Show path of the particle is hyperbolic

Question

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.

Answer 1

$\small (1-e^2)x^2+y^2 + 2 e p x = p^2$ 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.

Answer 2

Its easier to proceed as follows : $$y^2 - (e^2-1)x^2 + 2epx = p^2$$ $$y^2 - \left[ (e^2-1)x^2 - 2epx + \ldots \right] + \ldots= p^2$$ $$y^2 - \left[ (e^2-1)x^2 -2\cdot x\sqrt{e^2-1} \cdot \frac{ep}{\sqrt{e^2-1}} + \frac{e^2p^2}{e^2-1}\right] + \frac{e^2p^2}{e^2-1} = p^2$$

The middle term in the brackets is the $2ab$ term of $a^2-2ab+b^2$ and is written in such a way that it equals $2epx$ term in first statement. Can you take it from here?

Answer 3

I am surprized that no answer considers focus-directrix definition of a conic curve (see figure below).

Indeed, the given equation can be written under the form:

$$x^2+y^2=(ex-p)^2$$

Otherwise said:

$$\dfrac{\sqrt{x^2+y^2}}{|x-\tfrac{p}{e}|}=e \ \ \iff \ \ \dfrac{MO}{MH}=e$$

which is the definition of a hyperbola (with a focus situated at the origin) as the set of points $M$ such that the ratio of distances from a point, the focus $O$, and a vertical line (directrix $\Delta$ with equation $x=\tfrac{p}{e}$), is constant and equal to the eccentricity $e$.

For this figure, I have taken $e=2$ and $p=1$.

Remark: An advantage of this solution is that it works without any modification for cases $0<e<1$ (ellipse) and $e=1$ (parabola).