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

$$\sqrt{{(x-q)}^2+y^2}=\epsilon x.$$

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

$$\frac{r}{q\pm r\cos \phi }=\epsilon .$$

Solving this equation for the http://planetmath.org/node/6968polar radius $r$ yields the form

$r={\displaystyle \frac{\epsilon q}{1\mp \epsilon \cos \phi }}$(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.