-4x^2-4y^2-18z^2-4xy+8xz-8yz=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$- 4 x^{2} - 4 x y + 8 x z - 4 y^{2} - 8 y z - 18 z^{2} = 0$$
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
$$a_{11} = -4$$
$$a_{12} = -2$$
$$a_{13} = 4$$
$$a_{14} = 0$$
$$a_{22} = -4$$
$$a_{23} = -4$$
$$a_{24} = 0$$
$$a_{33} = -18$$
$$a_{34} = 0$$
$$a_{44} = 0$$
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
$$I_{1} = -26$$

     |-4  -2|   |-4  -4 |   |-4   4 |
I2 = |      | + |       | + |       |
     |-2  -4|   |-4  -18|   |4   -18|

$$I_{3} = \left|\begin{matrix}-4 & -2 & 4\\-2 & -4 & -4\\4 & -4 & -18\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}-4 & -2 & 4 & 0\\-2 & -4 & -4 & 0\\4 & -4 & -18 & 0\\0 & 0 & 0 & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & -2 & 4\\-2 & - \lambda - 4 & -4\\4 & -4 & - \lambda - 18\end{matrix}\right|$$

     |-4  0|   |-4  0|   |-18  0|
K2 = |     | + |     | + |      |
     |0   0|   |0   0|   | 0   0|

     |-4  -2  0|   |-4  -4   0|   |-4   4   0|
     |         |   |          |   |          |
K3 = |-2  -4  0| + |-4  -18  0| + |4   -18  0|
     |         |   |          |   |          |
     |0   0   0|   |0    0   0|   |0    0   0|

$$I_{1} = -26$$
$$I_{2} = 124$$
$$I_{3} = -24$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} - 26 \lambda^{2} - 124 \lambda - 24$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} + 26 \lambda^{2} + 124 \lambda + 24 = 0$$
$$\lambda_{1} = -6$$
$$\lambda_{2} = -10 - 4 \sqrt{6}$$
$$\lambda_{3} = -10 + 4 \sqrt{6}$$
entonces la forma canónica de la ecuación será
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- 6 \tilde x^{2} + \tilde y^{2} \left(-10 - 4 \sqrt{6}\right) + \tilde z^{2} \left(-10 + 4 \sqrt{6}\right) = 0$$
$$\frac{\tilde z^{2}}{\left(\frac{1}{\sqrt{10 - 4 \sqrt{6}}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{6}}{6}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{4 \sqrt{6} + 10}}\right)^{2}}\right) = 0$$
es la ecuación para el tipo cono imaginario
- está reducida a la forma canónica