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 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 $\frac{x^2}{9}-\frac{y^2}{16}=1$

For ellipse $e_1=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$
Foci $=(\pm 3,0)$
Let equation of hyperbola be $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ which passes through $(\pm 3,0)$
$\Rightarrow A^2=9,A=3,e_2=\frac{5}{3}$
Now,
$e_2^2=1+\frac{B^2}{A^2}$
$\frac{25}{9}=1+\frac{B^2}{9}\Rightarrow B^2=16$

$\therefore$ Required equation of hyperbola is$\frac{x^2}{9}-\frac{y^2}{16}=1$