Sr Examen
xz-4y+4z=0 forma canónica
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Solución
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$x z - 4 y + 4 z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = -2$$
$$a_{33} = 0$$
$$a_{34} = 2$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & \frac{1}{2} & 0\\0 & 0 & 0 & -2\\\frac{1}{2} & 0 & 0 & 2\\0 & -2 & 2 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & \frac{1}{2}\\0 & - \lambda & 0\\\frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = 1$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = -8$$
$$K_{3} = 0$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$- \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} + 4 \tilde z = 0$$
y
$$- \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} - 4 \tilde z = 0$$
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4}\right) = 0$$
y
$$2 \tilde z + \left(\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica
$$x z - 4 y + 4 z = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = \frac{1}{2}$$
$$a_{14} = 0$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = -2$$
$$a_{33} = 0$$
$$a_{34} = 2$$
$$a_{44} = 0$$
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
$$I_{1} = 0$$
|0 0| |0 0| | 0 1/2|
I2 = | | + | | + | |
|0 0| |0 0| |1/2 0 |$$I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 0 & 0\\\frac{1}{2} & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & \frac{1}{2} & 0\\0 & 0 & 0 & -2\\\frac{1}{2} & 0 & 0 & 2\\0 & -2 & 2 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & \frac{1}{2}\\0 & - \lambda & 0\\\frac{1}{2} & 0 & - \lambda\end{matrix}\right|$$
|0 0| |0 -2| |0 2|
K2 = | | + | | + | |
|0 0| |-2 0 | |2 0| |0 0 0 | |0 0 -2| | 0 1/2 0|
| | | | | |
K3 = |0 0 -2| + |0 0 2 | + |1/2 0 2|
| | | | | |
|0 -2 0 | |-2 2 0 | | 0 2 0|$$I_{1} = 0$$
$$I_{2} = - \frac{1}{4}$$
$$I_{3} = 0$$
$$I_{4} = 1$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \frac{\lambda}{4}$$
$$K_{2} = -8$$
$$K_{3} = 0$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - \frac{\lambda}{4} = 0$$
$$\lambda_{1} = - \frac{1}{2}$$
$$\lambda_{2} = \frac{1}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$- \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} + 4 \tilde z = 0$$
y
$$- \frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2} - 4 \tilde z = 0$$
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4}\right) = 0$$
y
$$2 \tilde z + \left(\frac{\tilde x^{2}}{4} - \frac{\tilde y^{2}}{4}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica