Find the eccentricity! - Part 5

By the properties of an ellipse, $PF_1 + PF_2 = 2a_1$ , and by the properties of a hyperbola, $PF_1 - PF_2 = 2a_2$ . These equations combine to $PF_1 = a_1 + a_2$ and $PF_2 = a_1 - a_2$ . Also by the properties of an ellipse and hyperbola, $F_1F_2 = 2a_1e_1 = 2a_2e_2$ .

By Pythagorean's Theorem on $\triangle PF_1F_2$ , $(a_1 + a_2)^2 + (a_1 - a_2)^2 = (2a_1e_1)^2$ . Substituting $a_2 = \frac{a_1e_1}{e_2}$ and rearranging eliminates $a_1$ and gives $e_2 = \frac{e_1}{\sqrt{2e_1^2 - 1}}$ .

For any ellipse, $e_1 < 1$ . For the circle to intersect the ellipse, its radius $e_1a_1$ must be larger than $b_1$ , so $e_1a_1 > b_1$ , or $e_1^2a_1^2 > a_1^2 - e_1^2a_1^2$ , which solves to $e_1 > \frac{\sqrt{2}}{2}$ . Therefore, the range of $e_1$ is $\frac{\sqrt{2}}{2} < e_1 < 1$ .

Since $e_2 = \frac{e_1}{\sqrt{2e_1^2 - 1}}$ , $S = e_1^2 + 9e_2^2 = e_1^2 + \frac{9e_1}{\sqrt{2e_1^2 - 1}} = \frac{18e_1^4 - 8e_1^2}{2e_1^2 - 1}$ . This has a minimum value when $S' = \frac{8e_1(9e_1^4 - 9e_1^2 + 2)}{(2e_1^2 - 1)^2} = 0$ , which for $\frac{\sqrt{2}}{2} < e_1 < 1$ is $e_1 = \sqrt{\frac{2}{3}}$ .

Therefore, the minimum value of $S$ is $S = \frac{18e_1^4 - 8e_1^2}{2e_1^2 - 1} = \frac{18(\sqrt{\frac{2}{3}})^4 - 8(\sqrt{\frac{2}{3}})^2}{2(\sqrt{\frac{2}{3}})^2 - 1} = \boxed{8}$ .