Proving ellipse focus-directrix implies equation On cuttheknot.org, a proof is given that the focus-directrix definition implies the equation definition (i.e. that an ellipse is a planar curve with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$). The first line of the proof states > Let $e$ be the ratio of distances and choose the system of coordinates so that the focus and the related directrix be $E(−ae,0)$ and $x=-\frac{a}{e}$. How this is possible, is not at all clear to me. How can we simply do this, if we do not know $a$? (Link to the full proof, go to 4 implies 2).

Given a focus $F$, directrix $d$ and eccentricity $e\in(0,1)$, the center of the ellipse, and so the origin of the coordinate system posited in the proof, can be found without explicitly computing the semi-major axis length $a$.

![focus/directrix ellipse](

Let $D$ be the intersection of the directrix $d$ and the perpendicular through $F$. ($\overline{DF}$ is, of course, the major axis of the ellipse.) One vertex $A$ is between $F$ and $D$. By definition, $(A-F)=e(D-A)$, so $$A=\frac1{1+e}F+\frac e{1+e}D.\tag1$$ The focus $F$ is between this point and the opposite vertex $A'$. For this other vertex we have $(A'-F)=e(A'-D)$, therefore $$A'=\frac1{1-e}F-\frac e{1-e}D.\tag2$$ The center of the ellipse is just the midpoint $$C=\frac12(A+A')={1\over1-e^2}F-{e^2\over1-e^2}D.\tag3$$ The semi-major axis length $a=\frac12\|A-A'\|$, but once you place the origin at $C$ and align the $x$-axis with the major axis ($\overline{DF}$), $F$ and $D$ will automatically have the correct coordinates.