4x^2=z^2+8x+16y+6z-3=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$4 x^{2} - 8 x - 16 y - z^{2} - 6 z + 3 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -4$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = -8$$
$$a_{33} = -1$$
$$a_{34} = -3$$
$$a_{44} = 3$$
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
$$I_{1} = 3$$

     |4  0|   |0  0 |   |4  0 |
I2 = |    | + |     | + |     |
     |0  0|   |0  -1|   |0  -1|

$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 0 & 0\\0 & 0 & -1\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & 0 & 0 & -4\\0 & 0 & 0 & -8\\0 & 0 & -1 & -3\\-4 & -8 & -3 & 3\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0 & 0\\0 & - \lambda & 0\\0 & 0 & - \lambda - 1\end{matrix}\right|$$

     |4   -4|   |0   -8|   |-1  -3|
K2 = |      | + |      | + |      |
     |-4  3 |   |-8  3 |   |-3  3 |

     |4   0   -4|   |0   0   -8|   |4   0   -4|
     |          |   |          |   |          |
K3 = |0   0   -8| + |0   -1  -3| + |0   -1  -3|
     |          |   |          |   |          |
     |-4  -8  3 |   |-8  -3  3 |   |-4  -3  3 |

$$I_{1} = 3$$
$$I_{2} = -4$$
$$I_{3} = 0$$
$$I_{4} = 256$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} + 4 \lambda$$
$$K_{2} = -80$$
$$K_{3} = -224$$
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
$$\lambda^{3} - 3 \lambda^{2} - 4 \lambda = 0$$
$$\lambda_{1} = 4$$
$$\lambda_{2} = -1$$
$$\lambda_{3} = 0$$
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
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$4 \tilde x^{2} - \tilde y^{2} + 16 \tilde z = 0$$
y
$$4 \tilde x^{2} - \tilde y^{2} - 16 \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{8}\right) = 0$$
y
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{2} - \frac{\tilde y^{2}}{8}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica