# Equations of Second Degree

(:keywords,equation of second degree,coordinate system,change of variables:)

A general equation of second degree is of the form

f(x, y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0.

Examples include, say, x^2 + (y - 2)^2 = 4 for a circle of radius 2  and center at (2, 0),  parabola y = x^2,   and hyperbola xy = 1. These are not just particular cases of the general equation. The general equation can be reduced to a few cases via a change of variables. To see how it is done, it is best to rewrite the equation in matrix form.

Let M = \pmatrix{A&\frac{B}{2}\\\frac{B}{2}&C} \xi = {x \choose y} d = {D \choose E}.  Then

f(x, y) = \xi ^T M \xi + d^T \xi + F,

where T denotes transpose that, in this case converts 2\times 1  column vectors into 1\times 2   row vectors.

If \xi = P\xi'  for some 2\times 2  matrix P, with \xi = {x' \choose y'},  then

f(x, y) = \xi' ^T P^T M P \xi' + d^T P\xi' + F,

so that the same function f(x, y) in the coordinate system x', y'  becomes

f'(x', y') = \xi' ^T M' \xi' + d'^T\xi' + F,

where M' = P^T M P  and d' = d^T P.

The purpose of this change of variables is to get rid of the Bxy  term. This requires to diagonalize matrix M.  To this end, we should be looking for eignevalues and eigenvectors of matrix M.

Let's consider an example taken from Brannan et al, p. 31.

The problem is to find a more revealing Cartesian system of coordinates for the equation

3x^2 -10xy +3y^2 + 14x -2y + 3 = 0.

For this equation, M = \pmatrix{3&-5\\-5&3} d = {14 \choose -2}   and transform matrix

P = \pmatrix{ \frac{1}{\sqrt 2}&\frac{1}{\sqrt 2} \\-\frac{1}{\sqrt 2}&\frac{1}{\sqrt 2} }.

The modified equation is that of a hyperbola

\frac{(y' - 3/\sqrt{2})^2}{4} - \frac{(x' + 1/\sqrt{2})^2}{1}= 1.

References

• D. A. Brannan et al, Geometry Cambridge University Press, 2002