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?
Best Answer
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 $$-\frac{1}{\sqrt{3}}(x-\frac{4}{3})=-\frac{4\sqrt{3}}{3}$$ $$x-\frac{4}{3}=4$$ $$FP=x=\frac{16}{3}$$