Show path of the particle is hyperbolic The equation for the path of the particle in cartesian coordinates is $$(1-e^2)x^2+y^2+2exp=p^2$$
where $p$ and $e$ are constant. Under the condition that $0<e<1$, I have shown that this equation describes an elliptical path. Now I'm looking at the situation when $e>1$. This means $0>(1-e^2)$, and dividing with this number gives $$x^2 + \frac{y^2}{(1-e^2)}+\frac{2exp}{(1-e^2)}=\frac{p^2}{(1-e^2)}$$
I complete the square by adding and reducing $(\frac{1}{2}\frac{2exp}{(1-e^2)})^2$. I have $$x^2+\bigg(\frac{exp}{(1-e^2)}\bigg)^2-\bigg(\frac{exp}{(1-e^2)}\bigg)^2 + \frac{y^2}{(1-e^2)}+\frac{2exp}{(1-e^2)}=\frac{p^2}{(1-e^2)}$$
and completing the square gives $$\bigg(x+\bigg(\frac{exp}{(1-e^2)}\bigg)\bigg)^2+\frac{y^2}{(1-e^2)}-\bigg(\frac{exp}{(1-e^2)}\bigg)^2=\frac{p^2}{(1-e^2)}$$
Because I divide $y^2$ with a negative number, I could change the sign in front of $y^2$. Then with a bit of algebra, the RHS can be made equal to one. At that point I would have an equation in ( at least resembling ) the form $$\frac{x^2 -y^2}{a} = 1$$
which describes a hyperbola. Is this correct? The difference between this and the elliptic case is that $(1-e^2)$ is now negative, and it seems natural that this would affect the sign of $y^2$, which in turn changes the path into a hyperbolic one.

$\small (1-e^2)x^2+y^2 + 2 e p x = p^2$ can be re-written as,
$\displaystyle \small x^2 + \frac{2 e p x}{1-e^2} + \frac{y^2}{1-e^2} = \frac{p^2}{1-e^2}$
$\displaystyle \small (x + \frac{e p}{1-e^2})^2 + \frac{y^2}{1-e^2} = \frac{p^2}{1-e^2} + \frac{e^2p^2}{(1-e^2)^2} = \frac{p^2}{(1-e^2)^2}$
$\displaystyle \small \frac{(x + e p / (1-e^2))^2}{(p / (1-e^2))^2} + \frac{y^2}{(p / \sqrt{1-e^2})^2} = 1$ for $0 \lt e \lt 1$ ...(i)
or $\displaystyle \small \frac{(x - e p / (e^2-1))^2}{(p / (e^2-1))^2} - \frac{y^2}{(p / \sqrt{e^2 -1})^2} = 1$ for $e \gt 1$ ...(ii)
In fact for (ii), we can confirm this is a hyperbola with eccentricity $e$.
Eccentricity of hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ is given by $\sqrt{1+\frac{b^2}{a^2}}$.
Here $a = \frac{p}{e^2-1}, b = \frac{p}{\sqrt{e^2-1}} \ $ so $\sqrt{1+\frac{b^2}{a^2}} = e$.
You can similarly confirm that for $0 \lt e \lt 1$, we get an ellipse, again with eccentricity $e$.
Please note $e = 1$ gives us parabola.
