14. Let $P$ be any point in the first quadrant on the ellipse $C: \frac{x^{2}}{5}+y^{2}=1$, and $F_{1}, F_{2}$ be the left and right foci, respectively. The lines $P F_{1}, P F_{2}$ intersect $C$ at $M, N$. If $\overrightarrow{P F_{1}}=\lambda_{1} \overrightarrow{F_{1} M}, \overrightarrow{P F_{2}}=\lambda_{2} \overrightarrow{F_{2} N}$, find the coordinates of point $P$ such that the slope of segment $M N$ is $-\frac{1}{9}$.
Let $P\left(x_{0}, y_{0}\right)$, from $\overrightarrow{O M}-\overrightarrow{O F_{1}}=\frac{1}{\lambda_{1}}\left(\overrightarrow{O F_{1}}-\overrightarrow{O P}\right) \Rightarrow \overrightarrow{O M}=\frac{\lambda_{1} \overrightarrow{O F_{1}}+\overrightarrow{O F_{1}}-\overrightarrow{O P}}{\lambda_{1}}$ $\Rightarrow M\left(\frac{-2 \lambda_{1}-2-x_{0}}{\lambda_{1}}, \frac{-y_{0}}{\lambda_{1}}\right)$, similarly we get $N\left(\frac{2 \lambda_{2}+2-x_{0}}{\lambda_{2}}, \frac{-y_{0}}{\lambda_{2}}\right)$. Substituting into the ellipse equation, we get $\left\{\begin{array}{l}\frac{\left(2 \lambda_{1}+2+x_{0}\right)^{2}}{5}+y_{0}^{2}=\lambda_{1}^{2}, \\ \frac{\left(2 \lambda_{2}+2-x_{0}\right)^{2}}{5}+y_{0}^{2}=\lambda_{2}^{2}\end{array} \Rightarrow\left\{\begin{array}{l}\frac{4\left(\lambda_{1}+1\right)^{2}+4 x_{0}\left(\lambda_{1}+1\right)}{5}=\lambda_{1}^{2}-1, \\ \frac{4\left(\lambda_{2}+1\right)^{2}-4 x_{0}\left(\lambda_{2}+1\right)}{5}=\lambda_{2}^{2}-1\end{array}\right.\right.$ $\Rightarrow\left\{\begin{array}{l}\frac{4\left(\lambda_{1}+1\right)+4 x_{0}}{5}=\lambda_{1}-1, \\ \frac{4\left(\lambda_{2}+1\right)-4 x_{0}}{5}=\lambda_{2}-1\end{array} \Rightarrow\left\{\begin{array}{l}\lambda_{1}=9+4 x_{0}, \\ \lambda_{2}=9-4 x_{0} .\end{array}\right.\right.$
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\text { Then } k_{M N}=\frac{-\frac{y_{0}}{\lambda_{2}}+\frac{y_{0}}{\lambda_{1}}}{\frac{2 \lambda_{2}+2-x_{0}}{\lambda_{2}}+\frac{2 \lambda_{1}+2+x_{0}}{\lambda_{1}}}=\frac{y_{0}\left(\lambda_{2}-\lambda_{1}\right)}{4 \lambda_{1} \lambda_{2}+2\left(\lambda_{1}+\lambda_{2}\right)+x_{0}\left(\lambda_{2}-\lambda_{1}\right)} \\
=\frac{-8 x_{0} y_{0}}{4\left(81-16 x_{0}^{2}\right)+36-8 x_{0}^{2}}=\frac{-8 x_{0} y_{0}}{-72 x_{0}^{2}+360}=-\frac{1}{9} \Rightarrow x_{0} y_{0}=5-x_{0}^{2}=5 y_{0}^{2} \\
\Rightarrow x_{0}=5 y_{0} \text {, combining with } x_{0}^{2}+5 y_{0}^{2}=5 \Rightarrow x_{0}=\frac{5 \sqrt{6}}{6}, y_{0}=\frac{\sqrt{6}}{6} . \text { Therefore, point } P\left(\frac{5 \sqrt{6}}{6}, \frac{\sqrt{6}}{6}\right) .
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