Q90 / 180 · COMEDK 2013 · Mathematics
If 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$1$
- Correct answerB$7$
- C$5$
- D$9$
Solution
The equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and equation of the hyperbola is $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ or $\frac{x^{2}}{\left(\frac{12}{5}\right)^{2}}-\frac{y}{\left(\frac{9}{5}\right)^{2}}=1$
$\therefore$ Eccentricity of ellipse, $e=\sqrt{1-\frac{b^{2}}{16}}$
and eccentricity of hyperbola,
$$
e^{\prime}=\sqrt{1+\frac{\left(\frac{9}{5}\right)^{2}}{\left(\frac{12}{5}\right)^{2}}}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}
$$
Since, foci of ellipse and hyperbola coincide.
$$
\begin{aligned}
& \therefore & 4 e &=\frac{12}{5} e^{\prime} \\
\Rightarrow & e &=\frac{3}{5} e^{\prime} \\
\Rightarrow \quad & \sqrt{1-\frac{b^{2}}{16}} &=\frac{3}{5} \times \frac{15}{12} \\
\Rightarrow & & 1-\frac{b^{2}}{16} &=\left(\frac{3}{4}\right)^{2} \Rightarrow b^{2}=7
\end{aligned}
$$
Hence, the correct answer is option: 'B'.