ax canonical form

Given line equation of 2-order:
$$a x = 0$$
This equation looks like:
$$a^{2} a_{22} + 2 a a_{12} x + 2 a a_{23} + a_{11} x^{2} + 2 a_{13} x + a_{33} = 0$$
where
$$a_{11} = 0$$
$$a_{12} = \frac{1}{2}$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}0 & \frac{1}{2}\\\frac{1}{2} & 0\end{matrix}\right|$$
$$\Delta = - \frac{1}{4}$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{0} a_{12} + a_{11} x_{0} + a_{13} = 0$$
$$a_{0} a_{22} + a_{12} x_{0} + a_{23} = 0$$
substitute coefficients
$$\frac{a_{0}}{2} = 0$$
$$\frac{x_{0}}{2} = 0$$
then
$$x_{0} = 0$$
$$a_{0} = 0$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'^{2} a_{22} + 2 a' a_{12} x' + a'_{33} + a_{11} x'^{2} = 0$$
where
$$a'_{33} = a_{0} a_{23} + a_{13} x_{0} + a_{33}$$
or
$$a'_{33} = 0$$
$$a'_{33} = 0$$
then equation turns into
$$a' x' = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = - \tilde a \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$a' = \tilde a \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = 0$$
then
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
substitute coefficients
$$x' = - \frac{\sqrt{2} \tilde a}{2} + \frac{\sqrt{2} \tilde x}{2}$$
$$a' = \frac{\sqrt{2} \tilde a}{2} + \frac{\sqrt{2} \tilde x}{2}$$
then the equation turns from
$$a' x' = 0$$
to
$$\left(- \frac{\sqrt{2} \tilde a}{2} + \frac{\sqrt{2} \tilde x}{2}\right) \left(\frac{\sqrt{2} \tilde a}{2} + \frac{\sqrt{2} \tilde x}{2}\right) = 0$$
simplify
$$- \frac{\tilde a^{2}}{2} + \frac{\tilde x^{2}}{2} = 0$$
$$\frac{\tilde a^{2}}{2} - \frac{\tilde x^{2}}{2} = 0$$
Given equation is degenerate hyperbole
$$- \frac{\tilde a^{2}}{\left(\sqrt{2}\right)^{2}} + \frac{\tilde x^{2}}{\left(\sqrt{2}\right)^{2}} = 0$$
- reduced to canonical form
The center of canonical coordinate system at point O

(0, 0)

Basis of the canonical coordinate system
$$\vec e_{1} = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_{2} = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$