# Steps to find the canonical form of conic section

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.

$(2-\sqrt5)X^2+(2+\sqrt5)Y^2=-4$

$-\frac{2-\sqrt5}{4}X^2-\frac{2+\sqrt5}{4}Y^2=1$

• Seem OK. Not checked the actual calculation – Shailesh Sep 26 '16 at 16:24
• @matt95 Not really. Swapping them amounts to swapping the $X$ and $Y$ axes. For a hyperbola, this means that you might end up with $Y^2/b^2-X^2/a^2+p=0$, which is still correct, but unconventional. – amd Sep 26 '16 at 18:10
• @matt95 If you’ve got a hyperbola, one eigenvalue is positive and the other negative. Since convention puts the minus sign on the $Y^2$ term, assign the negative eigenvalue to $\mu$. For an ellipse in standard position, the major axis conventionally coincides with the $X$-axis. The eigenvalues will have the same sign in that case, so you want the one with the smaller absolute value to go with $X^2$. – amd Sep 26 '16 at 18:18
• BTW, since the canonical-form matrix is diagonal, its determinant is the product of its diagonal elements, so $p=|A|/\lambda\mu$. – amd Sep 26 '16 at 18:19
• By “not really” in my comment above, I meant that it doesn’t really matter. – amd Sep 26 '16 at 18:25