Se da la ecuación de superficie de 2 grado:
$$4 x^{2} - 9 y^{2} + z^{2} + 36 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = -9$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 1$$
$$a_{34} = 0$$
$$a_{44} = 36$$
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
$$I_{1} = -4$$
|4 0 | |-9 0| |4 0|
I2 = | | + | | + | |
|0 -9| |0 1| |0 1|
$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & -9 & 0\\0 & 0 & 1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & 0 & 0 & 0\\0 & -9 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0 & 0\\0 & - \lambda - 9 & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|$$
|4 0 | |-9 0 | |1 0 |
K2 = | | + | | + | |
|0 36| |0 36| |0 36|
|4 0 0 | |-9 0 0 | |4 0 0 |
| | | | | |
K3 = |0 -9 0 | + |0 1 0 | + |0 1 0 |
| | | | | |
|0 0 36| |0 0 36| |0 0 36|
$$I_{1} = -4$$
$$I_{2} = -41$$
$$I_{3} = -36$$
$$I_{4} = -1296$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 4 \lambda^{2} + 41 \lambda - 36$$
$$K_{2} = -144$$
$$K_{3} = -1476$$
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
$$\lambda^{3} + 4 \lambda^{2} - 41 \lambda + 36 = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = -9$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$4 \tilde x^{2} + \tilde y^{2} - 9 \tilde z^{2} + 36 = 0$$
2 2 2
\tilde x \tilde y \tilde z
--------- + --------- - --------- = -1
2 2 2
/ 1 \ / 1\ / 1 \
|-----| \6 / |-----|
\2*1/6/ \3*1/6/
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica