5. Let the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ have one focus at $F$, and point $P$ be on the $y$-axis. The line $P F$ intersects the ellipse at points $M, N, \overrightarrow{P M}=$ $\lambda_{1} \overrightarrow{M F}, \overrightarrow{P N}=\lambda_{2} \overrightarrow{N F}$, then the real number $\lambda_{1}+\lambda_{2}=$ $\qquad$
$$
\begin{array}{c}
\text { 5. }-\frac{2 a^{2}}{b^{2}} . \\
\text { Solution: } \lambda_{1}+\lambda_{2}=\frac{x_{M}}{c-x_{M}}+\frac{x_{N}}{c-x_{N}}= \\
\frac{c\left(x_{M}+x_{N}\right)-2 x_{M} x_{N}}{c^{2}-c\left(x_{M}+x_{N}\right)+x_{M} x_{N}} \\
\text { From }\left\{\begin{array}{l}
y=k(x-c) \\
b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}
\end{array} \text { we get } x_{M}+x_{N}=\frac{2 a^{2} k^{2} c}{b^{2}+a^{2} k^{2}}\right. \text {, } \\
x_{M} x_{N}=\frac{a^{2} k^{2} c^{2}-a^{2} b^{2}}{b^{2}+a^{2} k^{2}}, \therefore \lambda_{1}+\lambda_{2}=-\frac{2 a^{2}}{b^{2}} .
\end{array}
$$
5. $-\frac{2 a^{2}}{b^{2}}$.