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Q.
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
Application of Derivatives
Solution:
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 $P M \perp O X$ and produce it to meet the ellipse at $Q$, then $A P Q$ is an isosceles triangle, let $S$ be its area, then
$S =2 \times \frac{1}{2} \times A M \times M P=(O A-O M) \times M P$
$=(a-a \cos \theta) \cdot b \sin \theta$
$\Rightarrow S =a b(\sin \theta-\sin \theta \cos \theta)=a b\left(\sin \theta-\frac{1}{2} \sin 2 \theta\right)$
On differentiating w.r.t. $\theta$, we get
$\frac{d S}{d \theta}=a b(\cos \theta-\cos 2 \theta)$
Again, differentiating w.r.t. $\theta$, we get
$\frac{d^2 S}{d \theta^2}=a b(-\sin \theta+2 \sin 2 \theta)$
For maxima or minima, put $\frac{d S}{d \theta}=0$
$\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}} =a b\left[-\sin \frac{2 \pi}{3}+2 \sin \left(2 \times \frac{2 \pi}{3}\right)\right] $
$=a b\left[-\sin \left(\pi-\frac{\pi}{3}\right)+2 \sin \left(\pi+\frac{\pi}{3}\right)\right] $
$ =a b\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] $
$ =a b\left(-\frac{\sqrt{3}}{2}-\frac{2 \sqrt{3}}{2}\right)$
$=a b\left(\frac{-3 \sqrt{3}}{2}\right)=\frac{-3 \sqrt{3} a b}{2} < 0$
$\therefore S$ is maximum, when $\theta=\frac{2 \pi}{3}$ and maximum value of
$S =a b\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) $
$ =a b\left[\sin \left(\pi-\frac{\pi}{3}\right)-\sin \left(\pi-\frac{\pi}{3}\right) \cos \left(\pi-\frac{\pi}{3}\right)\right] $
$ =a b\left(\sin \frac{\pi}{3}-\sin \frac{\pi}{3} \times\left(-\cos \frac{\pi}{3}\right)\right) $
$=a b\left(\sin \frac{\pi}{3}+\sin \frac{\pi}{3} \cos \frac{\pi}{3}\right)=a b\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}\right) $
$ =a b\left(\frac{2 \sqrt{3}+\sqrt{3}}{4}\right)=\frac{3 \sqrt{3}}{4} a b $ sq unit
Thus, maximum area of isosceles triangle is $\frac{3 \sqrt{3}}{4}$ ab sq unit.
