Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.5 AMC 10/12 Find the answer China

In a plane rectangular coordinate system $xOy$ , given parabola $\Gamma: y^2 = 2px$ ( $p > 0$ ), a line with inclination angle $\frac{\pi}{4}$ intersects $\Gamma$ at point $P(3, 2)$ and another point $Q$ . Then the area of $\triangle OPQ$ is ______.

Your answer

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

Solution

Since point $P(3, 2)$ is on $\Gamma$ , we have $2p = \frac{4}{3}$ .
The slope of the mentioned line is $1$ and it passes through point $P(3, 2)$ . Therefore, its equation is $y = x - 1$ , and it passes through point $A(1, 0)$ on the $x$ -axis. Substitute $x = y + 1$ into $y^2 = \frac{4}{3}x$ , eliminating $x$ and arranging it gives
$3y^2 - 4y - 4 = 0.$
The solution is $y_1 = 2$ , $y_2 = -\frac{2}{3}$ . Further, we can get
$S_{\triangle OPQ} = S_{\triangle OAP} + S_{\triangle OAQ} = \frac{1}{2}|OA|(|y_1| + |y_2|) = \frac{4}{3}.$

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