The maximum value of the area of the triangle with vertices  $\left(\mathrm{a},0\right), \left(\mathrm{acos} \mathrm{\theta },\mathrm{bsin} \mathrm{\theta }\right)$ and $\left(\mathrm{a} \cos \mathrm{\theta },-\mathrm{bsin} \mathrm{\theta }\right)$ is
 

$\mathrm{A}(\mathrm{a}, 0) \mathrm{B}(\mathrm{a} \mathrm{cos\theta }, \mathrm{b} \mathrm{sin\theta }) \mathrm{C}(\mathrm{a} \mathrm{cos\theta }, -\mathrm{b} \mathrm{sin\theta })$

area of $∆\mathrm{ABC}$

$$\begin{gathered} A=\frac{1}{2}\left|\begin{array}{cc}\mathrm{a}-\mathrm{a} \mathrm{cos\theta }& \mathrm{a}-\mathrm{a} \mathrm{cos\theta }\\ -\mathrm{b} \mathrm{sin\theta }& \mathrm{b} \mathrm{sin\theta }\end{array}\right| \\ =\frac{1}{2}\left|\mathrm{ab} \mathrm{sin\theta }-\mathrm{ab} \mathrm{sin\theta } \mathrm{cos\theta }+\mathrm{ab} \mathrm{sin\theta }-\mathrm{ab} \mathrm{sin\theta } \mathrm{cos\theta }\right| \\ =\mathrm{ab}\left|\sin \mathrm{\theta }-\sin \mathrm{\theta } \cos \mathrm{\theta }\right| \\ =\mathrm{ab} \mathrm{sin\theta }(1- \cos \mathrm{\theta }) \\ {A}^1=\mathrm{ab}\left(\cos \mathrm{\theta }(1-\mathrm{cos\theta })+\mathrm{sin\theta }\left(\mathrm{sin\theta }\right)\right) \end{gathered}$$

A is min or max

$$\begin{gathered} \Rightarrow A^1=0 \\ \mathrm{cos\theta }-{\cos }^2\mathrm{\theta }+1-{\cos }^2\mathrm{\theta }=0 \\ 2{\cos }^2\mathrm{\theta }-\mathrm{cos\theta }-1=0 \\ (2\mathrm{cos\theta }+1)(\mathrm{cos\theta }-1)=0 \\ \mathrm{cos\theta }=-\frac{1}{2} \\ \mathrm{\theta }=\frac{2\mathrm{\pi }}{3} \\ A=\mathbf{\left(}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}\mathbf{\right)}\left(1-(-\frac{1}{2})\right) \\ =\frac{3\sqrt{3}}{4}\mathrm{ab} \end{gathered}$$