# A hyperbola passes through the foci of the ellipse

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:

1. (1) $\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$

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

3. (3) $x^{2}-y^{2}=9$

4. (4) $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$

Correct Option: , 2

Solution:

$e_{1}=\sqrt{1-\frac{16}{25}}=\frac{3}{5} \quad$ foci $(\pm \mathrm{ae}, 0)$

Foci $=(\pm 3,0)$

Let equation of hyperbolabe $\frac{x^{2}}{A^{2}}-\frac{y^{2}}{B^{2}}=1$ Passes through $(\pm 3,0) A^{2}=9, A=3, e_{2}=\frac{5}{3}$

$e_{2}^{2}=1+\frac{B^{2}}{A^{2}}$

$\frac{25}{9}=1+\frac{B^{2}}{9} \Rightarrow B^{2}=16$

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