Let $\gamma$ be a generic conic in E2.
\begin{equation} \begin{pmatrix} 1 & x & y \end{pmatrix} \begin{pmatrix} a_{00} & a_{01} & a_{02} \\ a_{10} & a_{11} & a_{12} \\ a_{20} & a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} 1 \\ x \\ y \end{pmatrix} = 0, \end{equation} where the matrix is symmetrical.
I have read that in the case of a parabola, (i.e. the determinant of \begin{equation} A_0 = \begin{pmatrix} a_{11} & a_{12} \\ a_{21}=a_{12} & a_{22} \end{pmatrix} \end{equation}
is 0) the axis of the parabola is parallel to the eigen space of the eigenvalue $0$ of the matrix $A_0$.
I don't know how to prove this.
My attempt was to think of finding the Eigenspace of eigenvalue 0 as being equivalent to finding the kernel of $A_0$, that is, the solution of the system
\begin{equation} a_{11}x + a_{12} y = 0, a_{12}x + a_{22}y = 0 \end{equation} and that the vector found doing this is an axis of symmetry, but it doesn't seem to be a very efficient (if it works at all) or geometrically meaningful path.
For a parabola with equation $y=ax^2$, we have $A_0 =\begin{pmatrix} a & 0 \\ 0 & 0\end{pmatrix}$ and the eigenvector associated to eigenvalue $0$ is $(0,1)$, which is indeed parallel to the axis of the parabola.
Any parabola can be obtained from the above one with a rotation and translation. If $R$ is the rotation matrix, the $A_0$ matrix for the new parabola is $A_0'=RA_0R^{-1}$. The eigenvector of $A_0'$ associated to eigenvalue $0$ is then $R(0,1)$, which is of course parallel to the axis of the new parabola.