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

Answer 1

Answer: (D)

$3x^2+4y^2=12$
$\frac{x^2}{4}+\frac{y^2}{3}=1$ $x=2\cos \theta ,y=\sqrt{3}\sin \theta$
Let $P(2\cos \theta ,\sqrt{3\sin \theta })$
Equation of normal is $\frac{a^{2}x}{x_1}-\frac{b^{2}y}{y_1}=a^2-b^2$
$2x\sin \theta -\sqrt{3}\cos \theta y=\sin \theta \cos \theta$
Slope $\frac{2}{\sqrt{3}}\tan \theta =-2\therefore \tan \theta =-\sqrt{3}$
Equation of tangent is
it passes through (4, 4)
$3x\cos \theta +2\sqrt{3}\sin \theta y=6$
$12\cos \theta +8\sqrt{3}\sin \theta =6$
$\cos \theta =-\frac{1}{2},\sin \theta =\frac{\sqrt{3}}{2}\therefore \theta =120^{\circ }$
Hence point P is $\left(2\cos 120^{\circ },\sqrt{3}\sin 120^{\circ }\right)$
$P\left(1-,\frac{3}{2}\right),Q\left(4,4\right)$
$PQ=\frac{5\sqrt{5}}{2}$