Q90 / 160 · AP EAMCET 2015 · 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
- A$5$
- Correct answerB$7$
- C$9$
- D$1$
Solution
Equation of ellipse, $\frac{x^2}{16}+\frac{y^2}{b^2}=1$
Eccentricity,
$\therefore \quad e=\sqrt{1-\frac{b^2}{16}}=\frac{\sqrt{16-b^2}}{4}$
So, the focus will be $\left( \pm \sqrt{16-b^2}, 0\right)$ Also, the equation of hyperbola,
$\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25} \Rightarrow \frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{81}{25}\right)}=1$
$\therefore \quad e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{\frac{81}{25}}{\frac{144}{25}}}= \pm \frac{15}{12}$
So, the focus will be
$\left( \pm \frac{12}{5} \times \frac{15}{12}, 0\right) \text { i.e., }( \pm 3,0)$
On comparing the focus of ellipse with hyperbola, we get
$\begin{aligned}
& \sqrt{16-b^2}=3 \\
& 16-b^2=9 \\
& b^2=7
\end{aligned}$
Hence, the correct answer is option: 'B'.