5. If the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ contains a point $P(1,-1), F$ is the right focus, and there is a point $M$ on the ellipse such that $|M P|+$ $2|M F|$ is minimized, then the point $M$ is $\qquad$.
5. $\left(\frac{2}{3} \sqrt{6},-1\right)$ Hint: Since $e=\frac{1}{2}$, let the distance from point $M$ to the right directrix be $|M N|$, then $\frac{|M F|}{|M N|}=\frac{1}{2}$, so $|M N|=2|M F|$, thus $|M P|+2|M F|=|M P|+|M N|$. Therefore, draw a perpendicular from point $P$ to the directrix, its intersection with the ellipse is $M\left(\frac{2 \sqrt{6}}{3},-1\right)$