Question

If 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) 1

(b) 7

(c) 5

(d) 9

Answer 1

Eccentricity of hyperbola is $e=\sqrt{1+\frac{81}{144}}=\frac{15}{12}=\frac{5}{4}$

$\therefore \quad$ Foci $=( \pm a e, 0)=( \pm 3,0)$

Foci of ellipse $=\left( \pm a e^{\prime}, 0\right)=( \pm 4 e, 0)$

According to the given condition,

$\begin{array}{rlrl}4 e^{\prime}=3 \Rightarrow e^{\prime} =\frac{3}{4} \\\Rightarrow \sqrt{1-\frac{b^2}{16}} =\frac{3}{4} \\\Rightarrow & b^2 =7\end{array}$

So, The correct option of this question will be (B).