- #1
arestes
- 80
- 3
- TL;DR Summary
- Not sure about the accepted "canonical form" for a quadratic equation WITH linear term
Hello:
I'm not sure if there's an accepted canonical form for a quadratic equation in two (or more) variables:
$$ax^2+by^2+cxy+dx+ey+f=0$$
Is it the following form? (using the orthogonal matrix Q that diagonalizes the quadratic part):
$$ w^TDw+[d \ \ e]w+f=0$$
$$w^TDw+Lw+f=0$$
where
$$ w=\begin{pmatrix}
x' \\
y'
\end{pmatrix} = Q^T
\begin{pmatrix}
x \\
y
\end{pmatrix}
$$
and
$$ L=[d \ \ e] $$
Or is it a form with translated coordinates:
$$a(x'-m)^2+b(y'-n)^2+c(x'-m)(y'-n)+d(x'-m)+e(y'-n)+f'=0$$
with some to-be-determined constants m and n such that the linear terms vanish, which can be then used to change variables x'=x+m and y'=y+n.
I tried to find these m and n (expanding the binomials) but the simultaneous equations to satisfy in order to remove the linear terms are restricting and it seems to be impossible when $$c^2-4ab=0$$
Is it enough to leave the linear terms and call it "canonical form" just by diagonalizing the quadratic terms?
I'm not sure if there's an accepted canonical form for a quadratic equation in two (or more) variables:
$$ax^2+by^2+cxy+dx+ey+f=0$$
Is it the following form? (using the orthogonal matrix Q that diagonalizes the quadratic part):
$$ w^TDw+[d \ \ e]w+f=0$$
$$w^TDw+Lw+f=0$$
where
$$ w=\begin{pmatrix}
x' \\
y'
\end{pmatrix} = Q^T
\begin{pmatrix}
x \\
y
\end{pmatrix}
$$
and
$$ L=[d \ \ e] $$
Or is it a form with translated coordinates:
$$a(x'-m)^2+b(y'-n)^2+c(x'-m)(y'-n)+d(x'-m)+e(y'-n)+f'=0$$
with some to-be-determined constants m and n such that the linear terms vanish, which can be then used to change variables x'=x+m and y'=y+n.
I tried to find these m and n (expanding the binomials) but the simultaneous equations to satisfy in order to remove the linear terms are restricting and it seems to be impossible when $$c^2-4ab=0$$
Is it enough to leave the linear terms and call it "canonical form" just by diagonalizing the quadratic terms?