JEE PYQ: Circle Question 18
Question 18 - 2020 (05 Sep Shift 2)
Let $PQ$ be a diameter of the circle $x^2 + y^2 = 9$. If $\alpha$ and $\beta$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x + y = 2$ respectively, then the maximum value of $\alpha\beta$ is ______.
Show Answer
Answer: 7
Solution
$P(3\cos\theta, 3\sin\theta)$, $Q(-3\cos\theta, -3\sin\theta)$. $\alpha = \frac{|3\cos\theta + 3\sin\theta - 2|}{\sqrt{2}}$, $\beta = \frac{|3\cos\theta + 3\sin\theta + 2|}{\sqrt{2}}$. $\alpha\beta = \frac{|(3\cos\theta+3\sin\theta)^2 - 4|}{2} = \frac{|5 + 9\sin 2\theta|}{2}$. Max when $\sin 2\theta = 1$: $\frac{14}{2} = 7$.
BETA
