x^2-10*x*y+7*y^2+2*x-10*y=a forma canónica

Se da la ecuación de superficie de 2 grado:
$$- a + x^{2} - 10 x y + 2 x + 7 y^{2} - 10 y = 0$$
Esta ecuación tiene la forma:
$$a^{2} a_{33} + 2 a a_{13} x + 2 a a_{23} y + 2 a a_{34} + a_{11} x^{2} + 2 a_{12} x y + 2 a_{14} x + a_{22} y^{2} + 2 a_{24} y + a_{44} = 0$$
donde
$$a_{11} = 1$$
$$a_{12} = -5$$
$$a_{13} = 0$$
$$a_{14} = 1$$
$$a_{22} = 7$$
$$a_{23} = 0$$
$$a_{24} = -5$$
$$a_{33} = 0$$
$$a_{34} = - \frac{1}{2}$$
$$a_{44} = 0$$
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
$$I_{1} = 8$$

     |1   -5|   |7  0|   |1  0|
I2 = |      | + |    | + |    |
     |-5  7 |   |0  0|   |0  0|

$$I_{3} = \left|\begin{matrix}1 & -5 & 0\\-5 & 7 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & -5 & 0 & 1\\-5 & 7 & 0 & -5\\0 & 0 & 0 & - \frac{1}{2}\\1 & -5 & - \frac{1}{2} & 0\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -5 & 0\\-5 & 7 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$

     |1  1|   |7   -5|   | 0    -1/2|
K2 = |    | + |      | + |          |
     |1  0|   |-5  0 |   |-1/2   0  |

     |1   -5  1 |   |7    0     -5 |   |1   0     1  |
     |          |   |              |   |             |
K3 = |-5  7   -5| + |0    0    -1/2| + |0   0    -1/2|
     |          |   |              |   |             |
     |1   -5  0 |   |-5  -1/2   0  |   |1  -1/2   0  |

$$I_{1} = 8$$
$$I_{2} = -18$$
$$I_{3} = 0$$
$$I_{4} = \frac{9}{2}$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 8 \lambda^{2} + 18 \lambda$$
$$K_{2} = - \frac{105}{4}$$
$$K_{3} = 16$$
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
$$\lambda^{3} - 8 \lambda^{2} - 18 \lambda = 0$$
$$\lambda_{1} = 4 - \sqrt{34}$$
$$\lambda_{2} = 4 + \sqrt{34}$$
$$\lambda_{3} = 0$$
entonces la forma canónica de la ecuación será
$$\tilde a 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
y
$$- \tilde a 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde a + \tilde x^{2} \left(4 - \sqrt{34}\right) + \tilde y^{2} \left(4 + \sqrt{34}\right) = 0$$
y
$$- \tilde a + \tilde x^{2} \left(4 - \sqrt{34}\right) + \tilde y^{2} \left(4 + \sqrt{34}\right) = 0$$
$$- 2 \tilde a + \left(\frac{\tilde x^{2}}{\frac{1}{2} \frac{1}{-4 + \sqrt{34}}} - \frac{\tilde y^{2}}{\frac{1}{2} \frac{1}{4 + \sqrt{34}}}\right) = 0$$
y
$$2 \tilde a + \left(\frac{\tilde x^{2}}{\frac{1}{2} \frac{1}{-4 + \sqrt{34}}} - \frac{\tilde y^{2}}{\frac{1}{2} \frac{1}{4 + \sqrt{34}}}\right) = 0$$
es la ecuación para el tipo paraboloide hiperbólico
- está reducida a la forma canónica