Question

If $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and $e_2$ is the eccentricity of the hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ and $e_1{e}_2=1$, then $b^2$ is equal to

Answer 1

Given $e_1$ is the eccentricity of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$
$\Rightarrow e_1=\sqrt{\frac{16-b^2}{16}}$
Given $e_2$ is the eccentricity of the hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$
$e_2=\sqrt{\frac{9+b^2}{9}}$
Also given $e_1{e}_2=1$
$e_1{e}_2=\sqrt{\frac{16-b^2}{16}\times \frac{9+b^2}{9}}=1$
$\Rightarrow 7b^2-b^4=0$
$\Rightarrow b^2=7$