JEE Main 2022 (Online) 29th June Morning ShiftQuestion 1
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 :
Parabola
Solution
Concept. The right-angle condition fixes the focal chord as $x=1$; it is a focal chord of the ellipse too, so $ae=1$. Formulas used. (1) $y^2=4x$: point $(t^2,2t)$; focal chord: $ts=-1$.(2) $\angle PCQ=90^\circ\iff\vec{CP}\cdot\vec{CQ}=0$.(3) Ellipse: $b^2=a^2(1-e^2)=a^2-(ae)^2$.Step 1 - Use (2) with $C(3,0)$: $(t^2-3)(s^2-3)+4ts=0$. Step 2 - Use (1). $t^2s^2=1$, $ts=-1$: $1-3(t^2+s^2)+9-4=0\Rightarrow t^2+s^2=2$. With $t^2s^2=1$: $t^2=s^2=1$. So $P(1,2)$, $Q(1,-2)$ and $PQ$ is $x=1$. Step 3 - Ellipse focus. $x=1$ crosses the major axis at $(1,0)$, so $ae=1$ and by (3) $b^2=a^2-1$. Step 4 - P on E. $\dfrac1{a^2}+\dfrac4{a^2-1}=1\Rightarrow a^2-1+4a^2=a^4-a^2\Rightarrow a^4-6a^2+1=0$. $a^2=3\pm2\sqrt2$; need $a^2>1$: $a^2=3+2\sqrt2$.Step 5. $\dfrac1{e^2}=\dfrac{a^2}{(ae)^2}=a^2=3+2\sqrt2$. Answer: B