Q90 / 180 · COMEDK 2018 · Mathematics
If the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ coincide, then the value of $b^{2}$ is
- A25
- B9
- Correct answerC16
- D4
Solution
Eccentricity for $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is $b^{2}=a^{2}\left(1-e^{2}\right)$ and eccentricity for $\frac{x^{2}}{\frac{144}{25}}-\frac{y^{2}}{\frac{81}{25}}=1$ is
Again, foci $=a_{1} e_{1}=\frac{12}{5} \times \frac{15}{12}=3$
So, focus of hyperbola is $(3,0)=(a e, 0)$
and focus of ellipse is $(a e, 0)=(5 a, 0)$
As, this foci are same, so
$$
\begin{aligned}
& 5 a=3 \\
\therefore & e &=\frac{3}{5} \\
& \text { So, } \quad & e^{2} &=1-\frac{b^{2}}{a^{2}}=1-\frac{b^{2}}{25} \\
\Rightarrow & & \frac{b^{2}}{25} &=1-e^{2}=1-\frac{9}{25} \\
\Rightarrow & & \frac{b^{2}}{25} &=\frac{16}{25} \\
\Rightarrow & & b^{2} &=16
\end{aligned}
$$
Hence, the correct answer is option: 'C'.