Example 1. Find a point on the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$ such that the product of its distances to the two foci is 16.
Let the point to be found be $\mathrm{P}(\mathrm{x}, \mathrm{y}), \mathrm{F}_{1}$, $\mathrm{F}_{2}$ be the left and right foci of the ellipse.
Since $=5, \mathrm{~b}=3, \mathrm{c}=4, \mathrm{e}=\frac{4}{5}$,
it follows from formula (1) that
$$
\begin{array}{l}
\left|P F_{1}\right|=a+e x=5+\frac{4}{5} x, \\
\left|P F_{2}\right|=a-0 x=5-\frac{4}{5} x .
\end{array}
$$

According to the problem, $\left(5+\frac{4}{5} x\right)\left(5-\frac{4}{5} x\right)$ $=16$, solving this gives: $x= \pm \frac{15}{4}, y= \pm \frac{3 \sqrt{7}}{4}$.
Thus, the points to be found are:
$$
\begin{array}{l}
\left(\frac{15}{4}, \frac{3 \sqrt{7}}{4}\right),\left(\frac{15}{4},-\frac{3 \sqrt{7}}{4}\right) \\
\left(-\frac{15}{4}, \frac{3 \sqrt{7}}{4}\right),\left(-\frac{15}{4},-\frac{3 \sqrt{7}}{4}\right)
\end{array}
$$