Let $\triangle A B C$ be an isosceles triangle inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Let coordinate of $A$ be $(a, 0)$.
Area of $\triangle A B C, A=\frac{1}{2} \times A D \times B C$
$$
\begin{aligned}
&=\frac{1}{2} \times(a+x) \times 2 y=y(a+x) \\
&=\sqrt{b^{2}\left(1-\frac{x^{2}}{a^{2}}\right)}(a+x)=\frac{b}{a} \sqrt{a^{2}-x^{2}}(a+x)
\end{aligned}
$$



$\Rightarrow \frac{d A}{d x}=\frac{b}{a}\left[(a+x) \frac{1}{2} \times \frac{-2 x}{\sqrt{a^{2}-x^{2}}}+\sqrt{a^{2}-x^{2}}\right]$
$=\frac{b}{a}\left[\frac{-x(a+x)+\left(a^{2}-x^{2}\right)}{\sqrt{a^{2}-x^{2}}}\right]$
Put $\frac{d A}{d x}=0$
$\Rightarrow \quad 2 x^{2}+a x-a^{2}=0$
$\Rightarrow \quad 2 x^{2}+2 a x-a x-a^{2}=0$
$\Rightarrow \quad(2 x-a)(x+a)=0$
$\Rightarrow \quad x=\frac{a}{2},-a$
Since, $\frac{d^{2} A}{d x^{2}} < 0$, for $x=\frac{a}{2}$
$$
\begin{aligned}
\therefore \text { Maximum area } &=\frac{b}{a} \cdot\left(a+\frac{a}{2}\right) \sqrt{a^{2}-\frac{a^{2}}{4}} \\
&=\frac{b}{a} \cdot \frac{3}{2}(a) \frac{\sqrt{3} a}{2}=\frac{3 \sqrt{3} a b}{4} \text { sq units }
\end{aligned}
$$