JEE PYQ: Ellipse Question 22
Question 22 - 2019 (12 Apr Shift 1)
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) $\frac{5\sqrt{5}}{2}$ (b) $\frac{\sqrt{61}}{2}$ (c) $\frac{\sqrt{221}}{2}$ (d) $\frac{\sqrt{157}}{2}$
Show Answer
Answer: (a) $\frac{5\sqrt{5}}{2}$
Solution
Slope of tangent $= \frac{1}{2}$
$P(2\cos\theta, \sqrt{3}\sin\theta)$, tangent slope $= -\frac{\sqrt{3}}{2}\cot\theta = \frac{1}{2}$
$\tan\theta = -\sqrt{3} \Rightarrow \theta = \frac{2\pi}{3}$
$P = \left(-1, \frac{3}{2}\right)$, $PQ = \frac{5\sqrt{5}}{2}$
BETA
