Question

The maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with its vertex at one end of the major axis is

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

Answer 1

Answer: (A)

Let the equation of an ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then any point $P$ on the ellipse is $(a\cos \theta ,b\sin \theta )$.
From $P$, draw $PM\perp OX$ and produce it to meet the ellipse at $Q$, then $APQ$ is an isosceles triangle, let $S$ be its area, then
$S=2\times \frac{1}{2}\times AM\times MP=(OA-OM)\times MP$
$=(a-a\cos \theta )\cdot b\sin \theta$
$\Rightarrow S=ab(\sin \theta -\sin \theta \cos \theta )=ab\left(\sin \theta -\frac{1}{2}\sin 2\theta \right)$
On differentiating w.r.t. $\theta$, we get
$\frac{dS}{d\theta }=ab(\cos \theta -\cos 2\theta )$

Again, differentiating w.r.t. $\theta$, we get
$\frac{d^{2}S}{d{\theta }^2}=ab(-\sin \theta +2\sin 2\theta )$
For maxima or minima, put $\frac{dS}{d\theta }=0$
$\Rightarrow \cos \theta =\cos 2\theta$
$\Rightarrow 2\theta =2\pi -\theta$
$\Rightarrow \theta =\frac{2\pi }{3}$
At $\theta =\frac{2\pi }{3}$,
${\left(\frac{d^{2}S}{d{\theta }^2}\right)}_{\theta -\frac{2\pi }{3}}=ab\left[-\sin \frac{2\pi }{3}+2\sin \left(2\times \frac{2\pi }{3}\right)\right]$
$=ab\left[-\sin \left(\pi -\frac{\pi }{3}\right)+2\sin \left(\pi +\frac{\pi }{3}\right)\right]$
$=ab\left(-\sin \frac{\pi }{3}-2\sin \frac{\pi }{3}\right)$
$\left[\because \sin \left(\pi -\frac{\pi }{3}\right)=\sin \frac{\pi }{3},\sin \left(\pi +\frac{\pi }{3}\right)=-\sin \frac{\pi }{3}\right]$
$=ab\left(-\frac{\sqrt{3}}{2}-\frac{2\sqrt{3}}{2}\right)$
$=ab\left(\frac{-3\sqrt{3}}{2}\right)=\frac{-3\sqrt{3}ab}{2}<0$
$\therefore S$ is maximum, when $\theta =\frac{2\pi }{3}$ and maximum value of
$S=ab\left(\sin \frac{2\pi }{3}-\sin \frac{2\pi }{3}\cos \frac{2\pi }{3}\right)(\because \sin 2\theta =2\sin \theta \cos \theta )$
$=ab\left[\sin \left(\pi -\frac{\pi }{3}\right)-\sin \left(\pi -\frac{\pi }{3}\right)\cos \left(\pi -\frac{\pi }{3}\right)\right]$
$=ab\left(\sin \frac{\pi }{3}-\sin \frac{\pi }{3}\times \left(-\cos \frac{\pi }{3}\right)\right)$
$=ab\left(\sin \frac{\pi }{3}+\sin \frac{\pi }{3}\cos \frac{\pi }{3}\right)=ab\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2}\right)$
$=ab\left(\frac{2\sqrt{3}+\sqrt{3}}{4}\right)=\frac{3\sqrt{3}}{4}ab$ sq unit
Thus, maximum area of isosceles triangle is $\frac{3\sqrt{3}}{4}$ ab sq unit.