The foci of the ellipse x^216+y^2b^2=1 and the hyperbola x^2144-y^281=125 coincide.

Question

The foci of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{b^2}=1$ and the hyperbola $\dfrac{x^2}{144}-\dfrac{y^2}{81}=\dfrac{1}{25}$ coincide. Then the value of $b^2$ is:

Accepted Answer

Hyperbola: $\dfrac{x^2}{144/25}-\dfrac{y^2}{81/25}=1$, $e^2=1+\dfrac{81/25}{144/25}=\dfrac{225}{144}\Rightarrow e=\dfrac54$; foci at $ae=\dfrac{12}{5}\cdot\dfrac54=3$, i.e. $(\pm3,0)$.

For the ellipse with $a=4$: $ae=3\Rightarrow e=\tfrac34$. Then
$$b^2=a^2(1-e^2)=16\left(1-\frac{9}{16}\right)=7.$$

Solutions · 1