Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it China

In a plane rectangular coordinate system $xOy$ , points $A$ , $B$ are on the parabola $y^2 = 4x$ , satisfying $\vec{OA} \cdot \vec{OB} = -4$ , and point $F$ is the focus of the parabola. Then $S_{\triangle OFA} \cdot S_{\triangle OFB} = \_\_\_\_\_\_$ .

Solution

Let $F(1, 0)$ , $A(x_1, y_1)$ , $B(x_2, y_2)$ . Then $x_1 = \frac{y_1^2}{4}$ , $x_2 = \frac{y_2^2}{4}$ , and
$-4 = \vec{OA} \cdot \vec{OB} = x_1x_2 + y_1y_2 = \frac{1}{16}(y_1y_2)^2 + y_1y_2,$
from which we have $\frac{1}{16}(y_1y_2 + 8)^2 = 0$ , or $y_1y_2 = -8$ .
Therefore,
$\begin{align*} S_{\triangle OFA} \cdot S_{\triangle OFB} &= \left( \frac{1}{2} |OF| \cdot |y_1| \right) \cdot \left( \frac{1}{2} |OF| \cdot |y_2| \right) \\ &= \frac{1}{4} \cdot |OF|^2 \cdot |y_1 y_2| = 2. \end{align*}$
The answer is 2.

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