Mister Exam

Canonical form of a double hyperboloid

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

Detail solution (Invariants method)

Given equation of the surface of 2-order:
$$x^{2} - 2 x + y^{2} - 2 y - z^{2} + 2 z + 2 = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -1$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = -1$$
$$a_{33} = -1$$
$$a_{34} = 1$$
$$a_{44} = 2$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22} + a_{33}$$
     |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|

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

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

$$I_{1} = 1$$
$$I_{2} = -1$$
$$I_{3} = -1$$
$$I_{4} = -1$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + \lambda - 1$$
$$K_{2} = -1$$
$$K_{3} = -4$$
Because
I3 != 0

then by type of surface:
you need to
Make the characteristic equation for the surface:
$$- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0$$
or
$$\lambda^{3} - \lambda^{2} - \lambda + 1 = 0$$
$$\lambda_{1} = -1$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 1$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$- \tilde x^{2} + \tilde y^{2} + \tilde z^{2} + 1 = 0$$
$$- \frac{\tilde x^{2}}{\left(1^{-1}\right)^{2}} + \left(\frac{\tilde y^{2}}{\left(1^{-1}\right)^{2}} + \frac{\tilde z^{2}}{\left(1^{-1}\right)^{2}}\right) = -1$$
this equation is fora type two-sided hyperboloid
- reduced to canonical form