Question

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

• A. $\frac{\sqrt{6}}{65}$
• B. $\frac{4\sqrt{6}}{65}$
• C. $\frac{8\sqrt{6}}{65}$
• D. $\frac{48\sqrt{6}}{455}$

Answer 1

The correct option is D $\frac{48\sqrt{6}}{455}$


Chord of contact drawn from $P(8,27)$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ is $T=0$
$\Rightarrow 2x+3y=1\dots \left(i\right)$
Now, to get the equation of the pair of lines passing through origin and points $Q,R$.
Making equation of ellipse homogeneous using equation $\left(i\right),$ we get
$\frac{x^2}{4}+\frac{y^2}{9}=(2x+3y{)}^2$
$\Rightarrow 9x^2+4y^2=36(4x^2+12xy+9y^2)$
$\Rightarrow 135x^2+432xy+320y^2=0$
$\therefore \angle QOR=\varphi \Rightarrow \tan \varphi =\frac{2\sqrt{h^2-ab}}{a+b}$
$\Rightarrow \tan \varphi =\frac{2\sqrt{{216}^2-135\times 320}}{455}$
$\Rightarrow \tan \varphi =\frac{48\sqrt{6}}{455}$