Q:

Canonical form of a degenerate conic

I have the conic $\Gamma:x^2+y^2-2xy-6=0$, which is degenerate since
\begin{equation}
\begin{vmatrix}1&-1&0\\-1&1&0\\0&0&-6\end{vmatrix}=0
\end{equation}
and particular it represents two parallel lines as the rank of the complete matrix is 2(the second row is equal to the first one with inverted signs).
The exercise ask me the canonical form and gives the result $X^2-3=0$ but I've never did any exercise nor given any thought on how to write the canonical form of a degenerate conic. Can someone explain to me please? 

A:

You can easily detect the binomial theorem inside the formula, giving
$$
(x-y)^2-6=0
$$
$x-y=(1,-1)\dbinom{x}{y}$, and normalizing the first factor as part of an orthogonal matrix gives the equivalent form
$$
2X^2-6=0,\qquad 
\begin{bmatrix}X\\Y\end{bmatrix}
=\frac1{\sqrt2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}
·\begin{bmatrix}x\\y\end{bmatrix}
$$
Now divide by $2$.

