Trying to show that $r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right)$ is an ellipse I am trying to prove that planets move in ellipses,
I watched this video: https://www.youtube.com/watch?v=DurLVHPc1Iw
and read this: https://arxiv.org/pdf/1009.1738.pdf .
But both sources end up with this as an equation for the paths that planets follow :
$$
 r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right),
 $$
where $C=G\frac{m_{1}+m_{2}}{h^{2}}$ which is just a constant, as is $\omega$, and $e$ is the eccentricity, so this is an ellipse with the focus at the origin as long as $e<1$ and when I put the equation in graphing software it gives me ellipse.
But I don't understand why it is an ellipse? Can anyone show me how it is an ellipse? Either from the normal polar coordinates of a ellipse centred at the origin and shifting it (I couldn't get the algebra to work out when I added ae to the x coordinate) or just by explaining it conceptionally?

We have
$$
\begin{split}
r&= \dfrac{1}{C} \left(\dfrac{1}{1+e \cos(\theta+\omega)}\right) \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \cdot \dfrac{1/C}{1/C} \\
&= \dfrac{1/C}{1+ e\cos(\theta+\omega)}
\end{split}
$$
Write $P= 1/C$. Then we have
$$
\begin{split}
r= \dfrac{P}{1+e \cos(\theta+\omega)} \\
r+er \cos(\theta+\omega)= P \\
r= P - er \cos(\theta+\omega)
\end{split}
$$
Now let $x= r \cos \theta$. Then noting that in our case, $r^2= x^2+y^2$, we have
$$
\begin{split}
r&= P - er \cos(\theta+\omega) \\
r&= P - ex \\
r^2&= P^2 - 2Pex + e^2 x^2 \\
x^2+y^2&= P^2 - 2Pex + e^2 x^2 \\
x^2 - e^2x^2 +2Pex + y^2&= P^2 \\
(1-e^2)x^2 +2Pex + y^2&= P^2 \\
x^2 + \dfrac{2Pex}{1-e^2} + \dfrac{y^2}{1-e^2}&= \dfrac{P^2}{1-e^2}
\end{split}
$$
Now it is just a routine complete the square in $x$ and we have
$$
\left(x+ \dfrac{Pe}{1-e^2} \right)^2 + \dfrac{y^2}{1-e^2}= \dfrac{P^2}{(1-e^2)^2}
$$
Of course, this is the same as
$$
\dfrac{(x+Pe/(1-e^2))^2}{P^2/(1-e^2)^2} + \dfrac{y^2}{P^2/(1-e^2)} = 1
$$
So that we have an ellipse 'centered' at the point $(-Pe/(1-e^2), 0)$, semimajor axis length $P/(1-e^2)$, semiminor axis $P/\sqrt{1-e^2}$, and with foci at distance
$$
\sqrt{A^2 - B^2}= \sqrt{\dfrac{P^2}{(1-e^2)^2} - \dfrac{P^2}{1-e^2}}= \dfrac{P|e|}{1-e^2}
$$
