Example 33 (2003 National High School Competition Question) A line with an inclination angle of $60^{\circ}$ is drawn through the focus $F$ of the parabola $y^{2}=8(x+2)$. If this line intersects the parabola at points $A$ and $B$, and the perpendicular bisector of chord $AB$ intersects the $x$-axis at point $P$, then the length of segment $PF$ is ( ).
A. $\frac{16}{3}$
B. $\frac{8}{3}$
C. $\frac{16 \sqrt{3}}{3}$
D. $8 \sqrt{3}$
Solution: Choose A. Reason: By fully utilizing the given information, we know that the focus of the parabola coincides with the origin, so the equation of line $AB$ is $y = \tan 60^{\circ} \cdot x = \sqrt{3} x$.

Therefore, the x-coordinates of points $A$ and $B$ satisfy the equation $3 x^{2} - 8 x - 16 = 0$ (by substituting $y = \sqrt{3} x$ into $y^{2} = 8(x + 2)$). From this, we find that the x-coordinate of the midpoint of chord $AB$ is $x_{0} = \frac{1}{2}(x_{A} + x_{B}) = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$, and the y-coordinate is $y_{0} = \sqrt{3} x_{0} = \frac{4}{\sqrt{3}}$. Consequently, the equation of the perpendicular bisector is $y - \frac{4}{\sqrt{3}} = -\frac{1}{\sqrt{3}}(x - \frac{4}{3})$. Setting $y = 0$, we get the x-coordinate of point $P$ as $x = 4 + \frac{4}{3} = \frac{16}{3}$, which means $PF = \frac{16}{3}$.