Sr Examen
6x^2+5x+5z^2=6 forma canónica
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Solución
Ha introducido
[src]
2 2 -6 + 5*x + 5*z + 6*x = 0
$$6 x^{2} + 5 x + 5 z^{2} - 6 = 0$$
6*x^2 + 5*x + 5*z^2 - 6 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$6 x^{2} + 5 x + 5 z^{2} - 6 = 0$$
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
$$a_{11} = 6$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 5$$
$$a_{34} = 0$$
$$a_{44} = -6$$
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
$$I_{1} = 11$$
$$I_{3} = \left|\begin{matrix}6 & 0 & 0\\0 & 0 & 0\\0 & 0 & 5\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}6 & 0 & 0 & \frac{5}{2}\\0 & 0 & 0 & 0\\0 & 0 & 5 & 0\\\frac{5}{2} & 0 & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}6 - \lambda & 0 & 0\\0 & - \lambda & 0\\0 & 0 & 5 - \lambda\end{matrix}\right|$$
$$I_{1} = 11$$
$$I_{2} = 30$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 11 \lambda^{2} - 30 \lambda$$
$$K_{2} = - \frac{289}{4}$$
$$K_{3} = - \frac{845}{4}$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 11 \lambda^{2} + 30 \lambda = 0$$
$$\lambda_{1} = 6$$
$$\lambda_{2} = 5$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$6 \tilde x^{2} + 5 \tilde y^{2} - \frac{169}{24} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{2}{13} \sqrt{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{2}{13} \sqrt{6}}\right)^{2}} = 1$$
es la ecuación para el tipo cilindro elíptico
- está reducida a la forma canónica
$$6 x^{2} + 5 x + 5 z^{2} - 6 = 0$$
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
$$a_{11} = 6$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = \frac{5}{2}$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 5$$
$$a_{34} = 0$$
$$a_{44} = -6$$
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
$$I_{1} = 11$$
|6 0| |0 0| |6 0|
I2 = | | + | | + | |
|0 0| |0 5| |0 5|$$I_{3} = \left|\begin{matrix}6 & 0 & 0\\0 & 0 & 0\\0 & 0 & 5\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}6 & 0 & 0 & \frac{5}{2}\\0 & 0 & 0 & 0\\0 & 0 & 5 & 0\\\frac{5}{2} & 0 & 0 & -6\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}6 - \lambda & 0 & 0\\0 & - \lambda & 0\\0 & 0 & 5 - \lambda\end{matrix}\right|$$
| 6 5/2| |0 0 | |5 0 |
K2 = | | + | | + | |
|5/2 -6 | |0 -6| |0 -6| | 6 0 5/2| |0 0 0 | | 6 0 5/2|
| | | | | |
K3 = | 0 0 0 | + |0 5 0 | + | 0 5 0 |
| | | | | |
|5/2 0 -6 | |0 0 -6| |5/2 0 -6 |$$I_{1} = 11$$
$$I_{2} = 30$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 11 \lambda^{2} - 30 \lambda$$
$$K_{2} = - \frac{289}{4}$$
$$K_{3} = - \frac{845}{4}$$
Como
$$I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0$$
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
$$\lambda^{3} - 11 \lambda^{2} + 30 \lambda = 0$$
$$\lambda_{1} = 6$$
$$\lambda_{2} = 5$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0$$
$$6 \tilde x^{2} + 5 \tilde y^{2} - \frac{169}{24} = 0$$
$$\frac{\tilde x^{2}}{\left(\frac{\frac{1}{6} \sqrt{6}}{\frac{2}{13} \sqrt{6}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{2}{13} \sqrt{6}}\right)^{2}} = 1$$
es la ecuación para el tipo cilindro elíptico
- está reducida a la forma canónica