Your AI-Powered Personal Tutor
Question

If the foci of a hyperbola are same as that of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ and the eccentricity of the hyperbola is $\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point $\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to

$14 \sqrt{\frac{2}{5}}-\frac{4}{3}$
$7 \sqrt{\frac{2}{5}}+\frac{8}{3}$
$7 \sqrt{\frac{2}{5}}-\frac{8}{3}$
$14 \sqrt{\frac{2}{5}}-\frac{16}{3}$

Solution

Please login to view the detailed solution steps...

Go to DASH