Example 5 The hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ has an eccentricity of 2, and the distance from the origin to the line $AB$ is $\frac{\sqrt{3}}{2}$, where $A(0,-b), B(a, 0)$.
(1) Find the equation of the hyperbola;
(2) Let $F$ be the right focus of the hyperbola, and let the line $l$ pass through point $F$ and intersect the right branch of the hyperbola at two distinct points $P$ and $Q$. Let $M$ be the midpoint of segment $PQ$. If the projection of point $M$ on the line $x=-2$ is $N$, and $\overrightarrow{P N} \cdot \overrightarrow{Q N}=0,|\overrightarrow{P Q}|=10$, find the equation of the line $l$.
Solution: (1) Given the equation of $AB$ is $bx - ay - ab = 0$, and from $\left\{\begin{array}{l}\frac{c}{a}=2, \\ \frac{ab}{\sqrt{a^{2}+b^{2}}}=\frac{\sqrt{3}}{2},\end{array}\right.$ we get $a^{2}=1, b^{2}=3$,
$\therefore$ the equation of the hyperbola is $x^{2}-\frac{y^{2}}{3}=1$.
(2) The right focus $F(2,0)$, when $l \perp x$ axis, $|\overrightarrow{PQ}| = 6$ which does not meet the condition.

Assume the equation of $l$ is $y=k(x-2)$, substituting into $x^{2}-\frac{y^{2}}{3} = 1$ gives $(3-k^{2})x^{2} + 4k^{2}x - 4k^{2} - 3 = 0 \cdots \cdots$ (1), $3-k^{2} \neq 0$,
$\Delta=16k^{4} + 4(3-k^{2})(4k^{2} + 3) > 0$,
Let $P(x_{1}, y_{1}), Q(x_{2}, y_{2}), M(x_{0}, y_{0})$,
then $x_{1} + x_{2} = \frac{-4k^{2}}{3-k^{2}} > 0$,
$x_{1} x_{2} = \frac{4k^{2} + 3}{k^{2} - 3} > 0 \Rightarrow k^{2} > 3$
$\because \overrightarrow{PN} \cdot \overrightarrow{QN} = 0, \therefore PN \perp QN$.
Since $M$ is the midpoint of $PQ$, $|\overrightarrow{PQ}| = 10$,
$$
\begin{array}{l}
\therefore |PM| = |MN| = |MQ| = \frac{1}{2}|PQ| = 5. \\
|MN| = x_{0} + 2 = 5, \\
\therefore x_{0} = 3, \text{ and } x_{0} = \frac{x_{1} + x_{2}}{2} = \frac{2k^{2}}{k^{2} - 3} = 3,
\end{array}
$$
$\therefore k^{2} = 9$ satisfies (2), $\therefore k = \pm 3$,
The equation of the line $l$ is $y = \pm 3(x-2)$.