The Great Ellipse Eclipse

[S]uppose you have two congruent ellipses that, together, cover a unit circle (i.e., a circle with radius $1$). These ellipses can have any eccentricity you like, but they must be congruent to each other. What is the smallest possible area one of these ellipses can have, such that they completely cover the circle?

By symmetry (both rotational and optimization-al?), let's assume that we have two congruent ellipses that have parallel major axis along the $y$-axis and are centered at points $(c,0)$ and $(-c,0)$ for some $c > 0.$ That is, let the equations for the ellipses be \begin{align*} \frac{(x-c)^2}{a^2} + \frac{y^2}{b^2} &= 1 \\ \frac{(x+c)^2}{a^2} + \frac{y^2}{b^2} &= 1.\end{align*}

In order to ensure that these two ellipses cover the entire unit circle we would need to make sure that they cover the extreme points, $(\pm 1, 0)$ and $(0, \pm 1)$ (and for good measure that the origin is contained at least one of them). In order for the origin to be contained we can plug in $x=y=0$ and get $$\frac{c^2}{(1-c)^2} \leq 1,$$ or equivalently $2c - 1 \leq 0,$ or $c \leq \frac{1}{2}.$ In order for $(1,0)$ to be contained in the ellipse centered at $(c,0)$ we must have $$\frac{(1-c)^2}{a^2} + \frac{0}{b^2} = 1,$$ or equivalently $a = 1 - c.$ In order for (0,1) to be contained in either ellipse we must have $$\frac{c^2}{(1-c)^2} + \frac{1}{b^2} = 1$$ or equivalently $$b^2 = \frac{1}{1 - \frac{c^2}{(1-c)^2}} = \frac{(1-c)^2}{(1-c)^2 - c^2} = \frac{(1-c)^2}{1-2c},$$ that is, $b = \frac{1-c}{\sqrt{1-2c}}.$

The area of one of the ellipses is $$A(c) = \pi a(c)b(c) = \pi \frac{(1-c)^2}{\sqrt{1-2c}}.$$ Differentiating we get \begin{align*}\frac{\partial}{\partial c} A(c) &= \pi \left(\frac{(1-c)^2}{(1-2c)^{3/2}} + \frac{2(c-1)}{\sqrt{1-2c}} \right)\\ &= \pi\frac{ (1-c)^2 - 2 (1-2c) (1-c) }{(1-2c)^{3/2}}\\ &= \frac{\pi(1-c)(3c-1)}{(1-2c)^{3/2}}.\end{align*} So the unconstrained critical points of $A$ are $c = 1$ and $c = \frac{1}{3}.$ Since we want $c \leq \frac{1}{2},$ we can eliminate the critical point $c = 1,$ and instead need to just test the critical points $c \in \left\{ 0, \frac{1}{3}, \frac{1}{2} \right\}.$ We have $A(0) = \pi,$ and $\lim_{c \uparrow \frac{1}{2}} A(c) = +\infty$ and $$A\left(\frac{1}{3}\right) = \pi \frac{ \frac{4}{9} }{ \sqrt{\frac{1}{3} }} = \frac{4\pi \sqrt{3}}{9} \approx 2.418399....,$$ which is the smallest area of one ellipse of a congruent pair that can cover the unit circle. For completeness, those ellipses are \begin{align*} \frac{ (3x - 1)^2 }{4} + \frac{3y^2}{4} &= 1 \\ \frac{(3x+1)^2}{4} + \frac{3y^2}{4} &= 1\end{align*}