JEE Main 2014 — Parabola Question with Solution

The centre of the given ellipse $\left(0,0\right)$
The general equation of a tangent to the ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$  is  $y=mx\pm \sqrt{a^2{m}^2+b^2} ...\left(1\right)$

Given ellipse : $\frac{x^2}{6}+\frac{y^2}{2}=1$

A perpendicular line from the centre is $y=-\frac{x}{m} ...\left(2\right)$

Eliminating $m$,  from $\left(1\right)$ and $\left(2\right)$

$y=\left(\frac{-x}{y}\right)x\pm \sqrt{a^2\left(\frac{x^2}{y^2}\right)+b^2}$

$y^2=-x^2\pm \mathrm{y }\sqrt{a^2\left(\frac{x^2}{y^2}\right)+b^2}$

$\therefore x^2+y^2=\pm \sqrt{a^2{x}^2+b^2{y}^2}$

Squaring both sides,

${\left(x^2+y^2\right)}^2=a^2{x}^2+b^2{y}^2=6x^2+2y^2$