Solve this following


Let $P Q$ 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



Let $\mathrm{P}(3 \cos \theta, 3 \sin \theta)$

$Q(-3 \cos \theta,-3 \sin \theta)$

$\Rightarrow \alpha \beta=\frac{\left|(3 \cos \theta+3 \sin \theta)^{2}-4\right|}{2}$

$\Rightarrow \alpha \beta=\frac{5+9 \sin 2 \theta}{2} \leq 7$


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now