14. Given $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right)$ are two moving points on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=$ $1(a>b>0)$, $O$ is the origin, and $O A \perp$ $O B$. Find the minimum length of segment $A B$.
14. According to the problem, let's assume $A\left(r_{1} \cos \theta, r_{1} \sin \theta\right)$, $B\left(r_{2} \cos \left(\theta+\frac{\pi}{2}\right), r_{2} \sin \left(\theta+\frac{\pi}{2}\right)\right)$, then
$$
\begin{array}{l}
B\left(-r_{2} \sin \theta, r_{2} \cos \theta\right), A B^{2}=r_{1}^{2}+r_{2}^{2}, \\
\frac{r_{1}^{2} \cos ^{2} \theta}{a^{2}}+\frac{r_{1}^{2} \sin ^{2} \theta}{b^{2}}=1, r_{1}^{2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}, \\
\frac{r_{2}^{2} \sin ^{2} \theta}{a^{2}}+\frac{r_{2}^{2} \cos ^{2} \theta}{b^{2}}=1, r_{2}^{2}=\frac{a^{2} b^{2}}{b^{2} \sin ^{2} \theta+a^{2} \cos ^{2} \theta} .
\end{array}
$$
$$
\text { Hence } r_{1}^{2}+r_{2}^{2}=\frac{a^{2} b^{2}\left(a^{2}+b^{2}\right)}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)\left(b^{2} \sin ^{2} \theta+a^{2} \cos ^{2} \theta\right)}
$$
$$
\begin{array}{l}
=\frac{a^{2} b^{2}\left(a^{2}+b^{2}\right)}{\left(a^{4}+b^{4}\right) \sin ^{2} \theta \cdot \cos ^{2} \theta+a^{2} b^{2}\left(\sin ^{4} \theta+\cos ^{4} \theta\right)} \\
=\frac{a^{2} b^{2}\left(a^{2}+b^{2}\right)}{\left(a^{4}+b^{4}\right) \sin ^{2} \theta \cdot \cos ^{2} \theta+a^{2} b^{2}\left(1-2 \sin ^{2} \theta \cdot \cos ^{2} \theta\right)} \\
=\frac{a^{2} b^{2}\left(a^{2}+b^{2}\right)}{\left(a^{2}-b^{2}\right)^{2} \sin ^{2} \theta \cdot \cos ^{2} \theta+a^{2} b^{2}} \\
=\frac{4 a^{2} b^{2}\left(a^{2}+b^{2}\right)}{\left(a^{2}-b^{2}\right)^{2} \sin ^{2} 2 \theta+4 a^{2} b^{2}} \\
\geqslant \frac{4 a^{2} b^{2}\left(a^{2}+b^{2}\right)}{\left(a^{2}+b^{2}\right)^{2}} .
\end{array}
$$

When and only when $\theta=k \pi \pm \frac{\pi}{4}(k \in \mathbf{Z})$, the equality holds.
Therefore, the minimum length of segment $A B$ is $\frac{2 a b \sqrt{a^{2}+b^{2}}}{a^{2}+b^{2}}$.