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

Foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ are $(\pm 4e,0)$
Given hyperbola is $\frac{x^2}{(12/5{)}^2}-\frac{y^2}{(9/5{)}^2}=1\therefore b^2=a^2(e^2-1)\Rightarrow \frac{81}{25}=\frac{144}{25}(e^2-1)\Rightarrow e^2=\frac{25}{16}\Rightarrow e=\frac{5}{4}$
$\therefore Itsfociare(\pm ae,0)\Rightarrow (\pm \frac{12}{5}\times \frac{5}{4},0)\Rightarrow (\pm 3,0)$
Since foci of ellipse and hyperbola coincide $\Rightarrow 4e-3\Rightarrow e=\frac{3}{4}$. $\therefore b^2=a^2(1-e^2)=16(1-\frac{9}{16})=7$