Mathematics - Ellipse Question with Solution | TestHub

Given equation of ellipse is $$\frac{x^2}{16}+\frac{y^2}{b^2}=1$$ $$\begin{array}{rl}& \text{ eccentricity }=e=\sqrt{1-\frac{b^2}{16}}\\ & \text{ foci: }\pm ae=\pm 4\sqrt{1-\frac{b^2}{16}}\end{array}$$ Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ $$\begin{array}{rl}& \Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1\\ & \text{ eccentricity }=e=\sqrt{1+\frac{81}{25}\times \frac{25}{144}}=\sqrt{1+\frac{81}{144}}\\ & =\sqrt{\frac{225}{144}}=\frac{15}{12}\\ & \text{ foci: }\pm ae=\pm \frac{12}{5}\times \frac{15}{12}=\pm 3\\ & \end{array}$$ Since, foci of ellipse and hyperbola coincide & \therefore \pm 4 \sqrt{1-\frac{b^2}{16}}=\pm 3 \Rightarrow b^2=7 \\ $$