Question

If the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the value of $b^2$ is

• A. $25$
• B. $9$
• C. $16$
• D. $4$

Answer 1

Answer: (C)

$16$

For the given ellipse, $a=5,b=b$

So, $e=\sqrt{1-\frac{b^2}{25}}$

Foci $=(\pm ae,0)=(\pm 5\sqrt{1-\frac{b^2}{25}},0)$

For the hyperbola, $a=\frac{12}{5},b=\frac{9}{5}$

$e=\sqrt{1+\frac{81}{144}}=\frac{15}{12}$

Foci $=(\pm ae,0)=(\pm \frac{12}{5}\times \frac{15}{12},0)=(\pm 3,0)$

Since, the foci of ellipse and hyperbola are equal

$⟹3=5\sqrt{1-\frac{b^2}{25}}$

$⟹\frac{9}{25}=\frac{25-b^2}{25}$

$⟹b^2=16$