# If the normal to the ellipse

Question:

If the normal to the ellipse $3 x^{2}+4 y^{2}=12$ at a point $\mathrm{P}$ on it is parallel to the line, $2 x+y=4$ and the tangent to the ellipse at $\mathrm{P}$ passes through $\mathrm{Q}(4,4)$ then $\mathrm{PQ}$ is equal to :

1. (1) $\frac{5 \sqrt{5}}{2}$

2. (2) $\frac{\sqrt{61}}{2}$

3. (3) $\frac{\sqrt{221}}{2}$

4. (4) $\frac{\sqrt{157}}{2}$

Correct Option: 1,

Solution:

Slope of tangent on the line $2 x+y=4$ at point $P$ is $\frac{1}{2}$.

Given ellipse is,

$3 x^{2}+4 y^{2}=12 \Rightarrow \frac{x^{2}}{2^{2}}+\frac{y^{2}}{(\sqrt{3})^{2}}=1$

Let point $P(2 \cos \theta, \sqrt{3} \sin \theta)$

$\therefore$ equation of tangent on the ellipse, at $P$ is,

$\frac{x}{2} \cos \theta+\frac{y}{\sqrt{3}} \sin \theta=1$

$\Rightarrow m_{T}=-\frac{\sqrt{3}}{2} \cot \theta$

$\because$ both the tangents are parallel $\Rightarrow-\frac{\sqrt{3}}{2} \cot \theta=\frac{1}{2}$

$\frac{x}{2} \cos \theta+\frac{y}{\sqrt{3}} \sin \theta=1$

$\Rightarrow m_{T}=-\frac{\sqrt{3}}{2} \cot \theta$

$\because$ both the tangents are parallel $\Rightarrow-\frac{\sqrt{3}}{2} \cot \theta=\frac{1}{2}$

$\Rightarrow \tan \theta=-\sqrt{3} \Rightarrow \theta=\pi-\frac{\pi}{3}$ or $\theta=2 \pi-\frac{\pi}{3}$

Case-1: $\theta=\frac{2 \pi}{3}$, then point $P\left(-1, \frac{3}{2}\right)$ and $P Q=\frac{5 \sqrt{5}}{2}$

Case-2: $\theta=\frac{5 \pi}{3}$, then tangent does not pass through $Q(4,4)$.