Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.3 AIME Find the answer China

Let a line with the inclination angle of $60^\circ$ 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:

Pick one

Solution

It follows from the property of the focus of a parabola that $F = (0, 0)$ . Then the equation of the straight line through points $A$ and $B$ will be $y = \sqrt{3}x$ . Substitute it into the parabola equation, and then obtain
$3x^2 - 8x - 16 = 0.$
Let $E$ be the midpoint of the chord $AB$ , then the $x$ -coordinate of $E$ is $\frac{4}{3}$ . Then we have $|FE| = \frac{1}{\cos 60^\circ} \times \frac{4}{3} = \frac{8}{3}$ , $|PF| = 2|FE| = \frac{16}{3}$ . Answer: A.

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