Mister Exam

Canonical form of a double hyperboloid

For example, you have entered (calculator here):
x2−2x+y2−2y−z2+2z+2=0x^{2} - 2 x + y^{2} - 2 y - z^{2} + 2 z + 2 = 0

Detail solution (Invariants method)

Given equation of the surface of 2-order:
x2−2x+y2−2y−z2+2z+2=0x^{2} - 2 x + y^{2} - 2 y - z^{2} + 2 z + 2 = 0
This equation looks like:
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
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=0a_{13} = 0
a14=−1a_{14} = -1
a22=1a_{22} = 1
a23=0a_{23} = 0
a24=−1a_{24} = -1
a33=−1a_{33} = -1
a34=1a_{34} = 1
a44=2a_{44} = 2
The invariants of the equation when converting coordinates are determinants:
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|

substitute coefficients
I1=1I_{1} = 1
     |1  0|   |1  0 |   |1  0 |
I2 = |    | + |     | + |     |
     |0  1|   |0  -1|   |0  -1|

I3=∣10001000−1∣I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -1\end{matrix}\right|
I4=∣100−1010−100−11−1−112∣I_{4} = \left|\begin{matrix}1 & 0 & 0 & -1\\0 & 1 & 0 & -1\\0 & 0 & -1 & 1\\-1 & -1 & 1 & 2\end{matrix}\right|
I(λ)=∣1−λ0001−λ000−λ−1∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & - \lambda - 1\end{matrix}\right|
     |1   -1|   |1   -1|   |-1  1|
K2 = |      | + |      | + |     |
     |-1  2 |   |-1  2 |   |1   2|

     |1   0   -1|   |1   0   -1|   |1   0   -1|
     |          |   |          |   |          |
K3 = |0   1   -1| + |0   -1  1 | + |0   -1  1 |
     |          |   |          |   |          |
     |-1  -1  2 |   |-1  1   2 |   |-1  1   2 |

I1=1I_{1} = 1
I2=−1I_{2} = -1
I3=−1I_{3} = -1
I4=−1I_{4} = -1
I(λ)=−λ3+λ2+λ−1I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + \lambda - 1
K2=−1K_{2} = -1
K3=−4K_{3} = -4
Because
I3 != 0

then by type of surface:
you need to
Make the characteristic equation for the surface:
−I1λ2+I2λ−I3+λ3=0- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0
or
λ3−λ2−λ+1=0\lambda^{3} - \lambda^{2} - \lambda + 1 = 0
λ1=−1\lambda_{1} = -1
λ2=1\lambda_{2} = 1
λ3=1\lambda_{3} = 1
then the canonical form of the equation will be
(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~2+z~2+1=0- \tilde x^{2} + \tilde y^{2} + \tilde z^{2} + 1 = 0
−x~2(1−1)2+(y~2(1−1)2+z~2(1−1)2)=−1- \frac{\tilde x^{2}}{\left(1^{-1}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde z^{2}}{\left(1^{-1}\right)^{2}}\right) = -1
this equation is fora type two-sided hyperboloid
- reduced to canonical form
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