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Forma canónica/ 3*x^2+8/3*x*y+8/3*x*z+10/3*x+11/3*y^2+10/3*y+7/3*z^2-4/3z-5/3=0

3*x^2+8/3*x*y+8/3*x*z+10/3*x+11/3*y^2+10/3*y+7/3*z^2-4/3z-5/3=0 forma canónica

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Solución

Ha introducido [src] 
                      2                     2                    
  5      2   4*z   7*z    10*x   10*y   11*y    8*x*y   8*x*z    
- - + 3*x  - --- + ---- + ---- + ---- + ----- + ----- + ----- = 0
  3           3     3      3      3       3       3       3
3x2+8xy3+8xz3+10x3+11y23+10y3+7z23−4z3−53=03 x^{2} + \frac{8 x y}{3} + \frac{8 x z}{3} + \frac{10 x}{3} + \frac{11 y^{2}}{3} + \frac{10 y}{3} + \frac{7 z^{2}}{3} - \frac{4 z}{3} - \frac{5}{3} = 0 
3*x^2 + 8*x*y/3 + 8*x*z/3 + 10*x/3 + 11*y^2/3 + 10*y/3 + 7*z^2/3 - 4*z/3 - 5/3 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
3x2+8xy3+8xz3+10x3+11y23+10y3+7z23−4z3−53=03 x^{2} + \frac{8 x y}{3} + \frac{8 x z}{3} + \frac{10 x}{3} + \frac{11 y^{2}}{3} + \frac{10 y}{3} + \frac{7 z^{2}}{3} - \frac{4 z}{3} - \frac{5}{3} = 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=3a_{11} = 3
a12=43a_{12} = \frac{4}{3}
a13=43a_{13} = \frac{4}{3}
a14=53a_{14} = \frac{5}{3}
a22=113a_{22} = \frac{11}{3}
a23=0a_{23} = 0
a24=53a_{24} = \frac{5}{3}
a33=73a_{33} = \frac{7}{3}
a34=−23a_{34} = - \frac{2}{3}
a44=−53a_{44} = - \frac{5}{3}
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=9I_{1} = 9
     | 3   4/3 |   |11/3   0 |   | 3   4/3|
I2 = |         | + |         | + |        |
     |4/3  11/3|   | 0    7/3|   |4/3  7/3|

I3=∣3434343113043073∣I_{3} = \left|\begin{matrix}3 & \frac{4}{3} & \frac{4}{3}\\\frac{4}{3} & \frac{11}{3} & 0\\\frac{4}{3} & 0 & \frac{7}{3}\end{matrix}\right|
I4=∣34343534311305343073−235353−23−53∣I_{4} = \left|\begin{matrix}3 & \frac{4}{3} & \frac{4}{3} & \frac{5}{3}\\\frac{4}{3} & \frac{11}{3} & 0 & \frac{5}{3}\\\frac{4}{3} & 0 & \frac{7}{3} & - \frac{2}{3}\\\frac{5}{3} & \frac{5}{3} & - \frac{2}{3} & - \frac{5}{3}\end{matrix}\right|
I(λ)=∣3−λ434343113−λ043073−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}3 - \lambda & \frac{4}{3} & \frac{4}{3}\\\frac{4}{3} & \frac{11}{3} - \lambda & 0\\\frac{4}{3} & 0 & \frac{7}{3} - \lambda\end{matrix}\right|
     | 3   5/3 |   |11/3  5/3 |   |7/3   -2/3|
K2 = |         | + |          | + |          |
     |5/3  -5/3|   |5/3   -5/3|   |-2/3  -5/3|

     | 3   4/3   5/3 |   |11/3   0    5/3 |   | 3   4/3   5/3 |
     |               |   |                |   |               |
K3 = |4/3  11/3  5/3 | + | 0    7/3   -2/3| + |4/3  7/3   -2/3|
     |               |   |                |   |               |
     |5/3  5/3   -5/3|   |5/3   -2/3  -5/3|   |5/3  -2/3  -5/3|

I1=9I_{1} = 9
I2=23I_{2} = 23
I3=15I_{3} = 15
I4=−57I_{4} = -57
I(λ)=−λ3+9λ2−23λ+15I{\left(\lambda \right)} = - \lambda^{3} + 9 \lambda^{2} - 23 \lambda + 15
K2=−21K_{2} = -21
K3=−2053K_{3} = - \frac{205}{3}
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
λ3−9λ2+23λ−15=0\lambda^{3} - 9 \lambda^{2} + 23 \lambda - 15 = 0
λ1=5\lambda_{1} = 5
λ2=3\lambda_{2} = 3
λ3=1\lambda_{3} = 1
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
5x~2+3y~2+z~2−195=05 \tilde x^{2} + 3 \tilde y^{2} + \tilde z^{2} - \frac{19}{5} = 0
z~2(111995)2+(x~2(15511995)2+y~2(13311995)2)=1\frac{\tilde z^{2}}{\left(\frac{1}{\frac{1}{19} \sqrt{95}}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\frac{1}{5} \sqrt{5}}{\frac{1}{19} \sqrt{95}}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\frac{1}{3} \sqrt{3}}{\frac{1}{19} \sqrt{95}}\right)^{2}}\right) = 1
es la ecuación para el tipo elipsoide
- está reducida a la forma canónica
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