Question

A point $P(a,b)$ lies on the circle $x^2+y^2=4$ in the first quadrant. The maximum value of $\frac{ab}{2+a+b}$ can be expressed as $\sqrt{p}-q$ where $P$ and $Q$ are positive integers then

• A. $p=5$
• B. $q=1$
• C. $p=2$
• D. $q=2$

Answer 1

Answer: (B), (C)

$p=2$
$q=1$

$x^2+y^2=2^2$
Hence $P(a,b)=(2\cos \theta ,2\sin \theta )$ ...(parametric point on a circle).
Hence
$A=\frac{ab}{2+a+b}$
$=\frac{4\sin \theta \cos \theta }{2+2\sin \theta +2\cos \theta }$
$=\frac{2\sin \theta \cos \theta }{1+\sin \theta +\cos \theta }$
$=\frac{\sin 2\theta }{1+\sin \theta +\cos \theta }$
$A$ is maximum for $\theta =\frac{\pi }{4}$
By substituting we get
$A=\frac{1}{1+\sqrt{2}}$
Rationalizing
$A=\sqrt{2}-1$
$=\sqrt{p}-q$
By comparing coefficients, we get
$p=2$ and $q=1$