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 $\varepsilon$, is a conic section, which is an ellipse, a parabola or a hyperbola depending on whether $\varepsilon$ is less than, equal to or greater than 1.

For showing this, we choose the $y$-axis as the directrix and the point  $(q,\,0)$  as the focus.  The locus condition reads then

This is simplified to

If  $\varepsilon=1$,  we obtain the parabola

In the following, we thus assume that  $\varepsilon\neq 1$.

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

We take this point as the new origin (replacing $x$ by $x+x_{0}$); then the equation (1) changes to

From this we infer that the locus is

1. 1.

   in the case  $\varepsilon<1$  an ellipse with the semiaxes

   $$a\;=\;\frac{\varepsilon q}{1-\varepsilon^{2}},\quad b\;=\;\frac{\varepsilon q}{\sqrt{1-\varepsilon^{2}}}$$

   and with eccentricity $\varepsilon$;

2. 2.

   in the case  $\varepsilon>1$  a hyperbola with semiaxes

   $$a\;=\;\frac{\varepsilon q}{\varepsilon^{2}-1},\quad b\;=\;\frac{\varepsilon q}{\sqrt{\varepsilon^{2}-1}}$$

   and also now with the eccentricity $\varepsilon$.
   

Polar equation

Place 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 $\varphi$ be the polar coordinates of an arbitrary point of the conic.  Then the locus condition may be expressed as

Solving this equation for the polar radius $r$ yields the form

for the common polar equation of the conic.  The sign alternative ($\mp$) depends on whether the polar axis ($\varphi=0$) intersects the directrix or not.