xx+3yy-2xy forma canónica

Se da la ecuación de la línea de 2-o orden:
$$x^{2} - 2 x y + 3 y^{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0$$
donde
$$a_{11} = 1$$
$$a_{12} = -1$$
$$a_{13} = 0$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{33} = 0$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}1 & -1\\-1 & 3\end{matrix}\right|$$
$$\Delta = 2$$
Como
$$\Delta$$
no es igual a 0, entonces
hallamos el centro de coordenadas canónicas. Para eso resolvemos el sistema de ecuaciones
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
sustituimos coeficientes
$$x_{0} - y_{0} = 0$$
$$- x_{0} + 3 y_{0} = 0$$
entonces
$$x_{0} = 0$$
$$y_{0} = 0$$
Así pasamos a la ecuación en el sistema de coordenadas O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
donde
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
o
$$a'_{33} = 0$$
$$a'_{33} = 0$$
entonces la ecuación se transformará en
$$x'^{2} - 2 x' y' + 3 y'^{2} = 0$$
Hacemos el giro del sistema de coordenadas obtenido al ángulo de 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 - se define de la fórmula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
sustituimos coeficientes
$$\cot{\left(2 \phi \right)} = 1$$
entonces
$$\phi = \frac{\pi}{8}$$
$$\sin{\left(2 \phi \right)} = \frac{\sqrt{2}}{2}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{2}}{2}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
sustituimos coeficientes
$$x' = \tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}$$
entonces la ecuación se transformará de
$$x'^{2} - 2 x' y' + 3 y'^{2} = 0$$
en
$$3 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)^{2} - 2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right) + \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)^{2} = 0$$
simplificamos
$$- \frac{\sqrt{2} \tilde x^{2}}{2} - 2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + 2 \tilde x^{2} - \sqrt{2} \tilde x \tilde y + 4 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{\sqrt{2} \tilde y^{2}}{2} + 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + 2 \tilde y^{2} = 0$$
$$- \sqrt{2} \tilde x^{2} + 2 \tilde x^{2} + \sqrt{2} \tilde y^{2} + 2 \tilde y^{2} = 0$$
Esta ecuación es una elipsis degenerada
$$\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{2 - \sqrt{2}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\sqrt{2} + 2}}\right)^{2}} = 0$$
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O

(0, 0)

Base de las coordenadas canónicas
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}, \ \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)$$