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 $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$
$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} $