Example 6. Find the maximum value of the area of a triangle inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \quad(a>b>0)$.
Solve: Using the parametric equations of the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1 \quad(\mathrm{a}>\mathrm{b}>0)$:
$$
\left\{\begin{array}{l}
x=a \cos \theta, \\
y=b \sin \theta .
\end{array} \text { ( } \theta\right. \text { is the parameter) }
$$

As shown in Figure 8, there is an inscribed triangle $\triangle P, P_{2} P_{3}$ in the ellipse,
$\mathbf{P}_{1}\left(\mathrm{a} \cos \theta_{1}, \mathrm{~b} \sin \theta_{1}\right), \quad \mathbf{P}_{2}\left(\mathrm{a} \cos \theta_{2}, \mathrm{~b} \sin \theta_{2}\right)$
$\mathbf{P}_{3}\left(a \cos \theta_{3}, b \sin \theta_{3}\right)$.
$\because S \Delta p_{1} p_{2} D_{3}=\frac{1}{2}\left|\begin{array}{ll}a \cos \theta^{2} & b \sin \theta_{1} 1 \\ a \cos \theta_{2} & b \sin \theta_{2} 1 \\ a \cos \theta_{3} & b \sin \theta_{3} 1\end{array}\right|$ in absolute value $=\frac{1}{2} a b\left|\begin{array}{ll}\cos \theta_{1} & \sin \theta_{1} \\ \cos \theta_{2} & \sin \theta_{2} \\ \cos \theta_{3} & \sin \theta_{3}\end{array}\right|$ in absolute value.

The third-order determinant $\left\{\left.\begin{array}{lll}\cos \theta_{1} & \sin \theta_{1} & 1 \\ \cos \theta_{2} & \sin \theta_{2} & 1 \\ \cos \theta_{3} & \sin \theta_{3} & 1\end{array} \right\rvert\,\right.$
is exactly the area of the inscribed triangle in the unit circle. Its maximum value should be: the inscribed equilateral triangle in the unit circle, with an area of
$\frac{3}{4} \sqrt{3}$, hence
$S \Delta p_{1} p_{2} p_{3} \leqslant \frac{3}{8} \sqrt{3} a b$. Therefore, the maximum area of the inscribed triangle in the ellipse is $\frac{3}{8} \sqrt{3} \mathrm{ab}$.