Example 2. Find a point on the meadow $\frac{x^{2}}{45}+\frac{y^{2}}{20}=1$ such that the lines connecting it to the two foci are perpendicular. (Problem 13, Page 171)
Let $P$ be the point we are looking for. According to the ellipse equation, we can set $\mathbf{P}$
$$
\begin{array}{l}
(\sqrt{45} \cos \theta, \sqrt{20} \sin \theta) \\
\text { and }|\mathrm{C}_{1} \mathrm{O}|=|\mathrm{C}_{2} \mathrm{O}| \\
=\sqrt{45-20}=5 .
\end{array}
$$

Since $\mathrm{C}_{1} \mathrm{P} \perp \mathrm{C}_{2} \mathrm{P}$, we have $|\mathrm{C}_{1} \mathrm{D}| \cdot|\mathrm{C}_{2} \mathrm{D}|=|\mathrm{PD}|^{2}$.
$$
\text { Therefore, } \begin{aligned}
& (5+\sqrt{45} \cos \theta)(5-\sqrt{45} \cos \theta) \\
& =(\sqrt{20} \sin \theta)^{2},
\end{aligned}
$$

or $25-45 \cos ^{2} \theta=20 \sin ^{2} \theta$.
Thus, $\cos ^{2} \theta=\frac{1}{5}, \cos \theta= \pm \frac{\sqrt{5}}{5}$.
$$
\sin \theta= \pm \sqrt{1-\cos ^{2} \theta}= \pm \frac{2 \sqrt{5}}{5} \text {. }
$$

Therefore, the four points we are looking for are $P_{1}(3,4)$,
$$
P_{2}(-3,4), P_{3}(-3,-4), P_{4}(3,-4) \text {. }
$$