Eccentricity of ellipse

Question

An ellipse with focuses $F_1$ and $F_2$ passes through $A$, $B$, $C$ and $D$. If $ABF_1CDF_2$ is a regular hexagon, find the eccentricity of the ellipse.

Answer 1

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be an equation of the ellipse, $F_1(-c,0)$, $F_2(c,0)$ and $D$ placed in the first quadrant.

Thus, $DF_2=OF_2=c$, where $O$ is an origin.

Thus, since $\measuredangle F_1F_2D=60^{\circ}$, we see that $D\left(\frac{c}{2},\frac{c\sqrt3}{2}\right)$,

which since $\epsilon=\frac{c}{a}$ and $0<\epsilon<1$, gives $$\frac{c^2}{4a^2}+\frac{3c^2}{4b^2}=1$$ or $$\frac{c^2}{4a^2}+\frac{3c^2}{4(a^2-c^2)}=1$$ or $$\frac{\epsilon^2}{4}+\frac{3\epsilon^2}{4(1-\epsilon^2)}=1$$ or $$\epsilon^4-8\epsilon^2+4=0$$ or $$\epsilon^2=4-2\sqrt3$$ or $$\epsilon^2=(\sqrt3-1)^2$$ or $$\epsilon=\sqrt3-1.$$ Done!

Answer 2

Angle $\angle F_1AF_2$ is right and $\angle AF_1F_2=60$ so: $$\epsilon=\frac{F_1F_2}{F_1A+AF_2}=\frac{2c}{c(1+\sqrt3)}=\frac{2(\sqrt3-1)}{2}=\sqrt3-1$$