Question:

Step 1: Set up the geometric configuration and identify key relationships. Since point $B$ is symmetric to point $A$ about the origin $O$, we have $B = -A$. Let $F = (c, 0)$ be the right focus of the ellipse. We are given that $AF \perp BF$, which means $\angle AFB = 90°$. Step 2: Express distances using the focal radius formula. For a point on an ellipse, the distance to the right focus is given by the focal radius formula. For point $A$ on the ellipse: $$|AF| = a - ex_A$$ For point $B = (-x_A, -y_A)$, the distance to the right focus $F = (c, 0)$ is: $$|BF| = a + ex_A$$ Let $r_1 = |AF| = a - ex_A$ and $r_2 = |BF| = a + ex_A$. Step 3: Use the right angle property to establish a constraint. In right triangle $AFB$ with the right angle at $F$, the hypotenuse is $AB$. Since $O$ is the midpoint of $AB$ (as $A$ and $B$ are symmetric about $O$), the median from $F$ to the hypotenuse equals half the hypotenuse length. This gives us: $$|OF| = |OA|$$ Therefore: $$c^2 = |OA|^2$$ Step 4: Calculate $|OA|^2$ in terms of $x_A$. Since $A = (x_A, y_A)$ is on the ellipse: $$y_A^2 = b^2\left(1 - \frac{x_A^2}{a^2}\right)$$ Thus: $$|OA|^2 = x_A^2 + y_A^2 = x_A^2 + b^2 - \frac{b^2x_A^2}{a^2} = b^2 + x_A^2\left(\frac{a^2-b^2}{a^2}\right) = b^2 + \frac{c^2x_A^2}{a^2}$$ Step 5: Solve for $x_A^2$ using the constraint $c^2 = |OA|^2$. From $c^2 = b^2 + \frac{c^2x_A^2}{a^2}$: $$c^2 - b^2 = \frac{c^2x_A^2}{a^2}$$ Since $c^2 = a^2 - b^2$, we have $c^2 - b^2 = a^2 - 2b^2$. Therefore: $$x_A^2 = \frac{a^2(a^2-2b^2)}{c^2}$$ Step 6: Express the parameter $u = \frac{ex_A}{a}$ in terms of eccentricity. $$u^2 = \frac{e^2x_A^2}{a^2} = \frac{e^2(a^2-2b^2)}{c^2}$$ Since $b^2 = a^2 - c^2$: $$u^2 = \frac{e^2(a^2 - 2(a^2-c^2))}{c^2} = \frac{e^2(2c^2-a^2)}{c^2} = 2e^2 - 1$$ Thus $u = \sqrt{2e^2-1}$ (taking the positive root). Step 7: Relate the angle $\angle ABF$ to the eccentricity. In right triangle $AFB$: $$\tan(\angle ABF) = \frac{|AF|}{|BF|} = \frac{r_1}{r_2} = \frac{a-ex_A}{a+ex_A} = \frac{1-u}{1+u}$$ Let $\theta = \angle ABF \in \left[\frac{\pi}{6}, \frac{\pi}{4}\right]$, so $\tan\theta \in \left[\frac{1}{\sqrt{3}}, 1\right]$. Step 8: Find the eccentricity range by analyzing the tangent function. Let $v = \sqrt{2e^2-1}$. Then $\tan\theta = \frac{1-v}{1+v}$. **When $\theta = \frac{\pi}{4}$:** $\tan\theta = 1$ $$\frac{1-v}{1+v} = 1 \implies v = 0 \implies e = \frac{\sqrt{2}}{2}$$ **When $\theta = \frac{\pi}{6}$:** $\tan\theta = \frac{1}{\sqrt{3}}$ $$\frac{1-v}{1+v} = \frac{1}{\sqrt{3}} \implies \sqrt{3}(1-v) = 1+v$$ $$\sqrt{3} - \sqrt{3}v = 1 + v \implies v = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$$ Then: $$v^2 = (2-\sqrt{3})^2 = 7-4\sqrt{3}$$ $$2e^2 - 1 = 7-4\sqrt{3} \implies e^2 = 4-2\sqrt{3} = (\sqrt{3}-1)^2$$ $$e = \sqrt{3}-1$$ Step 9: State the final answer. As $\theta$ increases from $\frac{\pi}{6}$ to $\frac{\pi}{4}$, the eccentricity $e$ decreases from $\sqrt{3}-1$ to $\frac{\sqrt{2}}{2}$. Therefore, the range of eccentricity is: $$e \in \left[\