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 a e, 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 a e, 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$