Question

The value of $b^2$ in order that the foci of the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ and the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ coincide is

• A. 1
• B. 5
• C. 7
• D. 9

Answer 1

Answer: (C)

We have, Equation of hyperbola is
$\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\Rightarrow \frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{81}{25}\right)}=1$
$\Rightarrow \frac{x^2}{{\left(\frac{12}{5}\right)}^2}-\frac{y^2}{{\left(\frac{9}{5}\right)}^2}=1$
$\therefore$ Foci $=(\pm ae,0)$
$=\left(\pm \frac{12}{5}\sqrt{1+{\left(\frac{9/5}{12/5}\right)}^2},0\right)$
$=\left(\pm \frac{12}{5}\times 3,0\right)=(\pm 3,0)$
Equation of ellipse is
$\frac{x^2}{16}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{x^2}{4^2}+\frac{y^2}{b^2}=1$
$\therefore$ Foci $=(\pm ae,0)$
$=\left(\pm 4\sqrt{1-\frac{b^2}{4^2}},0\right)=\left(\pm \frac{4\sqrt{16-b^2}}{4},0\right)$
$=\left(\pm \sqrt{16-b^2},0\right)$
Since both have same foci
$\therefore \sqrt{16-b^2}=3$
$\Rightarrow 16-b^2=9$
$\Rightarrow b^2=7$