If the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\fr

Question

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

Professor PC Thomas Classes · Answer

Foci of hyperbola: $c_h^2 = a^2+b^2 = 144/25 + 81/25 = 9 \Rightarrow c_h = 3$. Ellipse foci $c_e^2 = 16 - b^2 = 9 \Rightarrow b^2 = 7$.