If the foci of the ellipse x^216+y^2b^2=1 coincide with the foci of the hyperbola x^2144-y…

Question

If the foci of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{b^2}=1$ coincide with the foci of the hyperbola $\dfrac{x^2}{144}-\dfrac{y^2}{81}=\dfrac{1}{25}$, then $b^2$ is equal to:

Accepted Answer

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

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

Solutions · 1