5. Let $A, B$ be moving points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$,

and $\overrightarrow{O A} \cdot \overrightarrow{O B}=0, O$ is the origin, then the distance from $O$ to the line $A B$ is $\qquad$
5. Solution: Let $A\left(x_{1}, y_{1}\right), B\left(-k y_{1}, k x_{1}\right)$, then we have $|O A|^{2}=x_{1}^{2}+y_{1}^{2}, |O B|^{2}=k^{2}|O A|^{2}$, and from $\frac{x_{1}^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1, \frac{y_{1}^{2}}{a^{2}}+\frac{x_{1}^{2}}{b^{2}}=\frac{1}{k^{2}}$ adding them together we get: $\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)$ $|O A|^{2}=\frac{k^{2}+1}{k^{2}}, |O A|^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \frac{k^{2}+1}{k^{2}}$, the distance from point $O$ to line $A B$ is $d^{2}=\frac{|O A|^{2}|O B|^{2}}{|A B|^{2}}=\frac{|O A|^{2}|O B|^{2}}{|O A|^{2}+|O B|^{2}}=$ $\frac{k^{2}|O A|^{2}}{1+k^{2}}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}, d=\frac{a b}{\sqrt{a^{2}+b^{2}}}$.