Track / Stage 4 / 329 of 340 #589 of 1964

Problem 589

AMC 12 late, AIME early

Geometry Difficulty 5.0 Multiple choice

4. Given $F_{1}, F_{2}$ are the left and right foci of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ , and there exist two points $A, B$ on the ellipse such that $\overrightarrow{F_{1} A}=3 \overrightarrow{F_{2} B}$ , then the range of the eccentricity of the ellipse is

Pick one

Official solution

Extend $A F_{1}$ to intersect the ellipse at point $C$ . By symmetry, we know $\left|F_{1} C\right|=\left|F_{2} B\right|$ . When $\left|F_{1} A\right|$ is maximized, $\left|F_{1} C\right|$ is minimized. Thus, $a+c>3(a-c) \Rightarrow 2 c>a \Rightarrow e \in\left(\frac{1}{2}, 1\right)$ , so the answer is $C$ .

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