Question

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, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

• A. $\frac{x^2}{9}–\frac{y^2}{4}=1$
• B. $\frac{x^2}{9}–\frac{y^2}{16}=1$
• C. $x^2–y^2=9$
• D. $\frac{x^2}{9}–\frac{y^2}{25}=1$

Answer 1

The correct option is B

$\frac{x^2}{9}–\frac{y^2}{16}=1$

Explanation for the correct option:

Step 1. Finding the foci of the ellipse:

$\frac{x^2}{25}+\frac{y^2}{16}=1$ by comparing with standard equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$a^2=25,b^2=16$

For the given ellipse,

Let $e_1$ is eccentricity of hyperbola

$\begin{array}{rcl}{\mathbf{e}}_{\mathbf{1}}& \mathbf{=}& \sqrt{\mathbf{1}\mathbf{–}\frac{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{a}}^{\mathbf{2}}}}\end{array}$

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

Foci $=(\pm 3,0)$ $\left[\because \mathrm{foci}=({\mathrm{ae}}_1,0)\right]$

Step 2. Let the equation of hyperbola be $\frac{x^2}{a^2}–\frac{y^2}{b^2}=1,$ passes through $(\pm 3,0)$

Let $e_2$ is eccentricity of hyperbola

$\begin{array}{rcl}{a}^2& =& 9\end{array}$

$\Rightarrow$$\begin{array}{rcl}a& =& \pm 3\end{array}$

$\begin{array}{rcl}\therefore e_2& =& \frac{5}{3}\end{array}$ $\left[\because \mathrm{given}{e}_1{e}_2=1\right]$

$\begin{array}{rcl}{e}_2^2& =& 1+\frac{b^2}{a^2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{25}{9}& =& 1+\frac{b^2}{9}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{25}{9}-1& =& \frac{b^2}{9}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{16}{9}& =& \frac{b^2}{9}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}16& =& b^2\end{array}$

$\therefore$The equation of the hyperbola is $\frac{\mathbf{}{\mathbf{x}}^{\mathbf{2}}}{\mathbf{9}}\mathbf{–}\mathbf{}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{16}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, the correct option is (B).