JEE Main 2021 — Ellipse Question with Solution

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the point $\left(\sqrt{\frac{3}{2}},1\right)$

So, $\frac{3}{2a^2}+\frac{1}{b^2}=1 ...\left(1\right)$ and $1-\frac{b^2}{a^2}=\frac{1}{3} ...\left(2\right)$

On solving equations $\left(1\right)$ and $\left(2\right)$, we get

$\Rightarrow a^2=3 \& b^2=2$

$\Rightarrow \frac{x^2}{3}+\frac{y^2}{2}=1 ...\left(3\right)$

We know that, focii of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\left(\pm ae, 0\right)$

Here, $a=\sqrt{3} \& e=\frac{1}{\sqrt{3}}$

$\therefore$ Its focus is $\left(1,0\right) \left(\because \alpha >0\right)$

Now, equation of circle is

$(x-1{)}^2+y^2=\frac{4}{3} ...\left(4\right)$

Solving $\left(3\right)$ and $\left(4\right)$ we get

$y=\pm \frac{2}{\sqrt{3}},x=1$

$\Rightarrow P{Q}^2=\sqrt{{\left(1-1\right)}^2+{\left(\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}\right)}^2}$

$={\left(\frac{4}{\sqrt{3}}\right)}^2=\frac{16}{3}$