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2x^2-y^2+2z^2-8x+6y-12z-10=0 forma canónica
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Solución
Ha introducido
[src]
2 2 2 -10 - y - 12*z - 8*x + 2*x + 2*z + 6*y = 0
$$2 x^{2} - 8 x - y^{2} + 6 y + 2 z^{2} - 12 z - 10 = 0$$
2*x^2 - 8*x - y^2 + 6*y + 2*z^2 - 12*z - 10 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$2 x^{2} - 8 x - y^{2} + 6 y + 2 z^{2} - 12 z - 10 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -4$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{24} = 3$$
$$a_{33} = 2$$
$$a_{34} = -6$$
$$a_{44} = -10$$
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
$$I_{1} = 3$$
$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & -1 & 0\\0 & 0 & 2\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 0 & 0 & -4\\0 & -1 & 0 & 3\\0 & 0 & 2 & -6\\-4 & 3 & -6 & -10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0 & 0\\0 & - \lambda - 1 & 0\\0 & 0 & 2 - \lambda\end{matrix}\right|$$
$$I_{1} = 3$$
$$I_{2} = 0$$
$$I_{3} = -4$$
$$I_{4} = 108$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - 4$$
$$K_{2} = -91$$
$$K_{3} = -88$$
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
$$\lambda^{3} - 3 \lambda^{2} + 4 = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 2$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \tilde x^{2} + 2 \tilde y^{2} + 2 \tilde z^{2} - 27 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{1}{\frac{1}{9} \sqrt{3}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{9} \sqrt{3}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{9} \sqrt{3}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica
$$2 x^{2} - 8 x - y^{2} + 6 y + 2 z^{2} - 12 z - 10 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -4$$
$$a_{22} = -1$$
$$a_{23} = 0$$
$$a_{24} = 3$$
$$a_{33} = 2$$
$$a_{34} = -6$$
$$a_{44} = -10$$
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
$$I_{1} = 3$$
|2 0 | |-1 0| |2 0|
I2 = | | + | | + | |
|0 -1| |0 2| |0 2|$$I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & -1 & 0\\0 & 0 & 2\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 0 & 0 & -4\\0 & -1 & 0 & 3\\0 & 0 & 2 & -6\\-4 & 3 & -6 & -10\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0 & 0\\0 & - \lambda - 1 & 0\\0 & 0 & 2 - \lambda\end{matrix}\right|$$
|2 -4 | |-1 3 | |2 -6 |
K2 = | | + | | + | |
|-4 -10| |3 -10| |-6 -10| |2 0 -4 | |-1 0 3 | |2 0 -4 |
| | | | | |
K3 = |0 -1 3 | + |0 2 -6 | + |0 2 -6 |
| | | | | |
|-4 3 -10| |3 -6 -10| |-4 -6 -10|$$I_{1} = 3$$
$$I_{2} = 0$$
$$I_{3} = -4$$
$$I_{4} = 108$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - 4$$
$$K_{2} = -91$$
$$K_{3} = -88$$
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
$$\lambda^{3} - 3 \lambda^{2} + 4 = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = 2$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \tilde x^{2} + 2 \tilde y^{2} + 2 \tilde z^{2} - 27 = 0$$
$$- \frac{\tilde x^{2}}{\left(\frac{1}{\frac{1}{9} \sqrt{3}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{9} \sqrt{3}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\frac{1}{2} \sqrt{2}}{\frac{1}{9} \sqrt{3}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo hiperboloide unilateral
- está reducida a la forma canónica