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

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

This is simplified to

$\displaystyle(1-\varepsilon^{2})x^{2}-2qx+y^{2}+q^{2}\;=\;0.$ | (1) |

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

$y^{2}\;=\;2qx-q^{2}.$ |

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

$x_{0}\;=\;\frac{q}{1-\varepsilon^{2}}.$ |

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

$\displaystyle(1-\varepsilon^{2})x^{2}+y^{2}\;=\;\frac{\varepsilon^{2}q^{2}}{1-% \varepsilon^{2}}.$ | (2) |

From this we infer that the locus is

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

$a\;=\;\frac{\varepsilon q}{1-\varepsilon^{2}},\qquad b\;=\;\frac{\varepsilon q% }{\sqrt{1-\varepsilon^{2}}}$ and with eccentricity $\varepsilon$;

- 2.
in the case $\varepsilon>1$ a hyperbola with semiaxes

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

$\frac{r}{q\pm r\cos{\varphi}}\;=\;\varepsilon.$ |

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

$\displaystyle r\;=\;\frac{\varepsilon q}{1\mp\varepsilon\cos{\varphi}}$ | (3) |

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.