Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.7 AMC 10/12 Find the answer China

Suppose points $F_1$ , $F_2$ are the foci of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ , $P$ is a point on the ellipse, and $|PF_1| : |PF_2| = 2 : 1$ . Then the area of $\triangle PF_1F_2$ is equal to ________.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

$|PF_1| + |PF_2| = 2a = 6$ by definition of an ellipse. Since $|PF_1| : |PF_2| = 2 : 1$ , then $|PF_1| = 4$ and $|PF_2| = 2$ . Notice that $|F_1F_2| = 2c = 2\sqrt{5}$ , and
$|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2.$
Then $\triangle PF_1F_2$ is a right triangle. So $S_{\triangle PF_1F_2} = \frac{1}{2} |PF_1| \cdot |PF_2| = 4$ .

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