Q:

Identify a quadric

Could you tell me how to identify a given quadric?
Given a conic section, I should find an orthonormal affine frame in $\mathbb{R}^2$ (with standard dot product) in which the equation has a canonical form.
Could you solve for example $\{(x,y) \in \mathbb{R}^2 : 73x^2 + 72xy + 52y^2 - 220x - 40y - 2300=0\}$ ? Or $x^2 - 6xy + 9y^2 + 2x - 5y -1=0$ ?
Thank you.
I would really appreciate a thorough explanation, because Kostrikin gives only the answers - there are no full solutions there.
For the second exmple I thought I could begin with a basis $(0,1), (1,0)$
$(x-3y)^2 + 2x-5y-1=0$ We set $\alpha = x-3y, \ \ \beta=y$, so $x=\alpha + 3\beta$.
$(0,0) + (x(1,0) + y(0,1)) = (0,0) + ((\alpha + 3\beta)(1,0) + \beta(0,1)) = (0,0) + \alpha(1,0) + \beta(3,1)$
So we have $\alpha^2 + 2 \alpha +  \beta -1 = 0$
$(\alpha + 1)^2 + \beta -2 = 0$
Then we set $\alpha_1 = \alpha + 1$, $\beta_1 = \beta -1$
The problem is that I don't know how to make the basis stay orthonormal.

A:

This solution uses projective geometry, in particular the fact that there is a unique conic up to isometry in the real projective plane, and that when embedding this into $\overline{\mathbb{R}^2}$, the type of conic it corresponds to in the restriction to $\mathbb{R}$ is characterized by how many ideal points it contains. We use the homogeneous coordinates in $\overline{\mathbb{R}^2}$ such that ideal points have $x_3=0$.

Let $q(x)=0$ be the homogenous equation of a projective conic in $\overline{\mathbb{R}^2}$ where $q$ is a quadratic form with $3\times 3$ symmetric matrix $M=(a_{ij})$. Show that whether the conic is an ellipse, a parabola, or a hyperbola in the restriction to $\mathbb{R}$ is determined by the sign of the upper-left $2\times 2$ minor of $M$.

Consider the ideal points of this conic: such points have $x_3=0$, and without loss of generality we may consider the representative vectors on the hyperplane $x_2=1$. Then
$$ q(x^*) = \begin{bmatrix} x_1 & 1 & 0 \end{bmatrix} \begin{bmatrix} a_{11}&a_{12}&a_{13} \\ a_{12}&a_{22}&a_{23} \\ a_{13}&a_{23}&a_{33} \end{bmatrix} \begin{bmatrix} x_1 \\ 1 \\ 0 \end{bmatrix} = a_{11}x_1^2+2a_{12}x_1+a_{22}.$$
The number of solutions to this quadratic is found by considering the sign of the discriminant $(2a_{12})^2-4a_{11}a_{22}$. Notice that this is precisely $-4$ times the upper-left $2\times 2$ minor of $M$. 
Explicitly: if the discriminant is positive, then there are two solutions to $q(x^*)=0$; if the discriminant is negative, there are none, and if it is zero, there is one. Since these solutions are the ideal points in the conic, this can be rewritten in the desired form: The conic is an ellipse precisely when the minor is positive, a parabola precisely when the minor is zero, and a hyperbola precisely when the minor is negative.

