JEE Main 2023 — Ellipse Question with Solution

Given,

The line $x=8$ is the directrix of the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the corresponding focus $\left(2,0\right)$,

So, equation of directrix is given by $x=\frac{a}{e}\Rightarrow 8=\frac{a}{e} \dots \left(1\right)$

And distance of corresponding focus is given by,

$ae=2 \dots \left(2\right)$

So, from both equation we get,

$8e=\frac{2}{e}$

$\Rightarrow e^2=\frac{1}{4}\Rightarrow e=\frac{1}{2}$

Hence, $a=4$

Now using eccentricity formula we get,

$b^2=a^2\left(1-e^2\right)$$=16\left(\frac{3}{4}\right)=12$

Now equation of tangent to ellipse at point $P\left(4\cos \theta ,\sqrt{12}\sin \theta \right)$ is given by,

$\frac{x\cos \theta }{4}+\frac{y\sin \theta }{2\sqrt{3}}=1$

Now given tangent passes through $\left(0,4\sqrt{3}\right)$,

So, $\sin \theta =\frac{1}{2}$$\Rightarrow \theta =30°$

Now, point $P\left(4\cos \theta ,\sqrt{12}\sin \theta \right)$ will be $P\left(2\sqrt{3},\sqrt{3}\right)$

And intersection of tangent with $x-$axis will give $\mathrm{Q}\left(\frac{8}{\sqrt{3}},0\right)$

Hence, by distance formula we get, ${\left(3PQ\right)}^2=39$