In my textbook there is a way of converting a quadratic equation to its canonical form using eigenvalues. I'm going to give you a rough description of the method:
The quadratic equation is of the form:
$a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_12xy+2a_{23}yz+2a_{31}xz+2a_{14}x+a_{24}y+a_{34}z+a_{44}=0$
Which can be express as following
$\begin{pmatrix}x & y & z\end{pmatrix}$$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}$$\begin{pmatrix} x\\ y\\ z \end{pmatrix}$$+2(a_{14}+a_{24}+a_{34})$$\begin{pmatrix} x\\ y\\ z \end{pmatrix}$$+a_{44}=0$
Let $A=\begin{pmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{pmatrix}$
because A is symmetric there exists an orthogonal matrix $P$ such that
$P^{-1}AP=$ $\begin{pmatrix} t_{1} & 0 & 0\\ 0 & t_{2} & 0\\ 0 & 0 & t_{3} \end{pmatrix}$, $t_{1},t_{2},t_{3}$ the eigenvalues of $A$
Now let $B=PB'$, $B'=$$\begin{pmatrix} x'\\ y'\\ z' \end{pmatrix}$
then $\begin{pmatrix}x' & y' & z'\end{pmatrix}$$\begin{pmatrix} t_{1} & 0 & 0\\ 0 & t_{2} & 0\\ 0 & 0 & t_{3} \end{pmatrix}$$\begin{pmatrix} x'\\ y'\\ z' \end{pmatrix}$$+2(a_{14}+a_{24}+a_{34})P$$\begin{pmatrix} x'\\ y'\\ z' \end{pmatrix}$$+a_{44}=0$$\Leftrightarrow$
$t_{1}x'^2+t_2y'^2+t_3z'^2$$+2(a_{14}+a_{24}+a_{34})P$$\begin{pmatrix} x'\\ y'\\ z' \end{pmatrix}$$+a_{44}=0$
My question now is the following, is there a specific order which determines which eigenvalue is number 1,2 etc? Because if for example an eigenvalue is zero this means that one of the $x^2,y^2,z^2$ is zeroed as well. How do you determine what eigenvalue to asign to each of the $x^2,y^2,z^2$?
The order of the eigenvalues is determined by how you want the figure to be oriented in space. To give you an example, consider a simple case: ellipse in 2D ($a_{33} = a_{23} = a_{31} = a_{34} = 0$) centered at the origin ($a_{14} = a_{24} = 0$ and $a_{44} = -1$) with major axis along the $x$ axis ($a_{11} = 1, a_{22} = 1 / b^2$, $a_{12} = 0$, $b<1$), that is
$$ x^2 + \frac{y^2}{b^2} = 1 \tag{1} $$
In this case
$$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1/b^2 \end{pmatrix} \tag{2} $$
This is a trivial example in the sense that the matrix is already diagonal,
$$ P = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}, ~~ T = \begin{pmatrix}1 & 0 \\ 0 & 1/b^2 \end{pmatrix} \tag{3} $$
But here's the idea, image that you decide to organize your eigenvalues as
$$ P_{\rm new} = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}, ~~ T_{\rm new} = \begin{pmatrix}1/b^2 & 0 \\ 0 & 1 \end{pmatrix} \tag{4} $$
In this case your coordinate system is defined as
$$ \begin{pmatrix}x' \\ y' \end{pmatrix} = P_{\rm new} \begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}y \\ x \end{pmatrix} \tag{5} $$
So that the ellipse is written as
$$ y'^2 + \frac{x'^2}{b^2} = 1 \tag{6} $$
Which is now an ellipse with major axis along the $y'$ axis! so changing the order of the eigenvalues we change the orientation of the figure in space