Discussion Forum

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 a hyperbola  passing through  the foci of the given ellipse  and $e_{1}e_{2}=1$ , then the equation of such a hyperbola among the following is 

Answer:

Explanation:

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

 Foci   =(0,$\pm 3$)

 $\Rightarrow$    $e_{1}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$ given $e_{1} e_{2}$ =1

             (where $e_{2}$ is eccentricity of hyperbola )

Let equation of hyperbola -$\frac{x^{2}}{a^{2}}+\frac{ y^{2}}{b^{2}}$=1

 its passes through (0, $\pm 3$)

$\therefore$     $b^{2}$=9

$\Rightarrow$     $ e_{2}=\sqrt{1+\frac{a^{2}}{b^{2}}}$

$\Rightarrow$    $ e_{2}^{2}=1+\frac{a^{2}}{b^{2}}$

$\Rightarrow$    $ \frac{1}{e_{2}^{2}}=1+\frac{a^{2}}{9}\Rightarrow \frac{25}{9}=1+\frac{a^{2}}{9}$

 $\Rightarrow$      $a^{2}$=16

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