The area of the greatest isosceles triangle that can be inscribed in an ellipse$\left({\mathrm{x}}^2/16\right)+\left({\mathrm{y}}^2/3\right)=1$ having its vertex coincident with an extremity of the major axis is

Let APQ be an isosceles  triangle inscribed in the ellipse $\left(\frac{{\mathrm{x}}^2}{16}\right)+\left(\frac{{\mathrm{y}}^2}{3}\right)=1$ with the Vertex A at the extremity (4, 0) of the major axis.

 If the coordinates of P are $(4\cos \mathrm{\theta },\sqrt{3}\sin \mathrm{\theta })$, then those of Q will be $(4\cos \mathrm{\theta },-\sqrt{3}\sin \mathrm{\theta })$
So the area of $\mathrm{△}\mathrm{APQ}$ is given by
$\mathrm{A}=\frac{1}{2}\mathrm{PQ}\times \mathrm{AD}=\frac{1}{2}\left(2\mathrm{PD}\right)\times \mathrm{AD}=\mathrm{PD}\times (\mathrm{OA}-\mathrm{OD})$
$$\begin{gathered} A=\left(b\sin \theta \right)(a-a\cos \theta ) \\ A=ab\sin \theta -ab\sin \theta \cos \theta \\ A=\frac{4\sqrt{3}}{2}(2\sin \theta -\sin 2\theta ) \\ Diff w.r.to \theta on both sides \\ \frac{dA}{d\theta }=2\sqrt{3}(2\cos \theta -2\cos 2\theta ) \\ \Rightarrow for max or min \frac{dA}{d\theta }=0 \\ \Rightarrow \cos \theta -2{\cos }^2\theta +1=0 \\ \Rightarrow 2{\cos }^2\theta -\cos \theta -1=0 \\ \Rightarrow (2\cos \theta +1)(\cos \theta -1)=0 \\ \Rightarrow \cos \mathrm{\theta }=-\frac{1}{2}\Rightarrow [\because \cos \mathrm{\theta }\ne 1\text{ as }\mathrm{\theta }\ne 0] \end{gathered}$$
$\Rightarrow \mathrm{\theta }=\frac{2\mathrm{\pi }}{3}$

Hence, maximum area
$$\begin{gathered} \mathrm{A}=\frac{1}{2}4\sqrt{3}\left[2\sin \frac{2\mathrm{\pi }}{3}-\sin \frac{4\mathrm{\pi }}{3}\right] \\ \Rightarrow A=2\sqrt{3}\left(2\right(\frac{\sqrt{3}}{2})+\frac{\sqrt{3}}{2}) \\ \Rightarrow A=9 \end{gathered}$$