137.2 ** Quadrilateral $P Q R S$ is inscribed in the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and is a rhombus.
(1) Draw a perpendicular from the origin $O$ to side $P Q$, with the foot of the perpendicular being $H$. Find the locus of $H$;
(2) Find the maximum value of $\frac{1}{|O P|}+\frac{1}{|O Q|}$;
(3) Find the maximum and minimum values of the area of $\triangle P O Q$;
(4) Find the maximum and minimum values of $|P Q|$.
(1) $P Q R S$ is an inscribed rhombus in an ellipse. Let $|O P|=\gamma_{1},|O Q|=\gamma_{2}$. Suppose the coordinates of point $P$ are $\left(\gamma_{1} \cos \theta, \gamma_{1} \sin \theta\right)$, then the coordinates of $Q$ are $\left(\gamma_{2} \cos \left(0+\frac{\pi}{2}\right), \gamma_{2} \sin \left(\theta+\frac{\pi}{2}\right)\right)$. Substituting into the ellipse equation, we get
$$
\frac{\left(\gamma_{1} \cos \theta\right)^{2}}{a^{2}}+\frac{\left(\gamma_{1} \sin \theta\right)^{2}}{b^{2}}=1, \frac{\left[\gamma_{1} \cos \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{a^{2}}+\frac{\left[\gamma_{2} \sin \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{b^{2}}=1,
$$

which simplifies to $\frac{1}{\gamma_{1}^{2}}=\frac{(\cos \theta)^{2}}{a^{2}}+\frac{(\sin \theta)^{2}}{b^{2}}, \frac{1}{\gamma_{2}^{2}}=\frac{\left[\cos \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{a^{2}}+\frac{\left[\sin \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{b^{2}}$,
thus
$$
\begin{array}{l}
\frac{1}{|O P|^{2}}+\frac{1}{|O Q|^{2}}=\frac{1}{\gamma_{1}^{2}}+\frac{1}{\gamma_{2}^{2}} \\
=\frac{(\cos \theta)^{2}}{a^{2}}+\frac{(\sin \theta)^{2}}{b^{2}}+\frac{\left[\cos \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{a^{2}}+\frac{\left[\sin \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{b^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} .
\end{array}
$$

In the right triangle $\triangle P O Q$, the altitude from $O$ to the hypotenuse $P Q$ is $|O H|$,
$$
|O H|=\frac{|O P||O Q|}{\sqrt{|O P|^{2}+|O Q|^{2}}}=\frac{\gamma_{1} \gamma_{2}}{\sqrt{\gamma_{1}^{2}+\gamma_{2}^{2}}}=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}},
$$

which is a constant. Therefore, the locus of point $H$ is a circle centered at the origin with radius $\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}$.
(2) According to the inequality $p+q \leqslant \sqrt{2\left(p^{2}+q^{2}\right)}$, we have $\frac{1}{|O P|}+\frac{1}{|O Q|} \leqslant$ $\sqrt{2\left(\frac{1}{|O P|^{2}}+\frac{1}{|O Q|^{2}}\right)}=\frac{\sqrt{2\left(a^{2}+b^{2}\right)}}{a b}$, with equality when $|O P|=|O Q|$.
(3) The area of $\triangle P O Q$ is
$$
\begin{aligned}
S & =\frac{1}{2}|O P||O Q|=\frac{1}{2} \gamma_{1} \cdot \gamma_{2}=\frac{a^{2} b^{2}}{2} \cdot \frac{1}{\sqrt{\left(a^{2}-c^{2} \cos ^{2} \theta\right)\left(a^{2}-c^{2} \sin ^{2} \theta\right)}} \\
& =\frac{a^{2} b^{2}}{2} \cdot \frac{1}{\sqrt{a^{4}-a^{2} c^{2}+\frac{1}{4} c^{2} \sin ^{2} 2 \theta}},
\end{aligned}
$$

When $\sin 2 \theta= \pm 1,|O P|=|O Q|$, the minimum area of $\triangle P O Q$ is $\frac{a^{2} b^{2}}{a^{2}+b^{2}}$;
When $\sin 2 \theta=0$, $O P$ and $O Q$ coincide with the $x$-axis and $y$-axis respectively, the maximum area of $\triangle P O Q$ is $\frac{1}{2} a b$.
(4) Since $\frac{1}{\gamma_{1}^{2}}+\frac{1}{\gamma_{2}^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$, and $\left(\gamma_{1}^{2}+\gamma_{2}^{2}\right)\left(\frac{1}{\gamma_{1}^{2}}+\frac{1}{\gamma_{2}^{2}}\right) \geqslant 4,|P Q|^{2}=\gamma_{1}^{2}+\gamma_{2}^{2}$, thus
$$
\begin{array}{c}
|P Q|^{2} \geqslant \frac{4 a^{2} b^{2}}{a^{2}+b^{2}},|P Q| \geqslant \frac{2 a b}{\sqrt{a^{2}+b^{2}}} . \\
\text { Also, } \frac{1}{\gamma_{1}^{2}}=\frac{(\cos \theta)^{2}}{a^{2}}+\frac{(\sin \theta)^{2}}{b^{2}}, \frac{1}{\gamma_{2}^{2}}=\frac{\left[\cos \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{a^{2}}+\frac{\left[\sin \left(\theta+\frac{\pi}{2}\right)\right]^{2}}{b^{2}}, \text { so } \\
\gamma_{1}^{2}+\gamma_{2}^{2}=\frac{a^{2} b^{2}}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}+\frac{a^{2} b^{2}}{a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta} \leqslant a^{2}+b^{2},
\end{array}
$$

Therefore, $|P Q| \leqslant \sqrt{a^{2}+b^{2}}$, which means $\frac{2 a b}{\sqrt{a^{2}+b^{2}}} \leqslant|P Q| \leqslant \sqrt{a^{2}+b^{2}}$.