Se da la ecuación de la línea de 2-o orden:
$$2 x^{2} - 4 x y + x + 2 y^{2} - 5 y + 2 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = -2$$
$$a_{13} = \frac{1}{2}$$
$$a_{22} = 2$$
$$a_{23} = - \frac{5}{2}$$
$$a_{33} = 2$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}2 & -2\\-2 & 2\end{matrix}\right|$$
$$\Delta = 0$$
Como
$$\Delta$$
es igual a 0, entonces
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
$$\cot{\left(2 \phi \right)} = 0$$
entonces
$$\phi = \frac{\pi}{4}$$
$$\sin{\left(2 \phi \right)} = 1$$
$$\cos{\left(2 \phi \right)} = 0$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
$$\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}$$
sustituimos coeficientes
$$x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}$$
$$y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}$$
entonces la ecuación se transformará de
$$2 x'^{2} - 4 x' y' + x' + 2 y'^{2} - 5 y' + 2 = 0$$
en
$$2 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 4 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) + 2 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right)^{2} - 5 \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 2 = 0$$
simplificamos
$$- 2 \sqrt{2} \tilde x + 4 \tilde y^{2} - 3 \sqrt{2} \tilde y + 2 = 0$$
$$2 \sqrt{2} \tilde x - 4 \tilde y^{2} + 3 \sqrt{2} \tilde y - 2 = 0$$
$$\left(2 \tilde y + \frac{3 \sqrt{2}}{4}\right)^{2} = 2 \sqrt{2} \tilde x - \frac{7}{8}$$
$$\left(\tilde y + \frac{3 \sqrt{2}}{8}\right)^{2} = \frac{\sqrt{2} \left(\tilde x - \frac{7 \sqrt{2}}{32}\right)}{2}$$
$$\tilde y'^{2} = \frac{\sqrt{2} \tilde x'}{2}$$
Esta ecuación es una parábola
- está reducida a la forma canónica
Centro de las coordenadas canónicas en Oxy
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{\sqrt{2}}{2} + 0 \frac{\sqrt{2}}{2}$$
$$y_{0} = 0 \frac{\sqrt{2}}{2} + 0 \frac{\sqrt{2}}{2}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
Centro de las coordenadas canónicas en el punto O
(0, 0)
Base de las coordenadas canónicas
$$\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$
$$\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)$$