Question

Let $PQ$ be a focal chord of the parabola $y^2=4x$ such that it subtends an angle of $\frac{\pi }{2}$ 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 :

• A. $1+\sqrt{2}$
• B. $3+2\sqrt{2}$
• C. $1+2\sqrt{3}$
• D. $4+5\sqrt{3}$

Answer 1

Answer: (B)

$PQ$ is focal chord

$m_{PR}\cdot m_{PQ}=-1$
$\frac{2t}{t^2-3}\times \frac{-2/t}{\frac{1}{t^2}-3}=-1$
${\left(t^2-1\right)}^2=0$
$\Rightarrow t=1$
$\Rightarrow P\&Q$ must be end point of latus rectum:
$P(1,2)\&Q(1,-2)$
$\therefore \frac{2b^2}{a}=4\&ae=1$
$\because$ We know that $b^2=a^2\left(1-e^2\right)$
$\therefore a=1+\sqrt{2}$
$\because e^2=1-\frac{b^2}{a^2}$
$\therefore e^2=3-2\sqrt{2}$
$\frac{1}{e^2}=3+2\sqrt{2}$