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4x^2-3y^2-z^2+6z-4x-20=0 forma canónica
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Solución
Ha introducido
[src]
2 2 2 -20 - z - 4*x - 3*y + 4*x + 6*z = 0
$$4 x^{2} - 4 x - 3 y^{2} - z^{2} + 6 z - 20 = 0$$
4*x^2 - 4*x - 3*y^2 - z^2 + 6*z - 20 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$4 x^{2} - 4 x - 3 y^{2} - z^{2} + 6 z - 20 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -2$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = -1$$
$$a_{34} = 3$$
$$a_{44} = -20$$
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
$$I_{1} = 0$$
$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & -3 & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & 0 & 0 & -2\\0 & -3 & 0 & 0\\0 & 0 & -1 & 3\\-2 & 0 & 3 & -20\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0 & 0\\0 & - \lambda - 3 & 0\\0 & 0 & - \lambda - 1\end{matrix}\right|$$
$$I_{1} = 0$$
$$I_{2} = -13$$
$$I_{3} = 12$$
$$I_{4} = -144$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 13 \lambda + 12$$
$$K_{2} = -13$$
$$K_{3} = 267$$
Como
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
$$\lambda^{3} - 13 \lambda - 12 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -3$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$4 \tilde x^{2} - \tilde y^{2} - 3 \tilde z^{2} - 12 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{6}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{1}{\frac{1}{6} \sqrt{3}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{3}}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica
$$4 x^{2} - 4 x - 3 y^{2} - z^{2} + 6 z - 20 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -2$$
$$a_{22} = -3$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = -1$$
$$a_{34} = 3$$
$$a_{44} = -20$$
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
$$I_{1} = 0$$
|4 0 | |-3 0 | |4 0 |
I2 = | | + | | + | |
|0 -3| |0 -1| |0 -1|$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & -3 & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & 0 & 0 & -2\\0 & -3 & 0 & 0\\0 & 0 & -1 & 3\\-2 & 0 & 3 & -20\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0 & 0\\0 & - \lambda - 3 & 0\\0 & 0 & - \lambda - 1\end{matrix}\right|$$
|4 -2 | |-3 0 | |-1 3 |
K2 = | | + | | + | |
|-2 -20| |0 -20| |3 -20| |4 0 -2 | |-3 0 0 | |4 0 -2 |
| | | | | |
K3 = |0 -3 0 | + |0 -1 3 | + |0 -1 3 |
| | | | | |
|-2 0 -20| |0 3 -20| |-2 3 -20|$$I_{1} = 0$$
$$I_{2} = -13$$
$$I_{3} = 12$$
$$I_{4} = -144$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 13 \lambda + 12$$
$$K_{2} = -13$$
$$K_{3} = 267$$
Como
I3 != 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
$$\lambda^{3} - 13 \lambda - 12 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = -3$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$4 \tilde x^{2} - \tilde y^{2} - 3 \tilde z^{2} - 12 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{1}{2 \frac{\sqrt{3}}{6}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{1}{\frac{1}{6} \sqrt{3}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{6} \sqrt{3}}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica