Sr Examen
3y^2+4z^2+3sqrt(6)x-2sqrt(3)y+13=0 forma canónica
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Solución
Ha introducido
[src]
2 2 ___ ___ 13 + 3*y + 4*z - 2*y*\/ 3 + 3*x*\/ 6 = 0
$$3 \sqrt{6} x + 3 y^{2} - 2 \sqrt{3} y + 4 z^{2} + 13 = 0$$
3*sqrt(6)*x + 3*y^2 - 2*sqrt(3)*y + 4*z^2 + 13 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$3 \sqrt{6} x + 3 y^{2} - 2 \sqrt{3} y + 4 z^{2} + 13 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = \frac{3 \sqrt{6}}{2}$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{24} = - \sqrt{3}$$
$$a_{33} = 4$$
$$a_{34} = 0$$
$$a_{44} = 13$$
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
$$I_{1} = 7$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 3 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & \frac{3 \sqrt{6}}{2}\\0 & 3 & 0 & - \sqrt{3}\\0 & 0 & 4 & 0\\\frac{3 \sqrt{6}}{2} & - \sqrt{3} & 0 & 13\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 3 - \lambda & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$
$$I_{1} = 7$$
$$I_{2} = 12$$
$$I_{3} = 0$$
$$I_{4} = -162$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 12 \lambda$$
$$K_{2} = \frac{149}{2}$$
$$K_{3} = \frac{99}{2}$$
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
$$\lambda^{3} - 7 \lambda^{2} + 12 \lambda = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$4 \tilde x^{2} + 3 \tilde y^{2} + 3 \sqrt{6} \tilde z = 0$$
y
$$4 \tilde x^{2} + 3 \tilde y^{2} - 3 \sqrt{6} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{2} \sqrt{6}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{2} \sqrt{6}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica
$$3 \sqrt{6} x + 3 y^{2} - 2 \sqrt{3} y + 4 z^{2} + 13 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = \frac{3 \sqrt{6}}{2}$$
$$a_{22} = 3$$
$$a_{23} = 0$$
$$a_{24} = - \sqrt{3}$$
$$a_{33} = 4$$
$$a_{34} = 0$$
$$a_{44} = 13$$
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
$$I_{1} = 7$$
|0 0| |3 0| |0 0|
I2 = | | + | | + | |
|0 3| |0 4| |0 4|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 3 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & \frac{3 \sqrt{6}}{2}\\0 & 3 & 0 & - \sqrt{3}\\0 & 0 & 4 & 0\\\frac{3 \sqrt{6}}{2} & - \sqrt{3} & 0 & 13\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 3 - \lambda & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$
| ___|
| 3*\/ 6 |
| 0 -------| | ___|
| 2 | | 3 -\/ 3 | |4 0 |
K2 = | | + | | + | |
| ___ | | ___ | |0 13|
|3*\/ 6 | |-\/ 3 13 |
|------- 13 |
| 2 | | ___|
| 3*\/ 6 | | ___|
| 0 0 -------| | 3*\/ 6 |
| 2 | | ___| | 0 0 -------|
| | | 3 0 -\/ 3 | | 2 |
| ___ | | | | |
K3 = | 0 3 -\/ 3 | + | 0 4 0 | + | 0 4 0 |
| | | | | |
| ___ | | ___ | | ___ |
|3*\/ 6 ___ | |-\/ 3 0 13 | |3*\/ 6 |
|------- -\/ 3 13 | |------- 0 13 |
| 2 | | 2 |
$$I_{1} = 7$$
$$I_{2} = 12$$
$$I_{3} = 0$$
$$I_{4} = -162$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 12 \lambda$$
$$K_{2} = \frac{149}{2}$$
$$K_{3} = \frac{99}{2}$$
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
$$\lambda^{3} - 7 \lambda^{2} + 12 \lambda = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 3$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$4 \tilde x^{2} + 3 \tilde y^{2} + 3 \sqrt{6} \tilde z = 0$$
y
$$4 \tilde x^{2} + 3 \tilde y^{2} - 3 \sqrt{6} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{2} \sqrt{6}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{3}{8} \sqrt{6}} + \frac{\tilde y^{2}}{\frac{1}{2} \sqrt{6}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica