((x^2)/4)-(y^2/1)=2z forma canónica

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

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

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

     |1/4  0|   |-1  0|   |0   -1|
K2 = |      | + |     | + |      |
     | 0   0|   |0   0|   |-1  0 |

     |1/4  0   0|   |-1  0   0 |   |1/4  0   0 |
     |          |   |          |   |           |
K3 = | 0   -1  0| + |0   0   -1| + | 0   0   -1|
     |          |   |          |   |           |
     | 0   0   0|   |0   -1  0 |   | 0   -1  0 |

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