If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola…
If the product of eccentricities of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=-1$ is 1 , then $b^2=$
$\frac{12}{25}$
$144$
$25$
$\frac{144}{25}$
Solution
Given the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and hyperbola
$\begin{aligned} & \frac{x^2}{9}-\frac{y^2}{16}=-1 \Rightarrow \frac{y^2}{16}-\frac{x^2}{9}=1 \\ & \because \text { eccentricity of ellipse }=\sqrt{1-\frac{b^2}{16}} \\ & \text { and eccentricity of hyperbola }=\sqrt{1+\frac{9}{16}} \\ & \text { Since, } \sqrt{1-\frac{b^2}{16}} \cdot \sqrt{1+\frac{9}{16}}=1 \\ & \Rightarrow 16-b^2=\frac{16 \times 16}{25} \Rightarrow b^2=\frac{144}{25}\end{aligned}$