(4) The parabola $y^{2}=2 p x(p>0)$ has a focus $F$, and a directrix $l$. $A$ and $B$ are two moving points on the parabola, and they satisfy $\angle A F B=\frac{\pi}{3}$. Let the midpoint of segment $A B$ be $M$, and the projection of $M$ on $l$ be $N$. Then the maximum value of $\frac{|M N|}{|A B|}$ is $\qquad$.
(4) 1 Hint: Method one Let $\angle A B F=\theta\left(0<\theta<\frac{2 \pi}{3}\right)$, then by the Law of Sines, we have
$$
\frac{|A F|}{\sin \theta}=\frac{|B F|}{\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}} .
$$

Therefore, $\frac{|A F|+|B F|}{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}=\frac{|A B|}{\sin \frac{\pi}{3}}$,
which means $\frac{|A F|+|B F|}{|A B|}=\frac{\sin \theta+\sin \left(\frac{2 \pi}{3}-\theta\right)}{\sin \frac{\pi}{3}}=2 \cos \left(\theta-\frac{\pi}{3}\right)$.
As shown in the figure, by the definition of a parabola and the midline theorem of a trapezoid, we get $|M N|=\frac{|A F|+|B F|}{2}$.
Thus, $\frac{|M N|}{|A B|}=\cos \left(\theta-\frac{\pi}{3}\right)$.
Therefore, when $\theta=\frac{\pi}{3}$, $\frac{|M N|}{|A B|}$ reaches its maximum value

of 1.
Method two By the definition of a parabola and the midline theorem of a trapezoid, we get $|M N|=\frac{|A F|+|B F|}{2}$.
In $\triangle A F B$, by the Law of Cosines, we have
$$
\begin{aligned}
|A B|^{2} & =|A F|^{2}+|B F|^{2}-2|A F| \cdot|B F| \cos \frac{\pi}{3} \\
& =(|A F|+|B F|)^{2}-3|A F| \cdot|B F| \\
& \geqslant(|A F|+|B F|)^{2}-3\left(\frac{|A F|+|B F|}{2}\right)^{2} \\
& =\left(\frac{|A F|+|B F|}{2}\right)^{2}=|M N|^{2} .
\end{aligned}
$$

Equality holds if and only if $|A F|=|B F|$. Hence, the maximum value of $\frac{|M N|}{|A B|}$ is 1.