Let a hyperbola passes through the focus of the ellipse The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then

Question

Let a hyperbola passes through the focus of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 .$ The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then [2006]

Smart Achievers · Accepted Answer

(1, 3)

For the given ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ $\Rightarrow e = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$

$\Rightarrow$ Eccentricity of hyperbola $= \frac{5}{3}$

Let the hyperbola be $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ then

$B^{2} = A^{2} (\frac{25}{9} - 1)$ $= \frac{16}{9} A^{2}$ $∴$ $\frac{x^{2}}{A^{2}} - \frac{9 y^{2}}{16 A^{2}} = 1$

As it passes through focus of ellipse i.e. $(3, 0)$

$∴$ $we get A^{2} = 9$ $\Rightarrow B^{2} = 16$

$∴$ Equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

Its focus is (5, 0) and vertex is (3, 0)

Solution

Q.
Let a hyperbola passes through the focus of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1 .$ The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then [2006]

1 the equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16} = 1$

2 the equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{25} = 1$

3 focus of hyperbola is $(5, 0)$

4 vertex of hyperbola is $(5 \sqrt{3}, 0)$

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Solution

Q. Let a hyperbola passes through the focus of the ellipse x225+y216=1. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then [2006]

1 the equation of hyperbola is x29-y216=1

2 the equation of hyperbola is x29-y225=1

3 focus of hyperbola is (5,0)