Consider the ellipse $S_{1} \equiv \frac{x^{2}}{9} + \frac{y^{2}}{8} - 1 = 0$ . Let $S_{2} = 0$ be a parabola on the right of y-axis confocal with S1 =0 having vertex at the centre of S1 = 0. Let P be the point of intersection of the parabola and directrix of the ellipse in the 1st quadrant. If tangent at point Q in 1st quadrant on S1 =0 and the line joining point P with the focus of S2 =0 intersect on the auxilliary circle of S1 =0 then the value of $9 \sin^{2} θ$ is_____( $θ$ is the eccentric angle of the point Q)

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

answer is 3.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Let Q be $(a \cos θ, a \sqrt{1 - e^{2}} \sin θ)$
Slope of tangent at point Q to the ellipse is $(- \sqrt{1 - e^{2}} \cot θ)$
But tangent at Q will be perpendicular to the line joining P to the focus of parabola.
So $(\frac{2 a - 0}{\frac{a}{e} - a e}) \times [- \sqrt{1 - e^{2}} \cot θ] = - 1 \Rightarrow \frac{2 e}{1 - e^{2}} \times \sqrt{1 - e^{2}} \cot θ = 1 \Rightarrow \tan θ = \frac{2 e}{\sqrt{1 - e^{2}}}$ $θ = \tan^{- 1} (\frac{2 e}{\sqrt{1 - e^{2}}})$

Question Image