Find the locus equation of points whose sum of distances to the fixed points $F_{1}(-c, 0), F_{2}(c, 0)$ is $2 \mathrm{a}$ $(\mathrm{a}>c>0)$
Let $M(x, y)$ be any point on the trajectory, and its distances to $\mathrm{F}_{1}, F_{1}$ are $r, r_{2} x$ respectively. $\mathrm{F}_{1}$ is the center with radius $r_{1}$, and $\mathrm{F}_{2}$ is the center with radius $\mathrm{r}_{2}$ of two circles (as shown in the figure). Therefore,
$$
\left\{\begin{array}{l}
(x+c)^{2}+y^{2}=r_{1}^{2}, \\
(x-c)^{2}+y^{2}=r_{2}^{2}, \\
r_{1}+r_{2}=2 a .
\end{array}\right.
$$

Eliminating the parameters $r_{1}, r_{2}$: (1) - (2) gives
$$
\mathrm{r}_{1}^{2}-\mathrm{r}_{2}^{2}=(\mathrm{x}+\mathrm{c})^{2}-(\mathrm{x}-\mathrm{c})^{2},
$$

which simplifies to $\left(r_{1}+r_{2}\right)\left(r_{1}-r_{2}\right)=4 c x$.

Thus, we have $\left\{\begin{array}{l}r_{1}+r_{2}=2 a, \\ r_{1}-r_{2}=\frac{2 c}{a} x .\end{array}\right.$
Therefore, $r_{1}=\frac{c x+a^{2}}{a}$.
Substituting (4) into (1) and setting $a^{2}-c^{2}=b^{2}$, we get
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 .
$$