Linear Algebra - How to find the axis of a parabola? Given the parabola $x^2+y^2-2xy+4x=0$ defined in $\mathbb{R}^2$, how can I find the axis?
The matrices associated to the curve are:
$$A=\begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}, B=\begin{bmatrix}2 \\ 0\end{bmatrix}, C=\begin{bmatrix}1 & -1 & 2\\ -1 & 1 & 0\\ 2 & 0 & 0\end{bmatrix}$$
My textbook says that we need to compute the characteristic polynomial $P_A(\lambda)=\lambda(\lambda-2)$ of the matrix A and then find the eigenspaces associated to the respective eigenvalues, but I can't understand why we would do that.
 A: Note that$$x^2+y^2-2xy+4x=(x-y)^2+2\bigl((x-y)+(x+y)\bigr).$$So, if you do the change of coordinates $X=x-y$ and $Y=x+y$, that curve becomes $X^2+2X+2Y=0$, which is equivalent to $(X+1)^2+2Y=1$, or $Y=1-(X+1)^2$. Can you take it from here?
A: Equation of a parabola can be written as $L_1^2= 4 A L_2$, provided $L_1$ and $ L_2$ are equations of lines which are non-parallel.
Further if $L_1$ and $L_2$ are perpendicular and normalized, then $4A$ is the length of 
latus rectum. Equation of axis is $L_1=0$, Equation of tangent at vertex is $L_2=0$. The focus is given by $(L_2=A, L_1=0)$. The equation the  directrix is $L_2=-A$. The equation of the latus rectum is $L_2=A$. A line is called normalized if written as $\frac{ax+by+c}{\sqrt{a^2+b^2}}.$
So in this problem as done by Santos above we have $(x-y+1)^2=(1-x-y)$. We re-write it as:
$$ \left ( \frac {x-y+1}{\sqrt{2}}\right)^2=\frac{1}{\sqrt{2}} \left(\frac{1-x-y}{\sqrt{2}}\right)$$
So the equation of axis is $$\frac{x-y+1}{\sqrt{2}}=0.$$ Tangent at vertex is $$\frac{1-x-y}{\sqrt{2}}=0.$$
The directrix is  $$\frac{1-x-y}{\sqrt{2}}=- ~\frac{1}{4\sqrt{2}}.$$
The length of the latus rectum of the parabola is $1/\sqrt{2}$.
A: Every real symmetric matrix is orthogonally diagonalizable. If you interpret such a matrix as representing a conic, those orthogonal eigenvectors give you the directions of the principal axes of the quadric. For an ellipse, these are the major and minor axes; for a hyperbola the transverse and conjugate axes; and for a parabola the directions of its axis and tangent at the vertex. In particular, a parabola’s axis direction is given by any eigenvector of zero. Basically, this says that there’s a rotation that leaves only $x^2$ as the only second-degree term.  
In this case, it’s obvious from inspection that $(1,1)^T$ is an eigenvector of $A$ with eigenvalue $0$ (add the two columns together), so that’s the parabola’s axis direction.  
If you’re familiar with homogeneous coordinates, another way to find the axis direction is to compute the parabola’s intersection with the line at infinity. The line at infinity is in fact tangent to every parabola, so the intersection point can be computed as $C^{-1}(0,0,1)^T$, i.e., as the last row/column of $C^{-1}$. This method uses the fact that the polar of a point on a conic is the tangent at that point.
