Sr Examen
2x^2+10y^2-2z^2+12xy-8yz+12x+4y+8z-1=0 forma canónica
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Solución
Ha introducido
[src]
2 2 2 -1 - 2*z + 2*x + 4*y + 8*z + 10*y + 12*x - 8*y*z + 12*x*y = 0
$$2 x^{2} + 12 x y + 12 x + 10 y^{2} - 8 y z + 4 y - 2 z^{2} + 8 z - 1 = 0$$
2*x^2 + 12*x*y + 12*x + 10*y^2 - 8*y*z + 4*y - 2*z^2 + 8*z - 1 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$2 x^{2} + 12 x y + 12 x + 10 y^{2} - 8 y z + 4 y - 2 z^{2} + 8 z - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 6$$
$$a_{13} = 0$$
$$a_{14} = 6$$
$$a_{22} = 10$$
$$a_{23} = -4$$
$$a_{24} = 2$$
$$a_{33} = -2$$
$$a_{34} = 4$$
$$a_{44} = -1$$
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
$$I_{1} = 10$$
$$I_{3} = \left|\begin{matrix}2 & 6 & 0\\6 & 10 & -4\\0 & -4 & -2\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 6 & 0 & 6\\6 & 10 & -4 & 2\\0 & -4 & -2 & 4\\6 & 2 & 4 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 6 & 0\\6 & 10 - \lambda & -4\\0 & -4 & - \lambda - 2\end{matrix}\right|$$
$$I_{1} = 10$$
$$I_{2} = -56$$
$$I_{3} = 0$$
$$I_{4} = 2304$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 10 \lambda^{2} + 56 \lambda$$
$$K_{2} = -66$$
$$K_{3} = -344$$
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
$$\lambda^{3} - 10 \lambda^{2} - 56 \lambda = 0$$
$$\lambda_{1} = 14$$
$$\lambda_{2} = -4$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$14 \tilde x^{2} - 4 \tilde y^{2} + \frac{24 \sqrt{14} \tilde z}{7} = 0$$
y
$$14 \tilde x^{2} - 4 \tilde y^{2} - \frac{24 \sqrt{14} \tilde z}{7} = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{6}{49} \sqrt{14}} - \frac{\tilde y^{2}}{\frac{3}{7} \sqrt{14}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{6}{49} \sqrt{14}} - \frac{\tilde y^{2}}{\frac{3}{7} \sqrt{14}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica
$$2 x^{2} + 12 x y + 12 x + 10 y^{2} - 8 y z + 4 y - 2 z^{2} + 8 z - 1 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 6$$
$$a_{13} = 0$$
$$a_{14} = 6$$
$$a_{22} = 10$$
$$a_{23} = -4$$
$$a_{24} = 2$$
$$a_{33} = -2$$
$$a_{34} = 4$$
$$a_{44} = -1$$
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
$$I_{1} = 10$$
|2 6 | |10 -4| |2 0 |
I2 = | | + | | + | |
|6 10| |-4 -2| |0 -2|$$I_{3} = \left|\begin{matrix}2 & 6 & 0\\6 & 10 & -4\\0 & -4 & -2\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 6 & 0 & 6\\6 & 10 & -4 & 2\\0 & -4 & -2 & 4\\6 & 2 & 4 & -1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 6 & 0\\6 & 10 - \lambda & -4\\0 & -4 & - \lambda - 2\end{matrix}\right|$$
|2 6 | |10 2 | |-2 4 |
K2 = | | + | | + | |
|6 -1| |2 -1| |4 -1| |2 6 6 | |10 -4 2 | |2 0 6 |
| | | | | |
K3 = |6 10 2 | + |-4 -2 4 | + |0 -2 4 |
| | | | | |
|6 2 -1| |2 4 -1| |6 4 -1|$$I_{1} = 10$$
$$I_{2} = -56$$
$$I_{3} = 0$$
$$I_{4} = 2304$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 10 \lambda^{2} + 56 \lambda$$
$$K_{2} = -66$$
$$K_{3} = -344$$
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
$$\lambda^{3} - 10 \lambda^{2} - 56 \lambda = 0$$
$$\lambda_{1} = 14$$
$$\lambda_{2} = -4$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$14 \tilde x^{2} - 4 \tilde y^{2} + \frac{24 \sqrt{14} \tilde z}{7} = 0$$
y
$$14 \tilde x^{2} - 4 \tilde y^{2} - \frac{24 \sqrt{14} \tilde z}{7} = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{6}{49} \sqrt{14}} - \frac{\tilde y^{2}}{\frac{3}{7} \sqrt{14}}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{\frac{6}{49} \sqrt{14}} - \frac{\tilde y^{2}}{\frac{3}{7} \sqrt{14}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica