Let the tangent and normal at the point $(3\sqrt{3},1)$ on the ellipse $\dfrac{x^2}{36}+\dfrac{y^2}{4}=1$ 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
Step-by-Step Solution
Key Concept: Find tangent and normal at $(3\sqrt{3},1)$ on the ellipse, get $y$-intercepts $A=(0,4)$ and $B=(0,-8)$. Circle on diameter $AB$: $x^2+(y-4)(y+8)=0$. Find $P$, $Q$ and tangent intersection.
$A=(0,4),B=(0,-8)$. Circle: $x^2+y^2+4y-32=0$. $P(2\sqrt{5},-6),Q(2\sqrt{5},2)$. Intersection at $(\frac{18}{\sqrt{5}},-2)$. $\alpha^2-\beta^2=\frac{324}{5}-4=\frac{304}{5}$.
Correct Answer: 3