MHT CET · Maths · Ellipse
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 \(\tan ^{-1} \frac{8 \sqrt{2}}{65}\)
- D None of these
Answer & Solution
Correct Answer
(D) None of these
Step-by-step Solution
Detailed explanation
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
\(
\begin{array}{l}
=\tan ^{-1} \frac{2 \sqrt{(216)^{2}-135 \cdot 320}}{455} \\
=\tan ^{-1} \frac{8 \sqrt{2916-2700}}{455} \\
=\tan ^{-1} \frac{8 \sqrt{216}}{455} \\
=\tan ^{-1} \frac{48 \sqrt{6}}{455}
\end{array}
\)
\(
\begin{aligned}
& 8 \cdot \frac{x}{4}+27 \cdot \frac{y}{9}=1 \\
\Rightarrow \quad & 2 x+3 y=1
\end{aligned}
\)
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
\(
\begin{array}{l}
=\tan ^{-1} \frac{2 \sqrt{(216)^{2}-135 \cdot 320}}{455} \\
=\tan ^{-1} \frac{8 \sqrt{2916-2700}}{455} \\
=\tan ^{-1} \frac{8 \sqrt{216}}{455} \\
=\tan ^{-1} \frac{48 \sqrt{6}}{455}
\end{array}
\)
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