Question

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

• A. $P\equiv (\frac{3}{2},1)$
• B. $P\equiv (\frac{\sqrt{3}}{2},\sqrt{3})$
• C. eccentric angle of $P$ is $\frac{\pi }{3}$
• D. eccentric angle of $P$ is $\frac{\pi }{6}$

Answer 1

Answer: (A), (D)

eccentric angle of $P$ is $\frac{\pi }{6}$

Any point $P(\alpha ,\beta )$ on the ellipse $\frac{x^2}{3}+\frac{y^2}{4}=1$ is $(\sqrt{3}\cos \theta ,2\sin \theta )$
As $P$ lies in first quadrant, thus
$\theta \in (0,\frac{\pi }{2})$


Line $PQ:y=2\sin \theta$
Line $PR:x=\sqrt{3}\cos \theta$
Line $OQ:y=x$ and
$Q\equiv (2\sin \theta ,2\sin \theta )$

$△=$ Area of trapezium $PQORP$
$=\frac{1}{2}(OR+PQ)PR=\frac{1}{2}(\sqrt{3}\cos \theta +(\sqrt{3}\cos \theta -2\sin \theta \left)\right)2\sin \theta =2\sqrt{3}\cos \theta \sin \theta -2{\sin }^2\theta =\sqrt{3}\sin 2\theta +\cos 2\theta -1=2\cos (2\theta -\frac{\pi }{3})-1$
Which is maximum when $\cos (2\theta -\frac{\pi }{3})$ is maximum
$\Rightarrow (2\theta -\frac{\pi }{3})=0$
$\Rightarrow \theta =\frac{\pi }{6}$
Hence, point $P$ is $(\frac{3}{2},1).$