5. Given $P$ as a moving point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, draw the perpendiculars from the foci $F_{1}$ and $F_{2}$ to the external angle bisector of $\angle F_{1} P F_{2}$. Then the locus of the feet of these perpendiculars is $\qquad$ .
5. $x^{2}+y^{2}=a^{2}$.

Assume without loss of generality that $F_{1} M$ is perpendicular to the external angle bisector $P M$ of $\angle F_{1} P F_{2}$ at point $M$, and intersects the extension of $P F_{2}$ at point $N$. Then $P N=P F_{1}$. Therefore, $F_{2} N=2 a$, and $O M$ is the midline of $\triangle N F_{1} F_{2}$, so $O M=a$.