Q96 / 160 · TS EAMCET 2013 · Mathematics
If the focii of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$ coincide, then $b^2$ is equal to
- A$4$
- Correct answerB$5$
- C$8$
- D$9$
Solution
Given, equation of ellipse is
$$
\frac{x^2}{25}+\frac{y^2}{16}=1
$$
and equation of hyperbola is
$$
\frac{x^2}{4}-\frac{y^2}{b^2}=1
$$
eccentricity of ellipse
$$
\begin{aligned}
& b^2=a^2\left(1-e^2\right) \\
& \Rightarrow \quad 16=25\left(1-e^2\right) \\
& \Rightarrow \quad e^2=1-\frac{16}{25}=\frac{9}{25} \\
& \Rightarrow \quad e= \pm \frac{3}{5} \\
&
\end{aligned}
$$
Focii of the ellipse $=( \pm a e, 0)=( \pm 3,0)$ which coincide with focii of the hyperbola.
Let $e_1$ be the eccentricity of the hyperbola.
$$
\begin{array}{ll}
\therefore & \pm a e_1= \pm 3 \\
\Rightarrow & e_1=\frac{3}{2}
\end{array}
$$
Now, $\quad b^2=a^2\left(e_1^2-1\right)$
$$
\begin{array}{cc}
\Rightarrow & b^2=4\left(\frac{9}{4}-1\right)=4 \times \frac{5}{4} \\
\Rightarrow & b^2=5
\end{array}
$$
Hence, the correct answer is option: 'B'.