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Let the tangent and normal at the point $\left ( 3\sqrt{3},1 \right )$ on the ellipse $\mathrm{\frac{x^{2}}{36}+\frac{y^{2}}{4}=1}$ 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 $\mathrm{x=2\sqrt{5}}$ intersect C at the points P and Q. If the tangents at the points P and Q on the circle intersect at the point $\left ( \alpha, \beta \right )$ , then $\left ( \alpha^{2}-\beta^{2} \right )$ is equal to
Option: 1
$\frac{305}{4}$

Option: 2
$60$

Option: 3
$\frac{314}{5}$

Option: 4
$61$

Answers (1)

best_answer

Given ellipse $\mathrm{\frac{x^{2}}{36}+\frac{y^{2}}{4}=1}$

$\mathrm{\frac{x}{4\sqrt{3}}+\frac{y}{4}=1}$

$\mathrm{y=4}$

$\begin{aligned} & \frac{x}{4}-\frac{4}{4 \sqrt{3}}=\frac{2}{\sqrt{3}} \\ \end{aligned}$

$\begin{aligned} & \mathrm{y}=-8 \\ \end{aligned}$

$\begin{aligned} & x^2+y^2+4 \mathrm{y}-32=0 \\ \end{aligned}$

$\mathrm{\begin{aligned} & \mathrm{~h} x+\mathrm{ky}+2(\mathrm{y}+\mathrm{k})-32=0 \\ \end{aligned}}$

$\begin{aligned} & \mathrm{k}=-2 \\ \end{aligned}$

$\begin{aligned} & \mathrm{hx}+2 \mathrm{k}-32=0 \\ \end{aligned}$

$\begin{aligned} & \mathrm{hx}=36 \\ & \alpha=\mathrm{h}=\frac{36}{2 \sqrt{5}} \\ \end{aligned}$

$\begin{aligned} & \beta=\mathrm{k}=-2 \\ & \alpha^2-\beta^2=\frac{304}{5} \end{aligned}$

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