Question

From point $P(8,27)$, tangents $PQ$ and $PR$ are drawn to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$, then the angle subtended by $QR$ at origin is

• A. ${\tan }^{-1}\frac{\sqrt{6}}{65}$
• B. ${\tan }^{-1}\frac{4\sqrt{6}}{65}$
• C. ${\tan }^{-1}\frac{8\sqrt{2}}{65}$
• D. ${\tan }^{-1}\frac{48\sqrt{6}}{455}$

Answer 1

Answer: (D)

${\tan }^{-1}\frac{48\sqrt{6}}{455}$

Chord of contact $S_1=0$

$\frac{8x}{4}+\frac{24y}{9}=1$

$2x+3y=1$

homogenizing

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{9}=1{(2x+3y)}^2$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{9}-1(4x^2+9y^2+12xy)=0$

$\Rightarrow x^2\left(\frac{-15}{4}\right)+\left(\frac{-80}{9}\right)y^2-12xy=0$

$\tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}$

$=\frac{2\sqrt{144-\frac{-15}{4}\times \frac{-80}{9}}}{-(\frac{15}{4}+\frac{80}{9})}=2\times \frac{\sqrt{\frac{332}{3}}}{\frac{455}{36}}=\frac{48\sqrt{6}}{455}$

$\theta ={\tan }^{-1}\left(\frac{48\sqrt{6}}{455}\right)$