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

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 $AB$ 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

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

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

$\begin{gathered} P(2\sqrt{5},-6)\, Q(2\sqrt{5},2) \ C:{x}^{2}+{y}^{2}+4y-32=0 \ P(2\sqrt{5},-6)\, Q(2\sqrt{5},2) \ T:x{x}_{1}+y{y}_{1}+2y+2{y}_{1}-32=0 \ {T}_{1}:2\sqrt{5}x-6y+2y-12-32=0 \ \, 2\sqrt{5}x-4y=44 \ {T}_{1}:\sqrt{5}x-2y=22.......(i) \ {T}_{2}:2\sqrt{5}x+2y+2y+4-32=0 \ 2\sqrt{5}x+4y=28 \ {T}_{2}:\sqrt{5}x+2y=14.......(ii) \end{gathered}$

From (i) and (ii), we get

$\begin{gathered} \alpha =\frac{18}{\sqrt{5}}\, \beta =-2 \ {\alpha }^{2}-{\beta }^{2}=\frac{304}{5} \end{gathered}$

Solutions · 1