Question

If PQ is a double ordinate of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and A is one end of the major axis then the maximum area of $\Delta APQ$ is

• A. $\frac{\sqrt{3}}{4}$ ab
• B. $\frac{3\sqrt{3}}{4}$ ab
• C. $\frac{\sqrt{3}}{2}$ ab
• D. $\frac{3\sqrt{3}}{2}$ ab

Answer 1

The correct option is B $\frac{3\sqrt{3}}{4}$ ab

Let $P\equiv (a\cos \theta ,b\sin \theta )$ and $Q\equiv (a\cos \theta ,-b\sin \theta )$
Height of $△APQ=a+a\cos \theta$
Area of $△APQ=\frac{1}{2}a(1+\cos \theta )\left(2b\sin \theta \right)$
$A=f\left(\theta \right)=ab(1+\cos \theta )\sin \theta$
$f^{\prime }\left(\theta \right)=ab(\cos 2\theta +\cos \theta )$

For maxima or minima,
$f^{\prime }\left(\theta \right)=0$
$\Rightarrow \theta =\frac{\pi }{3},\pi$ (not possible)

$\Rightarrow \theta =\frac{\pi }{3}$

$f^{\prime \prime }\left(\theta \right)=ab(-2\sin 2\theta -\sin \theta )$
$f^{\prime \prime }\left(\theta \right)<0$ at $\theta =\frac{\pi }{3}$
So, area is maximum at $\theta =\frac{\pi }{3}$
Maximum area $=\frac{3\sqrt{3}}{4}ab$