Let the tangent and normal at the point (3 3 , 1) on the ellipse x...

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

Observe that \left(3\sqrt{3}, 1\right)$lies on the given ellipse.</p><p>Equation of tangent to the ellipse at the given point is$T: \dfrac{3\sqrt{3}x}{36} + \dfrac{y}{4} = 1$</p><p>$⇒ \dfrac{\sqrt{3}x}{12} + \dfrac{y}{4} = 1$</p><p>Equation of normal passing through same point will be$N: \dfrac{x - 3\sqrt{3}}{\dfrac{3\sqrt{3}}{36}} = \dfrac{y - 1}{\dfrac{1}{4}}$</p><p>$⇒ \dfrac{12x - 36\sqrt{3}}{\sqrt{3}} = 4y - 4$</p><p>$⇒ 3x - 9\sqrt{3} = \sqrt{3}y - \sqrt{3}$</p><p>$⇒ 3x - \sqrt{3}y = 8\sqrt{3}$</p><p>Now tangent and normal intersects the$y -$axis at$A$and$B$respectively.</p><p>$A\left(0, 4\right)B\left(0, - 8\right)$</p><p>Equation of circle with$AB$as diameter will be</p><p>\left(x - {x}_{1}\right)\left(x - {x}_{2}\right) + \left(y - {y}_{1}\right)\left(y - {y}_{2}\right) = 0$

$\text{⇒}{x}^{2} + ( y - 4 ) ( y + 8 ) = 0$

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

$⇒ 20 + {y}^{2} + 4y - 32 = 0$

$⇒ {y}^{2} + 4y - 12 = 0$

$⇒ ( y + 6 ) ( y - 2 ) = 0$

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

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

$⇒ T: x{x}_{1} + y{y}_{1} + 2y + 2{y}_{1} - 32 = 0$

$⇒ {T}_{1}: 2\sqrt{5}x - 6y + 2y - 12 - 32 = 0$

$⇒ 2\sqrt{5}x - 4y = 44$

$⇒ {T}_{1}: \sqrt{5}x - 2y = 22 . . . . . . . . \left(i\right)$

$⇒ {T}_{2}: 2\sqrt{5}x + 2y + 2y + 4 - 32 = 0$

$⇒ 2\sqrt{5}x + 4y = 28$

$⇒ {T}_{2}: \sqrt{5}x + 2y = 14 . . . . . . . . . \left(ii\right)$

From \left(i\right)$and \left(ii\right)$

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

$⇒ {\alpha}^{2} - {\beta}^{2} = \dfrac{304}{5}$

Hence this is the required option.