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

Answer 1

The correct option is B

$\frac{x^2}{16}–\frac{y^2}{9}=-1$

Finding the equation of the hyperbola:

The eccentricity of $\frac{x^2}{16}+\frac{y^2}{25}=1$ is

Since, $\begin{array}{rcl}{e}_1& =& \sqrt{1–\frac{16}{25}}\\ & =& \frac{3}{5}\\ e_1{e}_2& =& 1\\ e_2& =& \frac{5}{3}\end{array}$

Clearly, the y-axis is the transverse axis of the ellipse.

⇒ Foci of the ellipse $(0,\pm 3)$

Let the hyperbola be $\frac{y^2}{b^2}–\frac{x^2}{a^2}=1...\left(i\right)$

given hyperbola passes through foci of the ellipse
$\Rightarrow b^2=9$ and also

$\begin{array}{rcl}{a}^2& =& b^2(e^2-1)\\ & =& 9\left(\frac{25}{9}-1\right)\\ & =& 16\end{array}$

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

Hence, the correct answer is option (B).