Question

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then , the value of $b^2$ is :

• A. $5$
• B. $7$
• C. $9$
• D. $1$

Answer 1

The correct option is B $7$

$\therefore$ For ellipse focus $=$$(\sqrt{16-b^2},0)$ .

$\therefore$ The value of $b^2$ $=$ $7$.