Se da la ecuación de superficie de 2 grado:
$$6 x^{2} + 4 x y - 4 x z + 4 x + y^{2} - 4 y + z^{2} - 2 z + 1 = 0$$
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
$$a_{11} = 6$$
$$a_{12} = 2$$
$$a_{13} = -2$$
$$a_{14} = 2$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = -2$$
$$a_{33} = 1$$
$$a_{34} = -1$$
$$a_{44} = 1$$
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
$$I_{1} = 8$$
|6 2| |1 0| |6 -2|
I2 = | | + | | + | |
|2 1| |0 1| |-2 1 |
$$I_{3} = \left|\begin{matrix}6 & 2 & -2\\2 & 1 & 0\\-2 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}6 & 2 & -2 & 2\\2 & 1 & 0 & -2\\-2 & 0 & 1 & -1\\2 & -2 & -1 & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}6 - \lambda & 2 & -2\\2 & 1 - \lambda & 0\\-2 & 0 & 1 - \lambda\end{matrix}\right|$$
|6 2| |1 -2| |1 -1|
K2 = | | + | | + | |
|2 1| |-2 1 | |-1 1 |
|6 2 2 | |1 0 -2| |6 -2 2 |
| | | | | |
K3 = |2 1 -2| + |0 1 -1| + |-2 1 -1|
| | | | | |
|2 -2 1 | |-2 -1 1 | |2 -1 1 |
$$I_{1} = 8$$
$$I_{2} = 5$$
$$I_{3} = -2$$
$$I_{4} = -8$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 8 \lambda^{2} - 5 \lambda - 2$$
$$K_{2} = -1$$
$$K_{3} = -46$$
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
$$\lambda^{3} - 8 \lambda^{2} + 5 \lambda + 2 = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = \frac{7}{2} - \frac{\sqrt{57}}{2}$$
$$\lambda_{3} = \frac{7}{2} + \frac{\sqrt{57}}{2}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + \tilde y^{2} \left(\frac{7}{2} - \frac{\sqrt{57}}{2}\right) + \tilde z^{2} \left(\frac{7}{2} + \frac{\sqrt{57}}{2}\right) + 4 = 0$$
2 2 2
\tilde x \tilde z \tilde y
--------- + ------------------------ - -------------------------- = -1
2 2 2
/ 1\ / 1 \ / 1 \
\2 / |---------------------| |-----------------------|
| ____________ | | ______________ |
| / ____ | | / ____ |
| / 7 \/ 57 | | / 7 \/ 57 |
| / - + ------ *1/2| | / - - + ------ *1/2|
\\/ 2 2 / \\/ 2 2 /
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica