Eclipsed Ellipse

The centre of the ellipse has coordinates $(0,\tfrac52)$ , and the foci are $F_1\;:\; \big(2e,\tfrac12(5-3e)\big)$ and $F_2\;:\; \big(-2e,\tfrac12(5 + 3e)\big)$ . If $P$ is the given point on the ellipse, then $F_1P \; = \; \sqrt{(2e-1)^2 + \tfrac14(1+3e)^2} \; = \; \tfrac12\sqrt{25e^2 - 10e + 5} \hspace{2cm} F_2P \; = \; \sqrt{(2e+1)^2 + \tfrac14(1-3e)^2} \; = \; \tfrac12\sqrt{25e^2 + 10e + 5}$ The distance between the points $(-2,4)$ and $2,1)$ is $5$ , so we need to solve $\begin{aligned} \sqrt{25e^2 - 10e + 5} + \sqrt{25e^2 + 10e + 5} & = \; 10 \\ 25e^2 + 10e + 5 & = \; 25e^2 - 10e + 5 - 20\sqrt{25e^2 - 10e + 5} + 100 \\ \sqrt{25e^2 - 10e + 5} & = \; 5 - e \\ 25e^2 - 10e + 5 & = \; e^2 - 10e + 25 \end{aligned}$ and hence $e^2 = \tfrac56$ , so that $e = \boxed{0.9128709292}$ .