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