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. $1$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

Given equation of 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 $(\pm ae,0)ie,\left(\pm \sqrt{16-b^2},0\right)$
Also, given equation of 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 the hyperbola are $(\pm ae,0)ie,(\pm 3,0)$
According to the given condition, foci of ellipse = foci of hyperbola
$\therefore \sqrt{16-b^2}=3$
$\Rightarrow b^2=7$