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

The given equation of the ellipse is $\cfrac{{{x^2}}}{{{a^2}}} + \cfrac{{{y^2}}}{{{b^2}}} = 1$, then any point $P$ on the ellipse is $(a\cos \theta ,\;b\sin \theta )$.

From $P$, draw $PQ$ parallel to $y$-axis to meet the ellipse again at $Q,$ then $PAQ$ is an isosceles triangle. Let $A$ be its area, then
$A = \cfrac{1}{2}PQ \cdot AL = \cfrac{1}{2}(2b\sin \theta )(a - a\cos \theta )$

$\Rightarrow A = (a - a\cos \theta ) \times b\sin 0 \Rightarrow A = ab(\sin \theta - \sin \theta \cos \theta )$

$\Rightarrow A = ab(\sin \theta - \cfrac{1}{2}\sin 2\theta )$ …(i)

Differentiating (i) w.r.t. $\theta$, we get $\cfrac{{dA}}{{d\theta }} = ab(\cos \theta - \cos 2\theta )$

(ii) For maxima/minima of$A,\,\,\,\cfrac{{dA}}{{d\theta }} = 0 \Rightarrow \cos \theta = \cos 2\theta$
$\Rightarrow \cos 2\theta = \cos (2\pi - \theta ) \Rightarrow 29 = 2\pi - \theta \Rightarrow \theta = \cfrac{{2\pi }}{3}$

Differentiating (ii) w.r. t. $\theta$, we get
$\cfrac{{{d^2}A}}{{d{\theta ^2}}} = ab( - \sin \theta + 2\sin 2\theta )$ …(iii)

Now, ${\left( {\cfrac{{{d^2}A}}{{d{\theta ^2}}}} \right)_{\theta = \cfrac{{2\pi }}{3}}} = ab\left( { - \sin \cfrac{{2\pi }}{3} + 2\sin \cfrac{{4\pi }}{3}} \right)$

$= ab\left( { - \cfrac{{\sqrt 3 }}{2} - \sqrt 3 } \right) = \cfrac{{ - 3\sqrt 3 }}{2}ab < 0.$

Hence we can say that $A$ is maximum when $\theta = \cfrac{{2\pi }}{3}.$

Also, the maximum area is $A = ab\left( {\sin \cfrac{{2\pi }}{3} - \cfrac{1}{2}\sin \cfrac{{4\pi }}{3}} \right)$

$= ab\left( {\cfrac{{\sqrt 3 }}{2} - \cfrac{1}{2}\left( { - \cfrac{{\sqrt 3 }}{2}} \right)} \right) = \cfrac{{3\sqrt 3 }}{4}ab$sq. units.