Example 2. Find the area of the smallest circumscribed parallelogram around the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Let the points of tangency of an external quadrilateral with the circle $x^{\prime 2}+y^{\prime 2}=a^{2}$ be $R(\cos \alpha, a \sin \alpha)$, $R^{\prime}(a \cos \beta, a \sin \beta)$, $Q(-a \cos \alpha$, $-a \sin \alpha)$, and $Q^{\prime}(-a \cos \beta, -a \sin \beta)$. Then the area of the external parallelogram is
$$
S^{\prime}=4 \cdot \frac{1}{2} a \cdot \frac{2 a}{\sin |\beta-\alpha|}=\frac{4 a^{2}}{\sin |\beta-\alpha|} .
$$

When $|\beta-\alpha|=\frac{\pi}{2}$, we get $S_{\text {min }}^{\prime}=4 a^{2}$. Under the transformation $x^{\prime}=x, y^{\prime}=-\frac{a}{b} y$, the minimum external parallelogram of the circle becomes the corresponding external parallelogram of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, with the area
$$
S_{\text {min }}=\frac{b}{a} \cdot 4 a^{2}=4 a b \text {. }
$$