Se da la ecuación de la línea de 2-o orden:
$$4 x_{1}^{2} + 4 x_{1} x_{2} + x_{2}^{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{2}^{2} + 2 a_{12} x_{1} x_{2} + 2 a_{13} x_{2} + a_{22} x_{1}^{2} + 2 a_{23} x_{1} + a_{33} = 0$$
donde
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{33} = 0$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}1 & 2\\2 & 4\end{matrix}\right|$$
$$\Delta = 0$$
Como
$$\Delta$$
es igual a 0, entonces
Hacemos el giro del sistema de coordenadas obtenido al ángulo de φ
$$x2' = - \tilde x1 \sin{\left(\phi \right)} + \tilde x2 \cos{\left(\phi \right)}$$
$$x1' = \tilde x1 \cos{\left(\phi \right)} + \tilde x2 \sin{\left(\phi \right)}$$
φ - se define de la fórmula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
sustituimos coeficientes
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
entonces
$$\phi = - \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{4}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{3}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{5}}{5}$$
sustituimos coeficientes
$$x2' = \frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}$$
$$x1' = \frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}$$
entonces la ecuación se transformará de
$$4 x1'^{2} + 4 x1' x2' + x2'^{2} = 0$$
en
$$\left(\frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}\right)^{2} + 4 \left(\frac{\sqrt{5} \tilde x1}{5} + \frac{2 \sqrt{5} \tilde x2}{5}\right) \left(\frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}\right) + 4 \left(\frac{2 \sqrt{5} \tilde x1}{5} - \frac{\sqrt{5} \tilde x2}{5}\right)^{2} = 0$$
simplificamos
$$5 \tilde x1^{2} = 0$$
$$\tilde x1^{2} = 0$$
$$\tilde x1^{2} = 0$$
$$\tilde x1'^{2} = 0$$
Esta ecuación es dos rectas paralelas
- está reducida a la forma canónica
donde se ha hecho la sustitución
$$\tilde x1' = \tilde x1$$
$$\tilde x2' = \tilde x2$$
Centro de las coordenadas canónicas en Oxy
$$x_{20} = - \tilde x1 \sin{\left(\phi \right)} + \tilde x2 \cos{\left(\phi \right)}$$
$$x_{10} = \tilde x1 \cos{\left(\phi \right)} + \tilde x2 \sin{\left(\phi \right)}$$
$$x_{20} = 0 \frac{2 \sqrt{5}}{5} + 0 \left(- \frac{\sqrt{5}}{5}\right)$$
$$x_{10} = 0 \left(- \frac{\sqrt{5}}{5}\right) + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{20} = 0$$
$$x_{10} = 0$$
Centro de las coordenadas canónicas en el punto O
(0, 0)
Base de las coordenadas canónicas
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$