Let $S$ and $S'$ be two foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If a circle described on $SS'$ as diameter intersects the ellipse in real and distinct points, then the eccentricity $e$ of the ellipse satisfies.
Step-by-Step Solution
Key Concept: The circle passing through the foci cuts the ellipse when the radius exceeds the semi-minor axis, giving a lower bound on eccentricity.
A circle with diameter $SS'$ (where $S, S'$ are the foci) has radius $r = ae$. For this circle to cut the ellipse, we need $r > b$, so $ae > b \Rightarrow e^2 > \frac{b^2}{a^2}$. Since $b^2 = a^2(1-e^2)$, we get $e^2 > 1 - e^2 \Rightarrow e^2 > \frac{1}{2} \Rightarrow e > \frac{1}{\sqrt{2}}$. Thus $e \in (\frac{1}{\sqrt{2}}, 1)$.
Correct Answer: 2