Question

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

• A. $1$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is C $7$

Given the equation of the ellipse is

$\frac{x^2}{16}-\frac{y^2}{b^2}=1$

Here $a^2=16$ $\Rightarrow$ $a=4$

$\therefore$ $e=\sqrt{1-\frac{b^2}{16}}=\frac{\sqrt{16-b^2}}{4}$

$\therefore$ Foci of ellipse are $(\pm ae,0)$ ie, $(\pm \sqrt{16-b^2},0)$

Also given the equation of the hyperbola is

$\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$

Here $a^2={\left(\frac{12}{5}\right)}^2,b^2={\left(\frac{9}{5}\right)}^2$

$\therefore$ $e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{81}{144}}=\frac{5}{4}$

$\therefore$ Foci of hyperbola are $(\pm ae,0)$ ie, $(\pm 3,0)$

According to the given condition

foci of ellipse$=$ foci of the hyperbola

$\therefore$ $\sqrt{16-b^2}=3$

$\Rightarrow$ $b^2=7$