JEE PYQ: Ellipse Question 10
Question 10 - 2020 (05 Sep Shift 1)
If the point P on the curve, $4x^2 + 5y^2 = 20$ is farthest from the point $Q(0, -4)$, then $PQ^2$ is equal to:
(a) 36 (b) 48 (c) 21 (d) 29
Show Answer
Answer: (a) 36
Solution
Ellipse $= \frac{x^2}{5} + \frac{y^2}{4} = 1$
Let a point on ellipse be $(\sqrt{5}\cos\theta, 2\sin\theta)$
$PQ^2 = 5\cos^2\theta + (2\sin\theta + 4)^2 = 21 + 16\sin\theta - \sin^2\theta$
$PQ^2$ is maximum when $\sin\theta = 1$
$PQ^2_{max} = 85 - 49 = 36$
BETA
