Question

If the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ intersect orthogonally, then the value of $b^2$ is

Answer 1

For hyperbola,
$e^2=1+\frac{b^2}{a^2}$
$\Rightarrow e^2=1+\frac{81}{144}=\frac{225}{144}$
$\Rightarrow e=\frac{15}{12}=\frac{5}{4}$
Also, $a^2=\frac{144}{25}\Rightarrow a=\frac{12}{5}$

If ellipse and hyperbola intersect orthogonally, then their foci coincide.
Hence, the foci are $(\pm ae,0)=(\pm 3,0)$

Now, for ellipse, $ae=3\Rightarrow a^2{e}^2=9$
$b^2=a^2(1-e^2)$
$=a^2-a^2{e}^2$
$=16-9=7$