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.

Title	simplest common equation of conics
Canonical name	SimplestCommonEquationOfConics
Date of creation	2015-03-12 8:24:02
Last modified on	2015-03-12 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