Sr Examen
2x^2+2y^2-5z^2+2xy+2*((2)^0.5)x-2*((2)^0.5)y+10z+74=0 forma canónica
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Solución
Ha introducido
[src]
2 2 2 ___ ___ 74 - 5*z + 2*x + 2*y + 10*z - 2*y*\/ 2 + 2*x*y + 2*x*\/ 2 = 0
$$2 x^{2} + 2 x y + 2 \sqrt{2} x + 2 y^{2} - 2 \sqrt{2} y - 5 z^{2} + 10 z + 74 = 0$$
2*x^2 + 2*x*y + 2*sqrt(2)*x + 2*y^2 - 2*sqrt(2)*y - 5*z^2 + 10*z + 74 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$2 x^{2} + 2 x y + 2 \sqrt{2} x + 2 y^{2} - 2 \sqrt{2} y - 5 z^{2} + 10 z + 74 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 1$$
$$a_{13} = 0$$
$$a_{14} = \sqrt{2}$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{24} = - \sqrt{2}$$
$$a_{33} = -5$$
$$a_{34} = 5$$
$$a_{44} = 74$$
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
$$I_{1} = -1$$
$$I_{3} = \left|\begin{matrix}2 & 1 & 0\\1 & 2 & 0\\0 & 0 & -5\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 1 & 0 & \sqrt{2}\\1 & 2 & 0 & - \sqrt{2}\\0 & 0 & -5 & 5\\\sqrt{2} & - \sqrt{2} & 5 & 74\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 1 & 0\\1 & 2 - \lambda & 0\\0 & 0 & - \lambda - 5\end{matrix}\right|$$
$$I_{1} = -1$$
$$I_{2} = -17$$
$$I_{3} = -15$$
$$I_{4} = -1125$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \lambda^{2} + 17 \lambda - 15$$
$$K_{2} = -103$$
$$K_{3} = -1350$$
Como
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
$$\lambda^{3} + \lambda^{2} - 17 \lambda + 15 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -5$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x^{2} + \tilde y^{2} - 5 \tilde z^{2} + 75 = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{15} \sqrt{3}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{15} \sqrt{3}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{1}{15} \sqrt{3}}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica
$$2 x^{2} + 2 x y + 2 \sqrt{2} x + 2 y^{2} - 2 \sqrt{2} y - 5 z^{2} + 10 z + 74 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 1$$
$$a_{13} = 0$$
$$a_{14} = \sqrt{2}$$
$$a_{22} = 2$$
$$a_{23} = 0$$
$$a_{24} = - \sqrt{2}$$
$$a_{33} = -5$$
$$a_{34} = 5$$
$$a_{44} = 74$$
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
$$I_{1} = -1$$
|2 1| |2 0 | |2 0 |
I2 = | | + | | + | |
|1 2| |0 -5| |0 -5|$$I_{3} = \left|\begin{matrix}2 & 1 & 0\\1 & 2 & 0\\0 & 0 & -5\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 1 & 0 & \sqrt{2}\\1 & 2 & 0 & - \sqrt{2}\\0 & 0 & -5 & 5\\\sqrt{2} & - \sqrt{2} & 5 & 74\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 1 & 0\\1 & 2 - \lambda & 0\\0 & 0 & - \lambda - 5\end{matrix}\right|$$
| ___| | ___|
| 2 \/ 2 | | 2 -\/ 2 | |-5 5 |
K2 = | | + | | + | |
| ___ | | ___ | |5 74|
|\/ 2 74 | |-\/ 2 74 | | ___ |
| 2 1 \/ 2 | | ___| | ___|
| | | 2 0 -\/ 2 | | 2 0 \/ 2 |
| ___| | | | |
K3 = | 1 2 -\/ 2 | + | 0 -5 5 | + | 0 -5 5 |
| | | | | |
| ___ ___ | | ___ | | ___ |
|\/ 2 -\/ 2 74 | |-\/ 2 5 74 | |\/ 2 5 74 |
$$I_{1} = -1$$
$$I_{2} = -17$$
$$I_{3} = -15$$
$$I_{4} = -1125$$
$$I{\left(\lambda \right)} = - \lambda^{3} - \lambda^{2} + 17 \lambda - 15$$
$$K_{2} = -103$$
$$K_{3} = -1350$$
Como
I3 != 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
$$\lambda^{3} + \lambda^{2} - 17 \lambda + 15 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -5$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x^{2} + \tilde y^{2} - 5 \tilde z^{2} + 75 = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{15} \sqrt{3}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{15} \sqrt{3}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\frac{1}{15} \sqrt{3}}\right)^{2}}\right) = -1$$
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica