Let PQ be a focal chord of the parabola $y^{2} = 4 x$ such that it subtends an angle of $\frac{π}{2}$ at the point (3,0). Let the line segment PQ be also a focal chord of the ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a^{2} > b^{2}$ . If e is the eccentricity of the ellipse E, then the value of $\frac{1}{e^{2}}$ is equal to:

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$1 + 2 \sqrt{3}$

$4 + 5 \sqrt{3}$

$1 + \sqrt{2}$

$3 + 2 \sqrt{2}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Question Image $\begin{array}{l} m_{PR} \cdot m_{PQ} = - 1 \Rightarrow \frac{2 t}{t^{2} - 3} \times \frac{- 2 / t}{\frac{1}{t^{2}} - 3} = - 1 \\ {(t^{2} - 1)}^{2} = 0 \Rightarrow t = 1 \\ \Rightarrow P & Q must be end point of latus rectum: \\ P (1, 2) & Q (1, - 2) \\ ∴ \frac{2 b^{2}}{a} = 4 & ae = 1 ∵ We know that b^{2} = a^{2} (1 - e^{2}) \\ \begin{aligned} ∴ a = 1 + \sqrt{2} \\ ∵ e^{2} = 1 - \frac{b^{2}}{a^{2}} \\ ∴ e^{2} = 3 - 2 \sqrt{2} \\ \frac{1}{e^{2}} = 3 + 2 \sqrt{2} \end{aligned} \end{array}$