Sr Examen
x^2-5y^2+z^2+4xy+2xz+4yz+3(2^(1/2))x-3(2^(1/2))z+12=0 forma canónica
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Solución
Ha introducido
[src]
2 2 2 ___ ___ 12 + x + z - 5*y - 3*z*\/ 2 + 2*x*z + 3*x*\/ 2 + 4*x*y + 4*y*z = 0
$$x^{2} + 4 x y + 2 x z + 3 \sqrt{2} x - 5 y^{2} + 4 y z + z^{2} - 3 \sqrt{2} z + 12 = 0$$
x^2 + 4*x*y + 2*x*z + 3*sqrt(2)*x - 5*y^2 + 4*y*z + z^2 - 3*sqrt(2)*z + 12 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$x^{2} + 4 x y + 2 x z + 3 \sqrt{2} x - 5 y^{2} + 4 y z + z^{2} - 3 \sqrt{2} z + 12 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = 1$$
$$a_{14} = \frac{3 \sqrt{2}}{2}$$
$$a_{22} = -5$$
$$a_{23} = 2$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = - \frac{3 \sqrt{2}}{2}$$
$$a_{44} = 12$$
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
$$I_{1} = -3$$
$$I_{3} = \left|\begin{matrix}1 & 2 & 1\\2 & -5 & 2\\1 & 2 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 2 & 1 & \frac{3 \sqrt{2}}{2}\\2 & -5 & 2 & 0\\1 & 2 & 1 & - \frac{3 \sqrt{2}}{2}\\\frac{3 \sqrt{2}}{2} & 0 & - \frac{3 \sqrt{2}}{2} & 12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 2 & 1\\2 & - \lambda - 5 & 2\\1 & 2 & 1 - \lambda\end{matrix}\right|$$
$$I_{1} = -3$$
$$I_{2} = -18$$
$$I_{3} = 0$$
$$I_{4} = 162$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 3 \lambda^{2} + 18 \lambda$$
$$K_{2} = -45$$
$$K_{3} = -189$$
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
$$\lambda^{3} + 3 \lambda^{2} - 18 \lambda = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -6$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$3 \tilde x^{2} - 6 \tilde y^{2} + 6 \tilde z = 0$$
y
$$3 \tilde x^{2} - 6 \tilde y^{2} - 6 \tilde z = 0$$
y
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica
$$x^{2} + 4 x y + 2 x z + 3 \sqrt{2} x - 5 y^{2} + 4 y z + z^{2} - 3 \sqrt{2} z + 12 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = 1$$
$$a_{14} = \frac{3 \sqrt{2}}{2}$$
$$a_{22} = -5$$
$$a_{23} = 2$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = - \frac{3 \sqrt{2}}{2}$$
$$a_{44} = 12$$
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
$$I_{1} = -3$$
|1 2 | |-5 2| |1 1|
I2 = | | + | | + | |
|2 -5| |2 1| |1 1|$$I_{3} = \left|\begin{matrix}1 & 2 & 1\\2 & -5 & 2\\1 & 2 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 2 & 1 & \frac{3 \sqrt{2}}{2}\\2 & -5 & 2 & 0\\1 & 2 & 1 & - \frac{3 \sqrt{2}}{2}\\\frac{3 \sqrt{2}}{2} & 0 & - \frac{3 \sqrt{2}}{2} & 12\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 2 & 1\\2 & - \lambda - 5 & 2\\1 & 2 & 1 - \lambda\end{matrix}\right|$$
| ___| | ___|
| 3*\/ 2 | | -3*\/ 2 |
| 1 -------| | 1 --------|
| 2 | |-5 0 | | 2 |
K2 = | | + | | + | |
| ___ | |0 12| | ___ |
|3*\/ 2 | |-3*\/ 2 |
|------- 12 | |-------- 12 |
| 2 | | 2 | | ___ |
| 3*\/ 2 |
| ___| |-5 2 0 | | 1 1 ------- |
| 3*\/ 2 | | | | 2 |
| 1 2 -------| | ___| | |
| 2 | | -3*\/ 2 | | ___|
| | |2 1 --------| | -3*\/ 2 |
K3 = | 2 -5 0 | + | 2 | + | 1 1 --------|
| | | | | 2 |
| ___ | | ___ | | |
|3*\/ 2 | | -3*\/ 2 | | ___ ___ |
|------- 0 12 | |0 -------- 12 | |3*\/ 2 -3*\/ 2 |
| 2 | | 2 | |------- -------- 12 |
| 2 2 |
$$I_{1} = -3$$
$$I_{2} = -18$$
$$I_{3} = 0$$
$$I_{4} = 162$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 3 \lambda^{2} + 18 \lambda$$
$$K_{2} = -45$$
$$K_{3} = -189$$
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
$$\lambda^{3} + 3 \lambda^{2} - 18 \lambda = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = -6$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$3 \tilde x^{2} - 6 \tilde y^{2} + 6 \tilde z = 0$$
y
$$3 \tilde x^{2} - 6 \tilde y^{2} - 6 \tilde z = 0$$
2 2
\tilde x \tilde y
--------- - --------- + 2*\tilde z = 0
1 1/2 y
2 2
\tilde x \tilde y
--------- - --------- - 2*\tilde z = 0
1 1/2 es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica