Question

The point ($\alpha ,\beta$) on the ellipse $4x^2+3y^2=12$, in the first quadrant, so that the area enclosed by the lines $y=x,y=\beta ,x=\alpha$ and the $x-$axis is maximum, is

• A. None of the above
• B. $(\sqrt{3},0)$
• C. $(\frac{3}{2},1)$
• D. $(\frac{1}{2},\sqrt{\frac{11}{3}})$

Answer 1

The correct option is C $(\frac{3}{2},1)$

Equation of the ellipse is $\frac{x^2}{3}+\frac{y^2}{4}=1$
Let point $P$ be$(\sqrt{3}\cos \theta ,2\sin \theta ),\theta \in$ $(0,\frac{\pi }{2})$
Clearly, line $PQ$ is$y=2\sin \theta$,
line $PR$ is$x=\sqrt{3}\cos \theta$and $OQ$ is $y=x$, and $Q$ is $(2\sin \theta ,2\sin \theta )$
$Z=$ area of the region $PQORP$ (trapezium)
$=\frac{1}{2}(OR+PQ)PR$
$=\frac{1}{2}(\sqrt{3}\cos \theta +(\sqrt{3}\cos \theta -2\sin \theta \left)\right)2\sin \theta$
$=(2\sqrt{3}\cos \theta \sin \theta -2si{n}^2\theta )$
$=(\sqrt{3}\sin 2\theta +\cos 2\theta -1)$
$=2\cos (2\theta -\frac{\pi }{3})-1$
Which is maximum when $\cos (2\theta -\frac{\pi }{3})=1$ or
$2\theta -\frac{\pi }{3}=0\Rightarrow \theta =\frac{\pi }{6}$
Hence, point $P\text{is}(\frac{3}{2},1)$