Let the tangent and normal at the point [(3 √3, 1)] on the ellipse [x^236+y^24=1] meet the…

Question

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 :

Accepted Answer

$$
\begin{aligned}
& \frac{x^2}{36}+\frac{y^2}{4}=1 \\
& T: \frac{3 \sqrt{3} x}{36}+\frac{y}{4}=1 \\
& T: \frac{\sqrt{3} x}{12}+\frac{y}{4}=1 \\
& N: \frac{x-3 \sqrt{3}}{\frac{3 \sqrt{3}}{36}}=\frac{y-1}{\frac{1}{4}}
\end{aligned}
$$

$$
\begin{aligned}
& \frac{12 x-36 \sqrt{3}}{\sqrt{3}}=4 y-4 \\
& 3 x-9 \sqrt{3}=\sqrt{3} y-\sqrt{3} \\
& N: 3 x-\sqrt{3} y=8 \sqrt{3} \\
& A(0,4) \quad B(0,-8) \\
& \text { C: } x^2+(y-4)(y+8)=0 \\
& \text { Line } x=2 \sqrt{5} \\
& 20+y^2+4 y-32=0 \\
& y^2+4 y-12=0 \\
& (y+6)(y-2)=0
\end{aligned}
$$

$$
\begin{aligned}
& P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\
& C: x^2+y^2+4 y-32=0 \\
& P(2 \sqrt{5},-6) \quad Q(2 \sqrt{5}, 2) \\
& T: x x_1+y y_1+2 y+2 y_1-32=0 \\
& T_1: 2 \sqrt{5} x-6 y+2 y-12-32=0 \\
& \quad 2 \sqrt{5} x-4 y=44 \\
& T_1: \sqrt{5} x-2 y=22 .......(i) \\
& T_2: 2 \sqrt{5} x+2 y+2 y+4-32=0 \\
& 2 \sqrt{5} x+4 y=28 \\
& T_2: \sqrt{5} x+2 y=14 .......(ii)
\end{aligned}
$$

From (i) and (ii), we get

$$
\begin{aligned}
& \alpha=\frac{18}{\sqrt{5}} \quad \beta=-2 \\
& \alpha^2-\beta^2=\frac{304}{5}
\end{aligned}
$$

Solutions · 1