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 the value of \(b^{2}\) is

Given equation of ellipse is
\(
\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1
\)
Here, \(a^{2}=16 \Rightarrow a=4\)
\(
e=\sqrt{1-\frac{b^{2}}{16}}=\frac{\sqrt{16-b^{2}}}{4}
\)
\(\therefore\) Foci of ellipse are \((\pm a e, 0) i e,\left(\pm \sqrt{16-b^{2}}, 0\right)\). Also, given equation of hyperbola is \(\frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25}\)
Here, \(\quad a^{2}=\left(\frac{12}{5}\right)^{2}, b^{2}=\left(\frac{9}{5}\right)^{2}\)
\(\therefore\)
\(
e=\sqrt{1+\frac{b^{2}}{a^{2}}}=\sqrt{1+\frac{81}{144}}=\frac{5}{4}
\)
\(\therefore\) Foci of the hyperbola are \((\pm a e, 0) i e,(\pm 3,0)\). According to the given condition, foci of ellipse \(=\) foci of hyperbola \(\therefore \sqrt{16-b^{2}}=3\)
\(\Rightarrow \quad b^{2}=7\)