Question

The foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}+\frac{y^2}{25}=\frac{1}{13}$ are the same.

The value of $b^2$ is ___ .

Answer 1

Consider the following hyperbola

$\frac{x^2}{144}-\frac{y^2}{25}=\frac{1}{13}\frac{13x^2}{144}-\frac{13y^2}{25}=1a^2=\frac{144}{13};b^2=\frac{25}{13}\text{Focal length,}c=\sqrt{a^2+b^2}c=\sqrt{\frac{144+25}{13}}=\sqrt{\frac{169}{13}}=\sqrt{13}$

Also, this is the focus of the ellipse
$\frac{x^2}{25}+\frac{y^2}{b^2}=1c^2=a^2-b^2\Rightarrow 13=25-b^2{b}^2=12$