设$$AB$$在准线$$l$$上的射影分别为$${A}'$$、$${B}'$$，  则$$\textbar M{M}'\textbar=\frac{\textbar A{A}'\textbar+\textbar B{B}'\textbar}{2}$$．  又因为$$\textbar A{A}'\textbar=\textbar AF\textbar$$，$$\textbar B{B}'\textbar=\textbar BF\textbar$$，  所以$$\textbar M{M}'\textbar=\frac{\textbar AF\textbar+\textbar BF\textbar}{2}$$．  因为$$\angle AFB=\frac{2}{3} \pi $$，  所以$$\textbar AB\textbar=\textbar FA{{\textbar}^{2}}+\textbar FB{{\textbar}^{2}}-2\textbar FA\textbar\cdot \textbar FB\textbar\cos \angle AFB$$  $$=\textbar FA{{\textbar}^{2}}+\textbar FB{{\textbar}^{2}}+\textbar FA\textbar\cdot \textbar FB\textbar$$  因此$$\textbar AB{{\textbar}^{2}}+\textbar M{M}'{{\textbar}^{2}}=\textbar AB{{\textbar}^{2}}+{{\left( \frac{\textbar AF\textbar+\textbar BF\textbar}{2} \right)}^{2}}$$  $$\geqslant \textbar AB{{\textbar}^{2}}+\textbar FA\textbar\cdot \textbar FB\textbar$$  $$={{(\textbar FA\textbar+\textbar FB\textbar)}^{2}}=4\textbar M{M}'{{\textbar}^{2}}$$，  则$$\textbar AB{{\textbar}^{2}}\geqslant 3\textbar M{M}'{{\textbar}^{2}}$$，即$$\frac{\textbar M{M}'\textbar}{\textbar AB\textbar}\leqslant \frac{\sqrt{3}}{3}$$．故选$$\text{B}$$． 
抛物线$${{y}^{2}}=4x$$的焦点为$$F$$，点$$A$$、$$B$$在抛物线上，且$$\angle AFB=\frac{2}{3} \pi $$，弦$$AB$$中点$$M$$在准线$$l$$上的射影为$${M}'$$，则$$\frac{\textbar M{M}'\textbar}{AB}$$的最大值为．
- $$\frac{4\sqrt{3}}{3}$$ 
- $$\frac{\sqrt{3}}{3}$$ 
- $$\frac{2\sqrt{3}}{3}$$ 
- $$\sqrt{3}$$ 