Q77 / 150 · MHT CET 2007 · Mathematics
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
- A1
- B5
- Correct answerC7
- D9
Solution
Given equation of ellipse is
$$
\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1
$$
Here, $a^{2}=16 \Rightarrow a=4$
$$
e=\sqrt{1-\frac{b^{2}}{16}}=\frac{\sqrt{16-b^{2}}}{4}
$$
$\therefore$ Foci of ellipse are $(\pm a e, 0) i e,\left(\pm \sqrt{16-b^{2}}, 0\right)$. Also, given equation of hyperbola is $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$
Here, $\quad a^{2}=\left(\frac{12}{5}\right)^{2}, b^{2}=\left(\frac{9}{5}\right)^{2}$
$\therefore$
$$
e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{81}{144}}=\frac{5}{4}
$$
$\therefore$ Foci of the hyperbola are $(\pm a e, 0) i e,(\pm 3,0)$. According to the given condition, foci of ellipse $=$ foci of hyperbola $\therefore \sqrt{16-b^{2}}=3$
$\Rightarrow \quad b^{2}=7$
Hence, the correct answer is option: 'C'.