Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.4 AMC 10/12 Find the answer

Given that point $P$ is on the ellipse $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$ , and the area of the triangle formed by point $P$ and the foci $F_{1}$ and $F_{2}$ is equal to $1$ . Find the coordinates of point $P$ .

Your answer

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

Solution

$F_{1}$ and $F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{5} + \frac{y^{2}}{4} = 1$ , respectively. We have $c = \sqrt{5-4} = 1$ . So, the coordinates of the foci are $F_{1}(-1,0)$ and $F_{2}(1,0)$ .

Let $P(x,y)$ be a point on the ellipse. By the formula for the area of a triangle, we know that $S = \frac{1}{2} \cdot 2c \cdot |y| = 1$ , which implies that $|y| = 1$ .

Substituting $|y| = 1$ into the equation of the ellipse, we have $\frac{x^{2}}{5} + \frac{1}{4} = 1$ . Solving for $x$ gives us $|x| = \frac{\sqrt{15}}{2}$ .

Therefore, the coordinates of point $P$ are $\boxed{( \frac{\sqrt{15}}{2}, 1 ), ( -\frac{\sqrt{15}}{2}, 1 ), ( -\frac{\sqrt{15}}{2}, -1 ), ( \frac{\sqrt{15}}{2}, -1 )}$ .

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