2x^2-7y^2-4z^2+4xy+20zy-16xz+60x-12y+12z-90=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$2 x^{2} + 4 x y - 16 x z + 60 x - 7 y^{2} + 20 y z - 12 y - 4 z^{2} + 12 z - 90 = 0$$
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
$$a_{11} = 2$$
$$a_{12} = 2$$
$$a_{13} = -8$$
$$a_{14} = 30$$
$$a_{22} = -7$$
$$a_{23} = 10$$
$$a_{24} = -6$$
$$a_{33} = -4$$
$$a_{34} = 6$$
$$a_{44} = -90$$
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
$$I_{1} = -9$$

     |2  2 |   |-7  10|   |2   -8|
I2 = |     | + |      | + |      |
     |2  -7|   |10  -4|   |-8  -4|

$$I_{3} = \left|\begin{matrix}2 & 2 & -8\\2 & -7 & 10\\-8 & 10 & -4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}2 & 2 & -8 & 30\\2 & -7 & 10 & -6\\-8 & 10 & -4 & 6\\30 & -6 & 6 & -90\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 2 & -8\\2 & - \lambda - 7 & 10\\-8 & 10 & - \lambda - 4\end{matrix}\right|$$

     |2   30 |   |-7  -6 |   |-4   6 |
K2 = |       | + |       | + |       |
     |30  -90|   |-6  -90|   |6   -90|

     |2   2   30 |   |-7  10  -6 |   |2   -8  30 |
     |           |   |           |   |           |
K3 = |2   -7  -6 | + |10  -4   6 | + |-8  -4   6 |
     |           |   |           |   |           |
     |30  -6  -90|   |-6  6   -90|   |30  6   -90|

$$I_{1} = -9$$
$$I_{2} = -162$$
$$I_{3} = 0$$
$$I_{4} = 52488$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 9 \lambda^{2} + 162 \lambda$$
$$K_{2} = -162$$
$$K_{3} = 20412$$
Como
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} + 9 \lambda^{2} - 162 \lambda = 0$$
$$\lambda_{1} = 9$$
$$\lambda_{2} = -18$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$9 \tilde x^{2} - 18 \tilde y^{2} + 36 \tilde z = 0$$
y
$$9 \tilde x^{2} - 18 \tilde y^{2} - 36 \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{1}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{1}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica