Conic Section
Let's start with a circle
$$
r^2 =a^2
$$
If we look at this on a rectangular grid it decomposes like
$$
x^2 + y^2 =a^2
$$
or
$$
\left(\frac{x}{a}\right)^2+ \left(\frac {y}{a}\right)^2 = 1
$$
Ellipse
We can deform this into an ellipse by scaling the $x$ and $y$ axes independently$$
\left(\frac{x}{a}\right)^2+ \left(\frac {y}{b}\right)^2 = 1
$$
With this form for the ellipse, the center and the major and minor axes are exposed clearly, but the focus is not. Let's look at another description of the ellipse, this time keeping the focus in focus. Begin again with the circle, $r=l$. Now, we deform it into an ellipse like so:
\begin{equation} \label{eq1}
r=\frac{l}{1+\epsilon\cos\theta}
\end{equation}
If $\epsilon = 0$ this is a circle, and if $\epsilon$ is a small number less than $1$, this describes an ellipse centered at the focus.
The distance $r$ goes through extrema which are exposed easily if we invent a new parameter $a$, and rewrite as
\begin{equation} \label{eq2}
r=\frac{a(1+\epsilon)(1-\epsilon)}{1+\epsilon\cos\theta}
\end{equation}
When $\theta =0$, the denominator is $1+\epsilon$, and $r$ achieves a minimum
$$
r_<=a(1-\epsilon)
$$
When $\theta =\pi$, the denominator is $1-\epsilon$, and $r$ achieves a maximum
$$
r_>=a(1+\epsilon)
$$
These are the apsidal distances. The semi-major axis, then, is
$$
a=\frac{r_> + r_<}{2}
$$
The focal distance is $a-r_<$, or
$$
\frac{r_> + r_<}{2} - r_< = \frac{r_> - r_<}{2} = \epsilon a \equiv c
$$
It's interesting to compare $a^2$ with $c^2$
$$
a^2 = \frac{1}{4} ({r_<}^2 + {r_>}^2 + 2 r_<r_>)
$$
$$
c^2 = \frac{1}{4} ({r_<}^2 + {r_>}^2 - 2r_<r_>)
$$
If we define a new parameter $b$ by
$$
b^2 \equiv a^2 - c^2 = r_<r_>
$$
then we have $a^2 = b^2 + c^2$, a right triangle.
Apparently, $a$ is the arithmetic average of the apsidal distances, and $b$ is their geometric average.
The form \eqref{eq1} for the ellipse handles the limiting case of the circle, as we have seen, but also dials smoothly into the hyperbolic case, scooping up the parabolic case along the way.
\begin{align*}
\epsilon &= 0 \text{ circle}\\
\epsilon &>0 \text{ ellipse}\\
\epsilon &= 1 \text{ parabola}\\
\epsilon &>1 \text{ hyperbola}\\
\end{align*}
Hyperbola
One might wonder how the analysis that began with \eqref{eq2} would go through for the hyperbolic case. For that scenario we would, since $\epsilon >1$, invent a parameter $a$ and rewrite \eqref{eq1} like so\begin{equation}
r=\frac{a(\epsilon+1)(\epsilon-1)}{1+\epsilon\cos\theta}
\end{equation}
Evidently, if $\theta = 0$, $r$ reaches a minimum of $r_<=a(\epsilon -1)$, while if $\cos\theta = -\frac{1}{\epsilon}$, $r$ goes bananas. When $\theta$ reaches $\pi$, $r=r_>=a(\epsilon+1)$. The midpoint between $r_<$ and $r_>$ is
$$
\frac{r_> - r_<}{2} = \frac {a (\epsilon +1) - a (\epsilon -1)}{2} = a
$$
while the focal distance is
$$
a + r_< = \frac {a (\epsilon +1) + a (\epsilon -1)}{2} = a \epsilon \equiv c
$$
Similar to the ellipse case, if we define a new parameter $b$ by
$$
b^2 \equiv c^2 -a^2 = r_<r_>
$$
then we have $a^2 + b^2 = c^2$, a right triangle.
Orbits
The motion of a mass in an inverse-square central force is described by a five-parameter family of curves. You probably won't be surprised to learn that these curves are in fact the conic sections reviewed above. After the plane (two parameters) and orientation (one parameter) are fixed, only two free parameters of the dynamics are available to describe the shape of the orbit, which will be some circle, ellipse, parabola, or hyperbola -- for instance the $a$ and $b$ used in the previous section.A quick way to relate the parameters controlling the dynamics, to the parameters controlling the orbit is to consider the energy equation.
$$
E = \frac{m}{2} \dot{r}^2 + \frac{l^2}{2mr^2} - \frac{k}{r}
$$
Since the turning points of the motion describe the apsidal distances, we can set $\dot{r}$ to $0$ and read off the two apsides.
$$E = \frac{l^2}{2mr^2} - \frac{k}{r}$$
$$Er^2 +kr - \frac{l^2}{2m} = 0$$
$$
r = -\frac{k}{2E} \pm \frac{\sqrt{k^2 + \frac{4El^2}{2m}}}{2E}
$$
$$
r = -\frac{k}{2E} \pm \frac{k}{2E}\sqrt{1 + \frac{2El^2}{mk^2}}
$$
Since the apsides are $a-c$ and $a+c$ (for the ellipse) we must have
$$ a= -\frac{k}{2E}$$
and
$$c= a\epsilon = -\frac{k}{2E}\sqrt{1 + \frac{2El^2}{mk^2}}$$
so that the major axis depends solely on the energy, and the focal distance depends on both the energy and the angular momentum.
