Question

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

Answer 1

Answer: 16

Given hyperbola $\frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{81}{25}\right)}=1$
Eccentricity of hyperbola $e=\frac{\sqrt{225/25}}{12/5}=\frac{15}{12}=\frac{5}{4}.$
Foci of hyperbola $=\left(\pm \frac{12}{5}\left(\frac{5}{4}\right),0\right)=\left(\pm 3,0\right)$
Given ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$
So, foci are $(-3,0),(3,0)$ Now for ellipse $b^2=25-9=16$ .