Se da la ecuación de la línea de 2-o orden:
$$x^{2} + 4 x y + 6 y^{2} + y = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = 6$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}1 & 2\\2 & 6\end{matrix}\right|$$
$$\Delta = 2$$
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
$$x_{0} + 2 y_{0} = 0$$
$$2 x_{0} + 6 y_{0} + \frac{1}{2} = 0$$
entonces
$$x_{0} = \frac{1}{2}$$
$$y_{0} = - \frac{1}{4}$$
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
$$a'_{33} = \frac{y_{0}}{2}$$
$$a'_{33} = - \frac{1}{8}$$
entonces la ecuación se transformará en
$$x'^{2} + 4 x' y' + 6 y'^{2} - \frac{1}{8} = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{5}{4}$$
entonces
$$\phi = - \frac{\operatorname{acot}{\left(\frac{5}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4 \sqrt{41}}{41}$$
$$\cos{\left(2 \phi \right)} = \frac{5 \sqrt{41}}{41}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}}$$
sustituimos coeficientes
$$x' = \tilde x \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}}$$
$$y' = - \tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}} + \tilde y \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}}$$
entonces la ecuación se transformará de
$$x'^{2} + 4 x' y' + 6 y'^{2} - \frac{1}{8} = 0$$
en
$$6 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}} + \tilde y \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}}\right)^{2} + 4 \left(- \tilde x \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}} + \tilde y \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}}\right) + \left(\tilde x \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}}\right)^{2} - \frac{1}{8} = 0$$
simplificamos
$$- \frac{25 \sqrt{41} \tilde x^{2}}{82} - 4 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}} \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}} + \frac{7 \tilde x^{2}}{2} - 10 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}} \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}} + \frac{20 \sqrt{41} \tilde x \tilde y}{41} + 4 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}} \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}} + \frac{25 \sqrt{41} \tilde y^{2}}{82} + \frac{7 \tilde y^{2}}{2} - \frac{1}{8} = 0$$
$$- \frac{\sqrt{41} \tilde x^{2}}{2} + \frac{7 \tilde x^{2}}{2} + \frac{\sqrt{41} \tilde y^{2}}{2} + \frac{7 \tilde y^{2}}{2} - \frac{1}{8} = 0$$
Esta ecuación es una elipsis
$$\frac{\tilde x^{2}}{\left(\frac{1}{2 \sqrt{2} \sqrt{\frac{7}{2} - \frac{\sqrt{41}}{2}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \sqrt{2} \sqrt{\frac{\sqrt{41}}{2} + \frac{7}{2}}}\right)^{2}} = 1$$
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O
(1/2, -1/4)
Base de las coordenadas canónicas
$$\vec e_1 = \left( \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{5 \sqrt{41}}{82}}, \ \sqrt{\frac{5 \sqrt{41}}{82} + \frac{1}{2}}\right)$$