Question

If a hyperbola passes through the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and its transverse and conjugate axes coincide with the major and minor axes of the ellipse and product of their eccentricities be 1, then the equation of hyperbola is

• A. $\frac{x^2}{9}-\frac{y^2}{25}=1$
• B. $\frac{x^2}{9}-\frac{y^2}{16}=1$
• C. $\frac{x^2}{16}-\frac{y^2}{25}=1$
• D. none of these.

Answer 1

Answer: (B)

Given ellipse is $\frac{x^{2}}{25} +\frac{y^{2}}{16} = 1 $
$ \therefore b^{2} = a^{2} \left(1-e^{2}\right) $
$\Rightarrow \frac{16}{25} = 1-e^{2} $
$ \Rightarrow e^{2}= \frac{9}{25}$
Let equation of hyperbola be
$ \frac{x^{2}}{a'^{2}}-\frac{y^{2}}{b'^{2}} = 1 \quad...\left(1\right)$
$ \therefore a'^{2}+b'^{2}= a'^{2}e'^{2} $
Since $e\times e' =1$
$\Rightarrow e' = \frac{5}{3} $
$ \Rightarrow a'^{2}+b'^{2} = \frac{25}{9}a'^{2} $
$ \Rightarrow 9b'^{2} = 16a'^{2}\quad...\left(2\right)$
also coordinates of focus of ellipse are $\left(\pm ae, 0\right) = \left( \pm3, 0\right)$
$\therefore $ hyperbola passes through $\left(\pm 3, 0\right)$
$ \therefore \frac{9}{a'^{2}} = 1 $
$ \Rightarrow a' = 9 $
$\Rightarrow a' = 3 \left(from \left(1\right)\right)$
from $ \left(ii\right) 9b'^{2}=16a'^{2}$
$ \Rightarrow b'^{2} = 16$
$\Rightarrow b' = 4$
$ \therefore $ equation of hyperbola is $\frac{x^{2}}{9} - \frac{y^{2}}{16 } = 1$