Sr Examen
sqrt(3)*x^2+2*x*y-sqrt(2)*y^2-2*x-2*sqrt(3)y=0 forma canónica
El profesor se sorprenderá mucho al ver tu solución correcta😉
Solución
Ha introducido
[src]
___ 2 ___ 2 ___ -2*x + \/ 3 *x - \/ 2 *y - 2*y*\/ 3 + 2*x*y = 0
$$\sqrt{3} x^{2} + 2 x y - 2 x - \sqrt{2} y^{2} - 2 \sqrt{3} y = 0$$
sqrt(3)*x^2 + 2*x*y - 2*x - sqrt(2)*y^2 - 2*sqrt(3)*y = 0
Solución detallada
Se da la ecuación de la línea de 2-o orden:
$$\sqrt{3} x^{2} + 2 x y - 2 x - \sqrt{2} y^{2} - 2 \sqrt{3} y = 0$$
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
$$a_{11} = \sqrt{3}$$
$$a_{12} = 1$$
$$a_{13} = -1$$
$$a_{22} = - \sqrt{2}$$
$$a_{23} = - \sqrt{3}$$
$$a_{33} = 0$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}\sqrt{3} & 1\\1 & - \sqrt{2}\end{matrix}\right|$$
$$\Delta = - \sqrt{6} - 1$$
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
$$\sqrt{3} x_{0} + y_{0} - 1 = 0$$
$$x_{0} - \sqrt{2} y_{0} - \sqrt{3} = 0$$
entonces
$$x_{0} = \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}}$$
$$y_{0} = - \frac{2 \sqrt{3}}{\sqrt{3} + 3 \sqrt{2}}$$
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
$$a'_{33} = - x_{0} - \sqrt{3} y_{0}$$
$$a'_{33} = - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}}$$
entonces la ecuación se transformará en
$$\sqrt{3} x'^{2} + 2 x' y' - \sqrt{2} y'^{2} - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}$$
entonces
$$\phi = \frac{\operatorname{acot}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right) \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}$$
$$\cos{\left(2 \phi \right)} = \frac{1}{\sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}$$
sustituimos coeficientes
$$x' = \tilde x \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}} + \tilde y \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}$$
entonces la ecuación se transformará de
$$\sqrt{3} x'^{2} + 2 x' y' - \sqrt{2} y'^{2} - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}} = 0$$
en
$$- \sqrt{2} \left(\tilde x \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}} + \tilde y \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}\right)^{2} + 2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}} + \tilde y \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}\right) + \sqrt{3} \left(\tilde x \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}\right)^{2} - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}} = 0$$
simplificamos
$$- \frac{\sqrt{2} \tilde x^{2}}{2} + 2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} + \frac{\sqrt{2} \tilde x^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{\sqrt{3} \tilde x^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{\sqrt{3} \tilde x^{2}}{2} - 2 \sqrt{3} \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} - 2 \sqrt{2} \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} + \frac{2 \tilde x \tilde y}{\sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} - \frac{\sqrt{3} \tilde y^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} - \frac{\sqrt{2} \tilde y^{2}}{2} - \frac{\sqrt{2} \tilde y^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} - 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} + \frac{\sqrt{3} \tilde y^{2}}{2} - \frac{\sqrt{6}}{\sqrt{3} + 3 \sqrt{2}} + \frac{3}{\sqrt{3} + 3 \sqrt{2}} = 0$$
$$\frac{\frac{\left(\sqrt{3} + 3 \sqrt{2}\right) \sqrt{2 \sqrt{6} + 9} \left(- \sqrt{2} \tilde x^{2} + \sqrt{3} \tilde x^{2} - \sqrt{2} \tilde y^{2} + \sqrt{3} \tilde y^{2}\right)}{2} + \frac{\left(\sqrt{3} + 3 \sqrt{2}\right) \left(4 \tilde x^{2} + \tilde x^{2} \sqrt{4 \sqrt{6} + 10} + \tilde x^{2} \sqrt{6 \sqrt{6} + 15} - 4 \sqrt{3} \tilde x \tilde y - 4 \sqrt{2} \tilde x \tilde y + 4 \tilde x \tilde y \sqrt{2 \sqrt{6} + 5} - \tilde y^{2} \sqrt{6 \sqrt{6} + 15} - \tilde y^{2} \sqrt{4 \sqrt{6} + 10} - 4 \tilde y^{2}\right)}{2} + \frac{\left(6 - 2 \sqrt{6}\right) \sqrt{2 \sqrt{6} + 9}}{2}}{\left(\sqrt{3} + 3 \sqrt{2}\right) \sqrt{2 \sqrt{6} + 9}} = 0$$
Esta ecuación es una hipérbola
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O
Base de las coordenadas canónicas
$$\vec e_1 = \left( \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}, \ \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}\right)$$
$$\sqrt{3} x^{2} + 2 x y - 2 x - \sqrt{2} y^{2} - 2 \sqrt{3} y = 0$$
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
$$a_{11} = \sqrt{3}$$
$$a_{12} = 1$$
$$a_{13} = -1$$
$$a_{22} = - \sqrt{2}$$
$$a_{23} = - \sqrt{3}$$
$$a_{33} = 0$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}\sqrt{3} & 1\\1 & - \sqrt{2}\end{matrix}\right|$$
$$\Delta = - \sqrt{6} - 1$$
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
$$\sqrt{3} x_{0} + y_{0} - 1 = 0$$
$$x_{0} - \sqrt{2} y_{0} - \sqrt{3} = 0$$
entonces
$$x_{0} = \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}}$$
$$y_{0} = - \frac{2 \sqrt{3}}{\sqrt{3} + 3 \sqrt{2}}$$
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
$$a'_{33} = - x_{0} - \sqrt{3} y_{0}$$
$$a'_{33} = - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}}$$
entonces la ecuación se transformará en
$$\sqrt{3} x'^{2} + 2 x' y' - \sqrt{2} y'^{2} - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}} = 0$$
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
$$\cot{\left(2 \phi \right)} = \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}$$
entonces
$$\phi = \frac{\operatorname{acot}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \right)}}{2}$$
$$\sin{\left(2 \phi \right)} = \frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right) \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}$$
$$\cos{\left(2 \phi \right)} = \frac{1}{\sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}$$
$$\cos{\left(\phi \right)} = \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}$$
$$\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}$$
sustituimos coeficientes
$$x' = \tilde x \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}$$
$$y' = \tilde x \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}} + \tilde y \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}$$
entonces la ecuación se transformará de
$$\sqrt{3} x'^{2} + 2 x' y' - \sqrt{2} y'^{2} - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}} = 0$$
en
$$- \sqrt{2} \left(\tilde x \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}} + \tilde y \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}\right)^{2} + 2 \left(\tilde x \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}} + \tilde y \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}\right) + \sqrt{3} \left(\tilde x \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}\right)^{2} - \frac{\sqrt{6} + 3}{\sqrt{3} + 3 \sqrt{2}} + \frac{6}{\sqrt{3} + 3 \sqrt{2}} = 0$$
simplificamos
$$- \frac{\sqrt{2} \tilde x^{2}}{2} + 2 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} + \frac{\sqrt{2} \tilde x^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{\sqrt{3} \tilde x^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{\sqrt{3} \tilde x^{2}}{2} - 2 \sqrt{3} \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} - 2 \sqrt{2} \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} + \frac{2 \tilde x \tilde y}{\sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} - \frac{\sqrt{3} \tilde y^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} - \frac{\sqrt{2} \tilde y^{2}}{2} - \frac{\sqrt{2} \tilde y^{2}}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} - 2 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}}} \sqrt{\frac{1}{2 \sqrt{\frac{1}{\frac{\sqrt{6}}{2} + \frac{5}{4}} + 1}} + \frac{1}{2}} + \frac{\sqrt{3} \tilde y^{2}}{2} - \frac{\sqrt{6}}{\sqrt{3} + 3 \sqrt{2}} + \frac{3}{\sqrt{3} + 3 \sqrt{2}} = 0$$
$$\frac{\frac{\left(\sqrt{3} + 3 \sqrt{2}\right) \sqrt{2 \sqrt{6} + 9} \left(- \sqrt{2} \tilde x^{2} + \sqrt{3} \tilde x^{2} - \sqrt{2} \tilde y^{2} + \sqrt{3} \tilde y^{2}\right)}{2} + \frac{\left(\sqrt{3} + 3 \sqrt{2}\right) \left(4 \tilde x^{2} + \tilde x^{2} \sqrt{4 \sqrt{6} + 10} + \tilde x^{2} \sqrt{6 \sqrt{6} + 15} - 4 \sqrt{3} \tilde x \tilde y - 4 \sqrt{2} \tilde x \tilde y + 4 \tilde x \tilde y \sqrt{2 \sqrt{6} + 5} - \tilde y^{2} \sqrt{6 \sqrt{6} + 15} - \tilde y^{2} \sqrt{4 \sqrt{6} + 10} - 4 \tilde y^{2}\right)}{2} + \frac{\left(6 - 2 \sqrt{6}\right) \sqrt{2 \sqrt{6} + 9}}{2}}{\left(\sqrt{3} + 3 \sqrt{2}\right) \sqrt{2 \sqrt{6} + 9}} = 0$$
Esta ecuación es una hipérbola
None
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O
___ ___
3 + \/ 6 -2*\/ 3
(---------------, ---------------)
___ ___ ___ ___
\/ 3 + 3*\/ 2 \/ 3 + 3*\/ 2 Base de las coordenadas canónicas
$$\vec e_1 = \left( \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}\right)$$
$$\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}}}, \ \sqrt{\frac{1}{2 \sqrt{\frac{1}{\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right)^{2}} + 1}} + \frac{1}{2}}\right)$$
Método de invariantes
Se da la ecuación de la línea de 2-o orden:
$$\sqrt{3} x^{2} + 2 x y - 2 x - \sqrt{2} y^{2} - 2 \sqrt{3} y = 0$$
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
$$a_{11} = \sqrt{3}$$
$$a_{12} = 1$$
$$a_{13} = -1$$
$$a_{22} = - \sqrt{2}$$
$$a_{23} = - \sqrt{3}$$
$$a_{33} = 0$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
sustituimos coeficientes
$$I_{1} = - \sqrt{2} + \sqrt{3}$$
$$I_{3} = \left|\begin{matrix}\sqrt{3} & 1 & -1\\1 & - \sqrt{2} & - \sqrt{3}\\-1 & - \sqrt{3} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + \sqrt{3} & 1\\1 & - \lambda - \sqrt{2}\end{matrix}\right|$$
$$I_{1} = - \sqrt{2} + \sqrt{3}$$
$$I_{2} = - \sqrt{6} - 1$$
$$I_{3} = - \sqrt{3} + \sqrt{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \sqrt{3} \lambda + \sqrt{2} \lambda - \sqrt{6} - 1$$
$$K_{2} = -4$$
Como
$$I_{2} < 0 \wedge I_{3} \neq 0$$
entonces por razón de tipos de rectas:
esta ecuación tiene el tipo : hipérbola
Formulamos la ecuación característica para nuestra línea:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
o
$$\lambda^{2} - \lambda \left(- \sqrt{2} + \sqrt{3}\right) - \sqrt{6} - 1 = 0$$
$$\lambda_{1} = - \frac{\sqrt{2 \sqrt{6} + 9}}{2} - \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}$$
$$\lambda_{2} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}$$
entonces la forma canónica de la ecuación será
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
o
$$\tilde x^{2} \left(- \frac{\sqrt{2 \sqrt{6} + 9}}{2} - \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right) + \tilde y^{2} \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}\right) + \frac{- \sqrt{3} + \sqrt{2}}{- \sqrt{6} - 1} = 0$$
$$\frac{\tilde x^{2}}{\frac{1}{1 + \sqrt{6}} \left(- \sqrt{2} + \sqrt{3}\right) \frac{1}{- \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}}} - \frac{\tilde y^{2}}{\frac{1}{1 + \sqrt{6}} \left(- \sqrt{2} + \sqrt{3}\right) \frac{1}{- \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}}} = 1$$
- está reducida a la forma canónica
$$\sqrt{3} x^{2} + 2 x y - 2 x - \sqrt{2} y^{2} - 2 \sqrt{3} y = 0$$
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
$$a_{11} = \sqrt{3}$$
$$a_{12} = 1$$
$$a_{13} = -1$$
$$a_{22} = - \sqrt{2}$$
$$a_{23} = - \sqrt{3}$$
$$a_{33} = 0$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22}$$
|a11 a12|
I2 = | |
|a12 a22|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
|a11 a13| |a22 a23|
K2 = | | + | |
|a13 a33| |a23 a33|sustituimos coeficientes
$$I_{1} = - \sqrt{2} + \sqrt{3}$$
| ___ |
|\/ 3 1 |
I2 = | |
| ___|
| 1 -\/ 2 |$$I_{3} = \left|\begin{matrix}\sqrt{3} & 1 & -1\\1 & - \sqrt{2} & - \sqrt{3}\\-1 & - \sqrt{3} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + \sqrt{3} & 1\\1 & - \lambda - \sqrt{2}\end{matrix}\right|$$
| ___ | | ___ ___|
|\/ 3 -1| |-\/ 2 -\/ 3 |
K2 = | | + | |
| -1 0 | | ___ |
|-\/ 3 0 |$$I_{1} = - \sqrt{2} + \sqrt{3}$$
$$I_{2} = - \sqrt{6} - 1$$
$$I_{3} = - \sqrt{3} + \sqrt{2}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \sqrt{3} \lambda + \sqrt{2} \lambda - \sqrt{6} - 1$$
$$K_{2} = -4$$
Como
$$I_{2} < 0 \wedge I_{3} \neq 0$$
entonces por razón de tipos de rectas:
esta ecuación tiene el tipo : hipérbola
Formulamos la ecuación característica para nuestra línea:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
o
$$\lambda^{2} - \lambda \left(- \sqrt{2} + \sqrt{3}\right) - \sqrt{6} - 1 = 0$$
$$\lambda_{1} = - \frac{\sqrt{2 \sqrt{6} + 9}}{2} - \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}$$
$$\lambda_{2} = - \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}$$
entonces la forma canónica de la ecuación será
$$\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0$$
o
$$\tilde x^{2} \left(- \frac{\sqrt{2 \sqrt{6} + 9}}{2} - \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right) + \tilde y^{2} \left(- \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}\right) + \frac{- \sqrt{3} + \sqrt{2}}{- \sqrt{6} - 1} = 0$$
$$\frac{\tilde x^{2}}{\frac{1}{1 + \sqrt{6}} \left(- \sqrt{2} + \sqrt{3}\right) \frac{1}{- \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}}} - \frac{\tilde y^{2}}{\frac{1}{1 + \sqrt{6}} \left(- \sqrt{2} + \sqrt{3}\right) \frac{1}{- \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{2 \sqrt{6} + 9}}{2}}} = 1$$
- está reducida a la forma canónica