6. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ (where $a>b>$ $0$), point $P$ is a point on it, $F_{1} 、 F_{2}$ are the foci of the ellipse, and the external angle bisector of $\angle F_{1} P F_{2}$ is $l$. Construct $F_{1} 、 F_{2}$ perpendicular to $l$ at $R 、 S$. When $P$ traverses the entire ellipse, the trajectory equation of $R 、 S$ is $\qquad$
6. $x^{2}+y^{2}=a^{2}$.

As shown in Figure 7, construct the symmetric point $F^{\prime}$ of $F_{2}$ with respect to $l$, $\angle F^{\prime} P S = \angle F_{2} P S$, and $l$ is the external angle bisector of $\angle F_{1} P F_{2}$. Then, $F^{\prime}$, $P$, and $F_{1}$ are collinear.

Let $S\left(x_{0}, y_{0}\right)$, $F^{\prime}\left(x_{1}, y_{1}\right)$, $F_{2}(c, 0)$, $F_{1}(-c, 0)$, and $F_{1} F^{\prime} = F_{1} P + P F^{\prime} = F_{1} P + P F_{2} = 2a$. Then, $\left(x_{1} + c\right)^{2} + y_{1}^{2} = (2a)^{2}$.
$$
\begin{array}{l}
\because x_{0} = \frac{x_{1} + c}{2}, y_{0} = \frac{y_{1}}{2}, \\
\therefore x_{1} = 2x_{0} - c, y_{1} = 2y_{0}, \\
\therefore (2x_{0})^{2} + (2y_{0})^{2} = (2a)^{2}. \\
\therefore x_{0}^{2} + y_{0}^{2} = a^{2}.
\end{array}
$$
Point $S$ satisfies the equation $x^{2} + y^{2} = a^{2}$.
Similarly, point $R$ also satisfies $x^{2} + y^{2} = a^{2}$.
Therefore, $x^{2} + y^{2} = a^{2}$.