Track / Stage 5 / 35 of 400 #635 of 2000

Problem 635

AIME late

Geometry Difficulty 5.1 Find the answer

2. In the Cartesian coordinate system $x O y$ , points $A$ and $B$ lie on the parabola $y^{2}=4 x$ , satisfying $\overrightarrow{O A} \cdot \overrightarrow{O B}=-4$ , $F$ is the focus of the parabola. Then $S_{\triangle O F A} \cdot S_{\triangle O F B}=$ $\qquad$ .

Your answer

Official solution

Let $A\left(\frac{y_{1}^{2}}{4}, y_{1}\right), B\left(\frac{y_{2}^{2}}{4}, y_{2}\right)$ , then $\overrightarrow{O A} \cdot \overrightarrow{O B}=\frac{y_{1}^{2} y_{2}^{2}}{16}+y_{1} y_{2}=-4$
$\Rightarrow y_{1}^{2} y_{2}^{2}+16 y_{1} y_{2}+64=0 \Rightarrow y_{1} y_{2}=-8 \text {. }$

Also, $F(1,0)$ , so $S_{\triangle O F A} \cdot S_{\triangle O F B}=\frac{1}{2}\left|y_{1}\right| \cdot \frac{1}{2}\left|y_{2}\right|=\frac{1}{4}\left|y_{1} y_{2}\right|=2$ .

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