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. The value of $b^2$ is

• A. $9$
• B. $1$
• C. $5$
• D. $7$

Answer 1

Answer: (D)

$7$

The equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$

Here, $a=\sqrt{\frac{144}{25}},b=\sqrt{\frac{81}{25}},e=\sqrt{1+\frac{81}{144}}=\frac{15}{12}=\frac{5}{4}$

$\therefore$ Foci $=(\pm 3,0)$

Also, focus of ellipse $=(3,0)\Rightarrow e=\frac{3}{4}$

$\therefore b^2=16(1-\frac{9}{16})=7$