Problem 325

Solve As shown in Figure 1, let the vertex A of the square inscribed in the ellipse in the first quadrant be
$$\left\{\begin{array}{l} x=a \cos \theta, \\ y=b \sin \theta \end{array}\right.$$
(obviously $0<\theta<\frac{x}{2}$)
From $x=x$, we have $a \cos \theta=b \sin \theta$, i.e., $\operatorname{tg} \theta=\frac{a}{b}$.
Thus, $\sin \theta=\frac{a}{\sqrt{a^{2}+b^{2}}}$.

Therefore, the side length of the square is $2 y=2 b \sin \theta=\frac{2 a b}{\sqrt{a^{2}+b^{2}}}$.