China Mathematical Competition

Define
$$P(|OP| \cos \theta, |OP| \sin \theta), \\ Q(|OQ| \cos(\theta \pm \frac{\pi}{2}), |OQ| \sin(\theta \pm \frac{\pi}{2})).$$
We have
$$\frac{1}{|OP|^2} = \frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}, \qquad \textcircled{1}$$
$$\frac{1}{|OQ|^2} = \frac{\sin^2\theta}{a^2} + \frac{\cos^2\theta}{b^2}. \qquad \textcircled{2}$$
Then
$$\frac{1}{|OP|^2} + \frac{1}{|OQ|^2} = \frac{1}{a^2} + \frac{1}{b^2}.$$
Therefore, $|OP| \times |OQ|$ reaches the minimum $\frac{2a^2b^2}{a^2+b^2}$ when $|OP| = |OQ| = \sqrt{\frac{2a^2b^2}{a^2+b^2}}$.