y^2+z+1=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$y^{2} + z + 1 = 0$$
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
$$a_{11} = 0$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = 0$$
$$a_{33} = 0$$
$$a_{34} = \frac{1}{2}$$
$$a_{44} = 1$$
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
$$I_{1} = 1$$

     |0  0|   |1  0|   |0  0|
I2 = |    | + |    | + |    |
     |0  1|   |0  0|   |0  0|

$$I_{3} = \left|\begin{matrix}0 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}0 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 0 & \frac{1}{2}\\0 & 0 & \frac{1}{2} & 1\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$

     |0  0|   |1  0|   | 0   1/2|
K2 = |    | + |    | + |        |
     |0  1|   |0  1|   |1/2   1 |

     |0  0  0|   |1   0    0 |   |0   0    0 |
     |       |   |           |   |           |
K3 = |0  1  0| + |0   0   1/2| + |0   0   1/2|
     |       |   |           |   |           |
     |0  0  1|   |0  1/2   1 |   |0  1/2   1 |

$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = 0$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2}$$
$$K_{2} = \frac{3}{4}$$
$$K_{3} = - \frac{1}{4}$$
Como
$$I_{2} = 0 \wedge I_{3} = 0 \wedge I_{4} = 0 \wedge I_{1} \neq 0 \wedge K_{3} \neq 0$$
entonces por razón de tipos de rectas:
hay que
entonces la forma canónica de la ecuación será
$$I_{1} \tilde x^{2} + \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
y
$$I_{1} \tilde x^{2} - \tilde y 2 \sqrt{\frac{\left(-1\right) K_{3}}{I_{1}}} = 0$$
$$\tilde x^{2} + \tilde y = 0$$
y
$$\tilde x^{2} - \tilde y = 0$$
$$\tilde x^{2} = \tilde y$$
y
$$\tilde x^{2} = - \tilde y$$
es la ecuación para el tipo cilindro parabólico
- está reducida a la forma canónica