13.7.17 $\star \star$ Given that $A$ and $B$ are two points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, $O$ is the center of the ellipse, and $O A \perp O B$.
(1) Prove that $\frac{1}{|O A|^{2}}+\frac{1}{|O B|^{2}}$ is a constant;
(2) Find the maximum and minimum values of $\frac{1}{|O A|}+\frac{1}{|O B|}$;
(3) Find the maximum and minimum values of the area of $\triangle A O B$.
(1) Taking the center of the ellipse $O$ as the pole and the $x$-axis as the polar axis, we establish a polar coordinate system. Then the polar equation of the ellipse is: $\frac{1}{\rho^{2}}=\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}$. Let's assume $A\left(\rho_{1}, \theta\right), B\left(\rho_{2}, \theta+\frac{\pi}{2}\right), \theta \in\left[0, \frac{\pi}{2}\right)$. Then $\frac{1}{|O A|^{2}}+\frac{1}{|O B|^{2}}=\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}+\frac{\sin ^{2} \theta}{a^{2}}+\frac{\cos ^{2} \theta}{b^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$ is a constant.
(2) $\frac{1}{|O A|}+\frac{1}{|O B|}=\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}+\sqrt{\frac{\sin ^{2} \theta}{a^{2}}+\frac{\cos ^{2} \theta}{b^{2}}}$
$$
\begin{array}{l}
=\frac{1}{a b}\left(\sqrt{a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta}+\sqrt{a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta}\right) \\
=\frac{1}{a b} \sqrt{a^{2}+b^{2}+2 \sqrt{a^{2} b^{2}\left(\sin ^{4} \theta+\cos ^{4} \theta\right)+\left(a^{4}+b^{4}\right) \sin ^{2} \theta \cos ^{2} \theta}}
\end{array}
$$
$$
=\frac{1}{a b} \sqrt{a^{2}+b^{2}+2 \sqrt{a^{2} b^{2}+\left(a^{2}-b^{2}\right)^{2} \frac{1}{4} \sin ^{2} 2 \theta}},
$$

Then, when $\theta=0$, the minimum value sought is $\frac{1}{a b} \sqrt{a^{2}+b^{2}+2 a b}=\frac{a+b}{a b}$; when $\theta=\frac{\pi}{4}$, the maximum value sought is
$$
\frac{1}{a b} \sqrt{a^{2}+b^{2}+2-\frac{1}{2}\left(a^{2}+b^{2}\right)}=\frac{\sqrt{2\left(a^{2}+b^{2}\right)}}{a b}.
$$
(3) The polar equation of the ellipse is rewritten as: $\rho^{2}=\frac{a^{2} b^{2}}{a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta}$. Therefore,
$$
\begin{aligned}
S_{\triangle A O B} & =\frac{1}{2} \rho_{1} \rho_{2}=\frac{1}{2} a^{2} b^{2} \sqrt{\frac{1}{\left(a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta\right)\left(a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta\right)}} \\
& =\frac{1}{2} a^{2} b^{2} \sqrt{\frac{1}{a^{2} b^{2}+\left(a^{2}-b^{2}\right)^{2} \cdot \frac{1}{4} \sin ^{2} 2 \theta}},
\end{aligned}
$$

Then, when $\theta=\frac{\pi}{4}$, the minimum area sought is $\frac{a^{2} b^{2}}{a^{2}+b^{2}}$; when $\theta=0$, the maximum area sought is $\frac{1}{2} a b$.