Q60 / 150 · MHT CET 2011 · Mathematics
From point $P(8,27)$ tangents $P Q$ and $P R$ are drawn to the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. Then, angle subtended by $Q R$ at origin is
- A$\tan ^{-1} \frac{2 \sqrt{6}}{65}$
- B$\tan ^{-1} \frac{4 \sqrt{6}}{65}$
- C$\tan ^{-1} \frac{8 \sqrt{2}}{65}$
- Correct answerDNone of these
Solution
Equation of chord of contact $Q R$ is
$$
\begin{aligned}
& 8 \cdot \frac{x}{4}+27 \cdot \frac{y}{9}=1 \\
\Rightarrow \quad & 2 x+3 y=1
\end{aligned}
$$
Now, equation of the pair of lines passing through origin and points $Q, R$ given by
$$
\begin{array}{l}
\quad\left(\frac{x^{2}}{4}+\frac{y^{2}}{9}\right)=(2 x+3 y)^{2} \\
\Rightarrow \quad 9 x^{2}+4 y^{2}=36\left(4 x^{2}+12 x y+9 y^{2}\right) \\
\Rightarrow \quad 135 x^{2}+432 x y+320 y^{2}=0
\end{array}
$$
$\therefore$ Required angle is
$$
\begin{array}{l}
=\tan ^{-1} \frac{2 \sqrt{(216)^{2}-135 \cdot 320}}{455} \\
=\tan ^{-1} \frac{8 \sqrt{2916-2700}}{455} \\
=\tan ^{-1} \frac{8 \sqrt{216}}{455} \\
=\tan ^{-1} \frac{48 \sqrt{6}}{455}
\end{array}
$$
Hence, the correct answer is option: 'D'.