Equation of Hyperbola
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Hyperbola

120736 A hyperbola with centre at \((0,0)\) has its transverse axis along \(\mathrm{X}\)-axis whose length is 12 . If \((8,2)\) is a point on the hyperbola, then its eccentricity is

1 \(\frac{8}{7}\)
2 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
3 \(\frac{3}{\sqrt{7}}\)
4 \(\frac{9}{7}\)
AP EAMCET-21.04.2019
Hyperbola

120737 If the foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the equation of hyperbola, if its eccentricity is 2 , is \(\qquad\)

1 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
2 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{16}-\frac{y^2}{144}=1\)
4 \(\frac{\mathrm{x}^2}{2}-\frac{\mathrm{y}^2}{12}=1\)
AP EAMCET-05.10.2021
Hyperbola

120738 The line \(y=x\) intersects the hyperbola \(\frac{x^2}{9}-\frac{y^2}{25}=1\) at the points \(P\) and \(Q\). The eccentricity of ellipse with PQ as major axis and minor axis of length \(\frac{5}{\sqrt{2}}\) is

1 \(\frac{\sqrt{5}}{3}\)
2 \(\frac{5}{\sqrt{3}}\)
3 \(\frac{5}{9}\)
4 \(\frac{2 \sqrt{2}}{3}\)
WB JEE-2013
Hyperbola

120739 Let \(e\) and \(e^{\prime}\) be the eccentricities of a hyperbola and its conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: J and K CET-2018], 1021. What will be the equation of the standard hyperbola where foci are \((0, \pm 10)\) and the length of the latus rectum is 30 ?,

1 0
2 1
3 2
4 None of these
J and K CET-2018]
Hyperbola

120740 Let \(L\left(x_1, 4\right)\) be the end of the Latus rectum of the hyperbola \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) lying in the first quadrant and let \(S\left(8, y_1\right)\) be the focus of the given hyperbopla. Then the length of its transverse axis is

1 \(2(\sqrt{17}-1)\)
2 \(4(\sqrt{17}-1)\)
3 \(2(\sqrt{17}+1)\)
4 \(4(\sqrt{17}+1)\)
AP EAMCET-06.07.2022
NEET DIGITAL LIBRARY
Hyperbola

120736 A hyperbola with centre at \((0,0)\) has its transverse axis along \(\mathrm{X}\)-axis whose length is 12 . If \((8,2)\) is a point on the hyperbola, then its eccentricity is

1 \(\frac{8}{7}\)
2 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
3 \(\frac{3}{\sqrt{7}}\)
4 \(\frac{9}{7}\)
AP EAMCET-21.04.2019
Hyperbola

120737 If the foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the equation of hyperbola, if its eccentricity is 2 , is \(\qquad\)

1 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
2 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{16}-\frac{y^2}{144}=1\)
4 \(\frac{\mathrm{x}^2}{2}-\frac{\mathrm{y}^2}{12}=1\)
AP EAMCET-05.10.2021
Hyperbola

120738 The line \(y=x\) intersects the hyperbola \(\frac{x^2}{9}-\frac{y^2}{25}=1\) at the points \(P\) and \(Q\). The eccentricity of ellipse with PQ as major axis and minor axis of length \(\frac{5}{\sqrt{2}}\) is

1 \(\frac{\sqrt{5}}{3}\)
2 \(\frac{5}{\sqrt{3}}\)
3 \(\frac{5}{9}\)
4 \(\frac{2 \sqrt{2}}{3}\)
WB JEE-2013
Hyperbola

120739 Let \(e\) and \(e^{\prime}\) be the eccentricities of a hyperbola and its conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: J and K CET-2018], 1021. What will be the equation of the standard hyperbola where foci are \((0, \pm 10)\) and the length of the latus rectum is 30 ?,

1 0
2 1
3 2
4 None of these
J and K CET-2018]
Hyperbola

120740 Let \(L\left(x_1, 4\right)\) be the end of the Latus rectum of the hyperbola \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) lying in the first quadrant and let \(S\left(8, y_1\right)\) be the focus of the given hyperbopla. Then the length of its transverse axis is

1 \(2(\sqrt{17}-1)\)
2 \(4(\sqrt{17}-1)\)
3 \(2(\sqrt{17}+1)\)
4 \(4(\sqrt{17}+1)\)
AP EAMCET-06.07.2022
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Hyperbola

120736 A hyperbola with centre at \((0,0)\) has its transverse axis along \(\mathrm{X}\)-axis whose length is 12 . If \((8,2)\) is a point on the hyperbola, then its eccentricity is

1 \(\frac{8}{7}\)
2 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
3 \(\frac{3}{\sqrt{7}}\)
4 \(\frac{9}{7}\)
AP EAMCET-21.04.2019
Hyperbola

120737 If the foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the equation of hyperbola, if its eccentricity is 2 , is \(\qquad\)

1 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
2 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{16}-\frac{y^2}{144}=1\)
4 \(\frac{\mathrm{x}^2}{2}-\frac{\mathrm{y}^2}{12}=1\)
AP EAMCET-05.10.2021
Hyperbola

120738 The line \(y=x\) intersects the hyperbola \(\frac{x^2}{9}-\frac{y^2}{25}=1\) at the points \(P\) and \(Q\). The eccentricity of ellipse with PQ as major axis and minor axis of length \(\frac{5}{\sqrt{2}}\) is

1 \(\frac{\sqrt{5}}{3}\)
2 \(\frac{5}{\sqrt{3}}\)
3 \(\frac{5}{9}\)
4 \(\frac{2 \sqrt{2}}{3}\)
WB JEE-2013
Hyperbola

120739 Let \(e\) and \(e^{\prime}\) be the eccentricities of a hyperbola and its conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: J and K CET-2018], 1021. What will be the equation of the standard hyperbola where foci are \((0, \pm 10)\) and the length of the latus rectum is 30 ?,

1 0
2 1
3 2
4 None of these
J and K CET-2018]
Hyperbola

120740 Let \(L\left(x_1, 4\right)\) be the end of the Latus rectum of the hyperbola \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) lying in the first quadrant and let \(S\left(8, y_1\right)\) be the focus of the given hyperbopla. Then the length of its transverse axis is

1 \(2(\sqrt{17}-1)\)
2 \(4(\sqrt{17}-1)\)
3 \(2(\sqrt{17}+1)\)
4 \(4(\sqrt{17}+1)\)
AP EAMCET-06.07.2022
For More Updates join with Us
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Hyperbola

120736 A hyperbola with centre at \((0,0)\) has its transverse axis along \(\mathrm{X}\)-axis whose length is 12 . If \((8,2)\) is a point on the hyperbola, then its eccentricity is

1 \(\frac{8}{7}\)
2 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
3 \(\frac{3}{\sqrt{7}}\)
4 \(\frac{9}{7}\)
AP EAMCET-21.04.2019
Hyperbola

120737 If the foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the equation of hyperbola, if its eccentricity is 2 , is \(\qquad\)

1 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
2 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{16}-\frac{y^2}{144}=1\)
4 \(\frac{\mathrm{x}^2}{2}-\frac{\mathrm{y}^2}{12}=1\)
AP EAMCET-05.10.2021
Hyperbola

120738 The line \(y=x\) intersects the hyperbola \(\frac{x^2}{9}-\frac{y^2}{25}=1\) at the points \(P\) and \(Q\). The eccentricity of ellipse with PQ as major axis and minor axis of length \(\frac{5}{\sqrt{2}}\) is

1 \(\frac{\sqrt{5}}{3}\)
2 \(\frac{5}{\sqrt{3}}\)
3 \(\frac{5}{9}\)
4 \(\frac{2 \sqrt{2}}{3}\)
WB JEE-2013
Hyperbola

120739 Let \(e\) and \(e^{\prime}\) be the eccentricities of a hyperbola and its conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: J and K CET-2018], 1021. What will be the equation of the standard hyperbola where foci are \((0, \pm 10)\) and the length of the latus rectum is 30 ?,

1 0
2 1
3 2
4 None of these
J and K CET-2018]
Hyperbola

120740 Let \(L\left(x_1, 4\right)\) be the end of the Latus rectum of the hyperbola \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) lying in the first quadrant and let \(S\left(8, y_1\right)\) be the focus of the given hyperbopla. Then the length of its transverse axis is

1 \(2(\sqrt{17}-1)\)
2 \(4(\sqrt{17}-1)\)
3 \(2(\sqrt{17}+1)\)
4 \(4(\sqrt{17}+1)\)
AP EAMCET-06.07.2022
NEET DIGITAL LIBRARY
Hyperbola

120736 A hyperbola with centre at \((0,0)\) has its transverse axis along \(\mathrm{X}\)-axis whose length is 12 . If \((8,2)\) is a point on the hyperbola, then its eccentricity is

1 \(\frac{8}{7}\)
2 \(\frac{2 \sqrt{2}}{\sqrt{7}}\)
3 \(\frac{3}{\sqrt{7}}\)
4 \(\frac{9}{7}\)
AP EAMCET-21.04.2019
Hyperbola

120737 If the foci of a hyperbola coincide with the foci of the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\), then the equation of hyperbola, if its eccentricity is 2 , is \(\qquad\)

1 \(\frac{x^2}{4}-\frac{y^2}{12}=1\)
2 \(\frac{x^2}{12}-\frac{y^2}{4}=1\)
3 \(\frac{x^2}{16}-\frac{y^2}{144}=1\)
4 \(\frac{\mathrm{x}^2}{2}-\frac{\mathrm{y}^2}{12}=1\)
AP EAMCET-05.10.2021
Hyperbola

120738 The line \(y=x\) intersects the hyperbola \(\frac{x^2}{9}-\frac{y^2}{25}=1\) at the points \(P\) and \(Q\). The eccentricity of ellipse with PQ as major axis and minor axis of length \(\frac{5}{\sqrt{2}}\) is

1 \(\frac{\sqrt{5}}{3}\)
2 \(\frac{5}{\sqrt{3}}\)
3 \(\frac{5}{9}\)
4 \(\frac{2 \sqrt{2}}{3}\)
WB JEE-2013
Hyperbola

120739 Let \(e\) and \(e^{\prime}\) be the eccentricities of a hyperbola and its conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
#[Qdiff: Hard, QCat: Numerical Based, examname: J and K CET-2018], 1021. What will be the equation of the standard hyperbola where foci are \((0, \pm 10)\) and the length of the latus rectum is 30 ?,

1 0
2 1
3 2
4 None of these
J and K CET-2018]
Hyperbola

120740 Let \(L\left(x_1, 4\right)\) be the end of the Latus rectum of the hyperbola \(\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) lying in the first quadrant and let \(S\left(8, y_1\right)\) be the focus of the given hyperbopla. Then the length of its transverse axis is

1 \(2(\sqrt{17}-1)\)
2 \(4(\sqrt{17}-1)\)
3 \(2(\sqrt{17}+1)\)
4 \(4(\sqrt{17}+1)\)
AP EAMCET-06.07.2022