JEE Main 2023 — Ellipse Question with Solution

Given that $\frac{x^2}{36}+\frac{y^2}{4}=1$

Observe that $\left(3\sqrt{3},1\right)$ lies on the given ellipse.

Equation of tangent to the ellipse at the given point is $T:\frac{3\sqrt{3}x}{36}+\frac{y}{4}=1$

$\Rightarrow \frac{\sqrt{3}x}{12}+\frac{y}{4}=1$

Equation of normal passing through same point will be $N:\frac{x-3\sqrt{3}}{\frac{3\sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}}$

$\Rightarrow \frac{12x-36\sqrt{3}}{\sqrt{3}}=4y-4$

$\Rightarrow 3x-9\sqrt{3}=\sqrt{3}y-\sqrt{3}$

$\Rightarrow 3x-\sqrt{3}y=8\sqrt{3}$

Now tangent and normal intersects the $y-$axis at $A$ and $B$ respectively.

$A\left(0,4\right) B\left(0,-8\right)$

Equation of circle with $AB$ as diameter will be

$\left(x-x_1\right)\left(x-x_2\right)+\left(y-y_1\right)\left(y-y_2\right)=0$

$\text{⇒}{x}^2+(y-4)(y+8)=0$

Line $x=2\sqrt{5}$ intersects the circle.

$\Rightarrow 20+y^2+4y-32=0$

$\Rightarrow y^2+4y-12=0$

$\Rightarrow (y+6)(y-2)=0$

Hence $P(2\sqrt{5},-6) Q(2\sqrt{5},2)$

Now let us find the equation of tangents to the circle at $P,Q$.

$\Rightarrow T:x{x}_1+y{y}_1+2y+2y_1-32=0$

$\Rightarrow T_1:2\sqrt{5}x-6y+2y-12-32=0$

$\Rightarrow 2\sqrt{5}x-4y=44$

$\Rightarrow T_1:\sqrt{5}x-2y=22........\left(\mathrm{i}\right)$

$\Rightarrow T_2:2\sqrt{5}x+2y+2y+4-32=0$

$\Rightarrow 2\sqrt{5}x+4y=28$

$\Rightarrow T_2:\sqrt{5}x+2y=14.........\left(\mathrm{ii}\right)$

From $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$\alpha =\frac{18}{\sqrt{5}} \beta =-2$

$\Rightarrow {\alpha }^2-{\beta }^2=\frac{304}{5}$

Hence this is the required option.