<s>[INST] 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.
 [/INST] 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 $1/\sqrt{2}$.
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