Ellipse and Normals

Standard bookwork tells us that the region $A$ is bounded by the evolute of the ellipse $\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} = 1$ , which has equation $\big(a|x|\big)^{\frac23} + \big(b|y|\big)^{\frac23} \; = \; (a^2 - b^2)^{\frac23}$ assuming that $a > b$ . Every point inside $A$ can have $4$ normals drawn from it to the ellipse. Since the region $A$ is a stretched version of an astroid, the area of $A$ is $\tfrac{1}{ab} \times \tfrac38\pi(a^2-b^2)^2$ and since this area is the same as that of the ellipse, we obtain $\begin{aligned} 3(a^2-b^2)^2 & =\; 8a^2b^2 \\ 3a^4e^4 & = \; 8a^4(1-e^2) \\ 3e^4 + 8e^2 - 8 & = \; 0 \end{aligned}$ so that $e^2 \; = \; \frac{2\sqrt{10} - 4}{3} \; = \; \frac{4}{2+\sqrt{10}}$ and hence $\boxed{e \; = \; \frac{2}{\sqrt{2+\sqrt{10}}}}$