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

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, \(\mathrm{e}^{\prime 2}=1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\)
\(
\Rightarrow \frac{9}{16}=1-\frac{\mathrm{b}^{2}}{16}
\)
\(
\therefore \mathrm{b}^{2}=7
\)