8x^2-2sqrt(6)xy-2sqrt(10)x+4sqrt(15)x+7y^2-4sqrt(10)y-2sqrt(15)+5 forma canónica

Se da la ecuación de la línea de 2-o orden:
$$8 x^{2} - 2 \sqrt{6} x y - 2 \sqrt{10} x + 4 \sqrt{15} x + 7 y^{2} - 4 \sqrt{10} y - 2 \sqrt{15} + 5 = 0$$
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
$$a_{11} = 8$$
$$a_{12} = - \sqrt{6}$$
$$a_{13} = - \sqrt{10} + 2 \sqrt{15}$$
$$a_{22} = 7$$
$$a_{23} = - 2 \sqrt{10}$$
$$a_{33} = 5 - 2 \sqrt{15}$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}8 & - \sqrt{6}\\- \sqrt{6} & 7\end{matrix}\right|$$
$$\Delta = 50$$
Como
$$\Delta$$
no es igual a 0, entonces
hallamos el centro de coordenadas canónicas. Para eso resolvemos el sistema de ecuaciones
$$a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0$$
$$a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0$$
sustituimos coeficientes
$$8 x_{0} - \sqrt{6} y_{0} - \sqrt{10} + 2 \sqrt{15} = 0$$
$$- \sqrt{6} x_{0} + 7 y_{0} - 2 \sqrt{10} = 0$$
entonces
$$x_{0} = - \frac{\sqrt{15}}{5} + \frac{7 \sqrt{10}}{50}$$
$$y_{0} = \frac{\sqrt{15}}{25} + \frac{\sqrt{10}}{5}$$
Así pasamos a la ecuación en el sistema de coordenadas O'x'y'
$$a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0$$
donde
$$a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}$$
o
$$a'_{33} = x_{0} \left(- \sqrt{10} + 2 \sqrt{15}\right) - 2 \sqrt{10} y_{0} - 2 \sqrt{15} + 5$$
$$a'_{33} = - 2 \sqrt{15} - 2 \sqrt{10} \left(\frac{\sqrt{15}}{25} + \frac{\sqrt{10}}{5}\right) + \left(- \sqrt{10} + 2 \sqrt{15}\right) \left(- \frac{\sqrt{15}}{5} + \frac{7 \sqrt{10}}{50}\right) + 5$$
entonces la ecuación se transformará en
$$8 x'^{2} - 2 \sqrt{6} x' y' + 7 y'^{2} - 2 \sqrt{15} - 2 \sqrt{10} \left(\frac{\sqrt{15}}{25} + \frac{\sqrt{10}}{5}\right) + \left(- \sqrt{10} + 2 \sqrt{15}\right) \left(- \frac{\sqrt{15}}{5} + \frac{7 \sqrt{10}}{50}\right) + 5 = 0$$
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
$$\cot{\left(2 \phi \right)} = - \frac{\sqrt{6}}{12}$$
entonces
$$\phi = - \frac{\operatorname{acot}{\left(\frac{\sqrt{6}}{12} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = - \frac{2 \sqrt{6}}{5}$$
$$\cos{\left(2 \phi \right)} = \frac{1}{5}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \frac{\sqrt{15}}{5}$$
$$\sin{\left(\phi \right)} = - \frac{\sqrt{10}}{5}$$
sustituimos coeficientes
$$x' = \frac{\sqrt{15} \tilde x}{5} + \frac{\sqrt{10} \tilde y}{5}$$
$$y' = - \frac{\sqrt{10} \tilde x}{5} + \frac{\sqrt{15} \tilde y}{5}$$
entonces la ecuación se transformará de
$$8 x'^{2} - 2 \sqrt{6} x' y' + 7 y'^{2} - 2 \sqrt{15} - 2 \sqrt{10} \left(\frac{\sqrt{15}}{25} + \frac{\sqrt{10}}{5}\right) + \left(- \sqrt{10} + 2 \sqrt{15}\right) \left(- \frac{\sqrt{15}}{5} + \frac{7 \sqrt{10}}{50}\right) + 5 = 0$$
en
$$7 \left(- \frac{\sqrt{10} \tilde x}{5} + \frac{\sqrt{15} \tilde y}{5}\right)^{2} - 2 \sqrt{6} \left(- \frac{\sqrt{10} \tilde x}{5} + \frac{\sqrt{15} \tilde y}{5}\right) \left(\frac{\sqrt{15} \tilde x}{5} + \frac{\sqrt{10} \tilde y}{5}\right) + 8 \left(\frac{\sqrt{15} \tilde x}{5} + \frac{\sqrt{10} \tilde y}{5}\right)^{2} - 2 \sqrt{15} - 2 \sqrt{10} \left(\frac{\sqrt{15}}{25} + \frac{\sqrt{10}}{5}\right) + \left(- \sqrt{10} + 2 \sqrt{15}\right) \left(- \frac{\sqrt{15}}{5} + \frac{7 \sqrt{10}}{50}\right) + 5 = 0$$
simplificamos
$$10 \tilde x^{2} + 5 \tilde y^{2} - 2 \sqrt{15} - \frac{32}{5} + 2 \sqrt{6} = 0$$
Esta ecuación es una elipsis
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{10} \sqrt{10}}{\frac{1}{\sqrt{- 2 \sqrt{6} + \frac{32}{5} + 2 \sqrt{15}}}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{\sqrt{- 2 \sqrt{6} + \frac{32}{5} + 2 \sqrt{15}}}}\right)^{2}} = 1$$
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O

     ____       ____    ____     ____ 
   \/ 15    7*\/ 10   \/ 10    \/ 15  
(- ------ + --------, ------ + ------)
     5         50       5        25   

Base de las coordenadas canónicas
$$\vec e_1 = \left( \frac{\sqrt{15}}{5}, \ - \frac{\sqrt{10}}{5}\right)$$
$$\vec e_2 = \left( \frac{\sqrt{10}}{5}, \ \frac{\sqrt{15}}{5}\right)$$