How to find the length if a segment given this problem? Let a line with the inclination angle of 60 degrees be drawn through the focus F of the parabola y^2 = 8(x+2). If the two intersection points of the line and the parabola are A and B, and the perpendicular bisector of the chord AB intersects the x-axis at the point P, then the length of the segment PF is?
My approach:
I try to find the focus of the parabola and got (0,0) but I am now confused on how to find their intersection points. Also on how to find the perpendicular bisector of the chord AB. Can somebody guide me how to solve this?
 A: 
The inclination angle of a line $(y-y_0)=k(x-x_0)$ is always $\arctan(k)$ so we have $AB:\ y=\sqrt{3}x$. Then to find points $A,B$ we are to solve $3x^2=8(x+2)$ $\Leftrightarrow$ $3x^2-8x-16=0$ but in this case there's no need to solve it, because we only need the middle of $AB$:
$M=\left(\frac{x_A+x_B}{2},\frac{y_A+y_B}{2}\right)$. By Vieta's theorem $x_A+x_B=\frac{8}{3}$ and $y_A+y_B=\sqrt{3}(x_A+x_B)=\frac{8\sqrt{3}}{3}$ and the perpendicular bisector equation is then
$$-\frac{1}{\sqrt{3}}(x-\frac{4}{3})=(y-\frac{4\sqrt{3}}{3})$$
Setting $y=0$ we get
$$-\frac{1}{\sqrt{3}}(x-\frac{4}{3})=-\frac{4\sqrt{3}}{3}$$
$$x-\frac{4}{3}=4$$
$$FP=x=\frac{16}{3}$$
A: The slope of the line is $\frac{\text{rise}}{\text{run}}$, or $\frac{\text{opp}}{\text{adj}}$ in a right triangle. This means the slope is equal to $\tan 60º = \sqrt{3}$, and the line is just $\sqrt{3}x$ since it passes through the origin.
Now substitute this into the equation of the parabola $y^2 = 8(x+2)$, which will give you a quadratic equation. This will give you the x-coordinates of A and B. Note that the perpendicular bisector of a line segment passes through the midpoint, and the product of the two slopes is always $-1$.
