Competition problem — Geometry — Maths Olympiad Prep

Library / /358 of 520

Geometry Difficulty 3.6 AMC 10/12 Find the answer

Given the parabola $y^{2}=4x$ with focus $F$ and its directrix intersects the $x$ -axis at point $H$ . If point $P$ is on the parabola and $|PH| = \sqrt{2}|PF|$ , then the $x$ -coordinate of point $P$ is __________.

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Analysis

This question examines the properties of a parabola. Let the distance from $P$ to the directrix be equal to $PM$ . According to the definition of a parabola, we have $|PF|=|PH|$ . Given that $|PH|= \sqrt{2}|PF|$ , it follows that $\triangle PMH$ is an isosceles right triangle. Let point $P$ be $( \frac{{y_{0}}^{2}}{4},y_{0} )$ . Since the equation of the directrix is $x=-1$ and $|PM|=|MH|$ , we have $\frac{{y_{0}}^{2}}{4}+1=y_{0}$ . Solving this, we get $y_{0}=±2$ , which leads to the result.

Solution

Let the distance from $P$ to the directrix be equal to $PM$ . According to the definition of a parabola, we have $|PF|=|PH|$ . Given that $|PH|= \sqrt{2}|PF|$ , it follows that $\triangle PMH$ is an isosceles right triangle. Let point $P$ be $( \frac{{y_{0}}^{2}}{4},y_{0} )$ .

Since the equation of the directrix is $x=-1$ and $|PM|=|MH|$ , we have $\frac{{y_{0}}^{2}}{4}+1=y_{0}$ .

Therefore, $y_{0}=±2$ ,

Thus, $P (1,±2)$ ,

Hence, the $x$ -coordinate of point $P$ is $1$ .

Therefore, the answer is $\boxed{1}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.