(5) For any two points $P, Q$ on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, if $O P \perp$ $O Q$, then the minimum value of the product $|O P| \cdot|O Q|$ is
(5) $\frac{2 a^{2} b^{2}}{a^{2}+b^{2}}$ Hint: Let
$$
\begin{array}{c}
P(|O P| \cos \theta,|O P| \sin \theta), \\
Q\left(|O Q| \cos \left(\theta \pm \frac{\pi}{2}\right),|O Q| \sin \left(\theta \pm \frac{\pi}{2}\right)\right) .
\end{array}
$$

Since $P$ and $Q$ are on the ellipse, we have
$$
\begin{array}{l}
\frac{1}{|O P|^{2}}=\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}, \\
\frac{1}{|O Q|^{2}}=\frac{\sin ^{2} \theta}{a^{2}}+\frac{\cos ^{2} \theta}{b^{2}},
\end{array}
$$
Adding (1) and (2) gives
$$
\frac{1}{|O P|^{2}}+\frac{1}{|O Q|^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} .
$$

Thus, when
$$
|O P|=|O Q|=\sqrt{\frac{2 a^{2} b^{2}}{a^{2}+b^{2}}}
$$

$|O P| \cdot|O Q|$ reaches its minimum value $\frac{2 a^{2} b^{2}}{a^{2}+b^{2}}$.