Affine classification of the 2-nd order curves

Affine classification of the 2-nd order curves


	Affine classification of the 2-nd order curves

The second order curve is defined by

$\begin{displaymath} a_{11} x^2 + a_{12} xy + a_{22} y^2 + b_1 x + b_2 y + c =0, \end{displaymath}$

where

$a_{11},\; a_{12}, \; a_{22}$

and

$b_1,\; b_2, \; c$

are real numbers.

The goal of this section is to classify all possible 2-nd order curves.

Consider the matrix of the 2-nd order curve,

$\begin{displaymath} A=\left( \begin{array}{cc} a_{11} & \frac{a_{12}}{2} \ \frac{a_{12}}{2} & a_{22} \ \end{array} \right) \end{displaymath}$

One can make A diagonal by the change of coordinates

$\begin{displaymath} \left( \begin{array}{c} x \ y \end{array}\right)= (\xi_1... ... \left( \begin{array}{c} \bar x \ \bar y \end{array}\right) \end{displaymath}$

where

$\xi_1$

and

$\xi_2$

are eigenvectors of A.

This change of coordinates corresponds to a rotation of the plane around the origin if we choose

$\xi_1, \;\; \xi_2$

so that

$\begin{displaymath} \vert \xi_1 \vert =1 ,\;\; \vert \xi_2 \vert =1 \mbox{ and } \det(\xi_1 \;\;\xi_2)=1, \end{displaymath}$

where

$\vert \nu \vert$

denotes the magnitude of the vector

$\nu = (\nu_1,\;\; \nu_2),$

$\begin{displaymath} \vert \nu \vert = \sqrt{\nu_1^2 + \nu_2^2}. \end{displaymath}$

In new coordinates

$(\bar x , \; \bar y )$

the equation for the 2-nd order curves is

$\begin{displaymath} \lambda_1 \bar{x}^2 + \lambda_2 \bar{y}^2 + \bar{b_1} \bar{x} + \bar{b_2} \bar{y} + c =0 \end{displaymath}$

where c is the same number as in the original equation of the curve.

$\begin{displaymath} (\bar{b_1} \;\; \bar{b_2} ) = (b_1 \;\; b_2 ) (\xi_1 \;\; \xi_2) \end{displaymath}$

In other words,

$\begin{eqnarray*} \bar{b_1} &=& b_1 \cdot \xi_{11} + b_2 \cdot \xi_{21} \ \bar{b_2} &=& b_1 \cdot \xi_{12} + b_2 \cdot \xi_{22} \end{eqnarray*}$

where

$\begin{displaymath} \xi_1 = \left( \begin{array}{c} \xi_{11} \ \xi_{21} \en... ...( \begin{array}{c} \xi_{12} \ \xi_{22} \end{array}\right) \end{displaymath}$

Now by shifting the origin of the frame of coordinates

$(\bar{x},\; \bar{y} )$

we can eliminate

$\bar{b_1},\;\; \bar{b_2},$

if both eigenvalues

$\lambda_1$

and

$\lambda_2$

are not equal to zero.That means we need to introduce new coordinates

$\begin{eqnarray*} \bar x &=& \tilde x + \ell_1 \ \bar y &=& \tilde y + \ell_2 \end{eqnarray*}$

with

$\begin{eqnarray*} \ell_1 &=& -\frac{\bar{b_1}}{2\cdot \lambda_1} \ \ell_2 &=& -\frac{\bar{b_2}}{2\cdot \lambda_2} \ \end{eqnarray*}$

After this shift the 2-nd order curve is given by

$\begin{displaymath} \lambda_1 \tilde{x}^2 + \lambda_2 \tilde{y}^2 + \bar c = 0 \end{displaymath}$

where

$\begin{displaymath} \bar c = \lambda_1 \ell_1^2 + \lambda_2 \ell_2^2 + \bar{b_1} \ell_1 + \bar{b_2} \ell_2 + c. \end{displaymath}$

Now we can give the complete affine classification of the 2-nd order curves.

**Table 1.1:** Affine classification of the 2-nd order curves
$\lambda_1 \lambda_2 >0$ and $\lambda_1 \cdot \bar c > 0$	imaginary ellipse
$\lambda_1 \lambda_2 >0$ and $\lambda_1 \cdot \bar c < 0$	ellipse
$\lambda_1 \lambda_2 <0$ and $\bar c\not= 0$	hyperbola
$\lambda_1 \lambda_2 <0$ and $\bar c = 0$	a pair of intersecting lines
$\lambda_1 \lambda_2 >0$ and $\bar c = 0$	a point
$\lambda_1 \not= 0$ and $\lambda_2 = 0,\; \bar b_2 \not= 0$	parabola
$\lambda_1 \not= 0$ and $\lambda_2 = 0,\; \bar b_2 = 0,\; \bar c = 0$	a pair of coinciding lines
$\lambda_1 > 0$ and $\lambda_2 =0 ,\; \bar b_2 = 0,\; \bar c < 0$	a pair of parallel lines
$\lambda_1 > 0$ and $\lambda_2 =0 ,\; \bar b_2 = 0, \; \bar c > 0$	a pair of imaginary parallel lines

The change of coordinates that brings the second order curve to its canonical form is depicted in Fig.1.1

**Figure 1.1:** Canonical change of coordinates
$\begin{figure}\begin{center} \psfig{file=images/secondOrdCurve.eps,scale=0.7} \end{center}\end{figure}$

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Sergey Nikitin 2004-11-04