Se da la ecuación de la línea de 2-o orden:
$$9 x^{2} + 24 x y - 20 x + 16 y^{2} + 110 y - 50 = 0$$
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
$$a_{11} = 9$$
$$a_{12} = 12$$
$$a_{13} = -10$$
$$a_{22} = 16$$
$$a_{23} = 55$$
$$a_{33} = -50$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}9 & 12\\12 & 16\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{7}{24}$$
entonces
$$\phi = - \frac{\operatorname{acot}{\left(\frac{7}{24} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{24}{25}$$
$$\cos{\left(2 \phi \right)} = \frac{7}{25}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{4}{5}$$
$$\sin{\left(\phi \right)} = - \frac{3}{5}$$
sustituimos coeficientes
$$x' = \frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}$$
$$y' = - \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}$$
entonces la ecuación se transformará de
$$9 x'^{2} + 24 x' y' - 20 x' + 16 y'^{2} + 110 y' - 50 = 0$$
en
$$16 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right)^{2} + 24 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) + 110 \left(- \frac{3 \tilde x}{5} + \frac{4 \tilde y}{5}\right) + 9 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right)^{2} - 20 \left(\frac{4 \tilde x}{5} + \frac{3 \tilde y}{5}\right) - 50 = 0$$
simplificamos
$$- 82 \tilde x + 25 \tilde y^{2} + 76 \tilde y - 50 = 0$$
$$82 \tilde x - 25 \tilde y^{2} - 76 \tilde y + 50 = 0$$
$$\left(5 \tilde y - \frac{38}{5}\right)^{2} = 82 \tilde x + \frac{2694}{25}$$
$$\left(\tilde y - \frac{38}{25}\right)^{2} = \frac{82 \tilde x}{25} + \frac{2694}{625}$$
$$\tilde y'^{2} = \frac{82 \tilde x}{25} + \frac{2694}{625}$$
Esta ecuación es una parábola
- está reducida a la forma canónica
Centro de las coordenadas canónicas en Oxy
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = \frac{0 \cdot 4}{5} + \frac{\left(-3\right) 0}{5}$$
$$y_{0} = \frac{\left(-3\right) 0}{5} + \frac{0 \cdot 4}{5}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
Centro de las coordenadas canónicas en el punto O
(0, 0)
Base de las coordenadas canónicas
$$\vec e_1 = \left( \frac{4}{5}, \ - \frac{3}{5}\right)$$
$$\vec e_2 = \left( \frac{3}{5}, \ \frac{4}{5}\right)$$