Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{4}=1$ meet the $y$-axis at the points $A$ and $B$ respectively. Let the circle $C$ be drawn taking $A B$ as a diameter and the line $x=2 \sqrt{5}$ intersect $C$ at the points $P$ and $Q$. If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha, \beta)$, then $\alpha^{2}-\beta^{2}$ is equal to :

$$ \begin{aligned} & \frac{x^2}{36}+\frac{y^2}{4}=1 \\\\ & T: \frac{3 \sqrt{3} x}{36}+\frac{y}{4}=1 \\\\ & T: \frac{\sqrt{3} x}{12}+\frac{y}{4}=1 \\\\ & N: \frac{x-3 \sqrt{3}}{\frac{3 \sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}} \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\\\ & 3 x-9 \sqrt{3}=\sqrt{3} y-\sqrt{3} \\\\ & N: 3 x-\sqrt{3} y=8 \sqrt{3} \\\\ & A(0,4) \quad B(0,-8) \\\\ & \text { C: } x^2+(y-4)(y+8)=0 \\\\ & \text { Line } x=2 \sqrt{5} \\\\ & 20+y^2+4 y-32=0 \\\\ & y^2+4 y-12=0 \\\\ & (y+6)(y-2)=0 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\ & C: x^2+y^2+4 y-32=0 \\\\ & P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\\\ & T: x x_1+y y_1+2 y+2 y_1-32=0 \\\\ & T_1: 2 \sqrt{5} x-6 y+2 y-12-32=0 \\\\ & \quad 2 \sqrt{5} x-4 y=44 \\\\ & T_1: \sqrt{5} x-2 y=22 .......(i) \\\\ & T_2: 2 \sqrt{5} x+2 y+2 y+4-32=0 \\\\ & 2 \sqrt{5} x+4 y=28 \\\\ & T_2: \sqrt{5} x+2 y=14 .......(ii) \end{aligned} $$ <br/><br/>From (i) and (ii), we get <br/><br/>$$ \begin{aligned} & \alpha=\frac{18}{\sqrt{5}} \quad \beta=-2 \\\\ & \alpha^2-\beta^2=\frac{304}{5} \end{aligned} $$