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 15/8 times the eccentricity of the ellipse, then the smaller focal distance of the point $(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}})$ on the hyperbola, is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

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

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} a = 3, b = 5 \\ F o c i l = (9 \pm b e) \\ = (0, \pm 4) \\ e_{H} = \frac{4}{5} \times \frac{15}{8} = \frac{3}{2} \\ E q^{n} o f H y p . \\ \frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1 \\ B . e_{H} = 4 \Rightarrow B = \frac{8}{3} \\ A^{2} = \frac{80}{9} \\ D i r e c t r i x y = \pm \frac{B}{e_{H}} = \pm \frac{16}{9} \\ S p = e p m (D e f o f c o n i c) \\ S p = 7 \sqrt{\frac{2}{5}} - \frac{8}{3} \\ A n s : - 3 \end{array}$