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sqrt3*x2*x3-(3*x3^2)/2 forma canónica
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Solución
Ha introducido
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2
3*x3 ___
- ----- + x2*x3*\/ 3 = 0
2
$$\sqrt{3} x_{2} x_{3} - \frac{3 x_{3}^{2}}{2} = 0$$
sqrt(3)*x2*x3 - 3*x3^2/2 = 0
Solución detallada
Se da la ecuación de la línea de 2-o orden:
$$\sqrt{3} x_{2} x_{3} - \frac{3 x_{3}^{2}}{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
donde
$$a_{11} = - \frac{3}{2}$$
$$a_{12} = \frac{\sqrt{3}}{2}$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 0$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}- \frac{3}{2} & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 0\end{matrix}\right|$$
$$\Delta = - \frac{3}{4}$$
Como
$$\Delta$$
no es igual a 0, entonces
hallamos el centro de coordenadas canónicas. Para eso resolvemos el sistema de ecuaciones
$$a_{11} x_{30} + a_{12} x_{20} + a_{13} = 0$$
$$a_{12} x_{30} + a_{22} x_{20} + a_{23} = 0$$
sustituimos coeficientes
$$\frac{\sqrt{3} x_{20}}{2} - \frac{3 x_{30}}{2} = 0$$
$$\frac{\sqrt{3} x_{30}}{2} = 0$$
entonces
$$x_{30} = 0$$
$$x_{20} = 0$$
Así pasamos a la ecuación en el sistema de coordenadas O'x'y'
$$a'_{33} + a_{11} x3'^{2} + 2 a_{12} x2' x3' + a_{22} x2'^{2} = 0$$
donde
$$a'_{33} = a_{13} x_{30} + a_{23} x_{20} + a_{33}$$
o
$$a'_{33} = 0$$
$$a'_{33} = 0$$
entonces la ecuación se transformará en
$$\sqrt{3} x2' x3' - \frac{3 x3'^{2}}{2} = 0$$
Hacemos el giro del sistema de coordenadas obtenido al ángulo de φ
$$x3' = - \tilde x2 \sin{\left(\phi \right)} + \tilde x3 \cos{\left(\phi \right)}$$
$$x2' = \tilde x2 \cos{\left(\phi \right)} + \tilde x3 \sin{\left(\phi \right)}$$
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
$$x3' = \tilde x2 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde x3 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}$$
$$x2' = \tilde x2 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \tilde x3 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}$$
entonces la ecuación se transformará de
$$\sqrt{3} x2' x3' - \frac{3 x3'^{2}}{2} = 0$$
en
$$- \frac{3 \left(\tilde x2 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde x3 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right)^{2}}{2} + \sqrt{3} \left(\tilde x2 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde x3 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right) \left(\tilde x2 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \tilde x3 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}\right) = 0$$
simplificamos
$$- \frac{3 \tilde x2^{2}}{4} + \frac{3 \sqrt{21} \tilde x2^{2}}{28} + \sqrt{3} \tilde x2^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - 3 \tilde x2 \tilde x3 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} + \frac{3 \sqrt{7} \tilde x2 \tilde x3}{7} - \frac{3 \tilde x3^{2}}{4} - \sqrt{3} \tilde x3^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \frac{3 \sqrt{21} \tilde x3^{2}}{28} = 0$$
$$- \frac{\sqrt{21} \tilde x2^{2}}{4} + \frac{3 \tilde x2^{2}}{4} + \frac{3 \tilde x3^{2}}{4} + \frac{\sqrt{21} \tilde x3^{2}}{4} = 0$$
Esta ecuación es una hipérbola degenerada
$$- \frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{- \frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{\sqrt{\frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} = 0$$
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O
Base de las coordenadas canónicas
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}, \ \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right)$$
$$\sqrt{3} x_{2} x_{3} - \frac{3 x_{3}^{2}}{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
donde
$$a_{11} = - \frac{3}{2}$$
$$a_{12} = \frac{\sqrt{3}}{2}$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 0$$
Calculemos el determinante
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
o, sustituimos
$$\Delta = \left|\begin{matrix}- \frac{3}{2} & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & 0\end{matrix}\right|$$
$$\Delta = - \frac{3}{4}$$
Como
$$\Delta$$
no es igual a 0, entonces
hallamos el centro de coordenadas canónicas. Para eso resolvemos el sistema de ecuaciones
$$a_{11} x_{30} + a_{12} x_{20} + a_{13} = 0$$
$$a_{12} x_{30} + a_{22} x_{20} + a_{23} = 0$$
sustituimos coeficientes
$$\frac{\sqrt{3} x_{20}}{2} - \frac{3 x_{30}}{2} = 0$$
$$\frac{\sqrt{3} x_{30}}{2} = 0$$
entonces
$$x_{30} = 0$$
$$x_{20} = 0$$
Así pasamos a la ecuación en el sistema de coordenadas O'x'y'
$$a'_{33} + a_{11} x3'^{2} + 2 a_{12} x2' x3' + a_{22} x2'^{2} = 0$$
donde
$$a'_{33} = a_{13} x_{30} + a_{23} x_{20} + a_{33}$$
o
$$a'_{33} = 0$$
$$a'_{33} = 0$$
entonces la ecuación se transformará en
$$\sqrt{3} x2' x3' - \frac{3 x3'^{2}}{2} = 0$$
Hacemos el giro del sistema de coordenadas obtenido al ángulo de φ
$$x3' = - \tilde x2 \sin{\left(\phi \right)} + \tilde x3 \cos{\left(\phi \right)}$$
$$x2' = \tilde x2 \cos{\left(\phi \right)} + \tilde x3 \sin{\left(\phi \right)}$$
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
$$x3' = \tilde x2 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde x3 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}$$
$$x2' = \tilde x2 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \tilde x3 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}$$
entonces la ecuación se transformará de
$$\sqrt{3} x2' x3' - \frac{3 x3'^{2}}{2} = 0$$
en
$$- \frac{3 \left(\tilde x2 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde x3 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right)^{2}}{2} + \sqrt{3} \left(\tilde x2 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} + \tilde x3 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right) \left(\tilde x2 \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \tilde x3 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}\right) = 0$$
simplificamos
$$- \frac{3 \tilde x2^{2}}{4} + \frac{3 \sqrt{21} \tilde x2^{2}}{28} + \sqrt{3} \tilde x2^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - 3 \tilde x2 \tilde x3 \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} + \frac{3 \sqrt{7} \tilde x2 \tilde x3}{7} - \frac{3 \tilde x3^{2}}{4} - \sqrt{3} \tilde x3^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}} \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}} - \frac{3 \sqrt{21} \tilde x3^{2}}{28} = 0$$
$$- \frac{\sqrt{21} \tilde x2^{2}}{4} + \frac{3 \tilde x2^{2}}{4} + \frac{3 \tilde x3^{2}}{4} + \frac{\sqrt{21} \tilde x3^{2}}{4} = 0$$
Esta ecuación es una hipérbola degenerada
$$- \frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{- \frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{\sqrt{\frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} = 0$$
- está reducida a la forma canónica
Centro de las coordenadas canónicas en el punto O
(0, 0)
Base de las coordenadas canónicas
$$\vec e_1 = \left( \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}\right)$$
$$\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{21}}{14}}, \ \sqrt{\frac{\sqrt{21}}{14} + \frac{1}{2}}\right)$$
Método de invariantes
Se da la ecuación de la línea de 2-o orden:
$$\sqrt{3} x_{2} x_{3} - \frac{3 x_{3}^{2}}{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
donde
$$a_{11} = - \frac{3}{2}$$
$$a_{12} = \frac{\sqrt{3}}{2}$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = - \frac{3}{2}$$
$$I_{3} = \left|\begin{matrix}- \frac{3}{2} & \frac{\sqrt{3}}{2} & 0\\\frac{\sqrt{3}}{2} & 0 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - \frac{3}{2} & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & - \lambda\end{matrix}\right|$$
$$I_{1} = - \frac{3}{2}$$
$$I_{2} = - \frac{3}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + \frac{3 \lambda}{2} - \frac{3}{4}$$
$$K_{2} = 0$$
Como
$$I_{3} = 0 \wedge I_{2} < 0$$
entonces por razón de tipos de rectas:
esta ecuación tiene el tipo : hipérbola degenerada
Formulamos la ecuación característica para nuestra línea:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
o
$$\lambda^{2} + \frac{3 \lambda}{2} - \frac{3}{4} = 0$$
$$\lambda_{1} = - \frac{\sqrt{21}}{4} - \frac{3}{4}$$
$$\lambda_{2} = - \frac{3}{4} + \frac{\sqrt{21}}{4}$$
entonces la forma canónica de la ecuación será
$$\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
o
$$\tilde x2^{2} \left(- \frac{3}{4} + \frac{\sqrt{21}}{4}\right) + \tilde x3^{2} \left(- \frac{\sqrt{21}}{4} - \frac{3}{4}\right) = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{- \frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{\sqrt{\frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} = 0$$
- está reducida a la forma canónica
$$\sqrt{3} x_{2} x_{3} - \frac{3 x_{3}^{2}}{2} = 0$$
Esta ecuación tiene la forma:
$$a_{11} x_{3}^{2} + 2 a_{12} x_{2} x_{3} + 2 a_{13} x_{3} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0$$
donde
$$a_{11} = - \frac{3}{2}$$
$$a_{12} = \frac{\sqrt{3}}{2}$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{33} = 0$$
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
$$I_{1} = - \frac{3}{2}$$
| ___|
| \/ 3 |
|-3/2 -----|
| 2 |
I2 = | |
| ___ |
|\/ 3 |
|----- 0 |
| 2 |$$I_{3} = \left|\begin{matrix}- \frac{3}{2} & \frac{\sqrt{3}}{2} & 0\\\frac{\sqrt{3}}{2} & 0 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - \frac{3}{2} & \frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2} & - \lambda\end{matrix}\right|$$
|-3/2 0| |0 0|
K2 = | | + | |
| 0 0| |0 0|$$I_{1} = - \frac{3}{2}$$
$$I_{2} = - \frac{3}{4}$$
$$I_{3} = 0$$
$$I{\left(\lambda \right)} = \lambda^{2} + \frac{3 \lambda}{2} - \frac{3}{4}$$
$$K_{2} = 0$$
Como
$$I_{3} = 0 \wedge I_{2} < 0$$
entonces por razón de tipos de rectas:
esta ecuación tiene el tipo : hipérbola degenerada
Formulamos la ecuación característica para nuestra línea:
$$- I_{1} \lambda + I_{2} + \lambda^{2} = 0$$
o
$$\lambda^{2} + \frac{3 \lambda}{2} - \frac{3}{4} = 0$$
$$\lambda_{1} = - \frac{\sqrt{21}}{4} - \frac{3}{4}$$
$$\lambda_{2} = - \frac{3}{4} + \frac{\sqrt{21}}{4}$$
entonces la forma canónica de la ecuación será
$$\tilde x2^{2} \lambda_{2} + \tilde x3^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0$$
o
$$\tilde x2^{2} \left(- \frac{3}{4} + \frac{\sqrt{21}}{4}\right) + \tilde x3^{2} \left(- \frac{\sqrt{21}}{4} - \frac{3}{4}\right) = 0$$
$$- \frac{\tilde x2^{2}}{\left(\frac{1}{\sqrt{- \frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} + \frac{\tilde x3^{2}}{\left(\frac{1}{\sqrt{\frac{3}{4} + \frac{\sqrt{21}}{4}}}\right)^{2}} = 0$$
- está reducida a la forma canónica