Question

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

• A. $\frac{5\sqrt{5}}{2}$ unit
• B. $\frac{\sqrt{157}}{2}$ unit
• C. $\frac{\sqrt{61}}{2}$ unit
• D. $\frac{\sqrt{221}}{2}$ unit

Answer 1

The correct option is A $\frac{5\sqrt{5}}{2}$ unit

Equation of ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$
Let point which is normal to the ellipse be $P(2\cos \theta ,\sqrt{3}\sin \theta )$
$\therefore$ Equation of normal: $2x\sec \theta -\sqrt{3}y\text{cosec}\theta =1$
Equation of normal parallel to the line, $2x+y=4$
Therefore $\frac{2\sec \theta }{\sqrt{3}\text{cosec}\theta }=-2$
$\Rightarrow \tan \theta =-\sqrt{3}\dots \left(1\right)$
Similarly equation of tangent at point $P:$
$\frac{x\cdot 2\cos \theta }{4}+\frac{y\cdot \sqrt{3}\sin \theta }{3}=1$
$\because$ tangent passes through point $Q(4,4)$
$\therefore 4\sqrt{3}\cos \theta +8\sin \theta =2\sqrt{3}\dots \left(2\right)$
Solving equation $\left(1\right)$ and $\left(2\right),$ we get
$\Rightarrow \theta =\frac{2\pi }{3}$
$\therefore P(-1,\frac{3}{2})\&Q(4,4)$
Hence $PQ=\frac{5\sqrt{5}}{2}$ unit