Given equation of the surface of 2-order:
x 2 + y 2 + z 2 − z = 0 x^{2} + y^{2} + z^{2} - z = 0 x 2 + y 2 + z 2 − z = 0 This equation looks like:
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 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 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 yz + 2 a 24 y + a 33 z 2 + 2 a 34 z + a 44 = 0 where
a 11 = 1 a_{11} = 1 a 11 = 1 a 12 = 0 a_{12} = 0 a 12 = 0 a 13 = 0 a_{13} = 0 a 13 = 0 a 14 = 0 a_{14} = 0 a 14 = 0 a 22 = 1 a_{22} = 1 a 22 = 1 a 23 = 0 a_{23} = 0 a 23 = 0 a 24 = 0 a_{24} = 0 a 24 = 0 a 33 = 1 a_{33} = 1 a 33 = 1 a 34 = − 1 2 a_{34} = - \frac{1}{2} a 34 = − 2 1 a 44 = 0 a_{44} = 0 a 44 = 0 The invariants of the equation when converting coordinates are determinants:
I 1 = a 11 + a 22 + a 33 I_{1} = a_{11} + a_{22} + a_{33} I 1 = a 11 + a 22 + a 33 |a11 a12| |a22 a23| |a11 a13|
I2 = | | + | | + | |
|a12 a22| |a23 a33| |a13 a33| I 3 = ∣ a 11 a 12 a 13 a 12 a 22 a 23 a 13 a 23 a 33 ∣ 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 3 = a 11 a 12 a 13 a 12 a 22 a 23 a 13 a 23 a 33 I 4 = ∣ 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 ∣ 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 4 = 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 I ( λ ) = ∣ a 11 − λ a 12 a 13 a 12 a 22 − λ a 23 a 13 a 23 a 33 − 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| I ( λ ) = a 11 − λ a 12 a 13 a 12 a 22 − λ a 23 a 13 a 23 a 33 − λ |a11 a14| |a22 a24| |a33 a34|
K2 = | | + | | + | |
|a14 a44| |a24 a44| |a34 a44| |a11 a12 a14| |a22 a23 a24| |a11 a13 a14|
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K3 = |a12 a22 a24| + |a23 a33 a34| + |a13 a33 a34|
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|a14 a24 a44| |a24 a34 a44| |a14 a34 a44| substitute coefficients
I 1 = 3 I_{1} = 3 I 1 = 3 |1 0| |1 0| |1 0|
I2 = | | + | | + | |
|0 1| |0 1| |0 1| I 3 = ∣ 1 0 0 0 1 0 0 0 1 ∣ I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right| I 3 = 1 0 0 0 1 0 0 0 1 I 4 = ∣ 1 0 0 0 0 1 0 0 0 0 1 − 1 2 0 0 − 1 2 0 ∣ I_{4} = \left|\begin{matrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & - \frac{1}{2}\\0 & 0 & - \frac{1}{2} & 0\end{matrix}\right| I 4 = 1 0 0 0 0 1 0 0 0 0 1 − 2 1 0 0 − 2 1 0 I ( λ ) = ∣ 1 − λ 0 0 0 1 − λ 0 0 0 1 − λ ∣ I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 1 - \lambda\end{matrix}\right| I ( λ ) = 1 − λ 0 0 0 1 − λ 0 0 0 1 − λ |1 0| |1 0| | 1 -1/2|
K2 = | | + | | + | |
|0 0| |0 0| |-1/2 0 | |1 0 0| |1 0 0 | |1 0 0 |
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K3 = |0 1 0| + |0 1 -1/2| + |0 1 -1/2|
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|0 0 0| |0 -1/2 0 | |0 -1/2 0 | I 1 = 3 I_{1} = 3 I 1 = 3 I 2 = 3 I_{2} = 3 I 2 = 3 I 3 = 1 I_{3} = 1 I 3 = 1 I 4 = − 1 4 I_{4} = - \frac{1}{4} I 4 = − 4 1 I ( λ ) = − λ 3 + 3 λ 2 − 3 λ + 1 I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - 3 \lambda + 1 I ( λ ) = − λ 3 + 3 λ 2 − 3 λ + 1 K 2 = − 1 4 K_{2} = - \frac{1}{4} K 2 = − 4 1 K 3 = − 1 2 K_{3} = - \frac{1}{2} K 3 = − 2 1 Because
I3 != 0 then by type of surface:
you need to
Make the characteristic equation for the surface:
− I 1 λ 2 + I 2 λ − I 3 + λ 3 = 0 - I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0 − I 1 λ 2 + I 2 λ − I 3 + λ 3 = 0 or
λ 3 − 3 λ 2 + 3 λ − 1 = 0 \lambda^{3} - 3 \lambda^{2} + 3 \lambda - 1 = 0 λ 3 − 3 λ 2 + 3 λ − 1 = 0 λ 1 = 1 \lambda_{1} = 1 λ 1 = 1 λ 2 = 1 \lambda_{2} = 1 λ 2 = 1 λ 3 = 1 \lambda_{3} = 1 λ 3 = 1 then the canonical form of the equation will be
( z ~ 2 λ 3 + ( x ~ 2 λ 1 + y ~ 2 λ 2 ) ) + I 4 I 3 = 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 ( z ~ 2 λ 3 + ( x ~ 2 λ 1 + y ~ 2 λ 2 ) ) + I 3 I 4 = 0 x ~ 2 + y ~ 2 + z ~ 2 − 1 4 = 0 \tilde x^{2} + \tilde y^{2} + \tilde z^{2} - \frac{1}{4} = 0 x ~ 2 + y ~ 2 + z ~ 2 − 4 1 = 0 z ~ 2 ( 1 2 ) 2 + ( x ~ 2 ( 1 2 ) 2 + y ~ 2 ( 1 2 ) 2 ) = 1 \frac{\tilde z^{2}}{\left(\frac{1}{2}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{1}{2}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{1}{2}\right)^{2}}\right) = 1 ( 2 1 ) 2 z ~ 2 + ( ( 2 1 ) 2 x ~ 2 + ( 2 1 ) 2 y ~ 2 ) = 1 this equation is fora type ellipsoid
- reduced to canonical form