Sr Examen
z^2-2sqrt(2)xz-2sqrt(6)x-2sqrt(3)z-1=0 forma canónica
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Solución
Ha introducido
[src]
2 ___ ___ ___ -1 + z - 2*x*\/ 6 - 2*z*\/ 3 - 2*x*z*\/ 2 = 0
$$- 2 \sqrt{2} x z - 2 \sqrt{6} x + z^{2} - 2 \sqrt{3} z - 1 = 0$$
-2*sqrt(2)*x*z - 2*sqrt(6)*x + z^2 - 2*sqrt(3)*z - 1 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$- 2 \sqrt{2} x z - 2 \sqrt{6} x + z^{2} - 2 \sqrt{3} z - 1 = 0$$
Esta ecuación tiene la forma:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
donde
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \sqrt{2}$$
$$a_{14} = - \sqrt{6}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = - \sqrt{3}$$
$$a_{44} = -1$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
sustituimos coeficientes
$$I_{1} = 1$$
$$I_{3} = \left|\begin{matrix}0 & 0 & - \sqrt{2}\\0 & 0 & 0\\- \sqrt{2} & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & - \sqrt{2} & - \sqrt{6}\\0 & 0 & 0 & 0\\- \sqrt{2} & 0 & 1 & - \sqrt{3}\\- \sqrt{6} & 0 & - \sqrt{3} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & - \sqrt{2}\\0 & - \lambda & 0\\- \sqrt{2} & 0 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = 1$$
$$I_{2} = -2$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + 2 \lambda$$
$$K_{2} = -10$$
$$K_{3} = -16$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
entonces por razón de tipos de rectas:
hay que
Formulamos la ecuación característica para nuestra superficie:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
o
$$\lambda^{3} - \lambda^{2} - 2 \lambda = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$2 \tilde x^{2} - \tilde y^{2} + 8 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{8} = -1$$
es la ecuación para el tipo cilindro hiperbólico
- está reducida a la forma canónica
$$- 2 \sqrt{2} x z - 2 \sqrt{6} x + z^{2} - 2 \sqrt{3} z - 1 = 0$$
Esta ecuación tiene la forma:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
donde
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = - \sqrt{2}$$
$$a_{14} = - \sqrt{6}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = - \sqrt{3}$$
$$a_{44} = -1$$
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
|a11 a12| |a22 a23| |a11 a13|
I2 = | | + | | + | |
|a12 a22| |a23 a33| |a13 a33|$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|$$
|a11 a14| |a22 a24| |a33 a34|
K2 = | | + | | + | |
|a14 a44| |a24 a44| |a34 a44| |a11 a12 a14| |a22 a23 a24| |a11 a13 a14|
| | | | | |
K3 = |a12 a22 a24| + |a23 a33 a34| + |a13 a33 a34|
| | | | | |
|a14 a24 a44| |a24 a34 a44| |a14 a34 a44|sustituimos coeficientes
$$I_{1} = 1$$
| ___|
|0 0| |0 0| | 0 -\/ 2 |
I2 = | | + | | + | |
|0 0| |0 1| | ___ |
|-\/ 2 1 |$$I_{3} = \left|\begin{matrix}0 & 0 & - \sqrt{2}\\0 & 0 & 0\\- \sqrt{2} & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & - \sqrt{2} & - \sqrt{6}\\0 & 0 & 0 & 0\\- \sqrt{2} & 0 & 1 & - \sqrt{3}\\- \sqrt{6} & 0 & - \sqrt{3} & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & - \sqrt{2}\\0 & - \lambda & 0\\- \sqrt{2} & 0 & 1 - \lambda\end{matrix}\right|$$
| ___| | ___|
| 0 -\/ 6 | |0 0 | | 1 -\/ 3 |
K2 = | | + | | + | |
| ___ | |0 -1| | ___ |
|-\/ 6 -1 | |-\/ 3 -1 | | ___ ___|
| ___| |0 0 0 | | 0 -\/ 2 -\/ 6 |
| 0 0 -\/ 6 | | | | |
| | | ___| | ___ ___|
K3 = | 0 0 0 | + |0 1 -\/ 3 | + |-\/ 2 1 -\/ 3 |
| | | | | |
| ___ | | ___ | | ___ ___ |
|-\/ 6 0 -1 | |0 -\/ 3 -1 | |-\/ 6 -\/ 3 -1 |
$$I_{1} = 1$$
$$I_{2} = -2$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + 2 \lambda$$
$$K_{2} = -10$$
$$K_{3} = -16$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
entonces por razón de tipos de rectas:
hay que
Formulamos la ecuación característica para nuestra superficie:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
o
$$\lambda^{3} - \lambda^{2} - 2 \lambda = 0$$
$$\lambda_{1} = 2$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$2 \tilde x^{2} - \tilde y^{2} + 8 = 0$$
$$\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{8} = -1$$
es la ecuación para el tipo cilindro hiperbólico
- está reducida a la forma canónica