3. 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 $A B$ intersects the $x$-axis at point $P$, then the length of segment $P F$ is
(A) $\frac{16}{3}$
(B) $\frac{8}{3}$
(C) $\frac{16 \sqrt{3}}{3}$
(D) $8 \sqrt{3}$
3.【Analysis and Solution】It is easy to know that the focus $F$ of this parabola coincides with the origin, so the equation of line $AB$ is $y=\sqrt{3} x$. Therefore, the x-coordinates of points $A$ and $B$ satisfy the equation $3 x^{2}-8 x-16=0$. From this, we find that the x-coordinate of the midpoint of chord $AB$ is $x_{0}=\frac{4}{3}$, and the y-coordinate is $y_{0}=\frac{4}{\sqrt{3}}$. Thus, the equation of the perpendicular bisector is $y-\frac{4}{\sqrt{3}}=-\frac{1}{\sqrt{3}}\left(x-\frac{4}{3}\right)$. Setting $y=0$, we get the x-coordinate of point $P$ as $x=4+\frac{4}{3}=\frac{16}{3}$, which means $PF=\frac{16}{3}$. Therefore, the answer is (A).