If foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\left(b^{2} < 16\right)$ and the hyperbola…
If foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\left(b^{2} < 16\right)$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}$ coincide,
then the value of $b^{2}$ is
$4$
$9$
$14$
$7$
Solution
Given hyperbola is, $\frac{\mathrm{x}^{2}}{144}-\frac{\mathrm{y}^{2}}{81}=\frac{1}{25}$
$\begin{array}{l}
\Rightarrow \frac{\mathrm{x}^{2}}{144 / 25}-\frac{\mathrm{y}^{2}}{81 / 25}=1 \\
\Rightarrow \mathrm{a}^{2}=\frac{144}{25}, \mathrm{~b}^{2}=\frac{81}{25}, \mathrm{e}=\sqrt{1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\sqrt{1+\frac{81}{144}}=\frac{15}{12}
\end{array}$
$\therefore$ foci of hyperbola are $(\pm \mathrm{ae}, 0)$ i.e $(\pm 3,0)$
Now given ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \mathrm{a}^{2}=16$
Assume eccentricity of this ellipse is e then its foci are $\left(\pm \mathrm{ae}^{\prime}, 0\right) \mathrm{i} . \mathrm{e}\left(\pm 4 \mathrm{e}^{\prime}, 0\right)$ Given foci of given hyperbola and ellipse coincide
$\Rightarrow 4 \mathrm{e}^{\prime}=3 \Rightarrow \mathrm{e}^{\prime}=\frac{3}{4}$
For ellipse, using eccentricity relationship, $\mathrm{e}^{\prime 2}=1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}$
$\Rightarrow \frac{9}{16}=1-\frac{\mathrm{b}^{2}}{16}$
$\therefore \mathrm{b}^{2}=7$