simplest common equation of conics
In the plane, the locus of the points having the ratio of their distances^{} from a certain point (the focus) and from a certain line (the directrix^{}) equal to a given constant $\epsilon $, is a conic section, which is an ellipse^{}, a parabola (http://planetmath.org/ConicSection) or a hyperbola^{} depending on whether $\epsilon $ is less than, equal to or greater than 1.
For showing this, we choose the $y$axis as the directrix and the point $(q,\mathrm{\hspace{0.17em}0})$ as the focus. The locus condition reads then
$$\sqrt{{(xq)}^{2}+{y}^{2}}=\epsilon x.$$ 
This is simplified to
$(1{\epsilon}^{2}){x}^{2}2qx+{y}^{2}+{q}^{2}=\mathrm{\hspace{0.33em}0}.$  (1) 
If $\epsilon =1$, we obtain the parabola
$${y}^{2}=\mathrm{\hspace{0.33em}2}qx{q}^{2}.$$ 
In the following, we thus assume that $\epsilon \ne 1$.
Setting $y:=0$ in (1) we see that the $x$axis cuts the locus in two points with the midpoint^{} of the segment connecting them having the abscissa^{}
$${x}_{0}=\frac{q}{1{\epsilon}^{2}}.$$ 
We take this point as the new origin (replacing $x$ by $x+{x}_{0}$); then the equation (1) changes to
$(1{\epsilon}^{2}){x}^{2}+{y}^{2}={\displaystyle \frac{{\epsilon}^{2}{q}^{2}}{1{\epsilon}^{2}}}.$  (2) 
From this we infer that the locus is

1.
in the case $$ an ellipse (http://planetmath.org/Ellipse2) with the semiaxes
$$a=\frac{\epsilon q}{1{\epsilon}^{2}},b=\frac{\epsilon q}{\sqrt{1{\epsilon}^{2}}}$$ and with eccentricity $\epsilon $;

2.
in the case $\epsilon >1$ a hyperbola (http://planetmath.org/Hyperbola2) with semiaxes
$$a=\frac{\epsilon q}{{\epsilon}^{2}1},b=\frac{\epsilon q}{\sqrt{{\epsilon}^{2}1}}$$ and also now with the eccentricity $\epsilon $.
equation
the origin into a focus of a conic section (and in the cases of ellipse and hyperbola, the abscissa axis through the other focus). As before, let $q$ be the distance of the focus from the corresponding directrix. Let $r$ and $\phi $ be the polar coordinates^{} of an arbitrary point of the conic. Then the locus condition may be expressed as
$$\frac{r}{q\pm r\mathrm{cos}\phi}=\epsilon .$$ 
Solving this equation for the http://planetmath.org/node/6968polar radius $r$ yields the form
$r={\displaystyle \frac{\epsilon q}{1\mp \epsilon \mathrm{cos}\phi}}$  (3) 
for the common polar equation of the conic. The sign alternative ($\mp $) depends on whether the polar axis ($\phi =0$) intersects the directrix or not.
Title  simplest common equation of conics 

Canonical name  SimplestCommonEquationOfConics 
Date of creation  20150312 8:24:02 
Last modified on  20150312 8:24:02 
Owner  pahio (2872) 
Last modified by  pahio (2872) 
Numerical id  11 
Author  pahio (2872) 
Entry type  Derivation 
Classification  msc 51N20 
Synonym  common equation of conics 
Related topic  ConicSection 
Related topic  QuadraticCurves 
Related topic  BodyInCentralForceField 