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

The correct option is C $7$

We are given Ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$

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

Foci of both the curves coincide.

The equation of Hyperbola can be written as $\frac{{25x}^2}{144}-\frac{{25y}^2}{81}=1\Rightarrow \frac{x^2}{{\left(\frac{12}{5}\right)}^2}-\frac{y^2}{{\left(\frac{9}{5}\right)}^2}=1$

We see that $a=\frac{12}{5}$, $b=\frac{9}{5}$

Now eccentricity $e=\frac{\sqrt{b^2+a^2}}{a}=\frac{\sqrt{\frac{144}{25}+\frac{81}{25}}}{\frac{12}{5}}=\frac{5}{4}$

So foci for hyperbola is $(ae,0)=\left(\frac{5}{4}\times \frac{12}{5},0\right)=(3,0)$

Now for the ellipse, focus is again $(ae,0)=(3,0)$

we have $a=4$ for ellipse, so eccentricity $e=\frac{\sqrt{16-b^2}}{4}$

so focus is $(ae,0)=(4\times \frac{\sqrt{16-b^2}}{4},0)=(\sqrt{16-b^2},0)$...................2

Since foci are equal then from 1 and 2 we have $\sqrt{16-b^2}=3\Rightarrow b^2=7$