9. Given the parabola $y^{2}=2 p x(p>0)$ with focus $F$, and $A$ and $B$ are two moving points on the parabola, satisfying $\angle A F B=120^{\circ}$. A perpendicular line $M N$ is drawn from the midpoint $M$ of chord $A B$ to the directrix of the parabola, with the foot of the perpendicular being $N$. Then the maximum value of $\frac{|M N|}{|A B|}$ is $\qquad$
9. $\frac{\sqrt{3}}{3}$.

According to the problem and the definition of a parabola, we have
$$
\begin{array}{l}
|A F|=\left|A A_{1}\right|,|B F|=\left|B B_{1}\right| \\
\Rightarrow|A F|+|B F|=\left|A A_{1}\right|+\left|B B_{1}\right|=2|M N| \\
\Rightarrow \frac{|M N|}{|A B|}=\frac{|A F|+|B F|}{2|A B|} .
\end{array}
$$

Notice that, in $\triangle A B F$,
$$
\begin{array}{l}
|A B|^{2}=|A F|^{2}+|B F|^{2}-2|A F||B F| \cos 120^{\circ} \\
\Rightarrow|A B|^{2}=|A F|^{2}+|B F|^{2}+|A F||B F| \\
\Rightarrow(|A F|+|B F|)^{2}-|A F||B F|=|A B|^{2} . \\
\text { Also, }|A F||B F| \leqslant \frac{(|A F|+|B F|)^{2}}{4}, \text { then } \\
|A B|^{2} \geqslant \frac{3}{4}(|A F|+|B F|)^{2} \\
\Rightarrow|A F|+|B F| \leqslant \frac{2 \sqrt{3}}{3}|A B| .
\end{array}
$$

Therefore, $\frac{|M N|}{|A B|}=\frac{|A F|+|B F|}{2|A B|} \leqslant \frac{\sqrt{3}}{3}$, with equality holding if and only if $|A F|=|B F|$.