Library / China Mathematical Competition /89 of 136

China Mathematical Competition

Geometry Difficulty 5.2 AIME, harder Prove it China

In a plane rectangular coordinate system $xOy$ , the focus of parabola $\Gamma: y^2 = 2px$ ( $p > 0$ ) is $F$ . Make a tangent line to $\Gamma$ passing through point $P$ (different from $O$ ) on $\Gamma$ and it intersects the $y$ -axis at point $Q$ . If $|FP| = 2$ , $|FQ| = 1$ , then the dot product of vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ is ______.

Solution

Let $P(\frac{t^2}{2p}, t)$ ( $t \neq 0$ ), and then the equation of the tangent line of $\Gamma$ is $yt = p(x + \frac{t^2}{2p})$ .
Let $x = 0$ , and we get $yt = \frac{t}{2}$ . The coordinates of $F$ are $(\frac{p}{2}, 0)$ , and thus
$|FP| = \sqrt{\left(\frac{p}{2} - \frac{t^2}{2p}\right)^2 + t^2} = \frac{p}{2} + \frac{t^2}{2p}, \\ |FQ| = \frac{\sqrt{p^2 + t^2}}{2}.$
Combining $|FP| = 2$ , $|FQ| = 1$ , we can get $p^2 + t^2 = 4p$ and $p^2 + t^2 = 4$ , respectively. Hence, $p = 1$ , $t^2 = 3$ .
Therefore, $\overrightarrow{OP} \cdot \overrightarrow{OQ} = \frac{t^2}{2} = \frac{3}{2}$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.