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

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 $\left(\right.-3,0\left.\right),\left(\right.3,0\left.\right)$ Now for ellipse $b^{2}=25-9=16$ .