If the foci of a hyperbola are the same as that of the ellipse x²/9 + y²/25 = 1 and the eccentricity of the hy…

Question

If the foci of a hyperbola are the 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:

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

Accepted Answer

A. $$7\sqrt{\frac{2}{5}} - \frac{8}{3}$$

Answer

B. $$14\sqrt{\frac{2}{5}} - \frac{4}{3}$$

Answer

C. $$14\sqrt{\frac{2}{5}} - \frac{16}{3}$$

Answer

D. $$7\sqrt{\frac{2}{5}} + \frac{8}{3}$$

Answer & step-by-step solution

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