Se da la ecuación de la línea de 2-o orden:
$$x^{2} + 4 x y - 20 x + 4 y^{2} + 10 y - 50 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = -10$$
$$a_{22} = 4$$
$$a_{23} = 5$$
$$a_{33} = -50$$
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
$$x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
φ - se define de la fórmula
$$\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}$$
sustituimos coeficientes
$$\cot{\left(2 \phi \right)} = - \frac{3}{4}$$
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
$$x' = \frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}$$
$$y' = - \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}$$
entonces la ecuación se transformará de
$$x'^{2} + 4 x' y' - 20 x' + 4 y'^{2} + 10 y' - 50 = 0$$
en
$$4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} + 4 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) + 10 \left(- \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right)^{2} - 20 \left(\frac{2 \sqrt{5} \tilde x}{5} + \frac{\sqrt{5} \tilde y}{5}\right) - 50 = 0$$
simplificamos
$$- 10 \sqrt{5} \tilde x + 5 \tilde y^{2} - 50 = 0$$
$$10 \sqrt{5} \tilde x - 5 \tilde y^{2} + 50 = 0$$
$$5 \tilde y^{2} = 10 \sqrt{5} \tilde x + 50$$
$$\tilde y^{2} = 2 \sqrt{5} \left(\tilde x + \sqrt{5}\right)$$
$$\tilde y'^{2} = 2 \sqrt{5} \tilde x'$$
Esta ecuación es una parábola
- está reducida a la forma canónica
Centro de las coordenadas canónicas en Oxy
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{2 \sqrt{5}}{5} + 0 \left(- \frac{\sqrt{5}}{5}\right)$$
$$y_{0} = 0 \left(- \frac{\sqrt{5}}{5}\right) + 0 \frac{2 \sqrt{5}}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
Centro de las coordenadas canónicas en el punto O
(0, 0)
Base de las coordenadas canónicas
$$\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ - \frac{\sqrt{5}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)$$