Let the hyperbola H: $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ pass through the point $(2 \sqrt{2}, - 2 \sqrt{2})$ . A parabola is drawn whose focus is same as the focus of H with positive abscissa and the directrix of the parabola passes through the other focus of H. If the length of the latus rectum of the parabola is e times the length of the latus rectum of H, where e is the eccentricity of H, then which of the following points lies on the parabola?

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$(2 \sqrt{3}, 3 \sqrt{2})$

$(3 \sqrt{3}, - 6 \sqrt{2})$

$(\sqrt{3}, - \sqrt{6})$

$(3 \sqrt{6}, 6 \sqrt{2})$

answer is B.

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Detailed Solution

$H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 F o c i : S (a e, 0), S^{'} (- a e, 0) F o o t o f d i r e c t r i x o f p a r a b o l a i s (- a e, 0) F o c u s o f p a r a b o l a i s (a e, 0) N o w, s e m i l a t u s r e c t u m o f p a r a b o l a = |S S^{'}| = 2 a e G i v e n, 4 a c = e (\frac{2 b^{2}}{a}) \Rightarrow b^{2} = 2 a^{2} . . . . . . (1) G i v e n, (2 \sqrt{2}, - 2 \sqrt{2}) l i e s o n H \Rightarrow \frac{1}{a^{2}} - \frac{1}{b^{2}} = \frac{1}{8} . . . . . . (2) F r o m (1) a n d (2) a^{2} = 4, b^{2} = 8 ∵ b^{2} = a^{2} (e^{2} - 1) ∴ e = \sqrt{3} \Rightarrow Equation of parabola is y^{2} = 8 \sqrt{3} x ∴ (3 \sqrt{3}, - 6 \sqrt{2}) lies on the parabola$