Sr Examen
5x^2+4xy+8y^2+3z^2+8x+14y+12z+5=0 forma canónica
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Solución
Ha introducido
[src]
2 2 2 5 + 3*z + 5*x + 8*x + 8*y + 12*z + 14*y + 4*x*y = 0
$$5 x^{2} + 4 x y + 8 x + 8 y^{2} + 14 y + 3 z^{2} + 12 z + 5 = 0$$
5*x^2 + 4*x*y + 8*x + 8*y^2 + 14*y + 3*z^2 + 12*z + 5 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
$$5 x^{2} + 4 x y + 8 x + 8 y^{2} + 14 y + 3 z^{2} + 12 z + 5 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{14} = 4$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{24} = 7$$
$$a_{33} = 3$$
$$a_{34} = 6$$
$$a_{44} = 5$$
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
$$I_{1} = 16$$
$$I_{3} = \left|\begin{matrix}5 & 2 & 0\\2 & 8 & 0\\0 & 0 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}5 & 2 & 0 & 4\\2 & 8 & 0 & 7\\0 & 0 & 3 & 6\\4 & 7 & 6 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 2 & 0\\2 & 8 - \lambda & 0\\0 & 0 & 3 - \lambda\end{matrix}\right|$$
$$I_{1} = 16$$
$$I_{2} = 75$$
$$I_{3} = 108$$
$$I_{4} = -1539$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 16 \lambda^{2} - 75 \lambda + 108$$
$$K_{2} = -21$$
$$K_{3} = -549$$
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
$$\lambda^{3} - 16 \lambda^{2} + 75 \lambda - 108 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 3$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + 4 \tilde y^{2} + 3 \tilde z^{2} - \frac{57}{4} = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{2}{57} \sqrt{57}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{2 \sqrt{57}}{57}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{2 \sqrt{57}}{57}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo elipsoide
- está reducida a la forma canónica
$$5 x^{2} + 4 x y + 8 x + 8 y^{2} + 14 y + 3 z^{2} + 12 z + 5 = 0$$
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
$$a_{11} = 5$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{14} = 4$$
$$a_{22} = 8$$
$$a_{23} = 0$$
$$a_{24} = 7$$
$$a_{33} = 3$$
$$a_{34} = 6$$
$$a_{44} = 5$$
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
$$I_{1} = 16$$
|5 2| |8 0| |5 0|
I2 = | | + | | + | |
|2 8| |0 3| |0 3|$$I_{3} = \left|\begin{matrix}5 & 2 & 0\\2 & 8 & 0\\0 & 0 & 3\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}5 & 2 & 0 & 4\\2 & 8 & 0 & 7\\0 & 0 & 3 & 6\\4 & 7 & 6 & 5\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}5 - \lambda & 2 & 0\\2 & 8 - \lambda & 0\\0 & 0 & 3 - \lambda\end{matrix}\right|$$
|5 4| |8 7| |3 6|
K2 = | | + | | + | |
|4 5| |7 5| |6 5| |5 2 4| |8 0 7| |5 0 4|
| | | | | |
K3 = |2 8 7| + |0 3 6| + |0 3 6|
| | | | | |
|4 7 5| |7 6 5| |4 6 5|$$I_{1} = 16$$
$$I_{2} = 75$$
$$I_{3} = 108$$
$$I_{4} = -1539$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 16 \lambda^{2} - 75 \lambda + 108$$
$$K_{2} = -21$$
$$K_{3} = -549$$
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
$$\lambda^{3} - 16 \lambda^{2} + 75 \lambda - 108 = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 3$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$9 \tilde x^{2} + 4 \tilde y^{2} + 3 \tilde z^{2} - \frac{57}{4} = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{2}{57} \sqrt{57}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{3 \frac{2 \sqrt{57}}{57}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2 \frac{2 \sqrt{57}}{57}}\right)^{2}}\right) = 1$$
es la ecuación para el tipo elipsoide
- está reducida a la forma canónica