Se da la ecuación de la línea de 2-o orden:
$$\sqrt{3} x^{2} + 2 x y + 10 y - 16 \sqrt{3} = 0$$
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
$$a_{11} = \sqrt{3}$$
$$a_{12} = 1$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 5$$
$$a_{33} = - 16 \sqrt{3}$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}\sqrt{3} & 1\\1 & 0\end{matrix}\right|$$
$$\Delta = -1$$
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
$$\sqrt{3} x_{0} + y_{0} = 0$$
$$x_{0} + 5 = 0$$
entonces
$$x_{0} = -5$$
$$y_{0} = 5 \sqrt{3}$$
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
$$a'_{33} = 5 y_{0} - 16 \sqrt{3}$$
$$a'_{33} = 9 \sqrt{3}$$
entonces la ecuación se transformará en
$$\sqrt{3} x'^{2} + 2 x' y' + 9 \sqrt{3} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{\sqrt{3}}{2}$$
entonces
$$\phi = \frac{\operatorname{acot}{\left(\frac{\sqrt{3}}{2} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{2 \sqrt{7}}{7}$$
$$\cos{\left(2 \phi \right)} = \frac{\sqrt{21}}{7}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}$$
sustituimos coeficientes
$$x' = \tilde x \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde y \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}$$
entonces la ecuación se transformará de
$$\sqrt{3} x'^{2} + 2 x' y' + 9 \sqrt{3} = 0$$
en
$$2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde y \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}\right) + \sqrt{3} \left(\tilde x \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}\right)^{2} + 9 \sqrt{3} = 0$$
simplificamos
$$\frac{3 \sqrt{7} \tilde x^{2}}{14} + 2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} + \frac{\sqrt{3} \tilde x^{2}}{2} - 2 \sqrt{3} \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} + \frac{2 \sqrt{21} \tilde x \tilde y}{7} - 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \frac{3 \sqrt{7} \tilde y^{2}}{14} + \frac{\sqrt{3} \tilde y^{2}}{2} + 9 \sqrt{3} = 0$$
$$\frac{\sqrt{3} \tilde x^{2}}{2} + \frac{\sqrt{7} \tilde x^{2}}{2} - \frac{\sqrt{7} \tilde y^{2}}{2} + \frac{\sqrt{3} \tilde y^{2}}{2} + 9 \sqrt{3} = 0$$
Esta ecuación es una hipérbola
$$\frac{\tilde x^{2}}{9 \sqrt{3} \frac{1}{\frac{\sqrt{3}}{2} + \frac{\sqrt{7}}{2}}} - \frac{\tilde y^{2}}{9 \sqrt{3} \frac{1}{- \frac{\sqrt{3}}{2} + \frac{\sqrt{7}}{2}}} = -1$$
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O
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(-5, 5*\/ 3 )
Base de las coordenadas canónicas
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}, \ \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right)$$