Question

If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ and $e_2$ is the eccentricity of the hyperbola passing through the foci of the ellipse and $e_1.e_2=1$, then the equation of the hyperbola is

• A. $\frac{16x^2}{81}-\frac{y^2}{9}=-1$
• B. $\frac{16x^2}{81}-\frac{y^2}{9}=1$
• C. $\frac{x^2}{9}-\frac{16y^2}{81}=-1$
• D. $\frac{x^2}{9}-\frac{16y^2}{81}=1$

Answer 1

The correct option is A $\frac{16x^2}{81}-\frac{y^2}{9}=-1$

$\frac{x^2}{16}+\frac{y^2}{25}=1$

$e_1=\frac{\sqrt{25-16}}{25}=\frac{3}{5}$
$e_1.e_2=1\Rightarrow e_2=\frac{5}{3}$

Equation of hyperbola :
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ (Assume)
$e_2^2=1+\frac{b^2}{a^2}\Rightarrow b^2=\frac{16}{9}{a}^2$
$\therefore \frac{x^2}{a^2}-\frac{y^2}{\frac{16}{9}{a}^2}=1$
Hyperbola passes through the foci of ellipse $(0,\pm 3)$
$\Rightarrow \frac{-9}{\frac{16}{9}{a}^2}=1\Rightarrow a^2=\frac{-81}{16}$
So, our assumption is wrong.

Therefore, equation of hyperbola is
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$
$\Rightarrow \frac{x^2}{\left(\frac{81}{16}\right)}-\frac{y^2}{\frac{16}{9}.\frac{81}{16}}=-1$
$\Rightarrow \frac{16x^2}{81}-\frac{y^2}{9}=-1$