Ellipse Question - None

Question

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

Accepted Answer

Given: Focal chord of $y^{2}=4 x$ is $P Q$ and it subtends an angle of $\frac{\pi}{2}$ at point $(3,0)$
Let parametric coordinates of point $P$ be $\left(t^{2}, 2 t\right)$ then parametric coordinates of point $Q$ be $\left(\frac{1}{t^{2}}, \frac{-2}{t}\right)$

$\because A P \perp A Q$
$\therefore \left(m_{A P}\right)\left(m_{A Q}\right) =-1$
$\Rightarrow \left(\frac{2 t}{t^{2}-3}\right)\left(\frac{\frac{-2}{t}}{\frac{1}{t^{2}}-3}\right) =-1$
$\Rightarrow \frac{-4 t^{2}}{\left(t^{2}-3\right)\left(1-3 t^{2}\right)} =-1$
$\Rightarrow 4 t^{2} =-3 t^{4}+10 t^{2}-3$
$\Rightarrow 3 t^{4}-6 t^{2}+3 =0$
$\Rightarrow \left(t^{2}-1\right)^{2} =0$
$\Rightarrow t =1$
$\therefore$ Coordinates of point $P$ and $Q$ are $(1,2)$ and $(1,-2)$ respectively.
Since, line segment $P Q$ is also a focal chord of ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2} \gt b^{2}$.
$\therefore {P}$ and $Q$ must be end points of latus rectum
$\Rightarrow \frac{2 b^{2}}{a}=4$ and $a e=1$
$\text { As we know, } b^{2}=a^{2}\left(1-e^{2}\right)$
$\Rightarrow b^{2}=a^{2}-a^{2} e^{2}$
$\Rightarrow b^{2}=a^{2}-1$
$\Rightarrow \frac{a^{2}-1}{a}=2$
$\Rightarrow a^{2}-2 a-1=0$
$\Rightarrow a=1+\sqrt{2}$
$\Rightarrow e=\frac{1}{a}=\frac{1}{1+\sqrt{2}}$
$\Rightarrow e^{2}=\frac{1}{3+2 \sqrt{2}}$
$\Rightarrow \frac{1}{e^{2}}=3+2 \sqrt{2}$

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