I have to find the canonical form of the conic:
$x^2+4xy+3y^2-2x+1=0$
I'm not sure if all these steps are correct, I would appreciate if you could check.
First of all I can say that this is an hyperbola since $\begin{vmatrix}A\end{vmatrix} \ne 0 $ and $\begin{vmatrix}Q\end{vmatrix} < 0$
\begin{align} \begin{vmatrix} A \end{vmatrix}= \begin{vmatrix} 1 & 2 & -1 \\ 2 & 3 & 0 \\ -1 & 0 & 1 \end{vmatrix} = -4 \end{align}
\begin{align} \begin{vmatrix} Q \end{vmatrix}= \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = -1 \end{align}
$Q$ being the matrix of the quadratic form and $A$ the matrix of the quadratic equation.
Now I know that the canonical form is like
$\lambda X^2 + \mu Y^2 + p = 0$
Eigenvalues of $Q$ are
\begin{align} \begin{vmatrix} 1-\lambda & 2 \\ 2 & 3-\lambda \end{vmatrix} = (1-\lambda)(3-\lambda)-4 = \lambda^2-4\lambda-1 = 0 \end{align}
We find that the solution to this equation is $\lambda = 2 - \sqrt5$, $\mu = 2 + \sqrt5$
Now I put the found solutions in the canonical form we had before, so now it becomes
$(2-\sqrt5)X^2+(2+\sqrt5)Y^2+p=0$
We need to find $p$ now. We know that the canonical form and the non-canonical form of the conic have the same matrix determinant, so
\begin{align} \begin{vmatrix} A' \end{vmatrix}= \begin{vmatrix} (2-\sqrt5) & 0 & 0 \\ 0 & (2+\sqrt5) & 0 \\ 0 & 0 & p \end{vmatrix} = -p = -4 \end{align}
Let's put $p$ in the quation and we get:
$(2-\sqrt5)X^2+(2+\sqrt5)Y^2+4=0$
We've now found the canonical form of the conic, we need one last step.
We divide by $p$ in this case so that the equation can result clearer to interpret.
$(2-\sqrt5)X^2+(2+\sqrt5)Y^2=-4$
$-\frac{2-\sqrt5}{4}X^2-\frac{2+\sqrt5}{4}Y^2=1$