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

Tangent at $(3\sqrt{3},1)$ is
 $$\begin{gathered} \frac{\mathrm{x}3\sqrt{3}}{36}+\frac{\mathrm{y}.1}{4}=1 \\ \mathrm{x}=0\Rightarrow \mathrm{y}=4 \\ \mathrm{A}=(0,4) \end{gathered}$$
$$\begin{gathered} Normal at (3\sqrt{3},1)\mathrm{is}\frac{36\mathrm{x}}{3\sqrt{3}}-\frac{4.\mathrm{y}}{1}=36-4 \\ \mathrm{x}=0\Rightarrow \mathrm{y}=-8 \\ \mathrm{B}=(0,-8) \end{gathered}$$
Circle on $\overline{\mathrm{AB}}$ as diameter is $(\mathrm{x}-0)(\mathrm{x}-0)+(\mathrm{y}-4)(\mathrm{y}+8)=0$
$\Rightarrow \mathrm{C}={\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{y}-32=0$
 Clearly PQ is the chord of contact of $(\mathrm{\alpha },\mathrm{\beta })$
 equation of PQ is $\mathrm{x}.\mathrm{\alpha }+\mathrm{y}.\mathrm{\beta }+2(\mathrm{y}+\mathrm{\beta })-32=0\Rightarrow \mathrm{\alpha }\mathrm{x}+(2+\mathrm{\beta })\mathrm{y}+2\mathrm{\beta }-32=0$ 
But $\overleftrightarrow{\mathrm{PQ}}$ is $\mathrm{x}=2\sqrt{5}$ 
Hence
 $$\begin{gathered} 2+\mathrm{\beta }=0,\frac{32-2\mathrm{\beta }}{\mathrm{\alpha }}=2\sqrt{5} \\ \Rightarrow \mathrm{\beta }=-2,\mathrm{\alpha }=\frac{36}{2\sqrt{5}} \\ \therefore {\mathrm{\alpha }}^2-{\mathrm{\beta }}^2=\frac{324}{5}-4=\frac{304}{5} \end{gathered}$$