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

Answer: (B)

7

eccentricity of ellipse is $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{b^2}{16}}$

eccentricity of hyperbola is $e^{\prime }=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}$

Since foci of ellipse and hyperbola coincide

Therefore, $4e=\frac{12}{5}{e}^{\prime }$

$\Rightarrow \sqrt{1-\frac{b^2}{16}}=\frac{3}{4}$

$\Rightarrow b^2=7$