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.

I'm guessing that the equation you got that "doesn't involve $r(\phi)$ anymore"
is equivalent to this:
$$(1-e^2)(x(\phi))^2 - (2e^2s) x(\phi) + y(\phi)^2 = e^2s^2. \tag 1$$
I'm going to "cheat" a little here: there happens to be a well-known formula,
$$r = \frac{a(1-e^2)}{1 - e \cos{\theta}}, \tag 2$$
for the polar coordinates $(r,\theta)$ of an ellipse with semimajor axis $a$
and eccentricity $e$, with one focus at the origin.
This gives me the idea that I should set a parameter 
$a = \frac{es}{1-e^2},$ because then I will have $es = a(1-e^2).$
The original equation then becomes Equation $(2)$,
and after substituting for $es$, Equation $(1)$ becomes
$$(1-e^2)(x(\phi))^2 - (2a(1-e^2)e)\, x(\phi) + (y(\phi))^2 = a^2(1-e^2)^2. $$
In preparation for the technique of completing the square, it's
helpful for the leading coefficient of the interesting squared variable
($(x(\phi))^2$ in this question) to be a square.
Typically we would want it to be $1$, but in this case
I have a strong hunch we will want to see something like 
$\left(\frac{x - x_0}{a}\right)^2$
in the final equation, so I'll try $\frac{1}{a^2}$ as the leading coefficient.
We get there by dividing both sides by $a^2(1-e^2)$:
$$\frac{x(\phi)^2}{a^2} - 2e\frac{x(\phi)}{a} + \frac{(y(\phi))^2}{a^2(1-e^2)}
 = 1 - e^2. $$
Now to complete the square, we use the fact that $(u+v)^2 = u^2 + 2uv + v^2$
for any $u$ and $v$; let $u^2$ be $\frac{x(\phi)^2}{a^2}$
and $2uv$ be $-2e\frac{x(\phi)}{a}$, that is, $u = \frac{x(\phi)}{a}$
and $v = -e$, so
$$\left(\frac{x(\phi)}{a} - e\right)^2 = 
   \frac{(x(\phi))^2}{a^2} - 2e\frac{x(\phi)}{a} + e^2.$$
So we can substitute $\left(\frac{x(\phi)}{a} - e\right)^2 - e^2$ for
$\frac{(x(\phi))^2}{a^2} - 2e\frac{x(\phi)}{a}.$ Do so:
$$\left(\frac{x(\phi)}{a} - e\right)^2 - e^2 + \frac{(y(\phi))^2}{a^2(1-e^2)}
 = 1 - e^2. $$
We can nicely cancel the constant $e^2$ term on each side:
$$\left(\frac{x(\phi)}{a} - e\right)^2 + \frac{(y(\phi))^2}{a^2(1-e^2)} = 1. $$
It should be reasonably clear already that this is an ellipse,
but if we set $b = a\sqrt{1-e^2}$ and $x_0 = ae,$
we can massage this equation into the form
$$
\left(\frac{x(\phi) - x_0}{a}\right)^2 + \left(\frac{y(\phi)}{b}\right)^2 = 1,
$$
which is the equation for an ellipse whose semi-major axis
(parallel to the $x$ axis) is $a$, whose semi-minor axis is $b$,
and whose center is at $(x_0,0)$.
