# 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 $\varepsilon$, is a conic section, which is an ellipse, a parabola (http://planetmath.org/ConicSection) 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. 1.

in the case  $\varepsilon<1$  an ellipse (http://planetmath.org/Ellipse2) with the semiaxes

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

and with eccentricity $\varepsilon$;

2. 2.

in the case  $\varepsilon>1$  a hyperbola (http://planetmath.org/Hyperbola2) with semiaxes

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

and also now with the eccentricity $\varepsilon$.

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 $\varphi$ 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 http://planetmath.org/node/6968polar 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.

Title simplest common equation of conics SimplestCommonEquationOfConics 2015-03-12 8:24:02 2015-03-12 8:24:02 pahio (2872) pahio (2872) 11 pahio (2872) Derivation msc 51N20 common equation of conics ConicSection QuadraticCurves BodyInCentralForceField