Question

Find 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.

Answer 1

Let the equation of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ then any point P on the ellipse is $(acos\theta ,bsin\theta )$.
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
$S=2\times \frac{1}{2}\times AM\times MP=(OA-OM)\times MP=(a-acos\theta ).bsin\theta$
$\Rightarrow S=ab(sin\theta -sin\theta cos\theta )=ab(sin\theta -\frac{1}{2}sin2\theta )$
On differentiating w.r.t. $\theta$, we get
$\frac{dS}{d\theta }=ab(cos\theta -cos2\theta )$
Again differentiating w.r.t. $\theta$, we get
$\frac{d^{2}S}{d{\theta }^2}=ab(-sin\theta +2sin2\theta )$
For maxima or miinima, put $\frac{dS}{d\theta }=0$
$\Rightarrow cos\theta =cos2\theta \Rightarrow 2\theta =2\pi -\theta$
$\Rightarrow \theta =\frac{2\pi }{3}$
At $\theta =\frac{2\pi }{3},(\frac{d^{2}S}{d{\theta }^2}{)}_{\theta =\frac{2\pi }{3}}=ab[-sin\frac{2\pi }{3}+2sin(2\times \frac{2\pi }{3})]$
$=ab[-sin(\pi -\frac{\pi }{3})+2sin(\pi +\frac{\pi }{3}\left)\right]=abb(-sin\frac{\pi }{3}-2sin\frac{\pi }{3})\left[\begin{array}{c}\because sin(\because -\frac{\pi }{3})=sin\frac{\pi }{3}\\ sin(\pi +\frac{\pi }{3})=\frac{-sin\pi }{3}\end{array}\right]=ab(-\frac{\sqrt{3}}{2}-\frac{2\sqrt{3}}{2})=ab\left(\frac{-3\sqrt{3}}{2}\right)=\frac{-3\sqrt{3}ab}{2}<0$
$\therefore$ S is maximum, when $0=\frac{2\pi }{3}$
and maximum value of $S=ab(sin\frac{2\pi }{3}-\frac{1}{2}.2sin\frac{2\pi }{3}cos\frac{2\pi }{3})$
$[\because sin2\theta =2sin\theta cos\theta ]=ab\left[sin\right(\pi -\frac{\pi }{3})-sin(\pi -\frac{\pi }{3}\left)cos\right(\pi -\frac{\pi }{3}\left)\right]$
$=ab(sin\frac{\pi }{3}-sin\frac{\pi }{3}\times (-cos\frac{\pi }{3}\left)\right)=ab(sin\frac{\pi }{3}+sin\frac{\pi }{3}cos\frac{\pi }{3})=ab(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2})=ab\left(\frac{2\sqrt{3}+\sqrt{3}}{4}\right)=\frac{3\sqrt{3}}{4}absqunit$
Thus, maximum area of isosceles triangle is $\frac{3\sqrt{3}}{4}absqunit.$