5x^2+12xy+10y^2-6x+4y-1=0 canonical form

Given line equation of 2-order:
$$5 x^{2} + 12 x y - 6 x + 10 y^{2} + 4 y - 1 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
where
$$a_{11} = 5$$
$$a_{12} = 6$$
$$a_{13} = -3$$
$$a_{22} = 10$$
$$a_{23} = 2$$
$$a_{33} = -1$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}5 & 6\\6 & 10\end{matrix}\right|$$
$$\Delta = 14$$
Because
$$\Delta$$
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
substitute coefficients
$$5 x_{0} + 6 y_{0} - 3 = 0$$
$$6 x_{0} + 10 y_{0} + 2 = 0$$
then
$$x_{0} = 3$$
$$y_{0} = -2$$
Thus, we have the equation in the coordinate system O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
where
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
or
$$a'_{33} = - 3 x_{0} + 2 y_{0} - 1$$
$$a'_{33} = -14$$
then equation turns into
$$5 x'^{2} + 12 x' y' + 10 y'^{2} - 14 = 0$$
Rotate the resulting coordinate system by an angle φ
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - determined from the formula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
substitute coefficients
$$\cot{\left(2 \phi \right)} = - \frac{5}{12}$$
then
$$\phi = - \frac{\operatorname{acot}{\left(\frac{5}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{12}{13}$$
$$\cos{\left(2 \phi \right)} = \frac{5}{13}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{3 \sqrt{13}}{13}$$
$$\sin{\left(\phi \right)} = - \frac{2 \sqrt{13}}{13}$$
substitute coefficients
$$x' = \frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}$$
$$y' = - \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}$$
then the equation turns from
$$5 x'^{2} + 12 x' y' + 10 y'^{2} - 14 = 0$$
to
$$10 \left(- \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right)^{2} + 12 \left(- \frac{2 \sqrt{13} \tilde x}{13} + \frac{3 \sqrt{13} \tilde y}{13}\right) \left(\frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}\right) + 5 \left(\frac{3 \sqrt{13} \tilde x}{13} + \frac{2 \sqrt{13} \tilde y}{13}\right)^{2} - 14 = 0$$
simplify
$$\tilde x^{2} + 14 \tilde y^{2} - 14 = 0$$
Given equation is ellipse
$$\frac{\tilde x^{2}}{\left(\frac{1}{\frac{1}{14} \sqrt{14}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{14} \sqrt{14}}{\frac{1}{14} \sqrt{14}}\right)^{2}} = 1$$
- reduced to canonical form
The center of canonical coordinate system at point O

(3, -2)

Basis of the canonical coordinate system
$$\vec e_1 = \left( \frac{3 \sqrt{13}}{13}, \ - \frac{2 \sqrt{13}}{13}\right)$$
$$\vec e_2 = \left( \frac{2 \sqrt{13}}{13}, \ \frac{3 \sqrt{13}}{13}\right)$$