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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?

Willie Taylor
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    Next time you ask a question, it would be better if you could show how you got the focus. Even though your answer is correct for this problem, if it were incorrect we would have a hard time helping you. – Toby Mak Jun 28 '20 at 08:16

2 Answers2

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

Alexey Burdin
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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 $-1$.

Toby Mak
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