PlanetMath: quadratic curves

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (210)
Orphanage
Unclass'd (1)
Unproven (473)
Corrections (40)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

quadratic curves

(Topic)

We want to determine the graphical representant of the general bivariate quadratic equation

$\displaystyle Ax^2+By^2+2Cxy+2Dx+2Ey+F = 0,$

(1)

where $A,\,B,\,C,\,D,\,E,\,F$ are known real numbers and

If $C \neq 0$ , we will rotate the coordinate system, getting new coordinate axes and , such that the equation (1) transforms into a new one having no more the mixed term . Let the rotation angle be $\alpha$ to the anticlockwise (positive) direction so that the - and -axes form the angles $\alpha$ and $\alpha+90^\circ$ with the original -axis, respectively. Then there is the connection

$\displaystyle x = x'\cos\alpha-y'\sin\alpha$
$\displaystyle y = x'\sin\alpha+y'\cos\alpha$

between the new and old coordinates (see rotation matrix). Substituting these expressions into (1) it becomes

$\displaystyle Mx'^2+Ny'^2+2Px'y'+2Gx'+2Hy'+F = 0,$

(2)

where

$\begin{align*}\begin{cases}M = A\cos^2\alpha+B\sin^2\alpha+C\sin2\alpha,\\ N = A... ...pha-C\sin2\alpha,\\ 2P = (B-A)\sin2\alpha+2C\cos2\alpha. \end{cases}\end{align*}$

(3)

It's always possible to determine $\alpha$ such that $(B-A)\sin2\alpha = -2C\cos2\alpha$ , i.e. that

$\displaystyle \tan2\alpha = \frac{2C}{A-B}$

for $A \neq B$ and $\alpha = 45^\circ$ for the case

. Then the term

vanishes in (2), which becomes, dropping out the apostrophes,

$\displaystyle Mx^2+Ny^2+2Gx+2Hy+F = 0.$

(4)

If none of the coefficients and equal zero, one can remove the first degree terms of (4) by first writing it as
$\displaystyle M\left(x+\frac{G}{M}\right)^2+N\left(y+\frac{H}{N}\right)^2 = \frac{G^2}{M}+\frac{H^2}{N}-F$
and then translating the origin to the point $\left(-\frac{G}{M},\,-\frac{H}{N}\right)$ , when we obtain the equation of the form

$\displaystyle Mx^2+Ny^2 = K.$ (5)

If and have the same sign, then in order that (5) could have a counterpart in the plane, the sign must be the same as the sign of ; then the counterpart is the ellipse
$\displaystyle \frac{x^2}{\left(\sqrt{\vert K/M\vert}\right)^2}+\frac{y^2}{\left(\sqrt{\vert K/N\vert}\right)^2} = 1.$
If and have opposite signs and $K \neq 0$ , then the curve (5) correspondingly is one of the hyperbolas
$\displaystyle \frac{x^2}{\left(\sqrt{\vert K/M\vert}\right)^2}-\frac{y^2}{\left(\sqrt{\vert K/N\vert}\right)^2} = \pm1,$
which for is reduced to a pair of intersecting lines.
If one of and , e.g. the latter, is zero, the equation (4) may be written

$\displaystyle M\left(x+\frac{G}{M}\right)^2+2Hy+F-\frac{G^2}{M} = 0$

i.e.

$\displaystyle M\left(x+\frac{G}{M}\right)^2+2H\left(y+\frac{MF-G^2}{2HM}\right) = 0.$

Translating now the origin to the point $\left(-\frac{G}{M},\,\frac{G^2-MF}{2HM}\right)$ the equation changes to

$\displaystyle Mx^2+2Hy = 0.$ (6)

For $H \neq 0$ , this is the equation $y = -\frac{M}{2H}x^2$ of a parabola, but for , of a double line .

The kind of the quadratic curve (1) can also be found out directly from this original form of the equation. Namely, from the formulae (3) between the old and the new coefficients one may derive the connection

$\displaystyle MN-P^2 = AB-C^2$