Sr Examen
4y^2+15z^2-10/2^(1/2)x+6/6^(1/2)y+3=0 forma canónica
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Solución
Ha introducido
[src]
2 2 ___ 3 + y + 4*y + 15*z - x*\/ 5 = 0
$$- \sqrt{5} x + 4 y^{2} + y + 15 z^{2} + 3 = 0$$
-sqrt(5)*x + 4*y^2 + y + 15*z^2 + 3 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$- \sqrt{5} x + 4 y^{2} + y + 15 z^{2} + 3 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = - \frac{\sqrt{5}}{2}$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{24} = \frac{1}{2}$$
$$a_{33} = 15$$
$$a_{34} = 0$$
$$a_{44} = 3$$
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
$$I_{1} = 19$$
$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 4 & 0\\0 & 0 & 15\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & - \frac{\sqrt{5}}{2}\\0 & 4 & 0 & \frac{1}{2}\\0 & 0 & 15 & 0\\- \frac{\sqrt{5}}{2} & \frac{1}{2} & 0 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 4 - \lambda & 0\\0 & 0 & 15 - \lambda\end{matrix}\right|$$
$$I_{1} = 19$$
$$I_{2} = 60$$
$$I_{3} = 0$$
$$I_{4} = -75$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 19 \lambda^{2} - 60 \lambda$$
$$K_{2} = \frac{111}{2}$$
$$K_{3} = \frac{305}{2}$$
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
$$\lambda^{3} - 19 \lambda^{2} + 60 \lambda = 0$$
$$\lambda_{1} = 15$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$15 \tilde x^{2} + 4 \tilde y^{2} + \sqrt{5} \tilde z = 0$$
y
$$15 \tilde x^{2} + 4 \tilde y^{2} - \sqrt{5} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{30} \sqrt{5}} + \frac{\tilde y^{2}}{\frac{1}{8} \sqrt{5}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{30} \sqrt{5}} + \frac{\tilde y^{2}}{\frac{1}{8} \sqrt{5}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica
$$- \sqrt{5} x + 4 y^{2} + y + 15 z^{2} + 3 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = - \frac{\sqrt{5}}{2}$$
$$a_{22} = 4$$
$$a_{23} = 0$$
$$a_{24} = \frac{1}{2}$$
$$a_{33} = 15$$
$$a_{34} = 0$$
$$a_{44} = 3$$
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
$$I_{1} = 19$$
|0 0| |4 0 | |0 0 |
I2 = | | + | | + | |
|0 4| |0 15| |0 15|$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 4 & 0\\0 & 0 & 15\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & - \frac{\sqrt{5}}{2}\\0 & 4 & 0 & \frac{1}{2}\\0 & 0 & 15 & 0\\- \frac{\sqrt{5}}{2} & \frac{1}{2} & 0 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 4 - \lambda & 0\\0 & 0 & 15 - \lambda\end{matrix}\right|$$
| ___ |
| -\/ 5 |
| 0 -------|
| 2 | | 4 1/2| |15 0|
K2 = | | + | | + | |
| ___ | |1/2 3 | |0 3|
|-\/ 5 |
|------- 3 |
| 2 | | ___ | | ___ |
| -\/ 5 | | -\/ 5 |
| 0 0 -------| | 0 0 -------|
| 2 | | 4 0 1/2| | 2 |
| | | | | |
K3 = | 0 4 1/2 | + | 0 15 0 | + | 0 15 0 |
| | | | | |
| ___ | |1/2 0 3 | | ___ |
|-\/ 5 | |-\/ 5 |
|------- 1/2 3 | |------- 0 3 |
| 2 | | 2 |$$I_{1} = 19$$
$$I_{2} = 60$$
$$I_{3} = 0$$
$$I_{4} = -75$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 19 \lambda^{2} - 60 \lambda$$
$$K_{2} = \frac{111}{2}$$
$$K_{3} = \frac{305}{2}$$
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
$$\lambda^{3} - 19 \lambda^{2} + 60 \lambda = 0$$
$$\lambda_{1} = 15$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$15 \tilde x^{2} + 4 \tilde y^{2} + \sqrt{5} \tilde z = 0$$
y
$$15 \tilde x^{2} + 4 \tilde y^{2} - \sqrt{5} \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{30} \sqrt{5}} + \frac{\tilde y^{2}}{\frac{1}{8} \sqrt{5}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{1}{30} \sqrt{5}} + \frac{\tilde y^{2}}{\frac{1}{8} \sqrt{5}}\right) = 0$$
es la ecuación para el tipo paraboloide elíptico
- está reducida a la forma canónica