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Problem 345

AMC 12 late, AIME early

Geometry Difficulty 4.6 Find the answer

7. Let $A_{1}$ and $A_{2}$ be the left and right vertices of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ . If there exists a point $P$ on the ellipse, different from $A_{1}$ and $A_{2}$ , such that $\overrightarrow{P O} \cdot \overrightarrow{P A_{2}}=0$ , where $O$ is the origin, then the range of the eccentricity $e$ of the ellipse is ( ).
(A) $\left(0, \frac{1}{2}\right)$
(B) $\left(0, \frac{\sqrt{2}}{2}\right)$
(C) $\left(\frac{1}{2}, 1\right)$
(D) $\left(\frac{\sqrt{2}}{2}, 1\right)$

Your answer

Official solution

7. D.

From the given, we know $\angle O P A_{2}=90^{\circ}$ .
Let $P(x, y)(x>0)$ .
The equation of the circle with $\mathrm{OA}_{2}$ as its diameter is
$\left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{a^{2}}{4} \text {, }$

Combining this with the ellipse equation, we get
$\left(1-\frac{b^{2}}{a^{2}}\right) x^{2}-a x+b^{2}=0 \text {. }$

From the given, we know that this equation has real roots in $(0, a)$ .
This leads to $0\frac{1}{2} \text {. }

Therefore, the range of $e$ is $\left(\frac{\sqrt{2}}{2}, 1\right)$ .

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