Showing that certain points lie on an ellipse I have the equation
$$r(\phi) = \frac{es}{1-e \cos{\phi}}$$
with $e,s>0$, $e<1$ and want to show that the points
$$ \begin{pmatrix}x(\phi)\\y(\phi)\end{pmatrix} = \begin{pmatrix}r(\phi)\cos{\phi}\\r(\phi)\sin{\phi}\end{pmatrix}$$
lie on an ellipse.
I have a basic and a specific problem which may be related. The basic problem is that I don't see how to get to the goal. 
I want to arrive at an ellipse equation $(x/a)^2 + (y/b)^2 = 1$ and am confused that we already have
$$\left(\frac{x(\phi)}{r(\phi))}\right)^2 + \left(\frac{y(\phi)}{r(\phi))}\right)^2 = 1$$
This can't be the right ellipse equation because it has the form of a circle equation and obviously, we don't have a circle here. But how should the equation I aim for look like then?
I started calculating anyway and arrived at a specific problem. Substituting $x(\phi)$ into $r(\phi)$, I got $r(\phi)=e(s+x(\phi))$ and using $r(\phi)^2 = x(\phi)^2 + y(\phi)^2$ I got an equation which doesn't involve $r(\phi)$ anymore but has $x(\phi)$ in addition to $x(\phi)^2, y(\phi)^2$. 
If I could eliminate the linear term somehow, the resulting equation is an ellipse equation. But I don't see a way how to accomplish this.

Your work seems correct. I suppose that from $x^2+y^2=e^2(s+x)^2$ you find:
$x^2(1-e^2)+y^2-2se^2x-e^2s^2=0$, and this is an equation of the form:
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
$$
that represents a conic section, and it is an ellipse if $B^2-4AC<0$, as it is easely verified since $0<e<1$.
