xx+4xy+4y2-20x+10y-50=0 forma canónica

Se da la ecuación de superficie de 2 grado:
$$x^{2} + 4 x y - 20 x + 10 y + 4 y_{2} - 50 = 0$$
Esta ecuación tiene la forma:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x y_{2} + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y y_{2} + 2 a_{24} y + a_{33} y_{2}^{2} + 2 a_{34} y_{2} + a_{44} = 0$$
donde
$$a_{11} = 1$$
$$a_{12} = 2$$
$$a_{13} = 0$$
$$a_{14} = -10$$
$$a_{22} = 0$$
$$a_{23} = 0$$
$$a_{24} = 5$$
$$a_{33} = 0$$
$$a_{34} = 2$$
$$a_{44} = -50$$
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
$$I_{1} = 1$$

     |1  2|   |0  0|   |1  0|
I2 = |    | + |    | + |    |
     |2  0|   |0  0|   |0  0|

$$I_{3} = \left|\begin{matrix}1 & 2 & 0\\2 & 0 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 2 & 0 & -10\\2 & 0 & 0 & 5\\0 & 0 & 0 & 2\\-10 & 5 & 2 & -50\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 2 & 0\\2 & - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$

     | 1   -10|   |0   5 |   |0   2 |
K2 = |        | + |      | + |      |
     |-10  -50|   |5  -50|   |2  -50|

     | 1   2  -10|   |0  0   5 |   | 1   0  -10|
     |           |   |         |   |           |
K3 = | 2   0   5 | + |0  0   2 | + | 0   0   2 |
     |           |   |         |   |           |
     |-10  5  -50|   |5  2  -50|   |-10  2  -50|

$$I_{1} = 1$$
$$I_{2} = -4$$
$$I_{3} = 0$$
$$I_{4} = 16$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + 4 \lambda$$
$$K_{2} = -179$$
$$K_{3} = -29$$
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
$$\lambda^{3} - \lambda^{2} - 4 \lambda = 0$$
$$\lambda_{1} = \frac{1}{2} - \frac{\sqrt{17}}{2}$$
$$\lambda_{2} = \frac{1}{2} + \frac{\sqrt{17}}{2}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde y2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde y2 \cdot 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{17}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{\sqrt{17}}{2}\right) + 4 \tilde y2 = 0$$
y
$$\tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{17}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{\sqrt{17}}{2}\right) - 4 \tilde y2 = 0$$
$$- 2 \tilde y2 + \left(\frac{\tilde x^{2}}{2 \frac{1}{- \frac{1}{2} + \frac{\sqrt{17}}{2}}} - \frac{\tilde y^{2}}{2 \frac{1}{\frac{1}{2} + \frac{\sqrt{17}}{2}}}\right) = 0$$
y
$$2 \tilde y2 + \left(\frac{\tilde x^{2}}{2 \frac{1}{- \frac{1}{2} + \frac{\sqrt{17}}{2}}} - \frac{\tilde y^{2}}{2 \frac{1}{\frac{1}{2} + \frac{\sqrt{17}}{2}}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica