Sr Examen
3*x^2-6*x*y+3*z^2+6*x*z-6*y*z-2*x-6*y-z-4=0 forma canónica
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Solución
Ha introducido
[src]
2 2 -4 - z - 6*y - 2*x + 3*x + 3*z - 6*x*y - 6*y*z + 6*x*z = 0
$$3 x^{2} - 6 x y + 6 x z - 2 x - 6 y z - 6 y + 3 z^{2} - z - 4 = 0$$
3*x^2 - 6*x*y + 6*x*z - 2*x - 6*y*z - 6*y + 3*z^2 - z - 4 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$3 x^{2} - 6 x y + 6 x z - 2 x - 6 y z - 6 y + 3 z^{2} - z - 4 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -3$$
$$a_{13} = 3$$
$$a_{14} = -1$$
$$a_{22} = 0$$
$$a_{23} = -3$$
$$a_{24} = -3$$
$$a_{33} = 3$$
$$a_{34} = - \frac{1}{2}$$
$$a_{44} = -4$$
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
$$I_{1} = 6$$
$$I_{3} = \left|\begin{matrix}3 & -3 & 3\\-3 & 0 & -3\\3 & -3 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & -3 & 3 & -1\\-3 & 0 & -3 & -3\\3 & -3 & 3 & - \frac{1}{2}\\-1 & -3 & - \frac{1}{2} & -4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & -3 & 3\\-3 & - \lambda & -3\\3 & -3 & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = 6$$
$$I_{2} = -18$$
$$I_{3} = 0$$
$$I_{4} = \frac{9}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} + 18 \lambda$$
$$K_{2} = - \frac{137}{4}$$
$$K_{3} = - \frac{39}{4}$$
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
$$\lambda^{3} - 6 \lambda^{2} - 18 \lambda = 0$$
$$\lambda_{1} = 3 - 3 \sqrt{3}$$
$$\lambda_{2} = 3 + 3 \sqrt{3}$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(3 - 3 \sqrt{3}\right) + \tilde y^{2} \left(3 + 3 \sqrt{3}\right) + \frac{\sqrt{2} \tilde z}{2} = 0$$
y
$$\tilde x^{2} \left(3 - 3 \sqrt{3}\right) + \tilde y^{2} \left(3 + 3 \sqrt{3}\right) - \frac{\sqrt{2} \tilde z}{2} = 0$$
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{-3 + 3 \sqrt{3}}} - \frac{\tilde y^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{3 + 3 \sqrt{3}}}\right) = 0$$
y
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{-3 + 3 \sqrt{3}}} - \frac{\tilde y^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{3 + 3 \sqrt{3}}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica
$$3 x^{2} - 6 x y + 6 x z - 2 x - 6 y z - 6 y + 3 z^{2} - z - 4 = 0$$
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
$$a_{11} = 3$$
$$a_{12} = -3$$
$$a_{13} = 3$$
$$a_{14} = -1$$
$$a_{22} = 0$$
$$a_{23} = -3$$
$$a_{24} = -3$$
$$a_{33} = 3$$
$$a_{34} = - \frac{1}{2}$$
$$a_{44} = -4$$
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
$$I_{1} = 6$$
|3 -3| |0 -3| |3 3|
I2 = | | + | | + | |
|-3 0 | |-3 3 | |3 3|$$I_{3} = \left|\begin{matrix}3 & -3 & 3\\-3 & 0 & -3\\3 & -3 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}3 & -3 & 3 & -1\\-3 & 0 & -3 & -3\\3 & -3 & 3 & - \frac{1}{2}\\-1 & -3 & - \frac{1}{2} & -4\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & -3 & 3\\-3 & - \lambda & -3\\3 & -3 & 3 - \lambda\end{matrix}\right|$$
|3 -1| |0 -3| | 3 -1/2|
K2 = | | + | | + | |
|-1 -4| |-3 -4| |-1/2 -4 | |3 -3 -1| |0 -3 -3 | |3 3 -1 |
| | | | | |
K3 = |-3 0 -3| + |-3 3 -1/2| + |3 3 -1/2|
| | | | | |
|-1 -3 -4| |-3 -1/2 -4 | |-1 -1/2 -4 |$$I_{1} = 6$$
$$I_{2} = -18$$
$$I_{3} = 0$$
$$I_{4} = \frac{9}{4}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 6 \lambda^{2} + 18 \lambda$$
$$K_{2} = - \frac{137}{4}$$
$$K_{3} = - \frac{39}{4}$$
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
$$\lambda^{3} - 6 \lambda^{2} - 18 \lambda = 0$$
$$\lambda_{1} = 3 - 3 \sqrt{3}$$
$$\lambda_{2} = 3 + 3 \sqrt{3}$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(3 - 3 \sqrt{3}\right) + \tilde y^{2} \left(3 + 3 \sqrt{3}\right) + \frac{\sqrt{2} \tilde z}{2} = 0$$
y
$$\tilde x^{2} \left(3 - 3 \sqrt{3}\right) + \tilde y^{2} \left(3 + 3 \sqrt{3}\right) - \frac{\sqrt{2} \tilde z}{2} = 0$$
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{-3 + 3 \sqrt{3}}} - \frac{\tilde y^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{3 + 3 \sqrt{3}}}\right) = 0$$
y
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{-3 + 3 \sqrt{3}}} - \frac{\tilde y^{2}}{\frac{1}{4} \sqrt{2} \frac{1}{3 + 3 \sqrt{3}}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica