# Canonical form of a degenerate conic

I have the conic $\Gamma:x^2+y^2-2xy-6=0$, which is degenerate since $$\begin{vmatrix}1&-1&0\\-1&1&0\\0&0&-6\end{vmatrix}=0$$ 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?

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 $2$.
• I think I have understood your explaination but still I want to be sure. One of the way to find the canonical form of a conic is through traslation+rotation. The traslation is written as $X=x-x_{0}\wedge Y=y-y_{0}$. The passage matrix you wrote is $\begin{bmatrix} \cos(\theta)&\sin(\theta)\\ -\sin(\theta)&\cos(\theta) \end{bmatrix}$ with $\theta=\frac{\pi}{4}$. By normalizing by dividing for $|(1,-1)|=\sqrt(2)$ and multiplying it for $(x,y)^T$ you get $(X,Y)^T$. – M.Giacchello Dec 5 '16 at 10:04
• You have to find an orthogonal vector. $[1,1]$ is orthogonal to $[1,-1]$, now normalize again. In the end you want to decompose $B=Q^TDQ$ with $Q$ orthogonal and $D$ diagonal, $B$ is your original symmetric matrix which has an eigen decomposition with simple real eigenvalues and orthogonal eigenvectors by the spectral theorem. – LutzL Dec 5 '16 at 10:08