Given directrix, eccentricity, and focus get center of ellipse Given
Directrix: $x=2$
Focus: $(0,0)$
Eccentricity: $0.8$
Find the semi major axis $a$.
I can write the cartesian equation $x^2+y^2=e^2(2-x)^2$ and work the center by manipulating it. However I've been looking for a formula for the semi major axis $a$, in terms of eccentricity and directrix when focus is fixed at $(0,0)$.  Any help?


My work:
$x^2+y^2=e^2(x-k)^2=e^2x^2-2e^2kx+e^2k^2 $
$(1-e^2)(x^2+2\frac{e^2k}{1-e^2}x)\cdots$
$\Rightarrow h=-\dfrac{e^2k}{1-e^2}$
I have it XD Is there a better more geometric/clever way?

Let the focus and directrix be $F=(f,0)$ and $x=d$; let $D:=(d,0)$ be the foot of the perpendicular from the focus to the directrix. The points on any conic satisfy 
$$\text{eccentricity}=\frac{\text{distance from focus}}{\text{distance from directrix}} \tag{1}$$
In particular, an endpoint $P:=(p,0)$ of the major axis satisfies
$$e = \frac{|PF|}{|PD|}=\frac{|p-f|}{|p-d|} \quad\to\quad \frac{p-f}{p-d} = \pm e \quad\to\quad p = \frac{f \pm d e}{1 \pm e} \tag{2}$$
The center $H := (h,0)$ is the midpoint of the vertices, so its $x$-coordinate is the average of the two $p$-values, namely,
$$h=\frac{f-de^2}{1-e^2} \tag{3}$$
(which agrees with OP's solution, with $f=0$ and $d=k$). The major radius is half the distance between the vertices, hence the absolute value of half the difference of the $p$-values:
$$a=\left|\frac{(f-d)e}{1-e^2}\right|\tag{4}$$
