Can You Eclipse via Ellipse

This Week's Fiddler

To cover a unit circle with two congruent ellipses, we can consider an ellipse like the following:

This sufficiently covers a semicircle (allowing two of them to cover the entire circle). It has points intersecting the circle at $(1,0)$, $(0,1)$, and $(0,-1)$. Let's say it is centered at $(c_x,0)$, with vertical axis semidiameter of length $a$ and horizontal axis semidiameter of length $b$. The equation of the ellipse is thus as follows:

$$\frac{(x-c_x)^2}{b^2}+\frac{y^2}{a^2}=1$$

Note that if $b=1$, we get a circle, and if $b=\frac{1}{2}$, we get two vertical lines. So the answer must lie within $b\in(\frac{1}{2},1]$, minimizing for the area.

Substituting the point $(1,0)$, we see that:

$$\frac{(1-c_x)^2}{b^2}=1 \\ c_x=1-b$$

Substituting the point $(0,1)$, we see that:

$$\frac{c_x^2}{b^2}+\frac{1}{a^2}=1 \\ \frac{(1-b)^2}{b^2}+\frac{1}{a^2}=1 \\ \frac{1-2b+b^2}{b^2}+\frac{1}{a^2}=1 \\ \frac{1-2b}{b^2}+\frac{1}{a^2}=0 \\ a^2=\frac{b^2}{2b-1} \\ a=\frac{b}{\sqrt{2b-1}}$$

Thus, the overall area is:

$$A=\pi ab=\pi\biggl(\frac{b^2}{\sqrt{2b-1}}\biggl)$$

Minimizing for $A$ over $b\in(\frac{1}{2},1]$, we find that when $b=\frac{2}{3}$, $A=\frac{4\pi}{3\sqrt3}$ is minimal.

This makes $A\approx0.7698\pi$, which is between $\frac{\pi}{2}$ and $\pi$ as expected.

See: Desmos visualization of the right ellipse as a function of $b$

Answer: 4*pi/3/sqrt(3)

Extra Credit

An ellipse with horizontal semidiameter length of $\cos{(\frac{\pi}{N})}$ and vertical semidiameter length of $\sin{(\frac{\pi}{N})}$ is sufficient to intersect with $(0,0)$, $(\cos{(\frac{\pi}{N})}, \sin{(\frac{\pi}{N})})$, and $(\cos{(\frac{-\pi}{N})}, \sin{(\frac{-\pi}{N})})$, successfully encapsulating a $\frac{1}{N}$ slice of the circle.

Thus, a sufficient area of one ellipse would be:

$$A=\pi\biggl(\cos{\bigl(\frac{\pi}{N}}\bigl)\biggl)\biggl(\sin{\bigl(\frac{\pi}{N}\bigl)}\biggl) \\ A=\frac{\pi}{2}\sin{\biggl(\frac{2\pi}{N}\biggl)}$$

Thus, for $N=3$, the answer would be $\frac{\pi}{2}\sin{(\frac{2\pi}{3})}=\frac{\sqrt3\pi}{4}$.

This makes $A\approx0.4330\pi$, which is above $\frac{\pi}{3}$ as expected.

Answer: sqrt(3)*pi/4 (for all N>2, pi/2*sin(2*pi/N))