JEE Math Practice 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

1. A.$$14 \sqrt{\frac{2}{5}}-\frac{4}{3}$$
2. B.$$7 \sqrt{\frac{2}{5}}+\frac{8}{3}$$
3. C.$$7 \sqrt{\frac{2}{5}}-\frac{8}{3}$$
4. D.$$14 \sqrt{\frac{2}{5}}-\frac{16}{3}$$