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 is vertex as at on end of the major axis.

Answer 1

Area $\Delta AP{P}^{\prime }=\frac{1}{2}\times p{p}^{\prime }\times AM$

$=\frac{1}{2}\times ab\left(2b\sin \theta \right)(a-a\cos \theta )$

$=ab[\sin \theta -\frac{1}{2}\sin 2\theta ]$

$\frac{2A}{2\theta }=0$

$\Rightarrow \theta =\frac{2\pi }{3}$

$\frac{2A}{2\theta }=ab\left[\frac{-3\sqrt{3}}{2}\right]<0$

A is maximum when $\theta =\frac{2\pi }{3}={120}^o$

Maximum valye of A$=a(\sin {120}^o-\frac{1}{2}\sin {240}^o)$

$=\frac{3\sqrt{3}}{4}ab$.