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(B) The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{4} = 1$. The point is $(3 \sqrt{3}, 1)$.The equation of the tangent at $(x_1, y_1)$ is

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$\frac{304}{5}$

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(B) The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{4} = 1$. The point is $(3 \sqrt{3}, 1)$.The equation of the tangent at $(x_1, y_1)$ is $\frac{x x_1}{36} + \frac{y y_1}{4} = 1$. Substituting $(3 \sqrt{3}, 1)$,we get $\frac{x (3 \sqrt{3})}{36} + \frac{y}{4} = 1$,which simplifies to $\frac{x \sqrt{3}}{12} + \frac{y}{4} = 1$.For the $y$-axis,set $x = 0$,so $\frac{y}{4} = 1 \implies y = 4$. Thus,$A = (0, 4)$.The slope of the tangent is $m = -\frac{\sqrt{3}/12}{1/4} = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}$. The slope of the normal is $m' = \sqrt{3}$.The equation of the normal is $y - 1 = \sqrt{3}(x - 3 \sqrt{3}) \implies y - 1 = x \sqrt{3} - 9 \implies y = x \sqrt{3} - 8$.For the $y$-axis,set $x = 0$,so $y = -8$. Thus,$B = (0, -8)$.The circle $C$ has $AB$ as diameter,where $A(0, 4)$ and $B(0, -8)$. The center is $(0, -2)$ and the radius is $r = \frac{4 - (-8)}{2} = 6$. The equation is $x^2 + (y + 2)^2 = 36$,or $x^2 + y^2 + 4y - 32 = 0$.The intersection with $x = 2 \sqrt{5}$ gives $(2 \sqrt{5})^2 + y^2 + 4y - 32 = 0 \implies 20 + y^2 + 4y - 32 = 0 \implies y^2 + 4y - 12 = 0$. Solving $(y + 6)(y - 2) = 0$,we get $y = 2$ and $y = -6$. So $P(2 \sqrt{5}, 2)$ and $Q(2 \sqrt{5}, -6)$.The tangent at $(x_0, y_0)$ to $x^2 + y^2 + 4y - 32 = 0$ is $x x_0 + y y_0 + 2(y + y_0) - 32 = 0$. For the intersection $(\alpha, \beta)$ of tangents at $P$ and $Q$,the chord of contact is $x(2 \sqrt{5}) + y(-2) + 2(y - 2) - 32 = 0 \implies x(2 \sqrt{5}) - 36 = 0 \implies x = \frac{18}{\sqrt{5}}$.Thus $\alpha = \frac{18}{\sqrt{5}}$ and $\beta = -2$. Then $\alpha^2 - \beta^2 = \frac{324}{5} - 4 = \frac{324 - 20}{5} = \frac{304}{5}$.

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