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Forma canónica/ xy+xz+yz+2*sqrt(2)*x+2*sqrt(2)*y-2*sqrt(2)*z-100=0

xy+xz+yz+2*sqrt(2)*x+2*sqrt(2)*y-2*sqrt(2)*z-100=0 forma canónica

El profesor se sorprenderá mucho al ver tu solución correcta😉

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Solución

Ha introducido [src] 
                               ___         ___         ___    
-100 + x*y + x*z + y*z - 2*z*\/ 2  + 2*x*\/ 2  + 2*y*\/ 2  = 0
xy+xz+22x+yz+22y−22z−100=0x y + x z + 2 \sqrt{2} x + y z + 2 \sqrt{2} y - 2 \sqrt{2} z - 100 = 0 
x*y + x*z + 2*sqrt(2)*x + y*z + 2*sqrt(2)*y - 2*sqrt(2)*z - 100 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
xy+xz+22x+yz+22y−22z−100=0x y + x z + 2 \sqrt{2} x + y z + 2 \sqrt{2} y - 2 \sqrt{2} z - 100 = 0
Esta ecuación tiene la forma:
a11x2+2a12xy+2a13xz+2a14x+a22y2+2a23yz+2a24y+a33z2+2a34z+a44=0a_{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
a11=0a_{11} = 0
a12=12a_{12} = \frac{1}{2}
a13=12a_{13} = \frac{1}{2}
a14=2a_{14} = \sqrt{2}
a22=0a_{22} = 0
a23=12a_{23} = \frac{1}{2}
a24=2a_{24} = \sqrt{2}
a33=0a_{33} = 0
a34=−2a_{34} = - \sqrt{2}
a44=−100a_{44} = -100
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
I1=a11+a22+a33I_{1} = a_{11} + a_{22} + a_{33}
     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

I3=∣a11a12a13a12a22a23a13a23a33∣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|
I4=∣a11a12a13a14a12a22a23a24a13a23a33a34a14a24a34a44∣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(λ)=∣a11−λa12a13a12a22−λa23a13a23a33−λ∣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
I1=0I_{1} = 0
     | 0   1/2|   | 0   1/2|   | 0   1/2|
I2 = |        | + |        | + |        |
     |1/2   0 |   |1/2   0 |   |1/2   0 |

I3=∣012121201212120∣I_{3} = \left|\begin{matrix}0 & \frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & 0 & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & 0\end{matrix}\right|
I4=∣01212212012212120−222−2−100∣I_{4} = \left|\begin{matrix}0 & \frac{1}{2} & \frac{1}{2} & \sqrt{2}\\\frac{1}{2} & 0 & \frac{1}{2} & \sqrt{2}\\\frac{1}{2} & \frac{1}{2} & 0 & - \sqrt{2}\\\sqrt{2} & \sqrt{2} & - \sqrt{2} & -100\end{matrix}\right|
I(λ)=∣−λ121212−λ121212−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & \frac{1}{2} & \frac{1}{2}\\\frac{1}{2} & - \lambda & \frac{1}{2}\\\frac{1}{2} & \frac{1}{2} & - \lambda\end{matrix}\right|
     |         ___|   |         ___|   |           ___|
     |  0    \/ 2 |   |  0    \/ 2 |   |  0     -\/ 2 |
K2 = |            | + |            | + |              |
     |  ___       |   |  ___       |   |   ___        |
     |\/ 2   -100 |   |\/ 2   -100 |   |-\/ 2    -100 |

     |                ___|   |                 ___ |   |                 ___ |
     |  0     1/2   \/ 2 |   |  0     1/2    \/ 2  |   |  0     1/2    \/ 2  |
     |                   |   |                     |   |                     |
     |                ___|   |                  ___|   |                  ___|
K3 = | 1/2     0    \/ 2 | + | 1/2     0     -\/ 2 | + | 1/2     0     -\/ 2 |
     |                   |   |                     |   |                     |
     |  ___    ___       |   |  ___     ___        |   |  ___     ___        |
     |\/ 2   \/ 2   -100 |   |\/ 2   -\/ 2    -100 |   |\/ 2   -\/ 2    -100 |

I1=0I_{1} = 0
I2=−34I_{2} = - \frac{3}{4}
I3=14I_{3} = \frac{1}{4}
I4=−452I_{4} = - \frac{45}{2}
I(λ)=−λ3+3λ4+14I{\left(\lambda \right)} = - \lambda^{3} + \frac{3 \lambda}{4} + \frac{1}{4}
K2=−6K_{2} = -6
K3=73K_{3} = 73
Como
I3 != 0

entonces por razón de tipos de rectas:
hay que
Formulamos la ecuación característica para nuestra superficie:
−I1λ2+I2λ−I3+λ3=0- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0
o
λ3−3λ4−14=0\lambda^{3} - \frac{3 \lambda}{4} - \frac{1}{4} = 0
λ1=1\lambda_{1} = 1
λ2=−12\lambda_{2} = - \frac{1}{2}
λ3=−12\lambda_{3} = - \frac{1}{2}
entonces la forma canónica de la ecuación será
(z~2λ3+(x~2λ1+y~2λ2))+I4I3=0\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0
x~2−y~22−z~22−90=0\tilde x^{2} - \frac{\tilde y^{2}}{2} - \frac{\tilde z^{2}}{2} - 90 = 0
−x~2(113010)2+(y~2(213010)2+z~2(213010)2)=−1- \frac{\tilde x^{2}}{\left(\frac{1}{\frac{1}{30} \sqrt{10}}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{\frac{1}{30} \sqrt{10}}\right)^{2}} + \frac{\tilde z^{2}}{\left(\frac{\sqrt{2}}{\frac{1}{30} \sqrt{10}}\right)^{2}}\right) = -1
es la ecuación para el tipo hiperboloide bilateral
- está reducida a la forma canónica
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