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 $\qquad$.
Let $|O P|=r_{1},|O Q|=r_{2}, P\left(r_{1} \cos \theta, r_{1} \sin \theta\right), Q\left(-r_{2} \sin \theta, r_{2} \cos \theta\right)$.
Substitute into the ellipse equation to get $\frac{r_{1}^{2} \cos ^{2} \theta}{a^{2}}+\frac{r_{1}^{2} \sin ^{2} \theta}{b^{2}}=1, \frac{r_{2}^{2} \sin ^{2} \theta}{a^{2}}+\frac{r_{2}^{2} \cos ^{2} \theta}{b^{2}}=1$ $\Rightarrow \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}} \geqslant \frac{2}{r_{1} r_{2}} \Rightarrow r_{1} r_{2} \geqslant \frac{2 a^{2} b^{2}}{a^{2}+b^{2}}$, equality holds when $r_{1}=r_{2}$.
Therefore, the minimum value of the product $|O P| \cdot|O Q|$ is $\frac{2 a^{2} b^{2}}{a^{2}+b^{2}}$.