Se da la ecuación de la línea de 2-o orden:
$$16 x^{2} - 72 x y + 144 x + 81 y^{2} + 144 y - 144 = 0$$
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
$$a_{11} = 16$$
$$a_{12} = -36$$
$$a_{13} = 72$$
$$a_{22} = 81$$
$$a_{23} = 72$$
$$a_{33} = -144$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}16 & -36\\-36 & 81\end{matrix}\right|$$
$$\Delta = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{65}{72}$$
entonces
$$\phi = \frac{\operatorname{acot}{\left(\frac{65}{72} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{72}{97}$$
$$\cos{\left(2 \phi \right)} = \frac{65}{97}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{9 \sqrt{97}}{97}$$
$$\sin{\left(\phi \right)} = \frac{4 \sqrt{97}}{97}$$
sustituimos coeficientes
$$x' = \frac{9 \sqrt{97} \tilde x}{97} - \frac{4 \sqrt{97} \tilde y}{97}$$
$$y' = \frac{4 \sqrt{97} \tilde x}{97} + \frac{9 \sqrt{97} \tilde y}{97}$$
entonces la ecuación se transformará de
$$16 x'^{2} - 72 x' y' + 144 x' + 81 y'^{2} + 144 y' - 144 = 0$$
en
$$81 \left(\frac{4 \sqrt{97} \tilde x}{97} + \frac{9 \sqrt{97} \tilde y}{97}\right)^{2} - 72 \left(\frac{4 \sqrt{97} \tilde x}{97} + \frac{9 \sqrt{97} \tilde y}{97}\right) \left(\frac{9 \sqrt{97} \tilde x}{97} - \frac{4 \sqrt{97} \tilde y}{97}\right) + 144 \left(\frac{4 \sqrt{97} \tilde x}{97} + \frac{9 \sqrt{97} \tilde y}{97}\right) + 16 \left(\frac{9 \sqrt{97} \tilde x}{97} - \frac{4 \sqrt{97} \tilde y}{97}\right)^{2} + 144 \left(\frac{9 \sqrt{97} \tilde x}{97} - \frac{4 \sqrt{97} \tilde y}{97}\right) - 144 = 0$$
simplificamos
$$\frac{1872 \sqrt{97} \tilde x}{97} + 97 \tilde y^{2} + \frac{720 \sqrt{97} \tilde y}{97} - 144 = 0$$
$$\left(\sqrt{97} \tilde y + \frac{360}{97}\right)^{2} = - \frac{1872 \sqrt{97} \tilde x}{97} + \frac{1484496}{9409}$$
$$\left(\tilde y + \frac{360 \sqrt{97}}{9409}\right)^{2} = - \frac{1872 \sqrt{97} \left(\tilde x - \frac{793 \sqrt{97}}{9409}\right)}{9409}$$
$$\tilde y'^{2} = - \frac{1872 \sqrt{97} \tilde x'}{9409}$$
Esta ecuación es una parábola
- está reducida a la forma canónica
Centro de las coordenadas canónicas en Oxy
$$x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}$$
$$y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}$$
$$x_{0} = 0 \frac{9 \sqrt{97}}{97} + 0 \frac{4 \sqrt{97}}{97}$$
$$y_{0} = 0 \frac{4 \sqrt{97}}{97} + 0 \frac{9 \sqrt{97}}{97}$$
$$x_{0} = 0$$
$$y_{0} = 0$$
Centro de las coordenadas canónicas en el punto O
(0, 0)
Base de las coordenadas canónicas
$$\vec e_1 = \left( \frac{9 \sqrt{97}}{97}, \ \frac{4 \sqrt{97}}{97}\right)$$
$$\vec e_2 = \left( - \frac{4 \sqrt{97}}{97}, \ \frac{9 \sqrt{97}}{97}\right)$$