions": "G: Ellipse;J: Line;H: Line;A: Point;P: Point;B: Point;E: Point;F: Point;O: Origin;Expression(G) = (x^2/4 + y^2/2 = 1);Expression(H) = (x = 3);RightVertex(G) = P;PointOnCurve(O, J);Intersection(J, G) = {A, B};Intersection(LineOf(P, A), H) = E;Intersection(LineOf(P, B), H) = F", "query_expressions": "Min(Abs(LineSegmentOf(E, F)))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[5, 42], [59, 61]], [[56, 58]], [[87, 94]], [[62, 65]], [[1, 4]], [[66, 69]], [[95, 98]], [[99, 102]], [[48, 55]], [[5, 42]], [[87, 94]], [[1, 46]], [[47, 58]], [[56, 71]], [[72, 104]], [[72, 104]]]", "query_spans": "[[[106, 119]]]", "process": "" }, { "text": "Given that the focus of the parabola $y^{2}=4x$ is $F$, and the point $P$ on the parabola has a horizontal coordinate of $2$, then $|PF|=$?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;P: Point;PointOnCurve(P, G);XCoordinate(P) = 2", "query_expressions": "Abs(LineSegmentOf(P, F))", "answer_expressions": "3", "fact_spans": "[[[2, 16], [26, 29]], [[2, 16]], [[20, 23]], [[2, 23]], [[30, 34]], [[26, 34]], [[30, 42]]]", "query_spans": "[[[45, 54]]]", "process": "The directrix of the parabola y^{2}=4x is given by the equation: x=-1. According to the definition of a parabola, |PF|=2-(-1)=3." }, { "text": "A line segment $AB$ of length $3$ has its endpoints $A$ and $B$ moving on the $x$-axis and $y$-axis respectively. A moving point $C(x, y)$ satisfies $\\overrightarrow{A C}=2 \\overrightarrow{C B}$. Then the trajectory equation of the moving point $C$ is?", "fact_expressions": "Length(LineSegmentOf(A,B)) = 3;Endpoint(LineSegmentOf(A,B)) = {A,B};A: Point;B: Point;PointOnCurve(A,xAxis) ;PointOnCurve(B,yAxis) ;C: Point;Coordinate(C) = (x1, y1);x1: Number;y1: Number;VectorOf(A,C) = 2*VectorOf(C,B)", "query_expressions": "LocusEquation(C)", "answer_expressions": "x^2 + y^2/4 = 1", "fact_spans": "[[[0, 13]], [[6, 23]], [[16, 19]], [[20, 23]], [[16, 36]], [[16, 36]], [[41, 50], [101, 104]], [[41, 50]], [[41, 50]], [[41, 50]], [[52, 97]]]", "query_spans": "[[[101, 111]]]", "process": "" }, { "text": "The line $ l $ intersects the hyperbola $ \\frac{x^{2}}{3}-\\frac{y^{2}}{2}=1 $ to form a chord of length $ 4 $, and its slope is $ 2 $. What is the $ y $-intercept of the line $ l $?", "fact_expressions": "l: Line;G: Hyperbola;Expression(G) = (x^2/3 - y^2/2 = 1);Length(InterceptChord(l,G))=4;Slope(l)=2", "query_expressions": "Intercept(l,yAxis)", "answer_expressions": "pm*sqrt(210)/3", "fact_spans": "[[[0, 5], [55, 56], [65, 70]], [[6, 44]], [[6, 44]], [[0, 53]], [[55, 62]]]", "query_spans": "[[[65, 81]]]", "process": "" }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$ are tangent to the circle $(x-3)^{2}+y^{2}=r^{2}(r>0)$. Find $r=?$", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/6 - y^2/3 = 1);H: Circle;Expression(H) = (y^2 + (x - 3)^2 = r^2);r: Number;r>0;IsTangent(Asymptote(G), H)", "query_expressions": "r", "answer_expressions": "sqrt(3)", "fact_spans": "[[[0, 38]], [[0, 38]], [[43, 72]], [[43, 72]], [[76, 79]], [[44, 72]], [[0, 74]]]", "query_spans": "[[[76, 81]]]", "process": "Test Analysis: One asymptote of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$ is $\\frac{\\sqrt{2}}{2}x-y=0$. The asymptotes of the hyperbola $\\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$ are tangent to the circle $(x-3)^{2}+y^{2}=r^{2}$ $(r>0)$, so $r=\\frac{|\\frac{\\sqrt{2}}{2}\\times3-0|}{\\sqrt{(\\frac{\\sqrt{2}}{2})^{2}+1^{2}}}\\sqrt{3}$." }, { "text": "The asymptotes of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{4}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/4 - y^2/4 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y = pm*x", "fact_spans": "[[[0, 38]], [[0, 38]]]", "query_spans": "[[[0, 46]]]", "process": "From the hyperbola equation, we know: a=2, b=2, so the equations of the asymptotes are: y=\\pm x" }, { "text": "The asymptotes of hyperbola $C $, centered at the origin and with coordinate axes as axes of symmetry, are both tangent to the circle $ (x-2)^{2}+y^{2}=1 $. What is the eccentricity of hyperbola $ C $?", "fact_expressions": "C: Hyperbola;G: Circle;O: Origin;Expression(G) = (y^2 + (x - 2)^2 = 1);Center(C) = O;SymmetryAxis(C)=axis;IsTangent(Asymptote(C),G)", "query_expressions": "Eccentricity(C)", "answer_expressions": "{2*sqrt(3)/3,2}", "fact_spans": "[[[14, 20], [52, 58]], [[27, 47]], [[3, 5]], [[27, 47]], [[0, 20]], [[6, 20]], [[14, 50]]]", "query_spans": "[[[52, 64]]]", "process": "Let y = kx be a tangent line to the circle (x−2)^2 + y^2 = 1, then \\frac{|2k|}{\\sqrt{k^2+1}} = 1, solving gives k = \\pm\\frac{\\sqrt{3}}{3}, so the tangent equations are y = \\pm\\frac{\\sqrt{3}}{3}x. These are also the asymptotes of a hyperbola. If the hyperbola is \\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1, then \\frac{b}{a} = \\frac{\\sqrt{3}}{3}, b = \\frac{\\sqrt{3}}{3}a, c = \\sqrt{a^{2}+b^{2}} = \\sqrt{a^2 + \\frac{1}{3}a^2} = \\frac{2\\sqrt{3}}{2}a, \\therefore e = \\frac{c}{a} = \\frac{2\\sqrt{3}}{\\frac{y}{a}} - \\frac{x^2}{b^2} = 1, then \\frac{a}{b} = \\frac{\\sqrt{3}}{3}, b = \\sqrt{3}a, c = \\sqrt{a^2 + b^2} = \\sqrt{a^{2}+3a^{2}} = 2a, \\therefore e = \\frac{c}{a} = 2." }, { "text": "Let the ellipse $\\frac{x^{2}}{6}+\\frac{y^{2}}{2}=1$ and the hyperbola $\\frac{x^{2}}{3}-y^{2}=1$ have common foci $F_{1}$, $F_{2}$, and let $P$ be a common point of the two curves. Then $\\cos \\angle F_{1} P F_{2}$ equals?", "fact_expressions": "G: Hyperbola;H: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (x^2/3 - y^2 = 1);Expression(H) = (x^2/6 + y^2/2 = 1);Focus(G)={F1,F2};Focus(H)={F1,F2};OneOf(Intersection(H,G))=P", "query_expressions": "Cos(AngleOf(F1, P, F2))", "answer_expressions": "1/3", "fact_spans": "[[[39, 67]], [[1, 38]], [[72, 79]], [[88, 91]], [[80, 87]], [[39, 67]], [[1, 38]], [[1, 87]], [[1, 87]], [[88, 102]]]", "query_spans": "[[[104, 134]]]", "process": "c^{2}=6-2=4\\Rightarrow c=2\\Rightarrow |F_{1}F_{2}|=2c=4, |PF_{1}|+|PF_{2}|=2\\sqrt{6}, |PF_{1}|-|PF_{2}|=2\\sqrt{3}, then |PF_{1}|=\\sqrt{6}+\\sqrt{3}, |PF_{2}|=\\sqrt{6}-\\sqrt{3}, |PF_{1}|+|PF_{2}|=3. |PF_{1}|^{2}+|PF_{2}|^{2}=|PF_{1}|+|PF_{2}|^{2}-2|PF_{1}|\\cdot|PF_{2}|=18, \\cos\\angle F_{1}PF_{2}=\\frac{|PF_{1}|^{2}+|PF_{2}|^{2}-|F_{1}F_{2}|^{2}}{2|PF_{1}||PF_{2}|}=\\frac{18-16}{6}=\\frac{1}{3}" }, { "text": "The length of the chord cut by the line $y=x-\\frac{1}{2}$ from the ellipse $x^{2}+4 y^{2}=4$ is?", "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2 + 4*y^2 = 4);Expression(H) = (y = x - 1/2)", "query_expressions": "Length(InterceptChord(H,G))", "answer_expressions": "2*sqrt(38)/5", "fact_spans": "[[[20, 39]], [[0, 19]], [[20, 39]], [[0, 19]]]", "query_spans": "[[[0, 46]]]", "process": "Mainly examines the geometric properties of ellipses, the positional relationship between a line and an ellipse, and the chord length formula. Substitute $ y = x - \\frac{1}{2} $ into $ x^{2} + 4y^{2} = 4 $ and simplify to obtain $ 5x^{2} - 4x - 3 = 0 $. Let the endpoints of the chord be $ A(x_{1}, y_{1}) $ and $ B(x_{2}, y_{2}) $, then $ \\begin{cases} x_{1} + x_{2} = \\frac{4}{5} \\\\ x_{1}x_{2} = -\\frac{3}{5} \\end{cases} $, so the chord length $ AB = \\sqrt{1 + 1^{2}} \\sqrt{(x_{1} + x_{2})^{2} -} \\frac{2\\sqrt{38}}{5} $" }, { "text": "It is known that $P$ is a point on the right branch of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ with $F_{1}$ and $F_{2}$ as the left and right foci, respectively, satisfying $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$ and $\\tan \\angle P F_{1} F_{2}=\\frac{1}{4}$. Then the eccentricity of this hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: 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;Tan(AngleOf(P, F1, F2)) = 1/4", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(17)/3", "fact_spans": "[[[29, 85], [200, 203]], [[32, 85]], [[32, 85]], [[2, 5]], [[7, 14]], [[15, 22]], [[32, 85]], [[32, 85]], [[29, 85]], [[6, 85]], [[6, 85]], [[2, 92]], [[96, 156]], [[158, 197]]]", "query_spans": "[[[200, 209]]]", "process": "Let $|\\overrightarrow{PF_{1}}|=m$, $|\\overrightarrow{PF_{2}}|=n$. Since $\\overrightarrow{PF_{1}}\\cdot\\overrightarrow{PF_{2}}=0$, in the right triangle $PF_{1}F_{2}$, by the Pythagorean theorem we have $m^{2}+n^{2}=4c^{2}$. Since $\\tan\\angle PF_{1}F_{2}=\\frac{1}{4}$, it follows that $\\frac{n}{m}=\\frac{1}{4}$. Solving gives $m=\\frac{8\\sqrt{17}}{17}c$, $n=\\frac{2\\sqrt{17}}{17}c$. By the definition of a hyperbola we get $m-n=\\frac{6\\sqrt{17}}{17}c=2a$. Therefore $e=\\frac{c}{a}=\\frac{\\sqrt{17}}{3}$." }, { "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)$. Draw a perpendicular line from point $F$ to one asymptote of the hyperbola $C$, with foot of perpendicular at $A$, intersecting the other asymptote at point $B$. If $2 \\overrightarrow{A F}=\\overrightarrow{F B}$, then the equation of the asymptotes of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;F: Point;B: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(C) = F;L1:Line;L2:Line;OneOf(Asymptote(C))=L1;OneOf(Asymptote(C))=L2;Negation(L1=L2);L:Line;PointOnCurve(F, L);IsPerpendicular(L1, L);FootPoint(L1, L)=A;Intersection(L,L2)=B;2*VectorOf(A, F) = VectorOf(F, B)", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y=pm*(sqrt(3)/3)*x", "fact_spans": "[[[5, 66], [77, 83], [161, 167]], [[13, 66]], [[13, 66]], [[96, 99]], [[1, 4], [72, 76]], [[108, 112]], [[13, 66]], [[13, 66]], [[5, 66]], [[1, 70]], [], [], [[77, 89]], [[77, 107]], [[77, 107]], [], [[71, 92]], [[71, 92]], [[71, 99]], [[71, 112]], [[114, 159]]]", "query_spans": "[[[161, 175]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, and $P Q$ is a chord of the hyperbola passing through $F_{1}$ and perpendicular to the $x$-axis. If $\\angle P F_{2} Q=90^{\\circ}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;a: Number;b: Number;P: Point;Q: Point;F1: Point;F2: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Focus(G) = {F1, F2};IsChordOf(LineSegmentOf(P, Q), G);PointOnCurve(F1, LineSegmentOf(P, Q));IsPerpendicular(xAxis, LineSegmentOf(P, Q));AngleOf(P, F2, Q) = ApplyUnit(90, degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1 + sqrt(2)", "fact_spans": "[[[18, 75], [105, 108], [145, 148]], [[21, 75]], [[21, 75]], [[81, 86]], [[81, 86]], [[2, 9], [89, 96]], [[10, 17]], [[21, 75]], [[21, 75]], [[18, 75]], [[2, 80]], [[81, 110]], [[81, 96]], [[81, 104]], [[114, 143]]]", "query_spans": "[[[145, 154]]]", "process": "" }, { "text": "The distance from a focus of the hyperbola $\\frac{y^{2}}{9}-\\frac{x^{2}}{16}=1$ to one of its asymptotes is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-x^2/16 + y^2/9 = 1)", "query_expressions": "Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))", "answer_expressions": "4", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 55]]]", "process": "" }, { "text": "Given that $P$ is a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and $M$, $N$ are points on the circles $(x+3)^{2}+y^{2}=1$ and $(x-3)^{2}+y^{2}=1$ respectively, then the minimum value of $PN + PM$ is?", "fact_expressions": "G: Ellipse;H: Circle;C:Circle;P: Point;N: Point;M: Point;Expression(G) = (x^2/25 + y^2/16 = 1);Expression(H) = (y^2 + (x + 3)^2 = 1);Expression(C) = (y^2 + (x - 3)^2 = 1);PointOnCurve(P, G);PointOnCurve(M, H);PointOnCurve(N, C)", "query_expressions": "Min(LineSegmentOf(P, M) + LineSegmentOf(P, N))", "answer_expressions": "8", "fact_spans": "[[[6, 45]], [[60, 80]], [[81, 101]], [[2, 5]], [[54, 57]], [[50, 53]], [[6, 45]], [[60, 80]], [[81, 101]], [[2, 49]], [[50, 104]], [[50, 104]]]", "query_spans": "[[[106, 121]]]", "process": "Let \\( F_{1}, F_{2} \\) be the left and right foci of the ellipse, then \\( |PM| + |PN| \\geqslant |PF_{1}| - 1 + |PF_{2}| - 1 = 2a - 2 = 8 \\). The equality holds when \\( P, F_{1}, M \\) are collinear and \\( P, F_{2}, N \\) are collinear. Therefore, the minimum value of \\( PN + PM \\) is 8." }, { "text": "Given that point $P$ is a point on the parabola $y^{2}=4x$, let $d_{1}$ denote the distance from point $P$ to the $y$-axis, and $d_{2}$ denote the distance from point $P$ to the line $l$: $3x-4y+12=0$. Then the minimum value of $d_{1}+d_{2}$ is?", "fact_expressions": "G: Parabola;l: Line;P: Point;Expression(G) = (y^2 = 4*x);Expression(l) = (3*x - 4*y + 12 = 0);PointOnCurve(P, G);d1:Number;d2:Number;Distance(P, yAxis) = d1;Distance(P, l) = d2", "query_expressions": "Min(d1 + d2)", "answer_expressions": "2", "fact_spans": "[[[7, 21]], [[50, 71]], [[2, 6], [26, 30], [45, 49]], [[7, 21]], [[50, 71]], [[2, 24]], [[37, 44]], [[74, 81]], [[26, 44]], [[45, 81]]]", "query_spans": "[[[83, 102]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$, and the circle $M$: $(x-a)^{2}+y^{2}=\\frac{b^{2}}{4}$. If an asymptote of the hyperbola $C$ is tangent to the circle $M$, then when $\\frac{a^{2}}{a^{2} b^{2}+1}-\\frac{4}{49} a^{2}$ attains its maximum value, what is the length of the real axis of $C$?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;M: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(M) = (y^2 + (-a + x)^2 = b^2/4);IsTangent(OneOf(Asymptote(C)),M);WhenMax(a^2/(a^2*b^2+1)-(4/49)*a^2)", "query_expressions": "Length(RealAxis(C))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 63], [104, 110], [181, 184]], [[10, 63]], [[10, 63]], [[64, 102], [117, 121]], [[10, 63]], [[10, 63]], [[2, 63]], [[64, 102]], [[104, 123]], [[125, 180]]]", "query_spans": "[[[181, 190]]]", "process": "] First, use the condition that a line is tangent to a circle to determine the relationship between a and b, then use the derivative to study the real axis length of the hyperbola when the function reaches its maximum value. A asymptote of the hyperbola has the equation: bx - ay = 0. Since the circle is tangent to the asymptote of the hyperbola, the distance from the center of the circle to the line equals the radius, namely: \\frac{ab}{\\sqrt{a^{2}+b^{2}}}=\\frac{ab}{c}=\\frac{b}{2}. From this it follows: c=2a, then c^{2}=4a^{2}, b^{2}=3a^{2}. Hence \\frac{a^{2}}{a^{2}b^{2}+1}-\\frac{4}{49}a^{2}=\\frac{a^{2}}{a^{2}\\times3a^{2}+1}-\\frac{4}{49}a^{2}=\\frac{3}{49}\\times\\frac{-12a^{6}+45a^{2}}{3a^{4}+1}. Let f(a)=\\frac{3}{49}\\times\\frac{-12a^{6}+45a^{2}}{3a^{4}+1} (a>0), then f'(a)=\\frac{9}{49}\\times\\frac{-6a(4a^{4}-1)(a^{4}+5)}{(3a^{4}+1)^{2}}. From the relationship between the monotonicity of the derivative and the original function, it follows that the function f(a) is monotonically increasing on the interval (0,\\frac{\\sqrt{2}}{2}) and monotonically decreasing on the interval (\\frac{\\sqrt{2}}{2},+\\infty). When a=\\frac{\\sqrt{2}}{2}, \\frac{a^{2}}{a^{2}b^{2}+1}-\\frac{4}{49}a^{2} reaches its maximum value, and at this point the real axis length of C is 2a=\\sqrt{2}." }, { "text": "Let the line $ l $ 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 $ \\triangle O A F $ ($ O $ being the origin) is $ 4 $, then what is the equation of the parabola?", "fact_expressions": "l: Line;Slope(l) = 2;G: Parabola;Expression(G) = (y^2 = a*x);a>0;a: Number;F: Point;PointOnCurve(F,l) = True;Focus(G) = F;Intersection(l,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, 13]], [[1, 13]], [[14, 33], [92, 95]], [[14, 33]], [[17, 33]], [[17, 33]], [[36, 39]], [[8, 39]], [[14, 39]], [[8, 52]], [[73, 76]], [[48, 52]], [[54, 90]]]", "query_spans": "[[[92, 100]]]", "process": "" }, { "text": "The equations of the asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2/16 - y^2/9 = 1)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=(pm*3/4)*x", "fact_spans": "[[[0, 39]], [[0, 39]]]", "query_spans": "[[[0, 47]]]", "process": "Given the hyperbola equation $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$, we can see that $a=4$, $b=3$; then the asymptotes of the hyperbola $\\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1$ are given by $y=\\pm\\frac{3}{4}x$." }, { "text": "Let $e_{1}$ and $e_{2}$ be the eccentricities of an ellipse and a hyperbola having common foci $F_{1}$ and $F_{2}$, respectively. Let $P$ be a common point of the two curves such that $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|=|\\overrightarrow{F_{1} F_{2}}|$. Then the value of $\\frac{e_{1} e_{2}}{\\sqrt{e_{1}^{2}+e_{2}^{2}}}$ is?", "fact_expressions": "G: Ellipse;H: Hyperbola;e1: Number;e2: Number;Eccentricity(G) = e1;Eccentricity(H) = e2;Focus(G) = {F1,F2};Focus(H) = {F1,F2};P: Point;OneOf(Intersection(G,H)) = P;Abs(VectorOf(P, F1) + VectorOf(P, F2)) = Abs(VectorOf(F1, F2));F1: Point;F2: Point", "query_expressions": "(e1*e2)/sqrt(e1^2 + e2^2)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[43, 45]], [[46, 49]], [[1, 8]], [[11, 18]], [[1, 53]], [[1, 53]], [[21, 49]], [[21, 49]], [[54, 57]], [[54, 67]], [[71, 155]], [[27, 34]], [[35, 42]]]", "query_spans": "[[[157, 209]]]", "process": "Problem Analysis: Let the equations of the ellipse and hyperbola be $\\frac{x^{2}}{a_{1}}+\\frac{y^{2}}{b_{1}}=1$, $\\frac{x^{2}}{a_{2}}-\\frac{y^{2}}{b_{2}}=1$, respectively. The common foci $F_{1}$, $F_{2}$ have coordinates $(-c,0)$, $(c,0)$, respectively. Let $O$ be the coordinate origin. Since $|\\overrightarrow{PF_{1}}+\\overrightarrow{PF_{2}}|=|\\overrightarrow{F_{1}F_{2}}|$, it follows that $2|PO|=|F_{1}F_{2}|$. Therefore, triangle $PF_{1}F_{2}$ is a right triangle with right angle at $P$. By the definitions of the ellipse and hyperbola, we obtain\n$$\n\\begin{cases}\n|PF_{1}|+|PF_{2}|=2a_{1}\\\\\n|PF_{1}|-|PF_{2}|=2a_{2}\n\\end{cases}\n$$\nThus, $|PF_{1}|=a_{1}+a_{2}$, $|PF_{2}|=a_{1}-a_{2}$. From $|PF_{1}|^{2}+|PF_{2}|^{2}=|F_{1}F_{2}|^{2}$, we get $a_{1}^{2}+a_{2}^{2}=2c^{2}$. Hence, $\\frac{1}{e_{1}^{2}}+\\frac{1}{e_{2}^{2}}=2$, so the value of $\\frac{e_{1}e_{2}}{\\sqrt{e_{1}^{2}+e_{2}^{2}}}$ is $\\frac{\\sqrt{2}}{2}$. Therefore, fill in $\\frac{\\sqrt{2}}{2}$." }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1$ with foci $F_{1}$, $F_{2}$, and $P$ a point on the ellipse, then the perimeter of $\\Delta F_{1} PF_{2}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G) = True", "query_expressions": "Perimeter(TriangleOf(F1, P, F2))", "answer_expressions": "4+2*sqrt(2)", "fact_spans": "[[[2, 39], [62, 64]], [[2, 39]], [[42, 49]], [[50, 57]], [[2, 57]], [[58, 61]], [[58, 67]]]", "query_spans": "[[[69, 95]]]", "process": "From the problem, we have $ a=2 $, $ b=c=\\sqrt{2} $. Since $ P $ is a point on the ellipse, $ |PF_{1}|+|PF_{2}|=2a=4 $. Therefore, the perimeter of $ \\triangle F_{1}PF_{2} $ is $ |PF_{1}|+|PF_{2}|+|F_{1}F_{2}|=4+2\\sqrt{2} $." }, { "text": "From the left focus $F$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$, draw two tangents to the circle $\\odot O$: $x^{2}+y^{2}=a^{2}$, and denote the points of tangency as $A$, $B$. Let $C$ be the left vertex of the hyperbola. If $\\angle A C B=120^{\\circ}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;O: Circle;A: Point;C: Point;B: Point;F: Point;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(O) = (x^2 + y^2 = a^2);LeftFocus(G) = F;l1: Line;l2: Line;TangentOfPoint(F, O) = {l1, l2};TangentPoint(l1,O)=A;TangentPoint(l2,O)=B;LeftVertex(G) = C;AngleOf(A, C, B) = ApplyUnit(120, degree)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 47], [103, 106], [144, 147]], [[4, 47]], [[4, 47]], [[55, 85]], [[95, 98]], [[110, 113]], [[99, 102]], [[51, 54]], [[1, 47]], [[55, 85]], [[1, 54]], [], [], [[0, 90]], [[0, 102]], [[0, 102]], [[103, 113]], [[115, 141]]]", "query_spans": "[[[144, 153]]]", "process": "From the given conditions, the asymptotes of the hyperbola are given by $ y = \\pm \\frac{b}{a}x $. Since $ \\angle ACB = 120^{\\circ} $, according to the characteristics of the graph, we have $ \\angle AFO = 30^{\\circ} $, so $ c = 2a $. Also, since $ b^{2} = c^{2} - a^{2} $, it follows that $ \\frac{b}{a} = \\sqrt{3} $, hence the eccentricity is $ 2 $." }, { "text": "Given that point $P$ is a moving point on the parabola $C$: $y^{2}=4x$, and point $Q$ is a moving point on the circle $x^{2}+(y-4)^{2}=1$. Let $d$ be the distance from point $P$ to the directrix of the parabola $C$. Then, the minimum value of $d+PQ$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);P: Point;PointOnCurve(P, C);G: Circle;Expression(G) = (x^2 + (y - 4)^2 = 1);Q: Point;PointOnCurve(Q, G);Distance(P, Directrix(C)) = d;d: Number", "query_expressions": "Min(d + LineSegmentOf(P, Q))", "answer_expressions": "sqrt(17)-1", "fact_spans": "[[[7, 25], [67, 73]], [[7, 25]], [[2, 6], [62, 66]], [[2, 31]], [[36, 56]], [[36, 56]], [[32, 35]], [[32, 60]], [[62, 82]], [[79, 82]]]", "query_spans": "[[[85, 98]]]", "process": "" }, { "text": "Let a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ be $F$, and let an endpoint of the imaginary axis be $B$. If the line $FB$ is perpendicular to an asymptote of the hyperbola, 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>0;b>0;OneOf(Focus(G)) = F;F: Point;OneOf(Endpoint(ImageinaryAxis(G))) = B;B: Point;IsPerpendicular(LineOf(F,B),OneOf(Asymptote(G))) = True", "query_expressions": "Eccentricity(G)", "answer_expressions": "(sqrt(5)+1)/2", "fact_spans": "[[[1, 57], [90, 93], [105, 108]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 66]], [[63, 66]], [[1, 78]], [[75, 78]], [[81, 101]]]", "query_spans": "[[[105, 114]]]", "process": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ has focus $F(c,0)$, $B(0,b)$. The line $FB$: $bx+cy-bc=0$ is perpendicular to the asymptote $y=\\frac{b}{a}x$, so $-\\frac{b}{c}\\cdot\\frac{b}{a}=-1$ $\\therefore b^{2}=ac$. Thus $c^{2}-a^{2}=ac$, i.e., $e^{2}-e-1=0$, so $e=\\frac{1+\\sqrt{5}}{2}$ or $e=\\frac{1-\\sqrt{5}}{2}$ (discarded)." }, { "text": "Given that point $F$ is the focus of the parabola $x^{2}=12 y$, and point $P$ is a point on the parabola, if there exists a point $M$ in the plane such that $\\overrightarrow{F M}=\\overrightarrow{M P}$, then what is the equation of the trajectory of point $M$?", "fact_expressions": "G: Parabola;F: Point;M: Point;P: Point;Expression(G) = (x^2 = 12*y);Focus(G) = F;PointOnCurve(P, G);VectorOf(F, M) = VectorOf(M, P)", "query_expressions": "LocusEquation(M)", "answer_expressions": "x^2=6*y-9", "fact_spans": "[[[7, 22], [31, 34]], [[2, 6]], [[47, 50], [99, 103]], [[26, 30]], [[7, 22]], [[2, 25]], [[26, 37]], [[53, 96]]]", "query_spans": "[[[99, 110]]]", "process": "Let M(x,y), and the coordinates of point P be (t, \\frac{1}{12}t^{2}). Since for the parabola x^{2}=12y, 2p=12 gives p=6, therefore F(0,3), thus \\overrightarrow{FM}=(x,y-3), \\overrightarrow{MP}=(t-x,\\frac{t^{2}}{12}-y). Also, since \\overrightarrow{FM}=\\overrightarrow{MP}, we have (x,y-3)=(t-x,\\frac{t^{2}}{12}-y), which yields \\begin{cases}x=t-x\\\\y-3=\\frac{t^{2}}{12}-y\\end{cases}. Eliminating the parameter t gives x^{2}=6y-9, so the trajectory equation of point M is x^{2}=6y-9." }, { "text": "The equation of the hyperbola that shares the same asymptotes as the hyperbola $x^{2}-\\frac{y^{2}}{4}=1$ and passes through the point $(2 , 2)$ is?", "fact_expressions": "G: Hyperbola;C:Hyperbola;H: Point;Expression(G) = (x^2 - y^2/4 = 1);Coordinate(H) = (2, 2);Asymptote(G) = Asymptote(C);PointOnCurve(H, C)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/3 - y^2/12 = 1", "fact_spans": "[[[1, 29]], [[50, 53]], [[39, 49]], [[1, 29]], [[39, 49]], [[0, 53]], [[38, 53]]]", "query_spans": "[[[50, 57]]]", "process": "Problem Analysis: A hyperbola shares the same asymptotes as $ x^{2}-\\frac{y^{2}}{4}=1 $. Let the equation of the desired hyperbola be $ x^{2}-\\frac{y^{2}}{4}=\\lambda $. Substituting the point $ (2,2) $, we obtain: $ \\lambda=3 $. Therefore, the answer should be: $ \\frac{x^{2}}{3}-\\frac{y^{2}}{12}=1 $" }, { "text": "Given the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, with left and right foci $F_{1}$ and $F_{2}$ respectively. If $M$ is a point on the right branch of the hyperbola such that $\\frac{|M F_{1}|}{|M F_{2}|}=3$, then the range of the eccentricity $e$ is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;M: 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(M, RightPart(G));Abs(LineSegmentOf(M, F1))/Abs(LineSegmentOf(M, F2)) = 3;Eccentricity(G)=e;e:Number", "query_expressions": "Range(e)", "answer_expressions": "(1, 2]", "fact_spans": "[[[2, 58], [59, 60], [89, 92]], [[5, 58]], [[5, 58]], [[84, 87]], [[67, 74]], [[75, 82]], [[5, 58]], [[5, 58]], [[2, 58]], [[59, 82]], [[59, 82]], [[84, 97]], [[100, 131]], [[89, 139]], [[136, 139]]]", "query_spans": "[[[136, 146]]]", "process": "Let the horizontal coordinate of point M be x. Since \\frac{|MF_{1}|}{|MF_{2}|}=3 and M lies on the right branch of the hyperbola (x\\geqslanta), according to the second definition of the hyperbola, we obtain 3e(x-\\frac{a^{2}}{c})=e(x+\\frac{a^{2}}{c}). \\therefore ex=2a. \\because x\\geqslanta," }, { "text": "The hyperbola $ C $: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 $ ($ a>0, b>0 $) has eccentricity $ 2 $. Its asymptotes are tangent to the circle $ (x-a)^{2}+y^{2}=\\frac{3}{4} $. Then the equation of this hyperbola is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;G: Circle;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y^2 + (-a + x)^2 = 3/4);Eccentricity(C) = 2;IsTangent(Asymptote(C),G)", "query_expressions": "Expression(C)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[0, 61], [70, 71], [110, 113]], [[8, 61]], [[8, 61]], [[75, 105]], [[8, 61]], [[8, 61]], [[0, 61]], [[75, 105]], [[0, 69]], [[70, 107]]]", "query_spans": "[[[110, 118]]]", "process": "From the given condition, \\frac{c}{a}=2, that is, c=2a, then b=\\sqrt{3}a. From the equation of the circle, its center is at (a,0) and radius r=\\frac{\\sqrt{3}}{2}. Take one asymptote of the hyperbola bx-ay=0, then \\frac{|ba|}{\\sqrt{a^{2}+b^{2}}}=\\frac{\\sqrt{3}}{2}, that is, \\frac{\\sqrt{3}a^{2}}{2a}=\\frac{\\sqrt{3}}{2}. Thus a=1, then b=\\sqrt{3}. Therefore, the required equation of the hyperbola is x^{2}-\\frac{y^{2}}{3}=1" }, { "text": "For any point $P$ on the hyperbola $\\frac{y^{2}}{3}-x^{2}=1$, draw perpendiculars from $P$ to the two asymptotes, with feet of perpendiculars denoted as $A$ and $B$. Then the minimum value of $|A B|$ is?", "fact_expressions": "G: Hyperbola;A: Point;B: Point;Expression(G) = (-x^2 + y^2/3 = 1);P:Point;PointOnCurve(P,G);L1:Line;L2:Line;Asymptote(G)={L1,L2};H1:Line;H2:Line;PointOnCurve(P,H1);PointOnCurve(P,H2);IsPerpendicular(H1,L1);IsPerpendicular(H2,L2);FootPoint(H1,L1)=A;FootPoint(H2,L2)=B", "query_expressions": "Min(Abs(LineSegmentOf(A, B)))", "answer_expressions": "3/2", "fact_spans": "[[[1, 29]], [[50, 53]], [[54, 57]], [[1, 29]], [[33, 36]], [[1, 36]], [], [], [[1, 41]], [], [], [[0, 44]], [[0, 44]], [[0, 44]], [[0, 44]], [[0, 57]], [[0, 57]]]", "query_spans": "[[[59, 72]]]", "process": "Problem Analysis: From the given conditions, points P, A, B, O are concyclic. To minimize |AB|, it suffices to minimize the diameter of the circle, i.e., the minimum diameter is a = \\sqrt{3}. Since the asymptotes of the hyperbola \\frac{y^{2}}{3} - x^{2} = 1 are y = \\pm\\sqrt{3}x, it follows that \\angle AOB = 120^{\\circ}. By the Law of Sines: \\frac{|AB|}{\\sin\\angle AOB} = 2R', so |AB| = 2R \\sin\\angle AOB = \\sqrt{3} \\sin 120^{\\circ} = \\sqrt{3} \\times \\frac{\\sqrt{3}}{2} = \\frac{3}{2}. Therefore, the answer should be \\frac{3}{2}." }, { "text": "If the equations of the asymptotes of a hyperbola are $y = \\pm 3x$, and it passes through the point $(2, -3\\sqrt{3})$, then what is the equation of the hyperbola?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (2, -3*sqrt(3));Expression(Asymptote(G)) = (y = pm*(3*x));PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2 - y^2/9 = 1", "fact_spans": "[[[1, 4], [49, 52]], [[27, 47]], [[27, 47]], [[1, 23]], [[1, 47]]]", "query_spans": "[[[49, 57]]]", "process": "" }, { "text": "The length of the chord cut by the line $y=x-1$ on the ellipse $\\frac{x^{2}}{2}+y^{2}=1$ is?", "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2/2 + y^2 = 1);Expression(H) = (y = x - 1)", "query_expressions": "Length(InterceptChord(H, G))", "answer_expressions": "4*sqrt(2)/3", "fact_spans": "[[[0, 27]], [[28, 37]], [[0, 27]], [[28, 37]]]", "query_spans": "[[[0, 44]]]", "process": "" }, { "text": "Given that $A$ and $B$ are the left and right vertices of the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and $P$ is a point on $C$. Let the slopes of lines $PA$ and $PB$ be $k_{1}$ and $k_{2}$, respectively. If $k_{1} k_{2}=-\\frac{4}{9}$, then what is the eccentricity of the ellipse?", "fact_expressions": "C: Ellipse;b: Number;a: Number;A: Point;P: Point;B: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftVertex(C) = A;RightVertex(C) = B;PointOnCurve(P, C);Slope(LineOf(P, A))=k1;Slope(LineOf(P, B)) = k2;k1:Number;k2:Number;k1*k2 = -4/9", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/3", "fact_spans": "[[[10, 67], [78, 81], [151, 153]], [[17, 67]], [[17, 67]], [[2, 5]], [[74, 77]], [[6, 9]], [[17, 67]], [[17, 67]], [[10, 67]], [[2, 73]], [[2, 73]], [[74, 84]], [[86, 121]], [[86, 121]], [[105, 112]], [[114, 121]], [[123, 149]]]", "query_spans": "[[[151, 159]]]", "process": "Let P(x_{0},y_{0}), A(-a,0), B(a,0), so k_{1}k_{2} = \\frac{y_{0}}{x_{0}+a} \\cdot \\frac{y_{0}}{x_{0}-a} = \\frac{y_{0}^{2}}{x_{0}^{2}-a^{2}}, and \\frac{x_{0}^{2}}{a^{2}} + \\frac{y_{0}^{2}}{b^{2}} = 1, so k_{1}k_{2} = \\frac{b^{2} - \\frac{b^{2}}{a^{2}}x_{0}^{2}}{x_{0}^{2}-a^{2}} = -\\frac{b^{2}}{a^{2}} = -\\frac{4}{9}, so b^{2} = a^{2} - c^{2} = \\frac{4}{9}a^{2}, so c^{2} = \\frac{5}{9}a^{2}, so e = \\frac{\\sqrt{5}}{3}." }, { "text": "Given the parabola $ C $: $ y^{2} = -4x $ with focus $ F $, and point $ A(-1,1) $, find the minimum value of the sum of the distances from a moving point $ P $ on curve $ C $ to points $ F $ and $ A $.", "fact_expressions": "C: Parabola;F: Point;P:Point;A:Point;Expression(C) = (y^2 = -4*x);Focus(C) = F;Coordinate(A) = (-1, 1);PointOnCurve(P,C)", "query_expressions": "Min(Distance(P,A)+Distance(P,F))", "answer_expressions": "2", "fact_spans": "[[[42, 47], [2, 22]], [[25, 28], [55, 59]], [[51, 54]], [[30, 40], [60, 64]], [[2, 22]], [[2, 28]], [[30, 40]], [[42, 54]]]", "query_spans": "[[[42, 75]]]", "process": "Problem Analysis: According to the parabola equation, find the focus coordinates and the directrix equation. Then, by the definition of a parabola, the sum of distances is minimized when points P, A, and the projection Q of P on the directrix are collinear, leading to the conclusion. \nSince the parabola equation is $ y^{2} = -4x $, \n$ \\therefore 2p = 4 $, giving the focus $ F(-1,0) $, and the directrix $ x = 1 $. \nLet $ Q $ be the projection of point $ P $ on the directrix $ l $ of the parabola, and $ A(-1,1) $. \nThen, by the definition of the parabola, the sum of the distance from point $ P $ to point $ (-1,1) $ and the distance from $ P $ to the focus of the parabola is minimized when points $ P $, $ Q $, and $ A $ are collinear. \n$ \\therefore $ The minimum value is $ 1 + 1 = 2 $." }, { "text": "The standard equation of the hyperbola with foci $(-2,0)$, $(2,0)$ and passing through the point $(2,3)$ is?", "fact_expressions": "G: Hyperbola;H: Point;I: Point;J: Point;Coordinate(H) = (-2, 0);Coordinate(I) = (2, 0);Coordinate(J) = (2, 3);Focus(G) = {H,I};PointOnCurve(J,G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[33, 36]], [[5, 13]], [[14, 21]], [[24, 32]], [[5, 13]], [[14, 21]], [[24, 32]], [[0, 36]], [[22, 36]]]", "query_spans": "[[[33, 43]]]", "process": "According to the definition of a hyperbola, let the standard equation of the hyperbola be given, and substitute the point coordinates into it to solve. [Detailed solution] From the given condition, c^{2}=4, so we can set the standard equation of the hyperbola as \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{4-a^{2}}=1. Substituting (2,3) into the equation gives \\frac{2}{a^{2}}-\\frac{3^{2}}{4-a^{2}}=1, solving yields a^{2}=1 or a^{2}=16 (discarded). Thus, b^{2}=4-a^{2}=3^{n}, so x^{2}-\\frac{y^{2}}{2}=1. Hence answer: y^{2}" }, { "text": "If one of the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an inclination angle of $60^{\\circ}$, then the range of $\\frac{b^{2}+1}{a}$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Inclination(OneOf(Asymptote(G))) = ApplyUnit(60, degree)", "query_expressions": "Range((b^2 + 1)/a)", "answer_expressions": "[2*sqrt(2),+oo]", "fact_spans": "[[[1, 57]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[1, 80]]]", "query_spans": "[[[82, 108]]]", "process": "The hyperbola \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 (a>0,b>0) has asymptotes y=\\pm\\frac{b}{a}x. Since the inclination angle of one asymptote is 60^{\\circ}, it follows that \\frac{b}{a}=\\sqrt{3}, i.e., b=\\sqrt{3}a. Therefore, \\frac{b^{2}+1}{a}=\\frac{3a^{2}+1}{a}=3a+\\frac{1}{a}\\geqslant2\\sqrt{3a\\cdot\\frac{1}{a}}=2\\sqrt{3}, with equality if and only if 3a=\\frac{1}{a}, i.e., a=\\frac{\\sqrt{3}}{3}. Hence, the range of \\underline{b^{2}+1} is [2\\sqrt{3},+\\infty)." }, { "text": "If the asymptotes of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ are tangent to the circle $(x-\\sqrt{3})^{2}+(y-1)^{2}=1$, 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;H: Circle;Expression(H) = ((x - sqrt(3))^2 + (y - 1)^2 = 1);IsTangent(Asymptote(G), H)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[1, 57], [98, 101]], [[1, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[4, 57]], [[62, 93]], [[62, 93]], [[1, 95]]]", "query_spans": "[[[98, 107]]]", "process": "" }, { "text": "The standard equation of an ellipse with eccentricity $e=\\frac{1}{2}$ and one focus at $F(0,-3)$ is?", "fact_expressions": "E: Ellipse;F: Point;e: Number;Coordinate(F) = (0, -3);Eccentricity(E) = e;e = 1/2;OneOf(Focus(E)) = F", "query_expressions": "Expression(E)", "answer_expressions": "x**2/27+y**2/36=1", "fact_spans": "[[[38, 40]], [[26, 37]], [[3, 18]], [[26, 37]], [[0, 40]], [[3, 18]], [[21, 40]]]", "query_spans": "[[[38, 46]]]", "process": "" }, { "text": "Let the ellipse $ C $: $ \\frac{x^{2}}{2} + y^{2} = 1 $ have left focus $ F $, and let the line $ l $: $ x - y + 2 = 0 $. A moving point $ P $ lies on the ellipse $ C $, and denote by $ d $ the distance from point $ P $ to the line $ l $. Then the maximum value of $ |P F| - d $ is?", "fact_expressions": "l: Line;C: Ellipse;P: Point;F: Point;Expression(C) = (x^2/2 + y^2 = 1);Expression(l)=(x-y+2=0);LeftFocus(C) = F;PointOnCurve(P, C);Distance(P, l) = d;d:Number", "query_expressions": "Max(Abs(LineSegmentOf(P, F))-d)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[42, 58], [78, 83]], [[1, 33], [65, 70]], [[61, 64], [73, 77]], [[38, 41]], [[1, 33]], [[42, 58]], [[1, 41]], [[61, 71]], [[73, 90]], [[87, 90]]]", "query_spans": "[[[92, 108]]]", "process": "" }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$, a line passing through $F_{1}$ intersects the ellipse at points $A$ and $B$. If $|F_{2} A|+| F_{2} B|=12$, then $|A B|$=?", "fact_expressions": "F1: Point;F2: Point;Focus(G) = {F1, F2};G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);H: Line;PointOnCurve(F1, H);A: Point;B: Point;Intersection(H, G) = {A, B};Abs(LineSegmentOf(F2, A)) + Abs(LineSegmentOf(F2, B)) = 12", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "8", "fact_spans": "[[[2, 9], [63, 70]], [[10, 17]], [[2, 61]], [[18, 56], [74, 76]], [[18, 56]], [[71, 73]], [[62, 73]], [[77, 80]], [[81, 84]], [[71, 86]], [[88, 113]]]", "query_spans": "[[[115, 124]]]", "process": "" }, { "text": "Given that point $P$ is a point on the ellipse $C$: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, and the product of the slopes of the lines joining $P$ to the upper and lower vertices of the ellipse is $-\\frac{4}{9}$, then the eccentricity of $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, C);Slope(LineSegmentOf(P, UpperVertex(C)))*Slope(LineSegmentOf(P, LowerVertex(C))) = -4/9", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/3", "fact_spans": "[[[7, 64], [72, 74], [103, 106]], [[14, 64]], [[14, 64]], [[2, 6], [68, 71]], [[14, 64]], [[14, 64]], [[7, 64]], [[2, 67]], [[68, 101]]]", "query_spans": "[[[103, 112]]]", "process": "" }, { "text": "The ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{t}=1$, the distance between the two foci is $6$, then $t$=?", "fact_expressions": "G: Ellipse;t: Number;Expression(G) = (x^2/25 + y^2/t = 1);Focus(G) = {F1,F2};F1: Point;F2: Point;Distance(F1,F2) = 6", "query_expressions": "t", "answer_expressions": "{16,34}", "fact_spans": "[[[0, 38]], [[51, 54]], [[0, 38]], [[0, 42]], [], [], [[0, 49]]]", "query_spans": "[[[51, 56]]]", "process": "" }, { "text": "Given that $A$ and $B$ are two moving points on the parabola $y^{2}=4x$, and $OA \\perp OB$, where $F$ is the focus of the parabola, then the minimum area of $\\triangle ABF$ is?", "fact_expressions": "G: Parabola;O: Origin;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);PointOnCurve(A, G);PointOnCurve(B, G);IsPerpendicular(LineSegmentOf(O, A), LineSegmentOf(O, B));Focus(G) = F", "query_expressions": "Min(Area(TriangleOf(A, B, F)))", "answer_expressions": "12", "fact_spans": "[[[10, 24], [48, 51]], [[32, 47]], [[2, 5]], [[6, 9]], [[55, 58]], [[10, 24]], [[2, 30]], [[2, 30]], [[32, 47]], [[48, 58]]]", "query_spans": "[[[60, 85]]]", "process": "Since $ S_{\\triangle ABF} = S_{\\triangle OAB} - (S_{\\triangle OFA} + S_{\\triangle OFB}) $, let $ A(t^{2}, 2t) $, $ B(m^{2}, 2m) $, from $ OA \\perp OB $, obtain the relationship between $ t $ and $ m $, use the AM-GM inequality to find the maximum of $ S_{\\triangle ABO} = \\frac{1}{2}|OA||OB| $, then solve accordingly. Let $ A(t^{2}, 2t) $, $ B(m^{2}, 2m) $, and $ OA \\perp OB $, so $ \\overrightarrow{OA} \\cdot \\overrightarrow{OB} = t^{2}m^{2} + 4tm = 0 $, thus $ tm = -4 $ or $ tm = 0 $ (discarded). Since $ tm < 0 $, $ t $ and $ m $ have opposite signs; without loss of generality, assume $ t > 0 $, $ m < 0 $. Then $ S_{\\triangle ABO} = \\frac{1}{2}|OA||OB| = \\frac{1}{2}\\sqrt{t^{4} + 4t^{2}} \\cdot \\sqrt{m^{4} + 4m^{2}} = \\frac{1}{2}|tm|\\sqrt{(t^{2} + 4)(m^{2} + 4)} \\geqslant 4\\sqrt{2|tm| + 8} = 16 $, equality holds if and only if $ |t| = |m| $, i.e., $ m = -t $. Therefore, the minimum area of $ \\triangle ABO $ is 16, at this time $ A(t^{2}, 2t) $, $ B(t^{2}, -2t) $. Thus, the inclination angle of line $ OA $ is $ 45^{\\circ} $, slope is 1. Hence, point $ A $ is the other intersection point (besides the origin) of the line $ y = x $ and the parabola $ y^{2} = 4x $: \n$$\n\\begin{cases}\ny = x \\\\\ny^{2} = 4x\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\nx = 0 \\\\\ny = 0\n\\end{cases},\n\\begin{cases}\nx = 4 \\\\\ny = 4\n\\end{cases}\n$$\nThus, $ A(4, 4) $, $ B(4, -4) $. Therefore, $ S_{\\triangle OFA} = S_{\\triangle OFB} = \\frac{1}{2} \\times 1 \\times 4 = 2 $. So $ S_{\\triangle ABF} = S_{\\triangle OAB} - (S_{\\triangle OFA} + S_{\\triangle OFB}) = 16 - 2 \\times 2 = 12 $." }, { "text": "Let the focus of the parabola $y^{2}=2 x$ be $F$. A line passing through point $F$ intersects the parabola at points $A$ and $B$ (point $A$ is in the first quadrant). If $\\overrightarrow{A F}=2 \\overrightarrow{F B}$, then $|A B|$=?", "fact_expressions": "G: Parabola;H: Line;A: Point;F: Point;B: Point;Expression(G) = (y^2 = 2*x);Focus(G) = F;PointOnCurve(F, H);Intersection(H, G) = {A, B};Quadrant(A) = 1;VectorOf(A, F) = 2*VectorOf(F, B)", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "9/4", "fact_spans": "[[[1, 15], [32, 35]], [[29, 31]], [[38, 41], [48, 52]], [[19, 22], [24, 28]], [[42, 45]], [[1, 15]], [[1, 22]], [[23, 31]], [[29, 47]], [[48, 57]], [[60, 105]]]", "query_spans": "[[[107, 116]]]", "process": "Let the equation of line AB be $ x = my + \\frac{1}{2} $. Substituting into the equation of the parabola yields $ y^{2} - 2my - 1 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ y_{1} + y_{2} = 2m $. Given $ |\\overrightarrow{AF}| = 2|\\overrightarrow{FB}| $, then $ y_{1} = -2y_{2} $. Substituting gives \n\\[\n\\begin{cases}\n-y_{2} = 2m \\\\\n-2y_{2}^{2} = -1\n\\end{cases}\n\\],\nobtaining \n\\[\n\\begin{cases}\nm = \\frac{\\sqrt{2}}{4} \\\\\ny = -\\frac{\\sqrt{2}}{2}\n\\end{cases}\n\\].\nThen $ |AB| = x_{1} + x_{2} + p = m(y_{1} + y_{2}) + 1 + 1 = \\frac{9}{4} $. The answer is: 9" }, { "text": "Given that the equation of the parabola is $y = a x^{2}$, what are the coordinates of its focus?", "fact_expressions": "G: Parabola;Expression(G) = (y = a*x^2);a: Number", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(0, 1/(4*a))", "fact_spans": "[[[2, 5], [22, 23]], [[2, 20]], [[9, 20]]]", "query_spans": "[[[22, 30]]]", "process": "Transform y = ax^{2} into x^{2} = \\frac{1}{a}y. When a > 0, the focus is (0, \\frac{1}{4a}); when a < 0, x^{2} = \\frac{1}{a}y = -(-\\frac{1}{a})y, then the focus is (0, -(-\\frac{1}{4a})), that is, (0, \\frac{1}{4a}). In summary, the focus is (0, \\frac{1}{4a})." }, { "text": "If the asymptotes of the hyperbola $C$: $x^{2}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ are given by $y=\\pm 7 x$, then $b=$?", "fact_expressions": "C: Hyperbola;b: Number;b>0;Expression(C)=(x^2-y^2/b^2 = 1);Expression(Asymptote(C))=(y=pm*7*x)", "query_expressions": "b", "answer_expressions": "7", "fact_spans": "[[[1, 43]], [[63, 66]], [[8, 43]], [[1, 43]], [[1, 61]]]", "query_spans": "[[[63, 68]]]", "process": "The asymptotes of the hyperbola are given by the equation $x^{2}-\\frac{y^{2}}{b^{2}}=0$, that is, $y=\\pm bx$. According to the problem, the equations of the asymptotes of this hyperbola are $y=\\pm7x$, and $b>0$, so we can conclude that $b=7$." }, { "text": "Given an ellipse $T$: $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ $(a > b > 0)$ with eccentricity $\\frac{\\sqrt{2}}{2}$, and right focus $F(1, 0)$. The three vertices of triangle $ABC$ lie on the ellipse $T$. Let the midpoints of its three sides $AB$, $BC$, $AC$ be $D$, $E$, $M$, respectively, and the slopes of the lines containing these three sides be $k_1$, $k_2$, $k_3$, respectively, where $k_1$, $k_2$, $k_3$ are all nonzero. Let $O$ be the origin. If the sum of the slopes of lines $OD$, $OE$, $OM$ is $1$, then $\\frac{1}{k_{1}} + \\frac{1}{k_{2}} + \\frac{1}{k_{3}} = $?", "fact_expressions": "T: Ellipse;Expression(T) = (x^2/a^2 + y^2/b^2 = 1);a: Number;b: Number;a > b;b > 0;Eccentricity(T) = sqrt(2)/2;F: Point;Coordinate(F) = (1, 0);RightFocus(T) = F;A: Point;B: Point;C: Point;PointOnCurve(Vertex(TriangleOf(A, B, C)), T) = True;MidPoint(LineSegmentOf(A, B)) = D;MidPoint(LineSegmentOf(B, C)) = E;MidPoint(LineSegmentOf(A, C)) = M;D: Point;E: Point;M: Point;k1: Number;k2: Number;k3: Number;Slope(OverlappingLine(LineSegmentOf(A, B))) = k1;Slope(OverlappingLine(LineSegmentOf(B, C))) = k2;Slope(OverlappingLine(LineSegmentOf(A, C))) = k3;Negation(k1 = 0);Negation(k2 = 0);Negation(k3 = 0);O: Origin;Slope(LineOf(O, D)) + Slope(LineOf(O, E)) + Slope(LineOf(O, M)) = 1", "query_expressions": "1/k2 + 1/k1 + 1/k3", "answer_expressions": "-2", "fact_spans": "[[[2, 54], [107, 112]], [[2, 54]], [[8, 54]], [[8, 54]], [[8, 54]], [[8, 54]], [[2, 79]], [[83, 91]], [[83, 91]], [[2, 91]], [[95, 100]], [[95, 100]], [[95, 100]], [[92, 113]], [[120, 151]], [[120, 151]], [[120, 151]], [[140, 143]], [[144, 147]], [[148, 151]], [[165, 170], [184, 189]], [[171, 176], [190, 195]], [[177, 182], [196, 201]], [[153, 182]], [[153, 182]], [[153, 182]], [[184, 207]], [[184, 207]], [[184, 207]], [[208, 211]], [[218, 246]]]", "query_spans": "[[[248, 299]]]", "process": "" }, { "text": "Given that the foci of ellipse $C$ are $F_{1}(-2 \\sqrt{2}, 0)$, $F_{2}(2 \\sqrt{2}, 0)$, and the length of the major axis is $6$. Let the line $y = x + 2$ intersect ellipse $C$ at points $A$ and $B$. Find the coordinates of the midpoint of segment $AB$.", "fact_expressions": "C: Ellipse;G: Line;B: Point;A: Point;F1: Point;F2: Point;Expression(G) = (y = x + 2);Coordinate(F1) = (-2*sqrt(2), 0);Coordinate(F2) = (2*sqrt(2), 0);Focus(C) = {F1,F2};Length(MajorAxis(C)) = 6;Intersection(G, C) = {A, B}", "query_expressions": "Coordinate(MidPoint(LineSegmentOf(A,B)))", "answer_expressions": "(-9/5,1/5)", "fact_spans": "[[[2, 7], [77, 82]], [[67, 76]], [[87, 90]], [[83, 86]], [[10, 33]], [[35, 57]], [[67, 76]], [[10, 33]], [[34, 57]], [[2, 57]], [[2, 65]], [[67, 92]]]", "query_spans": "[[[94, 108]]]", "process": "From the given conditions, the foci of the ellipse lie on the x-axis, where c=2\\sqrt{2}, a=3, thus b=1, ∴ its standard equation is: \\frac{x^{2}}{9}+y^{2}=1. Solving the system of equations \\begin{cases}\\frac{x^{2}}{2}+y^{2}=1\\\\y=x+2\\end{cases}, eliminating y gives 10x^{2}+36x+27=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), and let M(x_{0},y_{0}) be the midpoint of segment AB, then x_{1}+x_{2}=-\\frac{18}{5}, x_{0}=\\frac{x_{1}+x_{2}}{2}=-\\frac{9}{5} ∴ y_{0}=x_{0}+2=\\frac{1}{5}, that is, the coordinates of the midpoint of segment AB are (-\\frac{9}{5},\\frac{1}{5})." }, { "text": "Given that point $O$ is the coordinate origin, the parabola $y^{2}=3x$ intersects a line passing through the focus at points $A$ and $B$. Then $\\overrightarrow{OA} \\cdot \\overrightarrow{OB}$ equals?", "fact_expressions": "G: Parabola;H: Line;O: Origin;A: Point;B: Point;Expression(G) = (y^2 = 3*x);PointOnCurve(Focus(G),H);Intersection(H,G) = {A, B}", "query_expressions": "DotProduct(VectorOf(O, A), VectorOf(O, B))", "answer_expressions": "-27/16", "fact_spans": "[[[12, 26]], [[31, 33]], [[2, 6]], [[35, 38]], [[39, 42]], [[12, 26]], [[12, 33]], [[12, 44]]]", "query_spans": "[[[46, 98]]]", "process": "Let A(\\frac{y^{2}}{3},y_{1}), B(\\frac{y^{2}}{3},y_{2}). When the slope of line AB does not exist, y_{1}=p=\\frac{3}{2}, y_{2}=-p=\\frac{3}{2}, so \\overrightarrow{OA}\\cdot\\overrightarrow{OB}=(\\frac{y^{2}}{3},y_{1})(\\frac{y^{2}}{3},y_{2})=\\frac{1}{9}y_{1}^{2}y_{2}+y_{1}y_{2}=-\\frac{27}{16}. When the slope of line AB exists, let the equation be x=my+\\frac{3}{4} (m\\neq0), solving simultaneously with the parabola gives: y^{2}-3my-\\frac{9}{4}=0, so y_{1}y_{2}=-\\frac{9}{4}, \\therefore\\overrightarrow{OA}\\cdot\\overrightarrow{OB}=(\\frac{y_{1}^{2}}{3},y_{1})(\\frac{y_{2}^{2}}{3},y_{2})=\\frac{1}{9}y_{1}^{2}y_{2}^{2}+y_{1}y_{2}=-\\frac{27}{16}." }, { "text": "The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ passes through the point $(\\sqrt{3}, 2)$, and has eccentricity $3$. What is the length of its imaginary axis?", "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) = (sqrt(3), 2);PointOnCurve(H, G);Eccentricity(G)=3", "query_expressions": "Length(ImageinaryAxis(G))", "answer_expressions": "4*sqrt(5)", "fact_spans": "[[[0, 56], [85, 86]], [[3, 56]], [[3, 56]], [[58, 74]], [[3, 56]], [[3, 56]], [[0, 56]], [[58, 74]], [[0, 74]], [[0, 83]]]", "query_spans": "[[[85, 92]]]", "process": "From the given conditions, we have\n\\begin{cases}\\frac{3}{a}-\\frac{4}{b^{2}}=1,\\\\e^{2}=1+\\frac{b^{2}}{a}=9,\\\\a>0,b>0,\\end{cases}\nsolving which yields\n\\begin{cases}a=\\frac{\\sqrt{10}}{2},\\\\b=2\\sqrt{5}\\end{cases}.\nTherefore, the length of the imaginary axis of this hyperbola is $2b=4\\sqrt{5}$." }, { "text": "Given that the focus of the parabola $y = a x^{2} - 1$ is at the origin, what is the area of the triangle with vertices at the three intersection points of the parabola and the coordinate axes?", "fact_expressions": "G: Parabola;a: Number;O: Origin;Expression(G) = (y = a*x^2 - 1);Focus(G) = O;A: Point;B: Point;C: Point;Intersection(G, axis) = {A, B, C}", "query_expressions": "Area(TriangleOf(A, B, C))", "answer_expressions": "2", "fact_spans": "[[[2, 18], [29, 32]], [[5, 18]], [[22, 26]], [[2, 18]], [[2, 26]], [], [], [], [[29, 42]]]", "query_spans": "[[[28, 53]]]", "process": "" }, { "text": "Let $F_{1}$ and $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$ and $B$, such that $\\overrightarrow{A F}_{1} \\cdot \\overrightarrow{A F_{2}}=0$ and $\\frac{|A F_{2}|}{|B F_{2}|}=\\frac{1}{2}$. Find the eccentricity of the hyperbola $C$.", "fact_expressions": "l: Line;C: Hyperbola;b: Number;a: Number;A: Point;F2: Point;B: 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(F2, l);Intersection(l,RightPart(C))={A, B};DotProduct(VectorOf(A,F1),VectorOf(A,F2))=0;Abs(LineSegmentOf(A, F2))/Abs(LineSegmentOf(B, F2)) = 1/2", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(17)/3", "fact_spans": "[[[95, 100]], [[17, 78], [101, 107], [227, 233]], [[24, 78]], [[24, 78]], [[111, 114]], [[9, 16], [87, 94]], [[115, 118]], [[1, 8]], [[24, 78]], [[24, 78]], [[17, 78]], [[1, 85]], [[1, 85]], [[86, 100]], [[95, 120]], [[122, 181]], [[184, 225]]]", "query_spans": "[[[227, 239]]]", "process": "Using the definition of hyperbola to express respectively, then using the Pythagorean theorem and the definition of hyperbola to establish an equality relationship, find the eccentricity of the hyperbola. Let region, function, \\prime. According to the definition of hyperbola, region that is, obtain region, B.B's, obtain region, B in, B, that is, obtain function. According to the definition of hyperbola |AF_{1}|-|AF_{2}|=2a, that is given, obtain \\oversetfrown{CB}. Therefore, obtain e=\\frac{c}{a}=\\frac{\\sqrt{17}}{3}" }, { "text": "A line passing through the focus of the parabola $y=\\frac{1}{4} x^{2}$ intersects the parabola at points $A$ and $B$. The ordinate of the midpoint of $A$ and $B$ is $2$. What is the length of the chord $AB$?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2/4);H: Line;PointOnCurve(Focus(G), H);A: Point;B: Point;Intersection(H, G) = {A, B};YCoordinate(MidPoint(LineSegmentOf(A, B))) = 2;IsChordOf(LineSegmentOf(A, B), G)", "query_expressions": "Length(LineSegmentOf(A, B))", "answer_expressions": "6", "fact_spans": "[[[1, 25], [32, 35]], [[1, 25]], [[28, 30]], [[0, 30]], [[37, 40], [49, 52]], [[43, 46], [55, 58]], [[28, 48]], [[49, 68]], [[32, 75]]]", "query_spans": "[[[71, 80]]]", "process": "" }, { "text": "Through the point $P(1,-2)$, draw a chord $AB$ of the ellipse $3x^{2}+4y^{2}=24$ such that point $P$ bisects the chord $AB$. Then the equation of the line containing the chord $AB$ is?", "fact_expressions": "G: Ellipse;A: Point;B: Point;P: Point;Expression(G) = (3*x^2 + 4*y^2 = 24);Coordinate(P) = (1, -2);PointOnCurve(P, LineSegmentOf(A, B));IsChordOf(LineSegmentOf(A, B), G);MidPoint(LineSegmentOf(A, B)) = P", "query_expressions": "Expression(OverlappingLine(LineSegmentOf(A, B)))", "answer_expressions": "3*x - 8*y - 19 = 0", "fact_spans": "[[[13, 35]], [[37, 42]], [[37, 42]], [[2, 12], [45, 49]], [[13, 35]], [[2, 12]], [[0, 42]], [[13, 42]], [[45, 57]]]", "query_spans": "[[[60, 74]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=2, y_{1}+y_{2}=-4, k_{AB}=\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}, \\begin{cases}3x_{1}^{2}+4y_{1}^{2}=1\\\\3x_{2}^{2}+4y_{2}^{2}=1\\end{cases}. Subtracting the two equations yields 3(x_{1}+x_{2})+4(y_{1}+y_{2})\\times\\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=0, 3\\times2+4\\times(-4)\\times k_{AB}=0, k_{AB}=\\frac{3}{8}. The equation of line AB is y+2=\\frac{3}{8}(x-1), or 3x-8y-19=0." }, { "text": "The focal length of the ellipse $\\frac{x^{2}}{10-m}+\\frac{y^{2}}{m-2}=1$ is $4$, then $m$ equals?", "fact_expressions": "G: Ellipse;m: Number;Expression(G) = (x^2/(10 - m) + y^2/(m - 2) = 1);FocalLength(G) = 4", "query_expressions": "m", "answer_expressions": "{4,8}", "fact_spans": "[[[0, 42]], [[51, 54]], [[0, 42]], [[0, 49]]]", "query_spans": "[[[51, 57]]]", "process": "According to the problem, if the foci are on the x-axis, 10−m−m+2=4 ∴ m=4; if the foci are on the y-axis, m−2−10+m=4 ∴ m=8; in conclusion: m=4 or 8. This question examines the standard equation of an ellipse and tests students' understanding of concepts, ability to discuss by cases, and mathematical computation skills, and is a basic problem." }, { "text": "Given that a line $ l $ with slope $ k $ 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 $ k = $?", "fact_expressions": "l: Line;C: Parabola;A: Point;B: Point;k: Number;Expression(C) = (y^2 = 4*x);Slope(l)=k;PointOnCurve(Focus(C),l);Intersection(l,C)={A,B};Abs(LineSegmentOf(A, B))=8", "query_expressions": "k", "answer_expressions": "pm*1", "fact_spans": "[[[9, 14]], [[15, 34], [40, 46]], [[48, 51]], [[52, 55]], [[5, 8], [70, 73]], [[15, 34]], [[2, 14]], [[9, 37]], [[9, 57]], [[59, 68]]]", "query_spans": "[[[70, 75]]]", "process": "It is clear that the focus of the parabola $ C: y^{2} = 4x $ has coordinates $ (1, 0) $. If $ l $ is perpendicular to the $ x $-axis, then $ |AB| = 4 $, which does not satisfy the condition. Thus, assume the equation of the required 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 have $ x_{1} + x_{2} = \\frac{2k^{2} + 4}{k^{2}} $. Since $ AB $ passes through the focus, it follows that $ |AB| = x_{1} + x_{2} + p = \\frac{2k^{2} + 4}{k^{2}} + 2 = 8 $, so $ \\frac{2k^{2} + 4}{k^{2}} = 6 $. Solving gives $ k = \\pm 1 $." }, { "text": "The equation of a hyperbola centered at the origin, with one focus at $(5,0)$, and asymptotes given by the lines $y=\\pm \\frac{3}{4} x$ is?", "fact_expressions": "G: Hyperbola;O: Origin;Center(G) = O;H: Line;Expression(H) = (y = pm*((3/4)*x));Asymptote(G) = H;Coordinate(OneOf(Focus(G)))=(5,0)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16-y^2/9=1", "fact_spans": "[[[51, 54]], [[3, 7]], [[0, 54]], [[23, 46]], [[23, 46]], [[22, 54]], [8, 52]]", "query_spans": "[[[51, 58]]]", "process": "" }, { "text": "Given the parabola $C$: $y=\\frac{1}{4} x^{2}$ has focus $F$, and its directrix intersects the coordinate axes at point $E$. A line $l$ passing through $F$ intersects the parabola $C$ at points $A$ and $B$. If $3 \\overrightarrow{E F}=\\overrightarrow{E A}+2 \\overrightarrow{E B}$, then the slope $k$ of line $l$ is?", "fact_expressions": "l: Line;C: Parabola;E: Point;F: Point;A: Point;B: Point;k: Number;Expression(C) = (y = x^2/4);Focus(C) = F;Intersection(Directrix(C),axis)=E;PointOnCurve(F, l);Intersection(l, C) = {A, B};3*VectorOf(E, F) = VectorOf(E, A) + 2*VectorOf(E, B);Slope(l) = k", "query_expressions": "k", "answer_expressions": "pm*sqrt(2)/4", "fact_spans": "[[[58, 63], [153, 158]], [[2, 31], [39, 40], [64, 70]], [[48, 52]], [[35, 38], [54, 57]], [[72, 75]], [[76, 79]], [[161, 164]], [[2, 31]], [[2, 38]], [[39, 52]], [[53, 63]], [[58, 81]], [[83, 151]], [[153, 164]]]", "query_spans": "[[[161, 166]]]", "process": "" }, { "text": "Let $ P $ be an arbitrary point on the parabola $ y^2 = 4x $, and let $ Q $ be the projection of $ P $ onto the $ y $-axis. Given the point $ M(7,8) $, find the minimum value of the sum of the lengths $ |PM| $ and $ |PQ| $.", "fact_expressions": "P: Point;G: Parabola;Expression(G) = (y^2 = 4*x);PointOnCurve(P, G);Projection(P, yAxis) = Q;Q: Point;M: Point;Coordinate(M) = (7, 8)", "query_expressions": "Min(Abs(LineSegmentOf(P, M)) + Abs(LineSegmentOf(P, Q)))", "answer_expressions": "9", "fact_spans": "[[[0, 3], [24, 27]], [[4, 18]], [[4, 18]], [[0, 23]], [[24, 40]], [[37, 40]], [[41, 50]], [[41, 50]]]", "query_spans": "[[[52, 77]]]", "process": "The focus of the parabola y^{2}=4x is F(1,0), and the directrix is the line x=1. Therefore, |PQ|=|PF|-1. Connect MF, then the minimum value of |PM|+|PF| is |MF|=\\sqrt{(7-1)^{2}+8^{2}}=10. Therefore, the minimum value of |PM|+|PQ| is 10-1=9." }, { "text": "If a moving circle passes through the fixed point $A(-3,0)$ and is externally tangent to the fixed circle $C$: $(x-3)^{2}+y^{2}=4$, then what is the equation of the locus of the center $P$ of the moving circle?", "fact_expressions": "C: Circle;Z: Circle;A: Point;P: Point;Expression(C) = (y^2 + (x - 3)^2 = 4);Coordinate(A) = (-3, 0);Center(Z) = P;PointOnCurve(A, Z);IsOutTangent(Z, C)", "query_expressions": "LocusEquation(P)", "answer_expressions": "(x^2 - y^2/8 = 1)&(x <= -1)", "fact_spans": "[[[19, 42]], [[1, 3], [46, 48]], [[6, 15]], [[50, 53]], [[19, 42]], [[6, 15]], [[46, 53]], [[1, 15]], [[1, 44]]]", "query_spans": "[[[50, 60]]]", "process": "Let the radius of the moving circle P be r, then |PA| = r. Since the two circles are externally tangent, |PC| = 2 + r, thus |PC| - |PA| = 2. Then solve using the definition of a hyperbola. The center of the fixed circle is C(3,0), symmetric to A(-3,0) about the origin. Let the radius of the moving circle P be r, then |PA| = r. Because the two circles are externally tangent, |PC| = 2 + r, so |PC| - |PA| = 2 < |AC| = 6. Therefore, the locus of point P is the left branch of a hyperbola with foci at A and C. Thus, a = 1, c = 3, b^{2} = c^{2} - a^{2} = 8, so the equation of the locus is x^{2} - \\frac{y^{2}}{2} = 1 (x \\leqslant -1)." }, { "text": "Given that $F$ is the focus of the parabola $y^{2}=6 x$, $P$ is any point on the parabola, and point $B(4,3)$, find the minimum value of $|P B|+|P F|$.", "fact_expressions": "G: Parabola;B: Point;P: Point;F: Point;Expression(G) = (y^2 = 6*x);Focus(G) = F;PointOnCurve(P, G) = True;Coordinate(B) = (4, 3)", "query_expressions": "Min(Abs(LineSegmentOf(P, B)) + Abs(LineSegmentOf(P, F)))", "answer_expressions": "11/2", "fact_spans": "[[[6, 20], [28, 31]], [[38, 47]], [[24, 27]], [[2, 5]], [[6, 20]], [[2, 23]], [[24, 37]], [[38, 47]]]", "query_spans": "[[[49, 68]]]", "process": "" }, { "text": "Given that line $l$ is tangent to the parabola $x^{2}=2 p y (p>0)$ and the circle $x^{2}+(y+1)^{2}=1$ at points $A$ and $B$ ($A$, $B$ are distinct), point $F$ is the focus of the parabola. When $|A F|$ attains its minimum value, $p=$?", "fact_expressions": "l: Line;G: Parabola;Expression(G) = (x^2 = 2*(p*y));p: Number;p>0;H: Circle;Expression(H) = (x^2 + (y + 1)^2 = 1);A: Point;B: Point;TangentPoint(l, G) = A;TangentPoint(l, H) = B;Negation(A=B);F: Point;Focus(G) = F;WhenMin(Abs(LineSegmentOf(A, F)))", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 7]], [[10, 33], [83, 86]], [[10, 33]], [[105, 108]], [[13, 33]], [[34, 54]], [[34, 54]], [[55, 59], [65, 68]], [[61, 64], [70, 73]], [[2, 64]], [[2, 64]], [[65, 76]], [[78, 82]], [[78, 89]], [[90, 104]]]", "query_spans": "[[[105, 110]]]", "process": "Let A(x_{0},y_{0}), y = \\frac{x}{p}, so y_{x} = x_{0} = \\frac{x_{0}}{P} thus y - y_{0} = \\frac{x_{0}}{p}(x - x_{0}) so the equation of line l is x_{0}x = py + py_{0}, then \\frac{|p(y_{0}-1)|}{\\sqrt{x_{0}^{2}+p^{2}}} = 1, substituting x_{0}^{2} = 2py_{0}, we obtain \\begin{cases}x_{0}^{2}=4+4p,\\\\y_{0}=\\frac{2}{D}+2,\\end{cases}\\therefore|AF|=\\frac{2}{p}+2+\\frac{p}{2}\\geqslant4, equality holds if and only if p=2, therefore p=2." }, { "text": "The ellipse passes through the point $(3 , 0)$, and the eccentricity is $e=\\frac{\\sqrt{6}}{3}$. What is the equation of the ellipse?", "fact_expressions": "G: Ellipse;H: Point;Coordinate(H) = (3, 0);PointOnCurve(H, G);e: Number;Eccentricity(G) = e;e = sqrt(6)/3", "query_expressions": "Expression(G)", "answer_expressions": "{x^2/9 + y^2/3 = 1, x^2/9 + y^2/27 = 1}", "fact_spans": "[[[0, 2], [40, 42]], [[3, 13]], [[3, 13]], [[0, 13]], [[17, 39]], [[0, 39]], [[17, 39]]]", "query_spans": "[[[40, 47]]]", "process": "When the foci of the ellipse are on the x-axis, let the equation of the ellipse be: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0), then a=3, e=\\frac{c}{a}=\\frac{\\sqrt{6}}{3}, so c=\\sqrt{6}, b^{2}=3, so the equation of the ellipse is: \\frac{x^{2}}{9}+\\frac{y^{2}}{3}=1; when the foci of the ellipse are on the y-axis, let the equation of the ellipse be: \\frac{y^{2}}{a^{2}}+\\frac{x^{2}}{b^{2}}=1 (a>b>0), then b=3, e=\\frac{c}{a}=\\frac{\\sqrt{6}}{3}, so e^{2}=\\frac{c^{2}}{a^{2}}=\\frac{a^{2}-b^{2}}{a^{2}}=1-\\frac{b^{2}}{a^{2}}=\\frac{2}{3}, a^{2}=27, so the equation of the ellipse is: \\frac{y^{2}}{27}+\\frac{x^{2}}{9}=1. This problem examines finding the standard equation of an ellipse. The basic method is the undetermined coefficient method. The specific steps are: first determine the form, then determine the quantities, that is, first determine the location of the foci, and then establish a system of equations in terms of a and b based on the given conditions. After solving for a and b, the standard equation of the ellipse can be obtained. This is a basic problem." }, { "text": "Given that $F_{1}(0,-2)$, $F_{2}(0,2)$ are two foci of the ellipse $\\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}=1$ $(a>b>0)$, and the length of the major axis of the ellipse is $4 \\sqrt{2}$. Then the equation of this ellipse is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/b^2 + y^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;F1: Point;F2: Point;Coordinate(F1) = (0, -2);Coordinate(F2) = (0, 2);Focus(G) = {F1, F2};Length(MajorAxis(G)) = 4*sqrt(2)", "query_expressions": "Expression(G)", "answer_expressions": "y^2/8+x^2/4=1", "fact_spans": "[[[30, 82], [88, 90], [111, 113]], [[30, 82]], [[32, 82]], [[32, 82]], [[32, 82]], [[32, 82]], [[2, 15]], [[17, 29]], [[2, 15]], [[17, 29]], [[2, 86]], [[88, 107]]]", "query_spans": "[[[111, 118]]]", "process": "Since $ F_{1}(0,-2) $, $ F_{2}(0,2) $ are two foci of the ellipse $ \\frac{x^{2}}{b^{2}}+\\frac{y^{2}}{a^{2}}=1 $ $ (a>b>0) $, we have $ c=2 $. Since the length of the major axis of the ellipse is $ 4\\sqrt{2} $, we have $ 2a=4\\sqrt{2} $, hence $ a=2\\sqrt{2} $. Therefore, $ b^{2}=a^{2}-c^{2}=4 $. Thus, the equation of the ellipse is $ \\frac{y^{2}}{8}+\\frac{x^{2}}{4}=1 $." }, { "text": "The focus of the parabola $y^{2}=4 x$ is $F$, and a point $P$ lies on the parabola. If $|P F|=5$, then the area of $\\triangle POF$ is?", "fact_expressions": "G: Parabola;P: Point;O: Origin;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(P, G);Abs(LineSegmentOf(P, F)) = 5", "query_expressions": "Area(TriangleOf(P, O, F))", "answer_expressions": "2", "fact_spans": "[[[2, 16], [24, 27]], [[30, 33]], [[46, 61]], [[20, 23]], [[2, 16]], [[2, 23]], [[24, 33]], [[35, 44]]]", "query_spans": "[[[46, 66]]]", "process": "Given that the focus of the parabola $ y^2 = 4x $ is $ F $, the focus is $ F(1,0) $. Since $ |PF| = 5 $, by the definition of a parabola, the x-coordinate of point $ P $ is $ x_0 = 5 - 1 = 4 $. Substituting into $ y^2 = 4x $ gives the y-coordinate $ y_0 = 4 $. Therefore, the area of $ \\triangle POF $ is $ S = \\frac{1}{2} \\times 1 \\times 4 = 2 $." }, { "text": "A focus of the hyperbola $C$ is $(5,0)$, and an asymptote has equation $3x+4y=0$. Then the equation of hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;Coordinate(OneOf(Focus(C))) = (5, 0);Expression(OneOf(Asymptote(C)))=(3*x+4*y=0)", "query_expressions": "Expression(C)", "answer_expressions": "x^2/16-y^2/9=1", "fact_spans": "[[[0, 6], [42, 48]], [[0, 19]], [[0, 40]]]", "query_spans": "[[[42, 52]]]", "process": "Let the hyperbola equation be \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1. From the given conditions, c=5, and the asymptote equation is 3x+4y=0, then \\frac{b}{a}=\\frac{3}{4}, and a^{2}+b^{2}=c^{2}=5^{2}, \\therefore a=4, b=3. The hyperbola equation is \\frac{x^{2}}{16}-\\frac{y^{2}}{9}=1" }, { "text": "Given that $F$ is the focus of the parabola $y = x^{2}$, $M$ and $N$ are two points on this parabola, and $|MF| + |NF| = 3$, then what is the distance from the midpoint of segment $MN$ to the $x$-axis?", "fact_expressions": "G: Parabola;Expression(G) = (y = x^2);F: Point;Focus(G) = F;M: Point;N: Point;PointOnCurve(M, G);PointOnCurve(N, G);Abs(LineSegmentOf(M, F)) + Abs(LineSegmentOf(N, F)) = 3", "query_expressions": "Distance(MidPoint(LineSegmentOf(M, N)), xAxis)", "answer_expressions": "5/4", "fact_spans": "[[[6, 18], [31, 34]], [[6, 18]], [[2, 5]], [[2, 21]], [[22, 25]], [[26, 29]], [[22, 37]], [[22, 37]], [[38, 53]]]", "query_spans": "[[[55, 75]]]", "process": "According to the problem, the focus coordinates and directrix equation of the parabola can be obtained. Using the definition of a parabola, transform the distances from M and N to the focus into their distances to the directrix for calculation. The focus of the parabola is (0,\\frac{1}{4}), the directrix is y=-\\frac{1}{4}. Draw perpendiculars from M and N to the directrix respectively, then |MM'|=|MF|, |NN'|=|NF|, so |MM'|+|NN'|=|MF|+|NF|=3, thus the midline |PP'|=\\frac{|MM'|+|NN'|}{2}=\\frac{3}{2}, therefore the distance from the midpoint P of segment MN to the x-axis is |PP'|-\\frac{1}{4}=\\frac{3}{2}-\\frac{1}{4}=\\frac{5}{4}" }, { "text": "The eccentricity of the hyperbola $x^{2}-y^{2}=6$ is?", "fact_expressions": "G: Hyperbola;Expression(G) = (x^2 - y^2 = 6)", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[0, 18]], [[0, 18]]]", "query_spans": "[[[0, 24]]]", "process": "" }, { "text": "If a point on the parabola $y^{2}=2 p x(p>0)$ has distances of $5$ and $4$ to the focus and to the axis of symmetry of the parabola, respectively, then what is the equation of the parabola?", "fact_expressions": "G: Parabola;p: Number;F:Point;p>0;Expression(G) = (y^2 = 2*(p*x));PointOnCurve(F, G);Distance(F,Focus(G)) = 5;Distance(F,SymmetryAxis(G)) = 4", "query_expressions": "Expression(G)", "answer_expressions": "{y^2=4*x,y^2=16*x}", "fact_spans": "[[[1, 22], [51, 54], [30, 33]], [[4, 22]], [], [[4, 22]], [[1, 22]], [[1, 25]], [[1, 49]], [[1, 49]]]", "query_spans": "[[[51, 59]]]", "process": "Let the point be in the first quadrant. According to the problem, the ordinate of the point is 4, so the coordinates of the point are (\\frac{8}{p}, 4). By the definition of the parabola, we have \\frac{8}{p} + \\frac{p}{2} = 5. Rearranging gives p^{2} - 10p + 16 = 0. Solving this equation yields p = 2 or p = 8. Therefore, the equation of the parabola is y^{2} = 4x or y^{2} = 16x." }, { "text": "Given that a focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $(3 , 0)$, and a vertex is $(1,0)$, what is the equation of its asymptotes?", "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) = (3, 0);OneOf(Focus(G)) = H;Coordinate(OneOf(Vertex(G)))=(1,0)", "query_expressions": "Expression(Asymptote(G))", "answer_expressions": "y=pm*2*sqrt(2)*x", "fact_spans": "[[[2, 58], [91, 92]], [[5, 58]], [[5, 58]], [[64, 73]], [[5, 58]], [[5, 58]], [[2, 58]], [[64, 73]], [[2, 73]], [2, 70]]", "query_spans": "[[[91, 99]]]", "process": "Let the focal distance of the hyperbola be 2c. From the given conditions, the values of a and c can be known, then the value of b can be found according to b=\\sqrt{c^{2}-a^{2}}, and thus the result can be obtained. Let the focal distance of the hyperbola be 2c. From the problem, we know c=3, a=, so b=\\sqrt{c^{2}-a^{2}}=2\\sqrt{2}, therefore the asymptotes of the hyperbola are y=\\pm\\frac{b}{a}=\\pm2\\sqrt{2}x" }, { "text": "Given that $F$ is the focus of the parabola $y^{2}=4x$, and its directrix intersects the $x$-axis at point $P$. A line $l$ passing through $P$ intersects the parabola at points $A$ and $B$. Let the slopes of $FA$ and $FB$ be $m$ and $n$, respectively. Then $\\frac{m}{n}=$?", "fact_expressions": "l: Line;G: Parabola;F: Point;A: Point;B: Point;P:Point;m:Number;n:Number;Expression(G) = (y^2 = 4*x);Focus(G) = F;Intersection(Directrix(G), xAxis) = P;PointOnCurve(P, l);Intersection(l, G) = {A, B};Slope(LineSegmentOf(F, A)) = m;Slope(LineSegmentOf(F,B))=n", "query_expressions": "m/n", "answer_expressions": "-1", "fact_spans": "[[[44, 49]], [[6, 20], [24, 25], [50, 53]], [[2, 5]], [[55, 58]], [[59, 62]], [[34, 38], [40, 43]], [[84, 87]], [[88, 92]], [[6, 20]], [[2, 23]], [[24, 38]], [[39, 49]], [[44, 64]], [[66, 92]], [[66, 92]]]", "query_spans": "[[[94, 109]]]", "process": "\\because F(1,0), P(-1,0), let A(x_{1},y_{1}), B(x_{2},y_{2}), l: y = k(x - 1), from \\begin{cases} y = k(x + 1) \\\\ y^{2} = 4x \\end{cases} k^{2}x^{2} + (2k^{2} - 4)x + k^{2} = 0 \\Rightarrow x_{1} + x_{2} = \\frac{4 - 2k^{2}}{k^{2}}, x_{1} \\cdot x_{2} = 1, \\frac{-\\frac{x_{2} - 1}{k(x_{2} + 1)}}{+(x_{1} - x_{2}) - 1} = -1" }, { "text": "Given the hyperbola $3 x^{2}-y^{2}=9$, then the eccentricity of this hyperbola is equal to?", "fact_expressions": "G: Hyperbola;Expression(G) = (3*x^2 - y^2 = 9)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[2, 22], [25, 28]], [[2, 22]]]", "query_spans": "[[[25, 35]]]", "process": "The standard equation of the hyperbola is \\frac{x2}{3}-\\frac{y^{2}}{9}=1,\\thereforea=\\sqrt{3},c=2\\sqrt{3},\\thereforee=\\frac{c}{a}=2" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{5}=1$ $(a>0)$ has a focal distance of $10$, then what is the equation of the asymptotes of hyperbola $C$?", "fact_expressions": "C: Hyperbola;a: Number;Expression(C) = (-y^2/5 + x^2/a^2 = 1);a>0;FocalLength(C) = 10", "query_expressions": "Expression(Asymptote(C))", "answer_expressions": "y = pm*x/2", "fact_spans": "[[[2, 54], [64, 70]], [[10, 54]], [[2, 54]], [[10, 54]], [[2, 62]]]", "query_spans": "[[[64, 78]]]", "process": "From the problem, the foci of the hyperbola lie on the x-axis, so $ c = \\sqrt{a^{2} + 5} = 5 $, $ \\therefore a = 2\\sqrt{5} $, therefore the asymptotes of the hyperbola are $ y = \\pm\\frac{\\sqrt{5}}{2\\sqrt{5}}x = \\pm\\frac{1}{2}x $." }, { "text": "The focus of the parabola $C$: $y^{2}=2 p x$ has coordinates $F(\\frac{1}{2}, 0)$. Then the equation of the parabola $C$ is? If point $P$ moves on the parabola $C$, and point $Q$ moves on the line $x+y+5=0$, then the minimum value of $P Q$ equals?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;F: Point;Coordinate(F) = (1/2, 0);Focus(C) = F;P: Point;PointOnCurve(P, C);G: Line;Expression(G) = (x + y + 5 = 0);Q: Point;PointOnCurve(Q, G)", "query_expressions": "Expression(C);Min(LineSegmentOf(P, Q))", "answer_expressions": "y^2=2*x\n9*sqrt(2)/4", "fact_spans": "[[[0, 21], [49, 55], [66, 72]], [[0, 21]], [[8, 21]], [[27, 47]], [[27, 47]], [[0, 47]], [[61, 65]], [[61, 75]], [[81, 92]], [[81, 92]], [[76, 80]], [[76, 95]]]", "query_spans": "[[[49, 60]], [[97, 109]]]", "process": "" }, { "text": "The coordinates of the focus of the parabola $y^{2}=x$ are?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = x)", "query_expressions": "Coordinate(Focus(G))", "answer_expressions": "(1/4,0)", "fact_spans": "[[[0, 12]], [[0, 12]]]", "query_spans": "[[[0, 19]]]", "process": "In the parabola y^{2}=x, p=\\frac{1}{2}, \\therefore the coordinates of the focus are (\\frac{1}{4},0)" }, { "text": "The equation of the directrix of the parabola $y=a x^{2}$ is $y=4$. Then the value of $a$ is?", "fact_expressions": "G: Parabola;a: Number;Expression(G) = (y = a*x^2);Expression(Directrix(G)) = (y = 4)", "query_expressions": "a", "answer_expressions": "-1/16", "fact_spans": "[[[0, 14]], [[27, 30]], [[0, 14]], [[0, 25]]]", "query_spans": "[[[27, 34]]]", "process": "From y = ax^{2}, we get x^{2} = \\frac{1}{a}y, and its directrix equation is y = -\\frac{1}{4a}. Since the directrix equation of the parabola y = ax^{2} is y = 4, we have -\\frac{1}{4a} = 4. Solving for a gives a = -\\frac{1}{16}." }, { "text": "Given the ellipse $C$: $\\frac{x^{2}}{9}+\\frac{y^{2}}{7}=1$, the left focus of the ellipse is $F$. Point $A(-\\sqrt{2}, 1)$, and $P$ is a moving point on the ellipse $C$. Then the minimum value of the perimeter of $\\triangle P A F$ is?", "fact_expressions": "C: Ellipse;A: Point;P: Point;F: Point;Expression(C) = (x^2/9 + y^2/7 = 1);Coordinate(A) = (-sqrt(2), 1);LeftFocus(C) = F;PointOnCurve(P, C)", "query_expressions": "Min(Perimeter(TriangleOf(P, A, F)))", "answer_expressions": "4", "fact_spans": "[[[2, 44], [78, 83]], [[53, 71]], [[73, 77]], [[49, 52]], [[2, 44]], [[53, 71]], [[2, 52]], [[74, 87]]]", "query_spans": "[[[89, 115]]]", "process": "Let the right focus of the ellipse be M, a=3, b=\\sqrt{7}, c=\\sqrt{2}, point A(-\\sqrt{2},1) lies inside the ellipse, point F(-\\sqrt{2},0), and since P is a moving point on the ellipse C, |PF|+|PM|=6, |AF|=1, |AM|=\\sqrt{(2\\sqrt{2})^{2}+1^{2}}=3. The perimeter of \\Delta PAF is |PA|+|AF|+|FP|=|PA|+1+(6-|PM|)=7+(|PA|-|PM|)\\geqslant7-|AM|, with equality holding if and only if P lies at the intersection point of ray MA and the ellipse, so the minimum perimeter is 4." }, { "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, 14]], [[0, 14]]]", "query_spans": "[[[0, 21]]]", "process": "Since the equation of the parabola is x^{2}=4y, the focus lies on the y-axis, and the focus is (0,1)." }, { "text": "Given that an asymptote of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ passes through the point $(\\sqrt{2}, \\sqrt{6})$, what is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;I: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(I) = (sqrt(2), sqrt(6));PointOnCurve(I, OneOf(Asymptote(G)))", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[2, 58], [92, 95]], [[4, 58]], [[4, 58]], [[66, 89]], [[4, 58]], [[4, 58]], [[2, 58]], [[66, 89]], [[2, 89]]]", "query_spans": "[[[92, 101]]]", "process": "Find the asymptotes of the hyperbola, combine with the relationship among a, b, c, then use the eccentricity formula to compute the desired value. The hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ has asymptotes given by $y=\\pm\\frac{b}{a}x$. From the given condition, we have $\\frac{\\sqrt{2}b}{a}=\\sqrt{6}$, so $b=\\sqrt{3}a$, and thus the eccentricity of the hyperbola is $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=\\sqrt{1+3}=$" }, { "text": "The hyperbola $\\frac{x^{2}}{9}-\\frac{y^{2}}{16}=1$ has two foci $F_1$ and $F_2$. Point $P$ lies on the hyperbola. If $P F_1 \\perp P F_2$, then what is the distance from point $P$ to the $x$-axis?", "fact_expressions": "G: Hyperbola;P: Point;F2: Point;Expression(G) = (x^2/9 - y^2/16 = 1);F1: Point;Focus(G) = {F1,F2};PointOnCurve(P, G);IsPerpendicular(LineSegmentOf(P, F1),LineSegmentOf(P, F2))", "query_expressions": "Distance(P, xAxis)", "answer_expressions": "16/5", "fact_spans": "[[[0, 39], [64, 67]], [[59, 63], [91, 95]], [[53, 58]], [[0, 39]], [[45, 50]], [[0, 58]], [[59, 68]], [[70, 89]]]", "query_spans": "[[[91, 105]]]", "process": "" }, { "text": "If the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$ and the hyperbola $x^{2}-\\frac{y^{2}}{8}=1$ have the same foci $F_{1}$, $F_{2}$, and point $P$ is an intersection point of the two curves, then the value of $P F_{1} \\bullet PF_{2}$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/8 = 1);Expression(H) = (x^2/16 + y^2/7 = 1);Focus(G) = {F1, F2};Focus(H) = {F1, F2};OneOf(Intersection(G, H)) = P", "query_expressions": "LineSegmentOf(P, F2)*LineSegmentOf(P, F1)", "answer_expressions": "15", "fact_spans": "[[[40, 68]], [[1, 39]], [[91, 95]], [[74, 81]], [[83, 90]], [[40, 68]], [[1, 39]], [[1, 90]], [[1, 90]], [[91, 105]]]", "query_spans": "[[[107, 135]]]", "process": "Since the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$ and the hyperbola $x^{2}-\\frac{y^{2}}{8}=1$ have the same foci $F_{1}$, $F_{2}$, let $P$ be on the right branch of the hyperbola. Using the definitions of the ellipse and the hyperbola, we obtain: $|PF_{1}|+|PF_{2}|=2\\times4=8$ $\\textcircled{1}$, $|PF_{1}|-|PF_{2}|=2\\times1=2$ $\\textcircled{2}$. From $\\textcircled{1}$ $\\textcircled{2}$, we get $|PF_{1}|=5$, $|PF_{2}|=3$. $\\therefore |PF_{1}|\\cdot|PF_{2}|=15$." }, { "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}$ and $F_{2}$ respectively. A line passing through $F_{2}$ intersects the right branch of the hyperbola at points $P$ and $Q$, such that $P Q \\perp P F_{1}$ and $|P Q|=\\frac{8}{15}|P F_{1}|$. Find the eccentricity of the hyperbola.", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;P: Point;Q: 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(F2, H);Intersection(H, RightPart(G)) = {P, Q};IsPerpendicular(LineSegmentOf(P, Q), LineSegmentOf(P, F1));Abs(LineSegmentOf(P, Q)) = (8/15)*Abs(LineSegmentOf(P, F1))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(26)/4", "fact_spans": "[[[2, 58], [94, 97], [163, 166]], [[5, 58]], [[5, 58]], [[91, 93]], [[101, 104]], [[105, 108]], [[66, 73]], [[74, 81], [83, 90]], [[5, 58]], [[5, 58]], [[2, 58]], [[2, 81]], [[2, 81]], [[82, 93]], [[91, 110]], [[112, 131]], [[132, 161]]]", "query_spans": "[[[163, 172]]]", "process": "As shown in the figure, in the right triangle $ PF_{1}Q $, $ |QF_{1}| = \\sqrt{|PF_{1}|^{2} + |PQ|^{2}} = \\frac{17}{15}|PF_{1}| $, by the definition of the hyperbola we obtain: $ 2a = |PF_{1}| - |PF_{2}| = |QF_{1}| - |QF_{2}| $, from $ |PQ| = \\frac{8}{15}|PF_{1}| $, it follows that $ |PF_{2}| + |QF_{2}| = \\frac{8}{15}|PF_{1}| $, namely $ |PF_{1}| - 2a + \\frac{17}{15}|PF_{1}| - 2a = \\frac{8}{15}|PF_{1}| $, solving gives $ e = \\frac{\\sqrt{26}}{4} $" }, { "text": "Given that a chord passing through the focus $F$ of the parabola $y^{2}=4x$ intersects the parabola at points $A$ and $B$. Perpendiculars are drawn from $A$ and $B$ to the $y$-axis, with feet at $C$ and $D$ respectively. Then the minimum value of $|AC|+|BD|$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);F: Point;Focus(G) = F;L: LineSegment;IsChordOf(L,G) = True;PointOnCurve(F,L) = True;Intersection(L,G) = {A,B};A: Point;B: Point;L1: Line;L2: Line;PointOnCurve(A,L1) = True;IsPerpendicular(L1,yAxis) = True;PointOnCurve(B,L2) = True;IsPerpendicular(L2,yAxis) = True;FootPoint(L1,yAxis) = C;FootPoint(L2,yAxis) = D;C: Point;D: Point", "query_expressions": "Min(Abs(LineSegmentOf(A,C))+Abs(LineSegmentOf(B,D)))", "answer_expressions": "2", "fact_spans": "[[[3, 17], [26, 29]], [[3, 17]], [[20, 23]], [[3, 23]], [], [[2, 25]], [[2, 25]], [[2, 40]], [[31, 34], [42, 45]], [[35, 38], [46, 49]], [], [], [[41, 59]], [[41, 59]], [[41, 59]], [[41, 59]], [[41, 74]], [[41, 74]], [[65, 68]], [[71, 74]]]", "query_spans": "[[[77, 96]]]", "process": "" }, { "text": "Given that the distance from a point $M(1, m)$ on the parabola $y^{2}=2 p x$ to its focus is $5$, then the equation of the directrix of this parabola is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;M: Point;m: Number;Coordinate(M) = (1, m);PointOnCurve(M, G);Distance(M, Focus(G)) = 5", "query_expressions": "Expression(Directrix(G))", "answer_expressions": "x = -4", "fact_spans": "[[[2, 18], [31, 32], [44, 47]], [[2, 18]], [[5, 18]], [[21, 30]], [[21, 30]], [[21, 30]], [[2, 30]], [[21, 41]]]", "query_spans": "[[[44, 54]]]", "process": "Since the distance from point M(1, m) on the parabola y^{2}=2px to its focus is 5, we have 1+\\frac{p}{2}=5. Solving for p gives p=8. Therefore, the directrix equation of this parabola is x=-4." }, { "text": "Given the circle $O$: $x^{2}+y^{2}=9$, point $A(2,0)$, and a moving point $P$. The circle with segment $AP$ as diameter is internally tangent to circle $O$. What is the trajectory equation of the moving point $P$?", "fact_expressions": "O: Circle;G: Circle;A: Point;P: Point;Coordinate(A) = (2, 0);Expression(O) = (x^2 + y^2 = 9);IsDiameter(LineSegmentOf(A, P), G);IsInTangent(G, O)", "query_expressions": "LocusEquation(P)", "answer_expressions": "x^2/9 + y^2/5 = 1", "fact_spans": "[[[2, 23], [58, 62]], [[54, 55]], [[24, 33]], [[34, 38], [66, 69]], [[24, 33]], [[2, 23]], [[42, 55]], [[54, 62]]]", "query_spans": "[[[66, 76]]]", "process": "Problem Analysis: Let M be the midpoint of AP, N be the point of tangency, connect OM and MN, then |OM| + |MN| = |ON| = 3. Take A₁, the symmetric point of A with respect to the y-axis, and connect A₁P. Hence, |A₁P| + |AP| = 2|OM| + |ON| = 6. Therefore, the locus of point P is an ellipse with foci at A₁ and A, and major axis length 6. Here, a = 3, c = 2, b = \\sqrt{5}, so the equation of the locus of point P is \\frac{x^{2}}{9}+\\frac{y^{2}}{5}=1." }, { "text": "Let $P$ be a point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16} = 1$. If $F_{1}$ and $F_{2}$ are the two foci of the ellipse, then $|PF_{1}|+|PF_{2}|$=?", "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}", "query_expressions": "Abs(LineSegmentOf(P, F1)) + Abs(LineSegmentOf(P, F2))", "answer_expressions": "10", "fact_spans": "[[[5, 46], [69, 71]], [[1, 4]], [[51, 58]], [[61, 68]], [[5, 46]], [[1, 49]], [[51, 76]]]", "query_spans": "[[[78, 99]]]", "process": "" }, { "text": "Let $P$ be an arbitrary point on the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{16}=1$, and let point $Q(0, -4)$. Then the maximum value of $|P Q|$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/16 = 1);P: Point;PointOnCurve(P, G);Q: Point;Coordinate(Q) = (0, -4)", "query_expressions": "Max(Abs(LineSegmentOf(P, Q)))", "answer_expressions": "8", "fact_spans": "[[[5, 44]], [[5, 44]], [[1, 4]], [[1, 50]], [[52, 64]], [[52, 64]]]", "query_spans": "[[[66, 79]]]", "process": "" }, { "text": "Given that the distance from point $P(1,0)$ to one of the asymptotes of the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $\\frac{1}{2}$, then the eccentricity of the hyperbola $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;P: Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(P) = (1, 0);Distance(P,OneOf(Asymptote(C)))=1/2", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(5)/2", "fact_spans": "[[[12, 73], [98, 104]], [[19, 73]], [[19, 73]], [[2, 11]], [[19, 73]], [[19, 73]], [[12, 73]], [[2, 11]], [[2, 96]]]", "query_spans": "[[[98, 110]]]", "process": "Test analysis: One asymptote of the hyperbola is given by the equation bx - ay = 0, so \\frac{|b\\times1-a\\times0}{c}=\\frac{1}{2}, c=2b, a=\\sqrt{3}b, e=\\frac{2\\sqrt{3}}{3}" }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$. If point $P$ lies on the hyperbola and $\\overrightarrow{P F_{1}} \\cdot \\overrightarrow{P F_{2}}=0$, then $|\\overrightarrow{P F_{1}}+\\overrightarrow{P F_{2}}|$=?", "fact_expressions": "G: Hyperbola;P: Point;F2: Point;F1: Point;Expression(G) = (x^2 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P, F1), VectorOf(P, F2)) = 0", "query_expressions": "Abs(VectorOf(P, F1) + VectorOf(P, F2))", "answer_expressions": "2*sqrt(10)", "fact_spans": "[[[19, 47], [61, 64]], [[56, 60]], [[9, 16]], [[1, 8]], [[19, 47]], [[1, 53]], [[1, 53]], [[56, 65]], [[68, 127]]]", "query_spans": "[[[130, 185]]]", "process": "" }, { "text": "Given the parabola $ C $: $ y^{2} = 2x $, a line $ l $ passing through the point $ E(a, 0) $ intersects $ C $ at two distinct points $ P(x_{1}, y_{1}) $, $ Q(x_{2}, y_{2}) $, satisfying $ y_{1} y_{2} = -4 $. The segment with midpoint $ Q $ has endpoints $ M $, $ N $, where $ N $ lies on the $ x $-axis and $ M $ lies on $ C $. Then $ a = ? $ and the minimum value of $ |PM| $ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*x);E: Point;a: Number;Coordinate(E) = (a, 0);l: Line;PointOnCurve(E, l);P: Point;Q: Point;x1: Number;x2: Number;y1: Number;y2: Number;Coordinate(P) = (x1, y1);Coordinate(Q) = (x2, y2);Intersection(l, C) = {P, Q};y1*y2 = -4;M: Point;N: Point;MidPoint(H) = Q;Endpoint(H) = {M, N};PointOnCurve(N, xAxis);PointOnCurve(M, C);H:LineSegment;Negation(P=Q)", "query_expressions": "a;Min(Abs(LineSegmentOf(P, M)))", "answer_expressions": "2;4*sqrt(2)", "fact_spans": "[[[2, 20], [39, 42], [147, 150]], [[2, 20]], [[22, 32]], [[153, 156]], [[22, 32]], [[33, 38]], [[21, 38]], [[49, 66]], [[67, 84], [106, 109]], [[49, 66]], [[67, 84]], [[49, 66]], [[67, 84]], [[49, 66]], [[67, 84]], [[33, 84]], [[88, 104]], [[122, 125], [143, 146]], [[126, 130], [133, 136]], [[105, 115]], [[113, 130]], [[133, 142]], [[143, 151]], [[113, 115]], [[44, 84]]]", "query_spans": "[[[153, 158]], [[158, 171]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ has asymptotes with equations $y=\\pm 2 \\sqrt{2} x$, and the distance from point $A(1,2)$ to the right focus $F$ is $2 \\sqrt{2}$, then the equation of $C$ is?", "fact_expressions": "C: Hyperbola;b: Number;a: Number;A: Point;F:Point;a>0;b>0;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(A) = (1, 2);RightFocus(C)=F;Expression(Asymptote(C))=(y=pm*(2*sqrt(2)*x));Distance(A, F) = 2*sqrt(2)", "query_expressions": "Expression(C)", "answer_expressions": "x^2-y^2/8=1", "fact_spans": "[[[2, 62], [124, 127]], [[9, 62]], [[9, 62]], [[90, 99]], [[103, 106]], [[9, 62]], [[9, 62]], [[2, 62]], [[90, 99]], [[2, 106]], [[2, 89]], [[90, 122]]]", "query_spans": "[[[124, 132]]]", "process": "Let the semi-focal length of hyperbola C be c. Since the distance from point A(1,2) to the right focus is 2\\sqrt{2}, we have \\sqrt{(1-c)^{2}+(2-0)^{2}}=2\\sqrt{2}. Solving gives c=3 or c=-1 (discarded). Because \\frac{b}{a}=2\\sqrt{2}, it follows that e=\\frac{c}{a}=3, so a=1, b=2\\sqrt{2}. Therefore, the equation of hyperbola C is x^{2}-\\frac{y^{2}}{8}=1." }, { "text": "The ellipse with foci on the $x$-axis passes through the point $P(3,0)$ and has a focal distance of $2$. What is the eccentricity of the ellipse?", "fact_expressions": "G: Ellipse;P: Point;Coordinate(P) = (3, 0);PointOnCurve(P, G);FocalLength(G) = 2;PointOnCurve(Focus(G), xAxis)", "query_expressions": "Eccentricity(G)", "answer_expressions": "1/3", "fact_spans": "[[[9, 11], [30, 32]], [[12, 21]], [[12, 21]], [[9, 21]], [[9, 28]], [[0, 11]]]", "query_spans": "[[[30, 38]]]", "process": "Given that the ellipse with foci on the x-axis passes through the point P(3,0), then a=3, and the focal distance is 2, so c=1, thus the answer can be obtained. According to the conditions, let the standard equation of the ellipse be \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0). Since the ellipse passes through the point P(3,0), then a=3; and since the focal distance is 2, then c=1. Therefore, the eccentricity of the ellipse is e=\\frac{1}{3}." }, { "text": "Given that $F_{1}$ and $F_{2}$ are the two foci of the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, a line $l$ passing through $F_{1}$ intersects the ellipse at points $M$ and $N$. What is the perimeter of $\\Delta M F_{2} N$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};l: Line;PointOnCurve(F1, l);M: Point;N: Point;Intersection(l, G) = {M, N}", "query_expressions": "Perimeter(TriangleOf(M, F2, N))", "answer_expressions": "8", "fact_spans": "[[[18, 55], [76, 78]], [[18, 55]], [[2, 9], [62, 69]], [[10, 17]], [[2, 60]], [[70, 75]], [[61, 75]], [[79, 82]], [[83, 86]], [[70, 88]]]", "query_spans": "[[[90, 113]]]", "process": "According to the definition of an ellipse, the perimeter of \\triangle MF_{2}N = 4a = 8" }, { "text": "Given that the moving circle $P$ is externally tangent to the fixed circle $C$: $(x+2)^{2}+y^{2}=1$, and also tangent to the line $x=1$, what is the equation of the locus of the center $P$ of the moving circle?", "fact_expressions": "P: Circle;G: Line;C:Circle;P1:Point;Expression(C)=((x+2)^2+y^2=1);Expression(G) = (x = 1);IsOutTangent(P,C);IsTangent(P,G);Center(P)=P1", "query_expressions": "LocusEquation(P1)", "answer_expressions": "y^2=-8*x", "fact_spans": "[[[4, 7], [51, 53]], [[39, 46]], [[10, 33]], [[55, 58]], [[10, 33]], [[39, 46]], [[4, 36]], [[4, 48]], [[51, 58]]]", "query_spans": "[[[55, 65]]]", "process": "Let the distance from the center P of a circle to the line x=1 be equal to r, and let P(x,y). According to the given condition, we have PC = 1 + r, that is, \\sqrt{(x+2)^{2}+y^{2}} = 1 + r. Simplifying yields y^{2} = -8x." }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the inclination angle of one of its asymptotes is $60^{\\circ}$. Then the eccentricity of $C$ 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))) = ApplyUnit(60, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "2", "fact_spans": "[[[2, 63], [88, 91]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 63]], [[2, 86]]]", "query_spans": "[[[88, 97]]]", "process": "From the given conditions, $\\frac{b}{a}=\\tan60^{\\circ}=\\sqrt{3}$, the eccentricity $e=\\frac{c}{a}=\\sqrt{1+\\frac{b^{2}}{a^{2}}}=2$" }, { "text": "The standard equation of a parabola with vertex at the origin and directrix $y=-2$ is?", "fact_expressions": "G: Parabola;O: Origin;Vertex(G) = O;Expression(Directrix(G)) = (y = -2)", "query_expressions": "Expression(G)", "answer_expressions": "x^2 = 8*y", "fact_spans": "[[[18, 21]], [[3, 7]], [[0, 21]], [[8, 21]]]", "query_spans": "[[[18, 28]]]", "process": "From the given information, the standard equation of the parabola is x^{2}=2py. From -\\frac{p}{2}=-2, we get p=4. Therefore, the equation of the required parabola is x^{2}=8y." }, { "text": "A asymptote of the hyperbola $4 x^{2}-y^{2}=1$ is perpendicular to the line $x+t y+1=0$, then $t$=?", "fact_expressions": "G: Hyperbola;H: Line;t: Number;Expression(G) = (4*x^2 - y^2 = 1);Expression(H) = (t*y + x + 1 = 0);IsPerpendicular(OneOf(Asymptote(G)), H) = True", "query_expressions": "t", "answer_expressions": "pm*2", "fact_spans": "[[[0, 20]], [[27, 40]], [[44, 47]], [[0, 20]], [[27, 40]], [[0, 42]]]", "query_spans": "[[[44, 49]]]", "process": "Find the asymptotes of the hyperbola. The slope of the line $x + ty + 1 = 0$ is $-\\frac{1}{t}$. Using the condition for two lines to be perpendicular—the product of their slopes is $-1$ (when both slopes exist)—the required value can be calculated. [Solution] The standard equation of the hyperbola is $\\frac{x^{2}}{4} - y^{2} = 1$, from which the asymptotes of the hyperbola are found to be $y = \\pm 2x$. The slope of the line $x + ty + 1 = 0$ is $-\\frac{1}{t}$, while the slopes of the asymptotes are $\\pm 2$. When both slopes exist, the condition for two lines to be perpendicular is that the product of their slopes is $-1$. Thus, we obtain $-\\frac{1}{t} = \\frac{1}{2}$ or $-\\frac{1}{t} = -\\frac{1}{2}$, solving which gives $t = \\pm 2$." }, { "text": "It is known that the center of a hyperbola is at the origin and one focus is $F(\\sqrt{7}, 0)$. The line $y = x - 1$ intersects the hyperbola at points $M$ and $N$, and the x-coordinate of the midpoint of $MN$ is $-\\frac{2}{3}$. Find the equation of this hyperbola?", "fact_expressions": "G: Hyperbola;H: Line;M: Point;N: Point;F: Point;O:Origin;Center(G)=O;Expression(H) = (y = x - 1);OneOf(Focus(G))=F;Coordinate(F) = (sqrt(7), 0);Intersection(H,G)={M,N};XCoordinate(MidPoint(LineSegmentOf(M, N))) = -2/3", "query_expressions": "Expression(G)", "answer_expressions": "x^2/2-y^2/5=1", "fact_spans": "[[[2, 5], [43, 44], [85, 88]], [[33, 42]], [[47, 50]], [[51, 54]], [[16, 32]], [[8, 10]], [[2, 10]], [[33, 42]], [[2, 32]], [[16, 32]], [[33, 56]], [[57, 82]]]", "query_spans": "[[[85, 93]]]", "process": "" }, { "text": "Given the ellipse $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{9}=1$ $(a>3)$ with two foci $F_{1}$, $F_{2}$, and $| F_{1} F_{2}|=8$. The chord $AB$ passes through point $F_{1}$. What is the perimeter of $\\triangle A B F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/9 + x^2/a^2 = 1);a: Number;a>3;F1: Point;F2: Point;Focus(G) = {F1, F2};Abs(LineSegmentOf(F1, F2)) = 8;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G) = True;PointOnCurve(F1, LineSegmentOf(A, B)) = True", "query_expressions": "Perimeter(TriangleOf(A, B, F2))", "answer_expressions": "20", "fact_spans": "[[[2, 50]], [[2, 50]], [[4, 50]], [[4, 50]], [[56, 63], [99, 107]], [[64, 71]], [[2, 71]], [[73, 91]], [[94, 98]], [[94, 98]], [[2, 98]], [[94, 107]]]", "query_spans": "[[[109, 135]]]", "process": "" }, { "text": "Given that point $P$ is a point on the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{7}=1$, and the distance from $P$ to the right focus of the ellipse is $2$, then what is the distance from point $P$ to the left directrix of the ellipse?", "fact_expressions": "G: Ellipse;P: Point;Expression(G) = (x^2/16 + y^2/7 = 1);PointOnCurve(P, G);Distance(P, RightFocus(G)) = 2", "query_expressions": "Distance(P, LeftDirectrix(G))", "answer_expressions": "8", "fact_spans": "[[[7, 45], [53, 55], [72, 74]], [[2, 6], [49, 52], [67, 71]], [[7, 45]], [[2, 48]], [[49, 65]]]", "query_spans": "[[[67, 83]]]", "process": "Test analysis: From the equation of the ellipse, we know that $ a^{2}=16 $, $ b^{2}=7 $, $ \\therefore a=4 $, $ c=\\sqrt{a^{2}-b^{2}}=3 $, $ \\therefore e=\\frac{c}{a}=\\frac{3}{4} $. According to the definition of the ellipse, the distance from point $ P $ to the left focus of the ellipse is $ 2\\times4-2=6 $. Let $ d $ be the distance from point $ P $ to the left directrix of the ellipse. By the second definition of the ellipse, we have $ e=\\frac{6}{d}=\\frac{3}{4} $, solving gives $ d=8 $." }, { "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. If there exists a point $M$ on the ellipse $C$ such that the area of triangle $M F_{1} F_{2}$ is $\\sqrt{3} b^{2}$, then the range of the eccentricity $e$ of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;M: Point;F1: Point;F2: Point;e: Number;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(C) = F1;RightFocus(C) = F2;PointOnCurve(M, C);Area(TriangleOf(M, F1, F2)) = sqrt(3)*b^2;Eccentricity(C) = e", "query_expressions": "Range(e)", "answer_expressions": "[sqrt(3)/2, 1)", "fact_spans": "[[[2, 59], [85, 90], [138, 143]], [[120, 136]], [[9, 59]], [[93, 97]], [[68, 75]], [[76, 83]], [[147, 150]], [[9, 59]], [[9, 59]], [[2, 59]], [[2, 83]], [[2, 83]], [[85, 97]], [[98, 136]], [[138, 150]]]", "query_spans": "[[[147, 157]]]", "process": "Let M(x, y), then S_{\\triangle MF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}||y| = c|y| = \\sqrt{3}b^{2}, we obtain |y| = \\frac{\\sqrt{3}b^{2}}{c} \\in (0, b], and combining with b^{2} = a^{2} - c^{2}, the range of e = \\frac{c}{a} can be found. [Detailed solution] Let F_{1}(-c, 0), F_{2}(c, 0), M(x, y), then S_{\\Delta MF_{1}F_{2}} = \\frac{1}{2}|F_{1}F_{2}||y| = c|y|. If there exists a point M such that the area of triangle MF_{1}F_{2} is \\sqrt{3}b^{2}, then c|y| = \\sqrt{3}b^{2}, we obtain |y| = \\frac{\\sqrt{3}b^{2}}{c}. Since 0 < |y| \\leqslant b, it follows that 0 < \\frac{\\sqrt{3}b^{2}}{c} \\leqslant b, i.e., \\sqrt{3}b \\leqslant c, which gives c^{2} \\geqslant 3b^{2} = 3(a^{2} - c^{2}). Rearranging yields: 4c^{2} \\geqslant 3a^{2}, so e^{2} = \\frac{c^{2}}{a^{2}} \\geqslant \\frac{3}{4}, solving gives: e \\geqslant \\frac{\\sqrt{3}}{2}. Therefore, \\frac{\\sqrt{3}}{2} \\leqslant e < 1. Thus, the range of eccentricity e of ellipse C is" }, { "text": "Given the line $l$: $y = kx + 1$ intersects the ellipse $\\frac{x^{2}}{2} + y^{2} = 1$ at points $M$ and $N$, and $|MN| = \\frac{4\\sqrt{2}}{3}$, then $k = $?", "fact_expressions": "l: Line;G: Ellipse;M: Point;N: Point;Expression(G) = (x^2/2 + y^2 = 1);Expression(l)=(y=k*x+1);Intersection(l,G)={M,N};Abs(LineSegmentOf(M, N)) = 4*sqrt(2)/3;k:Number", "query_expressions": "k", "answer_expressions": "pm*1", "fact_spans": "[[[2, 18]], [[19, 46]], [[48, 51]], [[52, 55]], [[19, 46]], [[2, 18]], [[2, 57]], [[59, 87]], [[89, 92]]]", "query_spans": "[[[89, 94]]]", "process": "Let M(x_{1},y_{1}), N(x_{2},y_{2}) be given by \\begin{cases} y = kx + 1, \\\\ \\frac{x^{2}}{2} + y^{2} = 1, \\end{cases} eliminating y and simplifying yields (1 + 2k^{2})x^{2} + 4kx = 0, so x_{1} + x_{2} = -\\frac{4k}{1 + 2k^{2}}, x_{1}x_{2} = 0. Given MN = \\frac{4\\sqrt{2}}{3}, we have (x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} = \\frac{32}{9}, so (1 + k^{2})(x_{1} - x_{2})^{2} = \\frac{32}{9}, thus (1 + k^{2})[(x_{1} + x_{2})^{2} - 4x_{1}x_{2}] = \\frac{32}{9}, that is (1 + k^{2})\\left(\\frac{-4k}{1 + 2k^{2}}\\right)^{2} = \\frac{32}{9}, simplifying gives k^{4} + k^{2} - 2 = 0, so k^{2} = 1, hence k = \\pm 1." }, { "text": "The line $y = x + b$ and the curve $y = \\sqrt{x^{2} - 1}$ have exactly one common point; then the range of values for $b$ is?", "fact_expressions": "G: Line;b: Number;H: Curve;Expression(G) = (y = b + x);Expression(H) = (y = sqrt(x^2 - 1));NumIntersection(G, H) = 1", "query_expressions": "Range(b)", "answer_expressions": "[-1, 0) + [1, +oo) ", "fact_spans": "[[[0, 9]], [[41, 44]], [[10, 30]], [[0, 9]], [[10, 30]], [[0, 39]]]", "query_spans": "[[[41, 51]]]", "process": "According to the problem, the equation of the curve $ y = \\sqrt{x^2 - 1} $ can be rewritten as $ x^2 - y^2 = 1 $ ($ y \\geqslant 0 $). Note that one asymptote of the hyperbola $ x^2 - y^2 = 1 $ is $ y = x $. To make the line $ y = x + b $ intersect the curve $ y = \\sqrt{x^2 - 1} $ at exactly one point, there are two cases, as shown in the figure: (1) the line intersects the left part, in which case $ b \\geqslant 1 $; (2) the line intersects the right part, in which case $ -1 \\leqslant b < 0 $. Combining both cases, the range of $ b $ is $ b \\geqslant 1 $ or $ -1 \\leqslant b < 0 $, which in interval notation is $ [-1, 0) \\cup [1, +\\infty) $." }, { "text": "Given that the eccentricity of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{b^{2}}=1$ $(b>0)$ is $2$, then the distance from one of its foci to one of its asymptotes is?", "fact_expressions": "G: Hyperbola;b: Number;b>0;Expression(G) = (x^2/4 - y^2/b^2 = 1);Eccentricity(G) = 2", "query_expressions": "Distance(OneOf(Focus(G)),OneOf(Asymptote(G)))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 49], [59, 60]], [[5, 49]], [[5, 49]], [[2, 49]], [[2, 57]]]", "query_spans": "[[[59, 77]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the left and right foci of the hyperbola $x^{2}-\\frac{y^{2}}{9}=1$, respectively. If point $P$ lies on the hyperbola and $\\overrightarrow{PF_{1}} \\cdot \\overrightarrow{PF_{2}}=0$, then $|P F_{1}+P F_{2}|$=?", "fact_expressions": "G: Hyperbola;P: Point;F1: Point;F2: Point;Expression(G) = (x^2 - y^2/9 = 1);LeftFocus(G) = F1;RightFocus(G) = F2;PointOnCurve(P, G);DotProduct(VectorOf(P,F1),VectorOf(P,F2))=0", "query_expressions": "Abs(LineSegmentOf(P,F1)+LineSegmentOf(P,F2))", "answer_expressions": "2*sqrt(10)", "fact_spans": "[[[19, 47], [60, 63]], [[55, 59]], [[1, 8]], [[9, 16]], [[19, 47]], [[1, 53]], [[1, 53]], [[55, 64]], [[66, 124]]]", "query_spans": "[[[126, 147]]]", "process": "" }, { "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)$, and $P$ is an arbitrary point on the right branch of the hyperbola, if the minimum value of $\\frac{|PF_{1}|^{2}}{|PF_{2}|}$ is $8a$, then what is the range of values for 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));Min(Abs(LineSegmentOf(P, F1))^2/Abs(LineSegmentOf(P, F2))) = 8*a;Eccentricity(G) = e", "query_expressions": "Range(e)", "answer_expressions": "(1,3]", "fact_spans": "[[[20, 76], [87, 90], [127, 130]], [[23, 76]], [[23, 76]], [[83, 86]], [[2, 9]], [[10, 17]], [[150, 153]], [[23, 76]], [[23, 76]], [[20, 76]], [[2, 82]], [[2, 82]], [[83, 97]], [[99, 140]], [[143, 153]]]", "query_spans": "[[[150, 160]]]", "process": "" }, { "text": "Given that $P$ is an arbitrary point on the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 (a>b>0)$, the lines connecting $P$ to the two foci are perpendicular to each other, and the distances from $P$ to the two directrices are $6$ and $12$, respectively. Find the equation of the 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) = True;F1: Point;F2: Point;Focus(G) = {F1, F2};IsPerpendicular(LineSegmentOf(P, F1), LineSegmentOf(P, F2)) = True;L1: Line;L2: Line;Directrix(G) = {L1, L2};Distance(P, L1) = 6;Distance(P, L2) = 12", "query_expressions": "Expression(G)", "answer_expressions": "x^2/45+y^2/20=1", "fact_spans": "[[[6, 60], [105, 107]], [[6, 60]], [[8, 60]], [[8, 60]], [[8, 60]], [[8, 60]], [[2, 5], [66, 69], [81, 84]], [[2, 65]], [], [], [[6, 73]], [[6, 79]], [], [], [[6, 88]], [[6, 103]], [[6, 103]]]", "query_spans": "[[[105, 111]]]", "process": "" }, { "text": "Let $F$ be the right focus of the ellipse $C$: $\\frac{x^{2}}{a}+\\frac{y^{2}}{b}=1$ $(a>0, b>0)$, and let the maximum distance from a moving point on $C$ to $F$ be $d$. If there exists a point $P$ on the right directrix of $C$ such that $PF = d$, then the range of values for the eccentricity of the ellipse $C$ is?", "fact_expressions": "C: Ellipse;b: Number;a: Number;P: Point;F: Point;A: Point;a>0;b>0;Expression(C) = (y^2/b + x^2/a = 1);RightFocus(C) = F;PointOnCurve(A, C);Max(Distance(A, F))= d;PointOnCurve(P, RightDirectrix(C));LineSegmentOf(P, F) = d;d:Number", "query_expressions": "Range(Eccentricity(C))", "answer_expressions": "[1/2,1)", "fact_spans": "[[[5, 57], [62, 65], [85, 88], [111, 116]], [[12, 57]], [[12, 57]], [[95, 99]], [[1, 4], [71, 74]], [], [[12, 57]], [[12, 57]], [[5, 57]], [[1, 61]], [[62, 70]], [[62, 83]], [[85, 99]], [[102, 109]], [[80, 83]]]", "query_spans": "[[[111, 127]]]", "process": "Problem Analysis: Given that the maximum distance from a moving point on ellipse C to F is d, it follows that a + c = d. From PF = d, we obtain d ⩾ a²/c − c. Therefore, a + c ⩾ a²/c − c, which implies ac + c² ⩾ a² − c². Dividing both sides by a² and simplifying yields 2e² + e − 1 ⩽ 0, solving gives e ∈ [1/2, 1)." }, { "text": "The focus of the parabola $y^{2}=4x$ is used to draw a chord with inclination angle $\\alpha$ and length at most $8$. The line containing the chord intersects the circle $x^{2}+y^{2}=\\frac{3}{4}$. Find the range of possible values of $\\alpha$.", "fact_expressions": "G: Parabola;H: Circle;I: LineSegment;alpha: Number;Expression(G) = (y^2 = 4*x);Expression(H) = (x^2 + y^2 = 3/4);PointOnCurve(Focus(G), I);Inclination(I) = alpha;Negation(Length(I) > 8);IsChordOf(I, G);IsIntersect(OverlappingLine(I), H)", "query_expressions": "Range(alpha)", "answer_expressions": "[pi/4, pi/3] + [2*pi/3, 3*pi/4]", "fact_spans": "[[[1, 15]], [[52, 78]], [], [[25, 33], [84, 92]], [[1, 15]], [[52, 78]], [[0, 44]], [[0, 44]], [[0, 44]], [[0, 44]], [[0, 82]]]", "query_spans": "[[[84, 99]]]", "process": "" }, { "text": "Given the hyperbola $C$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, the right vertex of $C$ is $A$. A circle $A$ is drawn with center $A$ and radius equal to the semi-focal distance $c$. The circle $A$ intersects one asymptote of the hyperbola $C$ at points $M$ and $N$. If $|MN|=\\sqrt{3} 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;A: Point;RightVertex(C) = A;A1: Circle;Center(A1) = A;c: Number;HalfFocalLength(C) = c;Radius(A1) = c;M: Point;N: Point;Intersection(A1, OneOf(Asymptote(C))) = {M, N};Abs(LineSegmentOf(M, N)) = sqrt(3)*c", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)", "fact_spans": "[[[2, 63], [101, 107], [147, 150]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[68, 71], [74, 77]], [[2, 71]], [[91, 95], [96, 100]], [[73, 95]], [[84, 87]], [[2, 87]], [[84, 95]], [[115, 118]], [[119, 122]], [[96, 124]], [[127, 145]]]", "query_spans": "[[[147, 156]]]", "process": "Let the asymptote equation of hyperbola C be bx - ay = 0, and let d be the distance from point A(a,0) to the asymptote, then d = \\frac{ab}{\\sqrt{a^{2}+b^{2}}} = \\frac{ab}{c} \\therefore |MN| = 2\\sqrt{c^{2}-d^{2}} = 2\\sqrt{c^{2}-(\\frac{ab}{c})^{2}} = 2\\sqrt{\\frac{c}{\\because|MN|=\\sqrt{3}c\\therefore4(\\frac{c4-a2c2+a4}{c2})=3c^{2} simplifies to c^{4}-4a^{2}c^{2}+4a^{4}=0, i.e., e^{4}-4e^{2}+4=0, solving gives e=\\sqrt{2}. Hence, the eccentricity of C is \\sqrt{2}. 1 In problems where the hyperbola equation is unknown and the eccentricity is to be found: set up a homogeneous equation containing a, b, c, use b^{2}=c^{2}-a^{2} to eliminate b, then transform it into an equation in terms of e to solve." }, { "text": "Given that $F$ is the focus of the parabola $C$: $y^{2}=4 x$, two mutually perpendicular lines $l_{1}$ and $l_{2}$ are drawn through $F$. The line $l_{1}$ intersects $C$ at points $A$ and $B$, and the line $l_{2}$ intersects $C$ at points $D$ and $E$. Then the minimum value of $|A B|+4|D E|$ is?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 4*x);F: Point;Focus(C) = F;l1: Line;l2: Line;PointOnCurve(F, l1);PointOnCurve(F, l2);IsPerpendicular(l1, l2);A: Point;B: Point;Intersection(l1, C) = {A, B};D: Point;E: Point;Intersection(l2, C) = {D, E}", "query_expressions": "Min(Abs(LineSegmentOf(A, B)) + 4*Abs(LineSegmentOf(D, E)))", "answer_expressions": "36", "fact_spans": "[[[6, 25], [70, 73], [95, 98]], [[6, 25]], [[2, 5], [30, 33]], [[2, 28]], [[41, 50], [60, 69]], [[52, 59], [85, 94]], [[29, 59]], [[29, 59]], [[34, 59]], [[75, 78]], [[79, 82]], [[60, 84]], [[100, 103]], [[104, 107]], [[85, 109]]]", "query_spans": "[[[111, 131]]]", "process": "Let the equation of line $ l_{1} $ be $ y = k(x - 1) $. By solving the system of equations, we obtain $ |AB| = 4 + \\frac{4}{k^{2}} $ and $ |DE| = 4 + 4k^{2} $. Combining with the basic inequality, we can find the minimum value of $ |AB| + 4|DE| $, yielding the answer. From the problem, the parabola $ C: y^{2} = 4x $ has focus $ F(1, 0) $ and directrix $ x = -1 $. Let the equation of line $ l_{1} $ be $ y = k(x - 1) $, $ k \\neq 0 $. Solving the system \n\\[\n\\begin{cases}\ny^{2} = 4x \\\\\ny = k(x - 1)\n\\end{cases}\n\\]\ngives $ k^{2}x^{2} - (4 + 2k^{2})x + k^{2} = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 2 + \\frac{4}{k^{2}} $. By the definition of the parabola, $ |AB| = x_{1} + x_{2} + 2 = 4 + \\frac{4}{k^{2}} $. Since $ l_{1} \\perp l_{2} $, replacing $ k $ by $ -\\frac{1}{k} $ in the above expression gives $ |DE| = 4 + 4k^{2} $. Then $ |AB| + 4|DE| = 20 + 4\\left(4k^{2} + \\frac{1}{k^{2}}\\right) \\geqslant 20 + 8\\sqrt{4k^{2} \\cdot \\frac{1}{k^{2}}} = 36 $, with equality if and only if $ k = \\pm \\frac{\\sqrt{2}}{2} $. Thus, the minimum value of $ |AB| + 4|DE| $ is 36." }, { "text": "The foci of the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ are $F_{1}$, $F_{2}$, and $P$ is a point on the ellipse. Given that $P F_{1} \\perp P F_{2}$, what is the area of $\\Delta F_{1} P F_{2}$?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/25 + y^2/9 = 1);F1: Point;F2: Point;Focus(G) = {F1,F2};P: Point;PointOnCurve(P,G) = True;IsPerpendicular(LineSegmentOf(P,F1) , LineSegmentOf(P,F2)) = True", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "9", "fact_spans": "[[[0, 38], [61, 63]], [[0, 38]], [[41, 48]], [[49, 56]], [[0, 56]], [[57, 60]], [[57, 67]], [[70, 93]]]", "query_spans": "[[[95, 122]]]", "process": "" }, { "text": "A line $l$ passing through the point $M(0,1)$ intersects the ellipse $\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$ at points $A$ and $B$, $F$ is the right focus of the ellipse. When the perimeter of $\\triangle A B F$ is maximized, what is the area of $\\triangle A B F$?", "fact_expressions": "l: Line;G: Ellipse;M: Point;A: Point;B: Point;F: Point;Expression(G) = (x^2/8 + y^2/4 = 1);Coordinate(M) = (0, 1);PointOnCurve(M, l);Intersection(l, G) = {A, B};RightFocus(G) = F;WhenMax(Perimeter(TriangleOf(A,B,F)))", "query_expressions": "Area(TriangleOf(A, B, F))", "answer_expressions": "4*sqrt(10)/3", "fact_spans": "[[[11, 16]], [[17, 54], [69, 71]], [[1, 10]], [[55, 58]], [[59, 62]], [[65, 68]], [[17, 54]], [[1, 10]], [[0, 16]], [[11, 64]], [[65, 75]], [[76, 100]]]", "query_spans": "[[[101, 123]]]", "process": "Analysis: According to the definition and properties of the ellipse, the right focus is $ F(2,0) $, and let the left focus be $ F_{1}(-2,0) $. The perimeter is maximized if and only if points $ A $, $ B $, $ F $ are collinear. Then, using the two-point form, find the equation of the line, and further solve for the area. From the given conditions, the coordinates of the left and right foci of the ellipse are $ F_{1}(-2,0) $, $ F(2,0) $. By the definition of the ellipse, we have $ |AF| = 2\\sqrt{2} - |AF_{1}| $, $ |BF| = 2\\sqrt{2} - |BF_{1}| $. Therefore, the perimeter of $ \\triangle ABF $ is $ |AF| + |BF| + |AB| = 4\\sqrt{2} + |AB| - (|AF_{1}| + |BF_{1}|) $. Clearly, $ |AF_{1}| + |BF_{1}| \\geqslant |AB| $, and equality holds if and only if $ A $, $ B $, $ F_{1} $ are collinear, in which case the maximum perimeter is $ 8\\sqrt{2} $. At this time, the equation of the line is $ x - 2y - 2 = 0 $. Solving the system of equations\n$$\n\\begin{cases}\nx - 2y - 2 = 0 \\\\\n\\frac{x^{2}}{8} + \\frac{y^{2}}{4} = 1\n\\end{cases}\n$$\nwe obtain $ 3y^{2} + 4y - 2 = 0 $, so $ y_{1} + y_{2} = \\frac{4}{3} $. Thus, the area of $ \\triangle ABF $ is $ \\frac{1}{2} \\times 4 \\times \\frac{\\sqrt[2]{10}}{3} = \\frac{\\sqrt{4}}{3}, \\frac{4}{3} $." }, { "text": "If the line $ l $: $ y = (a - 1) x - 1 $ and the curve $ y^{2} = a x $ have exactly one common point, find the set of real values of $ a $?", "fact_expressions": "l: Line;G: Curve;a: Real;Expression(G) = (y^2 = a*x);Expression(l) = (y = x*(a - 1) - 1);NumIntersection(l, G) = 1", "query_expressions": "Range(a)", "answer_expressions": "{0, 1, 4/5}", "fact_spans": "[[[1, 21]], [[22, 35]], [[46, 51]], [[22, 35]], [[1, 21]], [[1, 43]]]", "query_spans": "[[[46, 57]]]", "process": "When a=0, the line l: y=-x-1 and the curve y^{2}=0 have exactly one common point, which satisfies the condition. When a\\neq0, solving the system of equations \\begin{cases}y=(a-1)x-\\\\x=\\frac{y^{2}}{a}\\end{cases} yields (a-1)y^{2}-ay-a=0. If a=1, then y=-1, and this has one common point with y^{2}=x, satisfying the condition; if a\\neq1, then A=a^{2}+4a(a-1)=0, solving gives a=\\frac{4}{5}, at which the line is tangent to the curve and has exactly one common point, satisfying the condition. Therefore, the set of real values of a is {0,1,\\frac{4}{5}}." }, { "text": "If the equation of an ellipse is $\\frac{x^{2}}{10-a}+\\frac{y^{2}}{a-2}=1$, and the eccentricity of this ellipse is $\\frac{\\sqrt{2}}{2}$, then the real number $a$=?", "fact_expressions": "G: Ellipse;a: Real;Expression(G) = (x^2/(10 - a) + y^2/(a - 2) = 1);Eccentricity(G) = sqrt(2)/2", "query_expressions": "a", "answer_expressions": "{14/3,22/3}", "fact_spans": "[[[1, 3], [49, 51]], [[78, 83]], [[1, 46]], [[49, 76]]]", "query_spans": "[[[78, 85]]]", "process": "When the foci are on the x-axis, the eccentricity of the ellipse is \\( e^2 = 1 - \\frac{b^{2}}{a^{2}} = 1 - \\frac{a-2}{10-a} = \\left(\\frac{\\sqrt{2}}{2}\\right)^{2} \\), solving gives \\( a = \\frac{14}{3} \\); when the foci are on the y-axis, the eccentricity of the ellipse is \\( e^{2} = 1 - \\frac{b^{2}}{a^{2}} = 1 - \\frac{10-a}{a-2} = \\left(\\frac{\\sqrt{2}}{2}\\right)^{2} \\), solving gives \\( a = \\frac{22}{3} \\)." }, { "text": "Given the parabola $C$: $y^{2}=2 px(p>0)$ with directrix $l$, a line passing through $M(1 , 0)$ with slope $\\sqrt{3}$ intersects $l$ at point $A$ and intersects $C$ at a point $B$. If $AM=MB$, then $p=?$", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 2*(p*x));p: Number;p>0;l: Line;Directrix(C) = l;M: Point;Coordinate(M) = (1, 0);G: Line;PointOnCurve(M, G);Slope(G) = sqrt(3);A: Point;Intersection(G, l) = A;B: Point;OneOf(Intersection(G, C)) = B;LineSegmentOf(A, M) = LineSegmentOf(M, B)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 27], [74, 77]], [[2, 27]], [[97, 100]], [[10, 27]], [[30, 33], [63, 66]], [[2, 33]], [[35, 45]], [[35, 45]], [[60, 62]], [[34, 62]], [[46, 62]], [[69, 72]], [[60, 72]], [[83, 86]], [[60, 86]], [[88, 95]]]", "query_spans": "[[[97, 102]]]", "process": "" }, { "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 $y=\\sqrt{2}x$. A line $l$ with slope $2$ passing through the left focus $F(-\\sqrt{3}, 0)$ intersects the hyperbola $C$ at points $A$ and $B$. Find the length of the chord $AB=?$", "fact_expressions": "C: Hyperbola;Expression(C) = (-y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a>0;b>0;Expression(OneOf(Asymptote(C))) = (y = sqrt(2)*x);F: Point;Coordinate(F) = (-sqrt(3), 0);LeftFocus(C) = F;PointOnCurve(F , l) = True;Slope(l) = 2;l: Line;Intersection(l, C) = {A, B};A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), C) = True", "query_expressions": "LineSegmentOf(A, B)", "answer_expressions": "10", "fact_spans": "[[[2, 63], [88, 89], [123, 129]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[2, 86]], [[92, 109]], [[92, 109]], [[88, 109]], [[87, 122]], [[110, 122]], [[117, 122]], [[117, 139]], [[130, 133]], [[134, 137]], [[123, 151]]]", "query_spans": "[[[146, 153]]]", "process": "From the given conditions: $\\frac{b}{a}=\\sqrt{2}$ and $c=\\sqrt{3}$, thus find the values of $a$, $b$, $c$, then find the equation of the hyperbola. Then assume the equation of line $l$, solve the system of equations of the line and the hyperbola, and use the chord length formula to find the length of the chord. \n$\\because$ Hyperbola $C: \\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1$ ($a>0$, $b>0$) has an asymptote with equation $y = \\sqrt{2}x$, \n$\\therefore \\frac{b}{a} = \\sqrt{2}$, i.e., $b = \\sqrt{2}a$. \n$\\because$ The left focus is $F(-\\sqrt{3}, 0)$, \n$\\therefore c = \\sqrt{3}$, \n$\\therefore c^{2} = a^{2} + b^{2} = 3a^{2} = 3$, \n$\\therefore a^{2} = 1$, $b^{2} = 2$, \n$\\therefore$ The equation of the hyperbola is $x^{2} - \\frac{y^{2}}{2} = 1$. \nThe equation of line $l$ is $y = 2(x + \\sqrt{3})$. \nLet $A(x_{1}, y_{1})$, $B(x_{2}, y_{2})$, \nfrom $\\begin{cases} y = 2(x + \\sqrt{3}) \\\\ x^{2} - \\frac{y^{2}}{2} = 1 \\end{cases}$" }, { "text": "Given that the line $y = x - 3m$ intersects the ellipse $C$: $\\frac{x^{2}}{a^{2}} + \\frac{y^{2}}{b^{2}} = 1$ $(a > b > 0)$ at points $P$ and $Q$, if the abscissa of the midpoint of $PQ$ is exactly $2m$, then what is the eccentricity of ellipse $C$?", "fact_expressions": "C: Ellipse;b: Number;a: Number;G: Line;m: Number;P: Point;Q: Point;a > b;b > 0;Expression(C) = (y^2/b^2 + x^2/a^2 = 1);Expression(G) = (y = -3*m + x);Intersection(G, C) = {P, Q};XCoordinate(MidPoint(LineSegmentOf(P, Q))) = 2*m", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(2)/2", "fact_spans": "[[[14, 71], [106, 111]], [[21, 71]], [[21, 71]], [[2, 13]], [[4, 13]], [[74, 77]], [[78, 81]], [[21, 71]], [[21, 71]], [[14, 71]], [[2, 13]], [[2, 83]], [[85, 104]]]", "query_spans": "[[[106, 117]]]", "process": "Let P(x_{1},y_{1}), Q(x_{2},y_{2}), substituting into the ellipse 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}}=0, which simplifies to \\frac{y_{1}-y_{2}}{x_{1}}-x_{2}\\cdot\\frac{y_{1}+y_{2}}{\\frac{x_{1}+x_{2}}{2}}=-\\frac{b^{2}}{a^{2}}. Since \\frac{x_{1}+x_{2}}{2}=2m, it follows that \\frac{y_{1}+y_{2}}{2}=\\frac{x_{1}-3m+x_{2}-3m}{2}=-m. Also, since k_{PQ}=\\frac{y_{1}}{x_{1}}, we have 1\\times\\frac{-m}{2m}=-\\frac{b^{2}}{a^{2}}, so \\frac{1}{2}=\\frac{b^{2}}{a^{2}}, thus e=\\frac{c}{a}=\\sqrt{\\frac{c^{2}}{a^{2}}}=\\sqrt{\\frac{a^{2}-b^{2}}{a^{2}}}=\\sqrt{1-\\frac{b^{2}}{a^{2}}}=\\sqrt{1-\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}. \\frac{1}{2}=\\frac{c^{2}}{a^{2}}-\\frac{b^{2}}{a^{2}}," }, { "text": "The parabola $y^{2}=4x$, line $l$ passes through the focus $F$ of the parabola and intersects the parabola at points $A$ and $B$. If $\\overrightarrow{BA}=4\\overrightarrow{BF}$, then the area of $\\triangle OAB$ ($O$ is the origin) is?", "fact_expressions": "l: Line;G: Parabola;O: Origin;A: Point;B: Point;F: Point;Expression(G) = (y^2 = 4*x);Focus(G) = F;PointOnCurve(F, l);Intersection(l,G)={A,B};VectorOf(B, A) = 4*VectorOf(B, F)", "query_expressions": "Area(TriangleOf(O,A,B))", "answer_expressions": "4*sqrt(3)/3", "fact_spans": "[[[15, 20]], [[0, 14], [33, 36], [22, 25]], [[115, 118]], [[38, 41]], [[43, 46]], [[28, 31]], [[0, 14]], [[22, 31]], [[15, 31]], [[15, 48]], [[50, 95]]]", "query_spans": "[[[97, 130]]]", "process": "According to the problem, first determine the equation of line AB, then solve simultaneously the equations of the line and the parabola, and use Vieta's formulas to find the area of $\\triangle C$. [Detailed Solution] From the given conditions: $AF = 3BF$, using the focal radius formula we have: \n$$\n\\frac{p}{1 - \\cos\\alpha} = \\frac{3p}{1 + \\cos\\alpha}\n$$ \nSolving gives: $\\cos\\alpha = \\frac{1}{2}$, $\\alpha = \\frac{\\pi}{3}$, thus the equation of line AB is: \n$$\ny = \\sqrt{3}(x - 1)\n$$ \nSolving simultaneously with the parabola equation yields: \n$$\n3y^{2} - 4\\sqrt{3}y - 12 = 0\n$$ \nThen: \n$$\n|y_{1} - y_{2}| = \\sqrt{\\left(\\frac{4\\sqrt{3}}{3}\\right)^{2} - 4 \\times (-4)} = \\frac{8}{\\sqrt{3}}\n$$ \nHence, the area $S$ of $\\triangle OAB$ is: \n$$\nS = \\frac{1}{2} \\times |OF| \\times |y_{1} - y_{2}| = \\frac{1}{2} \\times 1 \\times \\frac{8}{\\sqrt{3}} = \\frac{4}{3}\\sqrt{3}\n$$" }, { "text": "Given that the focus of the parabola $C$: $y^{2}=m x$ is the center of the circle $E$: $x^{2}+y^{2}+4 x-1=0$, then the equation of the directrix of the parabola $C$ is?", "fact_expressions": "C: Parabola;m: Number;E: Circle;Expression(C) = (y^2 = m*x);Expression(E) = (4*x + x^2 + y^2 - 1 = 0);Focus(C) = Center(E)", "query_expressions": "Expression(Directrix(C))", "answer_expressions": "x=2", "fact_spans": "[[[2, 20], [56, 62]], [[9, 20]], [[24, 51]], [[2, 20]], [[24, 51]], [[2, 54]]]", "query_spans": "[[[56, 69]]]", "process": "Convert the general form equation of a circle $ x^{2} + y^{2} + 4x - 1 = 0 $ into standard form to get $ (x+2)^{2} + y^{2} = 5 $, so the center is $ (-2, 0) $. Thus, the focus of the parabola $ C $ is $ (-2, 0) $, hence $ m = -8 $, and its directrix equation is $ x = 2 $." }, { "text": "Given the hyperbola $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. Let $P$ be a point on the left branch of the hyperbola $C$, such that $|P F_{1}|=|F_{1} F_{2}|$ and $\\angle P F_{1} F_{2}=120^{\\circ}$. Find the eccentricity of the hyperbola $C$.", "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, LeftPart(C)) = True;Abs(LineSegmentOf(P, F1)) = Abs(LineSegmentOf(F1, F2));AngleOf(P, F1, F2) = ApplyUnit(120, degree)", "query_expressions": "Eccentricity(C)", "answer_expressions": "(sqrt(3)+1)/2", "fact_spans": "[[[2, 61], [90, 96], [168, 174]], [[2, 61]], [[10, 61]], [[10, 61]], [[10, 61]], [[10, 61]], [[70, 77]], [[78, 85]], [[2, 85]], [[2, 85]], [[86, 89]], [[86, 103]], [[104, 129]], [[132, 166]]]", "query_spans": "[[[168, 180]]]", "process": "|PF_{1}|=|F_{1}F_{2}|, and \\angle PF_{1}F_{2}=120^{\\circ}, then |PF_{2}|=\\sqrt{3}|F_{1}F_{2}|=2\\sqrt{3}c, |PF_{1}|=|F_{1}F_{2}|=2c, hence |PF_{2}|-|PF_{1}|=2\\sqrt{3}c-2c=2a, thus e=\\frac{\\sqrt{3}+1}{2}" }, { "text": "Given the range of eccentricity of the ellipse $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ is $(0, \\frac{\\sqrt{2}}{2}]$, points $A$ and $B$ are the right focus and the upper vertex of the ellipse respectively, and the distance from the left focus to the line $AB$ is $2$. Then, what is the minimum value of the sum of distances from a point $P$ on the ellipse to the two foci?", "fact_expressions": "G: Ellipse;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);b: Number;a: Number;a > b;b > 0;Range(Eccentricity(G))=(0, sqrt(2)/2];A: Point;RightFocus(G) = A;B: Point;UpperVertex(G) = B;Distance(LeftFocus(G), LineOf(A, B)) = 2;A1: Point;LeftFocus(G) = A1;P: Point;PointOnCurve(P, G)", "query_expressions": "Min(Distance(P, A1) + Distance(P, A))", "answer_expressions": "4", "fact_spans": "[[[2, 54], [101, 103], [133, 135]], [[2, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[4, 54]], [[2, 89]], [[90, 94]], [[90, 111]], [[95, 98]], [[90, 111]], [[101, 131]], [], [[101, 116]], [[137, 141]], [[133, 141]]]", "query_spans": "[[[133, 156]]]", "process": "First obtain the equation of line AB, then from the distance from the left focus to the line we get \\frac{bc}{a}=1, and compute using the AM-GM inequality and a^{2}=b^{2}+c^{2}. Solution: Since the eccentricity of the ellipse is (0,\\frac{\\sqrt{2}}{2}], then 0<\\frac{c}{a}\\leqslant\\frac{\\sqrt{2}}{2}\\Rightarrow\\frac{c^{2}}{a^{2}}\\leqslant\\frac{1}{2}. Also a^{2}=b^{2}+c^{2}, so 2c^{2}\\leqslanta^{2}\\Rightarrow2c^{2}\\leqslantc^{2}+b^{2}\\Rightarrow c\\leqslant b. From the problem, the equation of line AB is: \\frac{x}{c}+\\frac{y}{b}=1. Thus the distance from the left focus to the line is: \\frac{bc}{\\sqrt{b^{2}+c^{2}}}=\\frac{bc}{a}. Also a^{2}=b^{2}+c^{2}\\geqslant2bc, so 1=\\frac{bc}{a}\\leqslant\\frac{b^{2}+c^{2}}{2a}, then a\\geqslant2, with equality if and only if b=c. Moreover, the sum of distances from any point P on the ellipse to the two foci is 2a\\geqslant4. Therefore, the minimum value of the sum of distances from a point P on the ellipse to the two foci is 4. Answer: 4" }, { "text": "Given that the line $l$ passing through the point $M(1,-1)$ intersects the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ at points $A$ and $B$, if point $M$ is the midpoint of $AB$, then the equation of line $l$ is?", "fact_expressions": "l: Line;G: Ellipse;A: Point;B: Point;M: Point;Expression(G) = (x^2/4 + y^2/3 = 1);Coordinate(M) = (1, -1);PointOnCurve(M,l);Intersection(l, G) = {A, B};MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Expression(l)", "answer_expressions": "3*x-4*y-7=0", "fact_spans": "[[[14, 19], [86, 91]], [[20, 57]], [[60, 63]], [[64, 67]], [[3, 13], [71, 75]], [[20, 57]], [[3, 13]], [[2, 19]], [[14, 69]], [[71, 84]]]", "query_spans": "[[[86, 96]]]", "process": "" }, { "text": "The hyperbola has its axes of symmetry coinciding with the coordinate axes, its center at the origin, foci at $(-2,0)$ and $(2,0)$, and passes through the point $P(-2,3)$. What is the standard equation of the hyperbola?", "fact_expressions": "G: Hyperbola;SymmetryAxis(G) = axis;O: Origin;Center(G) = O;F1: Point;F2: Point;Coordinate(F1) = (-2, 0);Coordinate(F2) = (2, 0);Focus(G) = {F1, F2};P: Point;Coordinate(P) = (-2, 3);PointOnCurve(P, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2-y^2/3=1", "fact_spans": "[[[0, 3], [57, 60]], [[0, 13]], [[17, 19]], [[0, 19]], [[25, 33]], [[34, 41]], [[0, 41]], [[0, 41]], [[0, 41]], [[45, 55]], [[45, 55]], [[0, 55]]]", "query_spans": "[[[57, 67]]]", "process": "Analysis: From the coordinates of the foci, the value of c can be obtained. Then, using the definition of the hyperbola, compute the distances from point P to the two foci to find the value of a. Next, use the relationship among a, b, and c to find b. Specifically, from the given conditions, c = 2, |2a = \\sqrt{(-2+2)^{2}+(0-3)^{2}} - \\sqrt{(2+2)^{2}+(0-3)^{2}}| = 2, ∴ a = 1, b = \\sqrt{3}, c = 2. Therefore, the standard equation of the hyperbola is x^{2} - \\frac{y^{2}}{3} = 1." }, { "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 $m=$?", "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": "m", "answer_expressions": "3", "fact_spans": "[[[1, 39]], [[76, 79]], [[1, 39]], [[1, 74]]]", "query_spans": "[[[76, 81]]]", "process": "Since the asymptote equations are $ y = \\pm \\frac{\\sqrt{3}}{2}x $, we have $ \\frac{\\sqrt{m}}{2} = \\frac{\\sqrt{3}}{2} $, solving gives $ m = 3 $." }, { "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. Point $A$ lies on the right branch of hyperbola $C$, the line $A F_{1}$ intersects the $y$-axis at point $B$, and $|F_{1} B|=3|A B|$, $A F_{1} \\perp A F_{2}$. Find the eccentricity of hyperbola $C$.", "fact_expressions": "C: Hyperbola;a: Number;b: Number;F1: Point;A: 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(A, RightPart(C));Intersection(LineOf(A,F1), yAxis) = B;Abs(LineSegmentOf(F1, B)) = 3*Abs(LineSegmentOf(A, B));IsPerpendicular(LineSegmentOf(A, F1), LineSegmentOf(A, F2))", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)+sqrt(6)", "fact_spans": "[[[0, 61], [91, 97], [173, 179]], [[8, 61]], [[8, 61]], [[70, 77]], [[86, 90]], [[78, 85]], [[121, 125]], [[8, 61]], [[8, 61]], [[0, 61]], [[0, 85]], [[0, 85]], [[86, 102]], [[103, 125]], [[127, 145]], [[147, 171]]]", "query_spans": "[[[173, 185]]]", "process": "" }, { "text": "The equation of the circle with center at the right focus of the hyperbola $x^{2}-\\frac{y^{2}}{3}=1$ and radius equal to its eccentricity is?", "fact_expressions": "G: Hyperbola;H: Circle;Expression(G) = (x^2 - y^2/3 = 1);RightFocus(G) = Center(H);Eccentricity(G)=Radius(H)", "query_expressions": "Expression(H)", "answer_expressions": "(x-2)^2+y^2=4", "fact_spans": "[[[1, 29]], [[44, 45]], [[1, 29]], [[0, 45]], [[0, 45]]]", "query_spans": "[[[44, 50]]]", "process": "" }, { "text": "The equation of the hyperbola that shares foci with the ellipse $\\frac{x^{2}}{25}+\\frac{y^{2}}{9}=1$ and has eccentricity $\\frac{4}{3}$ is?", "fact_expressions": "G: Hyperbola;H: Ellipse;Expression(H) = (x^2/25 + y^2/9 = 1);Eccentricity(G) = 4/3;Focus(G)=Focus(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9-y^2/7=1", "fact_spans": "[[[62, 65]], [[1, 39]], [[1, 39]], [[44, 65]], [[0, 65]]]", "query_spans": "[[[62, 70]]]", "process": "" }, { "text": "A focus of the hyperbola $8 kx^{2}-ky^{2}=8$ is $(0,3)$, then what is the value of $k$?", "fact_expressions": "G: Hyperbola;Expression(G) = (8*(k*x^2) - k*y^2 = 8);k: Number;H: Point;Coordinate(H) = (0, 3);OneOf(Focus(G)) = H", "query_expressions": "k", "answer_expressions": "-1", "fact_spans": "[[[0, 22]], [[0, 22]], [[37, 40]], [[28, 35]], [[28, 35]], [[0, 35]]]", "query_spans": "[[[37, 44]]]", "process": "" }, { "text": "If a chord of the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ is bisected by the point $(1,1)$, then what is the equation of the line containing this chord?", "fact_expressions": "G: Ellipse;H: LineSegment;I: Point;Expression(G) = (x^2/16 + y^2/4 = 1);Coordinate(I) = (1, 1);MidPoint(H)=I;IsChordOf(H,G)", "query_expressions": "Expression(OverlappingLine(H))", "answer_expressions": "x+4*y-5=0", "fact_spans": "[[[2, 40]], [], [[43, 51]], [[2, 40]], [[43, 51]], [[2, 53]], [[2, 42]]]", "query_spans": "[[[2, 68]]]", "process": "Let the two endpoints of the chord be A(x_{1},y_{1}) and B(x_{2},y_{2}). Then \n\\begin{cases}\\frac{x_{1}+x_{2}}{2}=1\\\\\\frac{y_{1}+y_{2}}{2}=1\\end{cases}, \ngiving \n\\begin{cases}x_{1}+x_{2}=2\\\\y_{1}+y_{2}=2\\end{cases}. \nSince points A and B lie on the ellipse \\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1, we have \n\\begin{cases}\\frac{x_{1}^{2}}{16}+\\frac{y_{1}^{2}}{4}=1\\\\\\frac{x_{2}^{2}}{16}+\\frac{y_{2}^{2}}{4}=1\\end{cases}. \nSubtracting these two equations yields \n\\frac{x_{2}^{2}-x_{1}^{2}}{16}+\\frac{y_{2}^{2}-y_{1}^{2}}{4}=0, \nwhich simplifies to \n\\frac{(x_{2}-x_{1})(x_{2}+x_{1})}{16}+\\frac{(y_{2}-y_{1})(y_{2}+y_{1})}{4}=0. \nThus, \n\\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{1}{4} \\cdot \\frac{x_{1}+x_{2}}{y_{1}+y_{2}}. \nGiven x_{1}+x_{2}=2 and y_{1}+y_{2}=2, we obtain \n\\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{1}{4} \\cdot \\frac{2}{2} = -\\frac{1}{4}. \nTherefore, the slope of line AB is k_{AB} = \\frac{y_{1}-y_{2}}{x_{1}-x_{2}} = -\\frac{1}{4}. \nHence, the equation of the line containing this chord is y-1 = -\\frac{1}{4}(x-1), or equivalently, x+4y-5=0." }, { "text": "Given that the line $l$: $x - y - m = 0$ passes through the focus of the parabola $C$: $y^{2} = 2 p x$ ($p > 0$), and intersects $C$ at points $A$ and $B$, if $|AB| = 6$, then the value of $p$ is?", "fact_expressions": "l: Line;Expression(l) = (-m + x - y = 0);m: Number;C: Parabola;Expression(C) = (y^2 = 2*p*x);p: Number;p>0;PointOnCurve(Focus(C), l) = True;Intersection(l, C) = {A, B};A: Point;B: Point;Abs(LineSegmentOf(A, B)) = 6", "query_expressions": "p", "answer_expressions": "3/2", "fact_spans": "[[[2, 18]], [[2, 18]], [[9, 18]], [[20, 46], [51, 54]], [[20, 46]], [[78, 81]], [[28, 46]], [[2, 49]], [[2, 65]], [[56, 59]], [[60, 63]], [[67, 76]]]", "query_spans": "[[[78, 85]]]", "process": "From \\begin{cases}x-y-m=0\\\\y^2=2px\\end{cases}, we obtain x^{2}-(2m+2p)x+m^{2}=0. Let A(x_{1},y_{1}), B(x_{2},y_{2}), then x_{1}+x_{2}=2m+2p. Since the line l: x-y-m=0 passes through the focus (\\frac{p}{2},0) of the parabola C: y^{2}=2px (p>0), \\therefore \\frac{p}{2}-0-m=0, solving gives m=\\frac{p}{2}. Also |AB|=(x_{1}+\\frac{p}{2})+(x_{2}+\\frac{p}{2})=x_{1}+x_{2}+p=2m+3p=4p=6, \\therefore p=\\frac{3}{2}. Thus the answer is \\frac{3}{5}." }, { "text": "Let $A(x_{1} , y_{1})$ , $B(x_{2} , y_{2})$ be two points on the parabola $y=2 x^{2}$, and line $l$ be the perpendicular bisector of $AB$. When the slope of line $l$ is $\\frac{1}{2}$, then the range of the $y$-intercept of line $l$ is?", "fact_expressions": "l: Line;G: Parabola;A: Point;B: Point;x1: Number;y1: Number;x2: Number;y2: Number;Expression(G) = (y = 2*x^2);Coordinate(A) = (x1, y1);Coordinate(B) = (x2, y2);PointOnCurve(A, G);PointOnCurve(B, G);PerpendicularBisector(LineSegmentOf(A, B)) = l;Slope(l) = 1/2", "query_expressions": "Range(Intercept(l, yAxis))", "answer_expressions": "(3/4, +oo)", "fact_spans": "[[[60, 65], [80, 85], [105, 110]], [[41, 55]], [[1, 20]], [[22, 40]], [[1, 20]], [[1, 20]], [[22, 40]], [[22, 40]], [[41, 55]], [[1, 20]], [[22, 40]], [[1, 59]], [[1, 59]], [[60, 77]], [[80, 102]]]", "query_spans": "[[[105, 125]]]", "process": "" }, { "text": "Given the parabola $C$: $x^{2}=-2 p y(p>0)$, the focus $F$ of the parabola coincides with one focus of $\\frac{y^{2}}{8}+\\frac{x^{2}}{4}=1$. A line passing through the focus $F$ intersects $C$ at two distinct points $A$ and $B$. The tangents to the parabola $C$ at points $A$ and $B$ intersect at point $M$, and the x-coordinate of $M$ is $2$. Then $|A B|$=?", "fact_expressions": "C: Parabola;Expression(C) = (x^2 = -2*p*y);p: Number;p>0;F: Point;Focus(C) = F;Q: Curve;Expression(Q) = (x^2/4 + y^2/8 = 1);OneOf(Focus(Q)) = F;PointOnCurve(F, G) = True;G: Line;Intersection(G, C) = {A, B};Negation(A = B);A: Point;B: Point;Intersection(TangentOnPoint(A, C), TangentOnPoint(B, C)) = M;M: Point;XCoordinate(MidPoint(M)) = 2", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "10", "fact_spans": "[[[2, 28], [88, 91], [105, 111]], [[2, 28]], [[9, 28]], [[9, 28]], [[31, 34], [81, 84]], [[2, 34]], [[35, 70]], [[35, 70]], [[31, 77]], [[78, 87]], [[85, 87]], [[85, 104]], [[93, 104]], [[93, 96], [112, 115]], [[97, 100], [116, 119]], [[105, 132]], [[128, 132], [134, 137]], [[134, 145]]]", "query_spans": "[[[147, 156]]]", "process": "From the given conditions, we have F(0, -2), so p = 4, and the equation of the parabola is x^{2} = -8y. Let the equation of line AB be y = kx - 2, A(x_{1}, y_{1}), B(x_{2}, y_{2}): y_{1} = -\\frac{x_{1}^{2}}{8}, y_{2} = -\\frac{x_{2}^{2}}{8}. From y = -\\frac{x^{2}}{8}, we get y' = -\\frac{x}{4}, so the slope at any point is -\\frac{x}{4}. Thus, the tangent line at point A is y - y_{1} = -\\frac{x_{1}}{4}(x - x_{1}), which simplifies to y = -\\frac{x_{1}}{4}x + \\frac{x_{1}^{2}}{8} \\textcircled{1}. Similarly, the tangent line at point B is y = -\\frac{x_{2}}{4}x + \\frac{x_{2}^{2}}{8} \\textcircled{2}. Solving \\textcircled{1} and \\textcircled{2} simultaneously gives x_{M} = \\frac{x_{1} + x_{2}}{2}. Since the x-coordinate of M is 2, we have x_{1} + x_{2} = 4. Substituting the equation of AB into the parabola yields x^{2} + 8kx - 16 = 0, so x_{1} + x_{2} = -8k = 4, hence k = -\\frac{1}{2}. Therefore, y_{1} + y_{2} = k(x_{1} + x_{2}) - 4 = -\\frac{1}{2} \\times 4 - 4 = -6. Thus, |AB| = 2p - y_{1} - y_{2} = 10." }, { "text": "If on the parabola $y^{2}=4 x$ there is a moving chord $A B$ of length $10$, then the shortest distance from the midpoint of $A B$ to the $y$-axis is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 4*x);A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);Length(LineSegmentOf(A, B)) = 10", "query_expressions": "Min(Distance(MidPoint(LineSegmentOf(A, B)), yAxis))", "answer_expressions": "4", "fact_spans": "[[[1, 15]], [[1, 15]], [[35, 40]], [[35, 40]], [[1, 33]], [[19, 33]]]", "query_spans": "[[[35, 55]]]", "process": "Let A(x_{1},y_{1}), B(x_{2},y_{2}), and the directrix of the parabola be x = -1. As shown in the figure, for any chord AB not passing through the focus, if E is the midpoint of AB, then GE = \\frac{AC + BD}{2} = \\frac{BF + AF}{2} > \\frac{AB}{2}. Therefore, among all chords of length AB = 10 in the parabola, only the chord passing through the focus has its midpoint closest to the directrix or the y-axis. It follows that: x_{1} + x_{2} + 2 = 10, and the midpoint of A, B has coordinates (\\frac{x_{1} + x_{2}}{2}, \\frac{y_{1} + y_{2}}{2}). Hence, the distance from the midpoint to the y-axis is \\frac{x_{1} + x_{2}}{2} = 4. The answer is: 4" }, { "text": "The length of the chord cut from the ellipse $x^{2}+4 y^{2}=16$ by the line $y=\\frac{1}{2} x+1$ is?", "fact_expressions": "G: Ellipse;H: Line;Expression(G) = (x^2 + 4*y^2 = 16);Expression(H) = (y = x/2 + 1)", "query_expressions": "Length(InterceptChord(H, G))", "answer_expressions": "sqrt(35)", "fact_spans": "[[[0, 20]], [[21, 42]], [[0, 20]], [[21, 42]]]", "query_spans": "[[[0, 49]]]", "process": "From \\begin{cases}x^{2}+4y^{2}=16\\\\y=\\frac{1}{2}x+1\\end{cases}, eliminating $y$ and simplifying yields $x^{2}+2x-6=0$. Let the intersection points of the line and the ellipse be $M(x_{1},y_{1})$, $N(x_{2},y_{2})$, then $x_{1}+x_{2}=-2$, $x_{1}x_{2}=-6$, so the chord length $|MN|=\\sqrt{1+k^{2}}|x_{1}-x_{2}|=\\sqrt{35}$. Hence, fill in $\\sqrt{35}$." }, { "text": "The line $y=2x+3$ intersects the curve $y=x^{2}$ at points $A$ and $B$. Then $|AB|$=?", "fact_expressions": "G: Line;H: Curve;A: Point;B: Point;Expression(G) = (y = 2*x + 3);Expression(H) = (y = x^2);Intersection(G, H) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "4*sqrt(5)", "fact_spans": "[[[0, 11]], [[12, 23]], [[26, 29]], [[30, 33]], [[0, 11]], [[12, 23]], [[0, 35]]]", "query_spans": "[[[37, 46]]]", "process": "" }, { "text": "If a hyperbola passes through the point $(6, \\sqrt{3})$ and its asymptotes are given by $y = \\pm \\frac{1}{3} x$, then what is the standard equation of this hyperbola?", "fact_expressions": "G: Hyperbola;H: Point;Coordinate(H) = (6, sqrt(3));PointOnCurve(H, G);Expression(Asymptote(G)) = (y = pm*(x/3))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/9-y^2=1", "fact_spans": "[[[1, 4], [24, 25], [55, 58]], [[6, 22]], [[6, 22]], [[1, 22]], [[24, 52]]]", "query_spans": "[[[55, 64]]]", "process": "From the given conditions, the asymptotes of the hyperbola are $ y = \\pm\\frac{1}{3}x $. Let the equation of the desired hyperbola be $ x^{2} - 9y^{2} = \\lambda $ ($ \\lambda \\neq 0 $). Since the point $ (6, \\sqrt{3}) $ lies on this hyperbola, we have $ \\lambda = 36 - 27 = 9 $. Therefore, the equation of the hyperbola is: $ x^{2} - 9y^{2} = 9 $, or $ \\frac{x^{2}}{9} - y^{2} = 1 $." }, { "text": "If the line $ m x - y - 2 m = 0 $ and the curve $ x = \\sqrt{1 + y^{2}} $ have two common points, then the range of real values for $ m $ is?", "fact_expressions": "G: Line;m: Real;H: Curve;Expression(G) = (-2*m + m*x - y = 0);Expression(H) = (x = sqrt(y^2 + 1));NumIntersection(G, H)=2", "query_expressions": "Range(m)", "answer_expressions": "(-oo,-1)+(1,+oo)", "fact_spans": "[[[1, 16]], [[45, 50]], [[17, 37]], [[1, 16]], [[17, 37]], [[1, 43]]]", "query_spans": "[[[45, 57]]]", "process": "The curve $ x = \\sqrt{1 + y^2} $, that is, $ x^2 - y^2 = 1 $ ($ x \\geqslant 1 $), has asymptotes: $ y = \\pm x $. The line $ mx - y - 2m = 0 $, that is, $ y = m(x - 2) $, passes through the fixed point $ (2, 0) $. In the same coordinate system, draw the graphs of both: from the graph it follows that $ m \\in (-\\infty, -1) \\cup (1, +\\infty) $." }, { "text": "Given that point $P(2, a)$ lies on the parabola $C$: $y^{2}=4 x$, then the distance from point $P$ to the focus of parabola $C$ is?", "fact_expressions": "P: Point;Coordinate(P) = (2, a);a: Number;C: Parabola;Expression(C) = (y^2 = 4*x);PointOnCurve(P, C) = True", "query_expressions": "Distance(P, Focus(C))", "answer_expressions": "3", "fact_spans": "[[[2, 12], [35, 39]], [[2, 12]], [[3, 12]], [[13, 32], [40, 46]], [[13, 32]], [[2, 33]]]", "query_spans": "[[[35, 54]]]", "process": "The directrix of the parabola $ C: y^{2} = 4x $ is $ x = -1 $. By the definition of a parabola, the distance $ d $ from point $ P(2, a) $ to the focus of parabola $ C $ is $ d = 2 - (-1) = 3 $. Therefore, the distance from point $ P $ to the focus of parabola $ C $ is 3. Hence, the answer is: 3" }, { "text": "Given that the equation $\\frac{x^{2}}{-2-\\lambda}+\\frac{y^{2}}{1+2 \\lambda}=-3$ represents an ellipse with foci on the $y$-axis, find the range of values for $\\lambda$.", "fact_expressions": "G: Ellipse;lambda: Number;Expression(G) = (x^2/(-lambda - 2) + y^2/(2*lambda + 1) = -3);PointOnCurve(Focus(G), yAxis)", "query_expressions": "Range(lambda)", "answer_expressions": "(-2, -1)", "fact_spans": "[[[70, 72]], [[74, 83]], [[2, 72]], [[61, 72]]]", "query_spans": "[[[74, 90]]]", "process": "According to the characteristics of the ellipse equation with foci on the y-axis, obtain the system of inequalities and solve it. \n$\\frac{x^{2}}{-2-\\lambda}+\\frac{y^{2}}{1+2\\lambda}=-3\\Rightarrow\\frac{y^{2}}{-3(1+2\\lambda)}+\\frac{x^{2}}{3(2+\\lambda)}=1=-3$ represents an ellipse with foci on the y-axis $|-\\frac{1}{2}|$, so we have \n$\\begin{cases}-3(1+2\\lambda)\\\\3(2+\\lambda)>0\\\\-3(1+2)\\end{cases}\\Rightarrow-2<\\lambda<-3(1+22)>3(2+\\lambda)$" }, { "text": "Given that the moving circle $M$ is tangent to the line $y=2$ and externally tangent to the fixed circle $C$: $x^{2}+(y+3)^{2}=1$, find the equation of the locus of the center $M$ of the moving circle.", "fact_expressions": "M: Circle;G: Line;Expression(G) = (y = 2);IsTangent(M, G);C: Circle;Expression(C) = (x^2 + (y + 3)^2 = 1);IsOutTangent(M, C);M1: Point;Center(M) = M1", "query_expressions": "LocusEquation(M1)", "answer_expressions": "x^2 = -12*y", "fact_spans": "[[[4, 7]], [[8, 15]], [[8, 15]], [[4, 17]], [[22, 46]], [[22, 46]], [[4, 48]], [[54, 57]], [[50, 57]]]", "query_spans": "[[[54, 64]]]", "process": "Let the radius of the moving circle be r, then the distance from M to the line y=2 is r, and MC=r+1. Hence, the distance from M to (0,-3) equals the distance from M to the line y=3, so the trajectory is a parabola, namely x^{2}=-12y." }, { "text": "Given that the right focus of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$ is $F(4,0)$, if a line passing through point $F$ with an inclination angle of $60^{\\circ}$ intersects the right branch of the hyperbola at exactly one point, then what is the range of values for $a$?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Line;F: Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Coordinate(F) = (4, 0);RightFocus(G) = F;PointOnCurve(F, H);Inclination(H) = ApplyUnit(60, degree);NumIntersection(H, RightPart(G)) = 1", "query_expressions": "Range(a)", "answer_expressions": "(0, 2]", "fact_spans": "[[[2, 58], [99, 102]], [[5, 58]], [[115, 118]], [[96, 98]], [[63, 71], [74, 78]], [[5, 58]], [[5, 58]], [[2, 58]], [[63, 71]], [[2, 71]], [[73, 98]], [[79, 98]], [[96, 113]]]", "query_spans": "[[[115, 125]]]", "process": "Test analysis: According to the problem, in order for the line to have exactly one intersection point with the right branch of the hyperbola, the slope $\\frac{b}{a}$ of the asymptote $y=\\frac{b}{a}x$ of the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0,b>0)$ must satisfy $\\frac{b}{a}\\geqslant\\tan60^{\\circ}$, that is, $(\\frac{b}{a})^{2}\\geqslant3$. Combining this with the right focus $F(4,0)$, which implies $c=4$, it follows that the range of values for $a$ is $0b>0)$, the left and right foci are $F_{1}$ and $F_{2}$ respectively, and $P$ is a moving point on the ellipse. If the maximum value of $\\angle F_{1} P F_{2}$ is $\\frac{2 \\pi}{3}$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;F1: Point;P: Point;F2: Point;Expression(G) = (y^2/b^2 + x^2/a^2 = 1);LeftFocus(G) =F1;RightFocus(G)=F2;PointOnCurve(P,G);Max(AngleOf(F1, P, F2)) = (2*pi)/3;a:Number;b:Number;a>b;b>0", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/2", "fact_spans": "[[[2, 4], [86, 88], [140, 142]], [[66, 73]], [[82, 85]], [[74, 81]], [[2, 57]], [[2, 81]], [[2, 81]], [[82, 92]], [[94, 138]], [[2, 57]], [[2, 57]], [[2, 57]], [[2, 57]]]", "query_spans": "[[[140, 148]]]", "process": "By the definition of an ellipse, |PF₁| + |PF₂| = 2a. By the law of cosines, |² = (|PF₁| + |PF₂|)² - |FF|F / |PF₁||P|² - 1 = (4b² - 1 = (22 / 2a²) - 1) / a F₁⊥|PF. Since the maximum value of ∠F₁PF₂ is 2π/3, then (2b² / a²) - 1 = cos(2π/3) = -1/2, we obtain b² / a² = 1/4. Therefore, the eccentricity of the ellipse is e = c/a = √(c² / a²) = √((a² - b²) / a²) = √(1 - b² / a²) = √3 / 2." }, { "text": "Given the curve $C$: $x^{2}+y^{2}=m$ has exactly three points whose distance to the line $12x+5y+26=0$ is $1$, then $m=?$", "fact_expressions": "C: Curve;Expression(C) = (x^2 + y^2 = m);m: Number;D1: Point;D2: Point;D3: Point;PointOnCurve(D1, C) ;PointOnCurve(D2, C) ;PointOnCurve(D3, C) ;Distance(D1, G) = 1;Distance(D2, G) = 1;Distance(D3, G) = 1;G: Line;Expression(G) = (12*x + 5*y + 26 = 0)", "query_expressions": "m", "answer_expressions": "9", "fact_spans": "[[[2, 24]], [[2, 24]], [[55, 58]], [], [], [], [[2, 29]], [[2, 29]], [[2, 29]], [[2, 53]], [[2, 53]], [[2, 53]], [[30, 47]], [[30, 47]]]", "query_spans": "[[[55, 60]]]", "process": "" }, { "text": "Let $F_{1}$ and $F_{2}$ be the two foci of the hyperbola $\\frac{x^{2}}{4}-y^{2}=1$, and let point $P$ lie on the hyperbola such that $\\angle F_{1} P F_{2}=60^{\\circ}$. 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);Focus(G) = {F1, F2};PointOnCurve(P, G);AngleOf(F1, P, F2) = ApplyUnit(60, degree)", "query_expressions": "Area(TriangleOf(P, F1, F2))", "answer_expressions": "sqrt(3)", "fact_spans": "[[[17, 45], [56, 59]], [[51, 55]], [[1, 8]], [[9, 16]], [[17, 45]], [[1, 50]], [[51, 60]], [[63, 96]]]", "query_spans": "[[[98, 128]]]", "process": "From the problem we have $ a^{2}=4 $, $ b^{2}=1 $, $ \\therefore c^{2}=4+1=5 $, $ \\therefore c=\\sqrt{5} $. From the problem we get \n\\[\n\\begin{cases}\n20=|PF_{1}|^{2}+|PF_{2}|^{2}-2|PF_{1}||PF_{2}|\\times\\frac{1}{2}=(|PF_{1}|-|PF_{2}|)^{2}+|PF_{1}||PF_{2}| \\\\\n\\therefore |PF_{1}||PF_{1}|-|PF_{2}||=4\n\\end{cases}\n\\]\nso $ S=\\frac{1}{2}\\cdot|PF_{1}||PF_{2}|\\cdot\\sin\\frac{\\pi}{3}=\\frac{1}{2}\\cdot4\\cdot\\frac{\\sqrt{3}}{2}=\\sqrt{3} $." }, { "text": "Through the right focus $F$ of the hyperbola $\\Gamma$: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{3 a^{2}}=1(a>0)$, draw a line with slope $k$ intersecting the right branch of the hyperbola at points $M$ and $N$. The perpendicular bisector of chord $MN$ intersects the $x$-axis at point $P$. Then $\\frac{|P F|}{|M N|}$=?", "fact_expressions": "Gamma: Hyperbola;a: Number;k:Number;H: Line;M: Point;N: Point;P: Point;F:Point;a>0;Expression(Gamma) = (x^2/a^2- y^2/(3*a^2) = 1);RightFocus(Gamma) = F;PointOnCurve(F, H);Slope(H) = k;Intersection(H,RightPart(Gamma))={M,N};IsChordOf(LineSegmentOf(M,N),Gamma);Intersection(PerpendicularBisector(LineSegmentOf(M,N)),xAxis)=P", "query_expressions": "Abs(LineSegmentOf(P, F))/Abs(LineSegmentOf(M, N))", "answer_expressions": "1", "fact_spans": "[[[1, 64], [82, 85]], [[14, 64]], [[75, 78]], [[79, 81]], [[89, 92]], [[94, 98]], [[119, 123]], [[68, 71]], [[14, 64]], [[1, 64]], [[1, 71]], [[0, 81]], [[72, 81]], [[79, 100]], [[79, 107]], [[101, 123]]]", "query_spans": "[[[125, 148]]]", "process": "Let $ l_{MN}: x = my + c $, ($ c = 2a, k = \\frac{1}{m} $), combining \n$$\n\\begin{cases}\n\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{3a^{2}} = 1 \\\\\nx = my + c\n\\end{cases}\n$$\nyields $ (a^{2} - b^{2}m^{2})y^{2} - 2b^{2}cmy - b^{4} = 0 $. Then the perpendicular bisector equation is: let $ y = 0 $, obtain $ x_{p} = \\frac{c^{a}}{a^{2} - b^{2}m^{2}} $, so \n$ |PF| = \\left| \\frac{c - b}{a^{2} - b^{2}m^{2}} - c \\right| = \\frac{b^{2}c(1 + m^{2})}{|a^{2} - b^{m}m^{2}|} $, \nhence $ \\frac{|PF|}{|MN|} = \\frac{c}{2a} = 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 $F_{1}$ and $F_{2}$ respectively, $P$ is a point on an asymptote of $C$, $O$ is the origin, and $|O P|=1$, the area of $\\Delta P F_{1} F_{2}$ is $\\sqrt{2} a$. Find the eccentricity of the hyperbola $C$.", "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, Asymptote(C)) = True;O: Origin;Abs(LineSegmentOf(O, P)) = 1;Area(TriangleOf(P, F1, F2)) = sqrt(2)*a", "query_expressions": "Eccentricity(C)", "answer_expressions": "sqrt(3)", "fact_spans": "[[[2, 63], [91, 94], [163, 169]], [[2, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[10, 63]], [[71, 78]], [[79, 86]], [[2, 86]], [[2, 86]], [[87, 90]], [[87, 100]], [[101, 104]], [[111, 120]], [[123, 161]]]", "query_spans": "[[[163, 175]]]", "process": "Let point $ P(x_{0},y_{0}) $ be a point in the first quadrant, then $ y_{0} = \\frac{b}{a}x_{0} $, $ S_{\\triangle PF_{1}F_{2}} = \\frac{1}{2} \\times 2c \\times \\frac{b}{a^{2}} + \\frac{2}{c^{2}} = 1 $, that is $ \\frac{3a}{a^{2}}\\left(\\frac{a}{b^{2}+1}\\right) = \\frac{2ac}{b^{2}c} = \\frac{2a^{2}}{b^{2}} = 1 $, thus $ \\frac{b^{2}}{a^{2}} = 2 $. Then $ y_{0} = \\frac{b}{a}x_{0} = \\frac{\\sqrt{2}a}{c} $, i.e., point $ P\\left(\\frac{\\sqrt{2}a^{2}}{bc}, \\frac{\\sqrt{2}a}{c}\\right) $." }, { "text": "The focus of the parabola $y=2 x^{2}$ is?", "fact_expressions": "G: Parabola;Expression(G) = (y = 2*x^2)", "query_expressions": "Focus(G)", "answer_expressions": "(0, 1/8)", "fact_spans": "[[[0, 14]], [[0, 14]]]", "query_spans": "[[[0, 19]]]", "process": "Problem Analysis: The parabola equation y = 2x^{2} is rewritten as x^{2} = \\frac{1}{2}y, \\therefore 2p = \\frac{1}{2}, \\therefore \\frac{p}{2} = \\frac{1}{4}, so the focus is (0, \\frac{1}{8})." }, { "text": "Point $P$ is a moving point on the hyperbola $\\frac{x^{2}}{25}-\\frac{y^{2}}{16}=1$ with foci $F_{1}$ and $F_{2}$. If point $I$ satisfies $\\overrightarrow{PI} \\cdot | \\overrightarrow{F_{1} F_{2}}|+\\overrightarrow{F_{1} I} \\cdot |\\overrightarrow{P F_{2}}|+\\overrightarrow{F_{2} I} \\cdot | \\overrightarrow{P F_{1}} |=\\overrightarrow{0}$, then the horizontal coordinate of point $I$ is?", "fact_expressions": "G: Hyperbola;P: Point;I: Point;F1: Point;F2: Point;Expression(G) = (x^2/25 - y^2/16 = 1);Focus(G)={F1,F2};PointOnCurve(P,G);VectorOf(F1, I)*Abs(VectorOf(P, F2)) + VectorOf(P, I)*Abs(VectorOf(F1, F2)) + VectorOf(F2, I)*Abs(VectorOf(P, F1)) = 0", "query_expressions": "XCoordinate(I)", "answer_expressions": "pm*5", "fact_spans": "[[[24, 64]], [[0, 4]], [[70, 74], [275, 279]], [[8, 15]], [[16, 23]], [[24, 64]], [[5, 64]], [[0, 68]], [[76, 272]]]", "query_spans": "[[[275, 285]]]", "process": "" }, { "text": "If the equation $\\frac{x^{2}}{m+1} + \\frac {y^{2}}{m+2}=1$ represents a hyperbola, then what is the range of values for $m$?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/(m + 1) + y^2/(m + 2) = 1)", "query_expressions": "Range(m)", "answer_expressions": "(-2, -1)", "fact_spans": "[[[48, 51]], [[54, 57]], [[2, 51]]]", "query_spans": "[[[54, 64]]]", "process": "" }, { "text": "Given that points $A$ and $B$ are two points on the hyperbola $x^{2}-\\frac{y^{2}}{2}=1$, and $O$ is the origin, if $\\overrightarrow{O A} \\cdot \\overrightarrow{O B}=0$, then what is the distance from point $O$ to the line $AB$?", "fact_expressions": "G: Hyperbola;B: Point;A: Point;O: Origin;Expression(G) = (x^2 - y^2/2 = 1);PointOnCurve(A, G);PointOnCurve(B, G);DotProduct(VectorOf(O, A), VectorOf(O, B)) = 0", "query_expressions": "Distance(O, LineOf(A,B))", "answer_expressions": "sqrt(2)", "fact_spans": "[[[11, 39]], [[7, 10]], [[2, 6]], [[44, 47], [107, 111]], [[11, 39]], [[2, 43]], [[2, 43]], [[52, 104]]]", "query_spans": "[[[107, 124]]]", "process": "" }, { "text": "Given that vertex $A$ of $\\triangle A B C$ is a focus of the ellipse $\\frac{x^{2}}{3}+y^{2}=1$, vertices $B$ and $C$ lie on the ellipse, and side $B C$ passes through the other focus of the ellipse, then the perimeter of $\\triangle A B C$ is?", "fact_expressions": "G: Ellipse;B: Point;C: Point;A: Point;Expression(G) = (x^2/3 + y^2 = 1);Negation(A=F);OneOf(Focus(G)) = A;PointOnCurve(B, G);PointOnCurve(C, G);F:Point;OneOf(Focus(G))=F;PointOnCurve(F,LineSegmentOf(B,C))", "query_expressions": "Perimeter(TriangleOf(A, B, C))", "answer_expressions": "4*sqrt(3)", "fact_spans": "[[[26, 53], [69, 71], [82, 84]], [[61, 64]], [[65, 68]], [[22, 25]], [[26, 53]], [[22, 90]], [[22, 58]], [[59, 72]], [[59, 72]], [], [[82, 90]], [[74, 90]]]", "query_spans": "[[[92, 114]]]", "process": "Solve according to the definition of an ellipse. It is easy to obtain that the perimeter of $\\triangle ABC$ is the sum of the distances from points $B$ and $C$ to the two foci, which is $2a+2a=4a=4\\sqrt{3}$." }, { "text": "If the focal length of the hyperbola $\\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1$ is $6$, then what is the value of $m$?", "fact_expressions": "G: Hyperbola;m: Number;Expression(G) = (x^2/4 - y^2/m = 1);FocalLength(G) = 6", "query_expressions": "m", "answer_expressions": "5", "fact_spans": "[[[1, 39]], [[49, 52]], [[1, 39]], [[1, 46]]]", "query_spans": "[[[49, 56]]]", "process": "\\because the hyperbola \\frac{x^2}{4}-\\frac{y^{2}}{m}=1, \\therefore a=2, b=\\sqrt{m}; \\because the focal distance of the hyperbola \\frac{x^{2}}{4}-\\frac{y^{2}}{m}=1 is 6, \\therefore 6=2\\sqrt{4+m}, solving gives m=5" }, { "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. If a point $P$ on the ellipse satisfies $P F_{2} \\perp x$-axis and $|P F_{1}|=2|P F_{2}|$, then the eccentricity of the ellipse is?", "fact_expressions": "G: Ellipse;b: Number;a: Number;P: 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;PointOnCurve(P,G);IsPerpendicular(LineSegmentOf(P,F2),xAxis);Abs(LineSegmentOf(P,F1))=2*Abs(LineSegmentOf(P,F2))", "query_expressions": "Eccentricity(G)", "answer_expressions": "sqrt(3)/3", "fact_spans": "[[[2, 54], [80, 82], [134, 136]], [[4, 54]], [[4, 54]], [[84, 88]], [[71, 78]], [[63, 70]], [[4, 54]], [[4, 54]], [[2, 54]], [[2, 78]], [[2, 78]], [[80, 88]], [[90, 108]], [[109, 131]]]", "query_spans": "[[[134, 142]]]", "process": "Let |PF_{2}| = m, then |PF_{1}| = 2|PF_{2}| = 2m. By the definition of the ellipse, we have: |PF_{1}| + |PF_{2}| = 3m = 2a, so m = \\frac{2a}{3}. Thus, |PF_{1}| = \\frac{4a}{3}, |PF_{2}| = \\frac{2a}{3}. Since PF_{2} \\perp x-axis, triangle PF_{1}F_{2} is a right triangle. By the Pythagorean theorem: |PF_{1}|^{2} = |PF_{2}|^{2} + |F_{1}F_{2}|^{2}, that is, (\\frac{4a}{3})^{2} = (\\frac{2a}{3})^{2} + (2c)^{2}, which gives (\\frac{c}{a})^{2} = \\frac{1}{3}. Therefore, the eccentricity e = \\frac{c}{a} = \\frac{\\sqrt{3}}{3}." }, { "text": "If $AB$ is a moving chord of the parabola $y^2 = 2px$ $(p > 0)$, and $|AB| = a$ $(a > 2p)$, then the shortest distance from the midpoint $M$ of $AB$ to the $y$-axis is?", "fact_expressions": "G: Parabola;Expression(G) = (y^2 = 2*(p*x));p: Number;p>0;A: Point;B: Point;IsChordOf(LineSegmentOf(A, B), G);a: Number;Abs(LineSegmentOf(A, B)) = a;a>2*p;M: Point;MidPoint(LineSegmentOf(A, B)) = M", "query_expressions": "Min(Distance(M, yAxis))", "answer_expressions": "(a-p)/2", "fact_spans": "[[[7, 28]], [[7, 28]], [[10, 28]], [[10, 28]], [[1, 6]], [[1, 6]], [[1, 31]], [[34, 50]], [[34, 50]], [[34, 50]], [[61, 64]], [[53, 64]]]", "query_spans": "[[[61, 76]]]", "process": "" }, { "text": "Given that point $P(x_1, y_1)$ satisfies $\\sqrt{(x-1)^{2}+y^{2}}+\\sqrt{(x+1)^{2}+y^{2}}=4$, then the maximum value of $|O P|$ ($O$ is the origin) is equal to?", "fact_expressions": "P: Point;O: Origin;C: Curve;Coordinate(P) = (x1, y1);Expression(C) = (sqrt(y^2 + (x - 1)^2) + sqrt(y^2 + (x + 1)^2) = 4);PointOnCurve(P, C);x1: Number;y1: Number", "query_expressions": "Max(Abs(LineSegmentOf(O, P)))", "answer_expressions": "2", "fact_spans": "[[[2, 14]], [[75, 78]], [[16, 65]], [[2, 14]], [[16, 65]], [[2, 65]], [[3, 14]], [[3, 14]]]", "query_spans": "[[[67, 92]]]", "process": "Problem Analysis: According to the problem, the locus of point P is the set of points whose sum of distances to two fixed points (1,0) and (-1,0) is 4. This locus is an ellipse with the standard equation \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1, and the maximum value of |OP| is a=2." }, { "text": "Given that point $P$ is an arbitrary point on the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0, b>0)$, draw lines through point $P$ parallel to the asymptotes of the hyperbola, intersecting the two asymptotes at points $M$ and $N$, respectively. If $|P M| \\cdot|P N|=b^{2}$, then the eccentricity of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;P: Point;M: Point;N: Point;l1: Line;l2: Line;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);PointOnCurve(P, G);L1:Line;L2:Line;PointOnCurve(P,L1);PointOnCurve(P,L2);Asymptote(G)={l1,l2};IsParallel(L1,l1);IsParallel(L2,l2);Intersection(L1,l2)=M;Intersection(L2,l1)=N;Abs(LineSegmentOf(P, M))*Abs(LineSegmentOf(P, N)) = b^2", "query_expressions": "Eccentricity(G)", "answer_expressions": "2*sqrt(3)/3", "fact_spans": "[[[7, 63], [75, 78], [134, 137]], [[10, 63]], [[10, 63]], [[2, 6], [70, 74]], [[96, 99]], [[100, 103]], [], [], [[10, 63]], [[10, 63]], [[7, 63]], [[2, 68]], [], [], [[69, 86]], [[69, 86]], [[75, 94]], [[69, 86]], [[69, 86]], [[69, 105]], [[69, 105]], [[107, 131]]]", "query_spans": "[[[134, 143]]]", "process": "Test Analysis: Let P(a,0), then the asymptotes of the hyperbola are y=\\pm\\frac{b}{a}x. The equations of lines passing through point P and parallel to the two asymptotes of the hyperbola are y=\\pm\\frac{b}{a}(x-a)\\begin{matrix}m&y=\\frac{s}{x-0}\\\\y=\\frac{a-a}{b}&x\\end{matrix}= Given c^{2}=a^{2}+b^{2}=4b^{2}, c=2b, \\therefore e=\\frac{c}{a}=\\frac{2b}{\\sqrt{3}b}=\\frac{2\\sqrt{3}}{3}," }, { "text": "Given the parabola $C$: $y^{2}=2px$ ($p>0$) with directrix $l$, a line passing through $M(1,0)$ with slope $\\sqrt{3}$ intersects $l$ at $A$ and intersects $C$ at a point $B$. If $\\overrightarrow{AM}=\\overrightarrow{MB}$, then $p=$?", "fact_expressions": "C: Parabola;p: Number;G:Line;M: Point;A: Point;B: Point;l: Line;p>0;Expression(C) = (y^2 = 2*p*x);Coordinate(M) = (1, 0);Directrix(C)=l;PointOnCurve(M,G);Slope(G)=sqrt(3);Intersection(G,l)=A;OneOf(Intersection(G,C))=B;VectorOf(A, M) = VectorOf(M, B)", "query_expressions": "p", "answer_expressions": "2", "fact_spans": "[[[2, 26], [71, 74]], [[130, 133]], [[57, 59]], [[34, 42]], [[66, 69]], [[80, 83]], [[29, 32], [60, 63]], [[9, 26]], [[2, 26]], [[34, 42]], [[2, 32]], [[33, 59]], [[43, 59]], [[57, 69]], [[57, 83]], [[85, 128]]]", "query_spans": "[[[130, 135]]]", "process": "Problem Analysis: From the given conditions, we have $ l: x = -\\frac{p}{2} $, and the equation of the line is $ y = \\sqrt{3}(x - 1) $. Then $ A\\left(-\\frac{p}{2}, \\sqrt{3}\\left(-\\frac{p}{2} - 1\\right)\\right) $. Since $ \\overrightarrow{AM} = \\overrightarrow{MB} $, it follows that $ B\\left(2 + \\frac{p}{2}, \\sqrt{3}\\left(\\frac{p}{2} + 1\\right)\\right) $. Substituting $ B\\left(2 + \\frac{p}{2}, \\sqrt{3}\\left(\\frac{p}{2} + 1\\right)\\right) $ into $ y^2 = 2px $, we get $ 3\\left(\\frac{p}{2} + 1\\right)^2 = 2p\\left(2 + \\frac{p}{2}\\right) $. Solving this equation yields $ p = 2 $. Hence, fill in $ 2 $." }, { "text": "Given the ellipse $\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$, draw a line $l$ through the point $(1,0)$ with an inclination angle of $45^{\\circ}$, intersecting the ellipse at points $A$ and $B$, and let $O$ be the origin. Then the area of $\\triangle A O B$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/4 + y^2/3 = 1);H: Point;Coordinate(H) = (1, 0);l: Line;PointOnCurve(H, l) = True;Inclination(l) = ApplyUnit(45, degree);Intersection(l, G) = {A, B};A: Point;O: Origin;B: Point", "query_expressions": "Area(TriangleOf(A, O, B))", "answer_expressions": "6*sqrt(2)/7", "fact_spans": "[[[2, 39], [73, 75]], [[2, 39]], [[41, 49]], [[41, 49]], [[67, 72]], [[40, 72]], [[50, 72]], [[67, 85]], [[76, 79]], [[86, 89]], [[80, 83]]]", "query_spans": "[[[96, 118]]]", "process": "The line $ l $ passing through the point $ (1,0) $ with an inclination angle of $ 45^{\\circ} $ has the equation $ y = x - 1 $. Solving the system \n\\[\n\\begin{cases}\n\\frac{x^{2}}{4} + \\frac{y^{2}}{3} = 1 \\\\\ny = x - 1\n\\end{cases}\n\\]\nyields $ 7x^{2} - 8x - 8 = 0 $, where $ \\triangle = 64 + 4 \\times 7 \\times 8 > 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $. Then $ x_{1} + x_{2} = \\frac{8}{7} $, $ x_{1}x_{2} = -\\frac{8}{7} $. \n\\[\n|AB| = \\sqrt{(1+1)\\left[\\left(\\frac{8}{7}\\right)^{2} + 4 \\times \\frac{8}{7}\\right]} = \\frac{24}{7}\n\\]\nThe distance from the origin $ O(0,0) $ to the line $ y = x - 1 $ is $ d = \\frac{|-1|}{\\sqrt{2}} = \\frac{\\sqrt{2}}{2} $. Therefore, the area $ S $ of $ \\triangle AOB $ is \n\\[\nS = \\frac{1}{2} \\times d \\times |AB| = \\frac{1}{2} \\times \\frac{\\sqrt{2}}{2} \\times \\frac{24}{7} = \\frac{6\\sqrt{2}}{7}\n\\]" }, { "text": "Given that the hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ $(a>0 , b>0)$ and the ellipse $\\frac{x^{2}}{16}+\\frac{y^{2}}{9}=1$ have the same foci, and the eccentricity of the hyperbola is twice the eccentricity of the ellipse, then the equation of the hyperbola is?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;H: Ellipse;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);Expression(H) = (x^2/16 + y^2/9 = 1);Focus(G) = Focus(H);Eccentricity(G)=2*Eccentricity(H)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/4-y^2/3=1", "fact_spans": "[[[2, 59], [106, 109], [125, 128]], [[5, 59]], [[5, 59]], [[60, 98], [115, 117]], [[5, 59]], [[5, 59]], [[2, 59]], [[60, 98]], [[2, 104]], [[106, 123]]]", "query_spans": "[[[125, 133]]]", "process": "" }, { "text": "The focal distance of a hyperbola is $4$ times the distance between its two directrices. What is the eccentricity of this hyperbola?", "fact_expressions": "G: Hyperbola;L1:Line;L2:Line;Directrix(G)={L1,L2};FocalLength(G)=4*Distance(L1,L2)", "query_expressions": "Eccentricity(G)", "answer_expressions": "2", "fact_spans": "[[[0, 3], [21, 24]], [], [], [[0, 10]], [[0, 18]]]", "query_spans": "[[[21, 31]]]", "process": "According to the focal length and the distance between the two directrices, set up an equation and then solve for the answer. [Detailed solution] Since the equations of the directrices of the hyperbola are: $ x = \\pm\\frac{a^{2}}{c} $, it follows that $ 2c = 4\\times\\frac{2a^{2}}{c} $, that is, $ c^{2} = 4a^{2} $, then $ e^{2} = \\frac{c^{2}}{a^{2}} = 4 $, $ e = 2 $." }, { "text": "Given that the center of an ellipse is at the origin, one focus is $F(-2\\sqrt{3}, 0)$, and the length of the major axis is twice the length of the minor axis, then the standard equation of this ellipse is?", "fact_expressions": "G: Ellipse;O: Origin;Center(G) = O;F: Point;Coordinate(F) = (-2*sqrt(3), 0);OneOf(Focus(G)) = F;Length(MajorAxis(G)) = 2*Length(MinorAxis(G))", "query_expressions": "Expression(G)", "answer_expressions": "x^2/16+y^2/4=1", "fact_spans": "[[[2, 4], [53, 55]], [[7, 9]], [[2, 9]], [[15, 36]], [[15, 36]], [[2, 36]], [[2, 50]]]", "query_spans": "[[[53, 62]]]", "process": "" }, { "text": "Given that the ellipse $a x^{2}+y^{2}=4$ and $\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1$ have equal focal distances, then $a$=?", "fact_expressions": "G: Ellipse;C:Ellipse;a: Number;Expression(G) = (a*x^2 + y^2 = 4);Expression(C)=(x^2/12 + y^2/9 = 1);FocalLength(G)=FocalLength(C)", "query_expressions": "a", "answer_expressions": "{4/7,4}", "fact_spans": "[[[2, 21]], [[22, 58]], [[65, 68]], [[2, 21]], [[22, 58]], [[2, 63]]]", "query_spans": "[[[65, 70]]]", "process": "" }, { "text": "Given that a hyperbola with asymptotes $x \\pm 2y = 0$ passes through the point $(4,1)$, then the standard equation of this hyperbola is?", "fact_expressions": "G: Hyperbola;Expression(Asymptote(G)) = (pm*2*y + x = 0);H: Point;Coordinate(H) = (4, 1);PointOnCurve(H, G)", "query_expressions": "Expression(G)", "answer_expressions": "x^2/12-y^2/3=1", "fact_spans": "[[[21, 24], [37, 40]], [[2, 24]], [[26, 34]], [[26, 34]], [[21, 34]]]", "query_spans": "[[[37, 47]]]", "process": "Let the hyperbola equation be x^{2}-4y^{2}=\\lambda. Substituting the point (4,1), we find the asymptotes of the hyperbola are x\\pm2y=0. Then let the hyperbola equation be: x^{2}-4y^{2}=\\lambda. Substituting the point (4,1), we get \\lambda=12. Hence, the hyperbola equation is: \\frac{x^{2}}{12}-\\frac{y^{2}}{3}=1" }, { "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 line to the asymptote $y=\\frac{b}{a}x$, with foot of perpendicular at $M$. This line intersects the left and right branches of the hyperbola at points $A$ and $B$ respectively. Find the range of values for the eccentricity of the hyperbola?", "fact_expressions": "G: Hyperbola;b: Number;a: Number;F: Point;A: Point;B: Point;M:Point;a>0;b>0;Expression(G) = (-y^2/b^2 + x^2/a^2 = 1);RightFocus(G) = F;L:Line;OneOf(Asymptote(G))=L;Expression(L)=(y=(b/a)*x);L1:Line;IsPerpendicular(L,L1);FootPoint(L,L1)=M;PointOnCurve(F, L1);Intersection(L1,LeftPart(G)) = A;Intersection(L1,RightPart(G)) = B", "query_expressions": "Range(Eccentricity(G))", "answer_expressions": "(sqrt(2),+oo)", "fact_spans": "[[[1, 57], [97, 100], [121, 124]], [[4, 57]], [[4, 57]], [[61, 64]], [[110, 113]], [[114, 117]], [[92, 95]], [[4, 57]], [[4, 57]], [[1, 57]], [[1, 64]], [[68, 85]], [[1, 85]], [[68, 85]], [], [[0, 88]], [[0, 95]], [[0, 88]], [[0, 119]], [[0, 119]]]", "query_spans": "[[[121, 134]]]", "process": "Since line AB intersects both the left and right branches of the hyperbola, line AB must intersect the asymptote y=-\\frac{b}{a}x in the second quadrant. Therefore, the slope of line AB must be greater than the slope of the asymptote y=-\\frac{b}{a}x, that is, -\\frac{a}{b}>-\\frac{b}{a}, which implies b^{2}>a^{2}. Since b^{2}=c^{2}-a^{2}, it follows that c^{2}>2a^{2}. The eccentricity of the hyperbola e=\\frac{c}{a}>\\sqrt{2}. Therefore, the range of values for the eccentricity of the hyperbola is (\\sqrt{2},+\\infty)." }, { "text": "The equation of a parabola with vertex at the origin, focus on the $x$-axis, and passing through the point $(-4,-8)$ is?", "fact_expressions": "G: Parabola;H: Point;O: Origin;Coordinate(H) = (-4, -8);PointOnCurve(Focus(G), xAxis);PointOnCurve(H, G);Vertex(G) = O", "query_expressions": "Expression(G)", "answer_expressions": "y^2 = -16*x", "fact_spans": "[[[28, 31]], [[17, 27]], [[3, 5]], [[17, 27]], [[6, 31]], [[16, 31]], [[0, 31]]]", "query_spans": "[[[28, 35]]]", "process": "According to the problem, let the equation of the parabola be y^{2}=mx, m\\neq0. Then (-8)^{2}=m\\cdot(-4), solving gives m=-16, so the required parabola equation is y^{2}=-16x. Hence the answer is: y^{2}=-16x" }, { "text": "Given that the foci of the ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{2}=1$ are $F_{1}$ and $F_{2}$ respectively, and point $P$ lies on the ellipse. If $|P F_{1}|=4$, then the area of triangle $F_{1} P F_{2}$ is?", "fact_expressions": "G: Ellipse;Expression(G) = (x^2/9 + y^2/2 = 1);F1: Point;F2: Point;Focus(G) = {F1, F2};P: Point;PointOnCurve(P, G);Abs(LineSegmentOf(P, F1)) = 4", "query_expressions": "Area(TriangleOf(F1, P, F2))", "answer_expressions": "2*sqrt(3)", "fact_spans": "[[[2, 39], [66, 68]], [[2, 39]], [[45, 52]], [[53, 60]], [[2, 60]], [[61, 65]], [[61, 69]], [[71, 84]]]", "query_spans": "[[[86, 109]]]", "process": "" }, { "text": "Let $F$ be the focus of the parabola $C$: $y^{2}=8 x$. A line passing through $F$ with an inclination angle of $30^{\\circ}$ intersects $C$ at points $A$ and $B$. Then $|A B|$=?", "fact_expressions": "C: Parabola;Expression(C) = (y^2 = 8*x);F: Point;Focus(C) = F;G: Line;PointOnCurve(F, G);Inclination(G) = ApplyUnit(30, degree);A: Point;B: Point;Intersection(G, C) = {A, B}", "query_expressions": "Abs(LineSegmentOf(A, B))", "answer_expressions": "32", "fact_spans": "[[[5, 24], [54, 57]], [[5, 24]], [[1, 4], [29, 32]], [[1, 27]], [[50, 52]], [[28, 52]], [[33, 52]], [[58, 61]], [[62, 65]], [[50, 67]]]", "query_spans": "[[[69, 78]]]", "process": "From $ y^{2} = 8x $, we get $ 2p = 8 $, $ p = 4 $, then $ F(2,0) $. Therefore, the equation of the line passing through $ A $ and $ B $ is $ y = \\frac{\\sqrt{3}}{3}(x - 2) $. Solving the system \n\\[\n\\begin{cases}\ny^{2} = 8x \\\\\ny = \\frac{\\sqrt{3}}{3}(x - 2)\n\\end{cases}\n\\]\nyields $ x^{2} - 28x + 4 = 0 $. Let $ A(x_{1}, y_{1}) $, $ B(x_{2}, y_{2}) $, then $ x_{1} + x_{2} = 28 $, and since $ |AB| = x_{1} + x_{2} + p = 28 + 4 = 32 $." } ]