Q43 / 120 · VITEEE 2016 · 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 value of $b^{2}$ is
- A1
- B5
- Correct answerC7
- D9
Solution
Given ellipse : $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$
Now $b^{2}=a^{2}\left(1-e^{2}\right)$
$$
\begin{array}{l}
\Rightarrow b^{2}=16\left(1-e^{2}\right), \Rightarrow \frac{b^{2}}{16}=1-e^{2} \\
\Rightarrow e^{2}=1-\frac{b^{2}}{16}=\frac{16-b^{2}}{16} \Rightarrow e=\frac{\sqrt{16-b^{2}}}{4} \\
\text { Foci }=(\pm a e, 0)=\left(\pm \sqrt{16-b^{2}}, 0\right)
\end{array}
$$
Given hyperbola : $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$
$$
\Rightarrow \frac{x^{2}}{\left(\frac{12}{5}\right)^{2}}-\frac{y^{2}}{\left(\frac{9}{5}\right)^{2}}=1
$$
Now, $b^{2}=a^{2}\left(e^{2}-1\right)$
$$
\begin{array}{l}
\Rightarrow\left(\frac{9}{5}\right)^{2}=\left(\frac{12}{5}\right)^{2}\left(e^{2}-1\right) \\
\Rightarrow\left(\frac{9}{12}\right)^{2}=e^{2}-1 \\
\Rightarrow e^{2}=1+\frac{81}{144}=\frac{144+81}{144} \\
\Rightarrow e=\frac{15}{12}=\frac{5}{4} \\
\text { Foci }=(\pm a e, o)=(\pm 3,0)
\end{array}
$$
Since foci of the given ellipse and hyperbola coincide, therefore
$$
\begin{array}{l}
\sqrt{16-b^{2}}=3 \Rightarrow 16-b^{2}=9 \\
\therefore b^{2}=7
\end{array}
$$
Hence, the correct answer is option: 'C'.