If the foci of the ellipse (x^2)/(16)+(y^2)/(b^2)=1 coincide with...

Given equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{b^2}=1$$\begin{aligned}& \text{ eccentricity }=e=\sqrt{1-\frac{b^2}{16}}\\ & \text{ foci: } \pma e=\pm4 \sqrt{1-\frac{b^2}{16}} \end{aligned}$Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}\begin{aligned}& \Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1 \\ & \text{ eccentricity }=e=\sqrt{1+\frac{81}{25} \times \frac{25}{144}}=\sqrt{1+\frac{81}{144}}\\ & =\sqrt{\frac{225}{144}}=\frac{15}{12}\\ & \text{ foci: } \pma e=\pm \frac{12}{5} \times \frac{15}{12}=\pm3 \\ & \end{aligned}$Since, foci of ellipse and hyperbola coincide & \therefore\pm$4 \sqrt{1-\frac{b^2}{16}}=\pm3 \Rightarrow b²=7 \\$