Mister Exam

Canonical form of a elliptical paraboloid

For example, you have entered (calculator here):
$$x^{2} - 6 x + y^{2} + 6 y - 4 z + 18 = 0$$

Detail solution (Invariants method)

Given equation of the surface of 2-order:
$$x^{2} - 6 x + y^{2} + 6 y - 4 z + 18 = 0$$
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
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = -3$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = 3$$
$$a_{33} = 0$$
$$a_{34} = -2$$
$$a_{44} = 18$$
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
$$I_{1} = 2$$
     |1  0|   |1  0|   |1  0|
I2 = |    | + |    | + |    |
     |0  1|   |0  0|   |0  0|

$$I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 0\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}1 & 0 & 0 & -3\\0 & 1 & 0 & 3\\0 & 0 & 0 & -2\\-3 & 3 & -2 & 18\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & - \lambda\end{matrix}\right|$$
     |1   -3|   |1  3 |   |0   -2|
K2 = |      | + |     | + |      |
     |-3  18|   |3  18|   |-2  18|

     |1   0  -3|   |1  0   3 |   |1   0   -3|
     |         |   |         |   |          |
K3 = |0   1  3 | + |0  0   -2| + |0   0   -2|
     |         |   |         |   |          |
     |-3  3  18|   |3  -2  18|   |-3  -2  18|

$$I_{1} = 2$$
$$I_{2} = 1$$
$$I_{3} = 0$$
$$I_{4} = -4$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 2 \lambda^{2} - \lambda$$
$$K_{2} = 14$$
$$K_{3} = -8$$
Because
$$I_{3} = 0 \wedge I_{2} \neq 0 \wedge I_{4} \neq 0$$
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
$$\lambda^{3} - 2 \lambda^{2} + \lambda = 0$$
$$\lambda_{1} = 1$$
$$\lambda_{2} = 1$$
$$\lambda_{3} = 0$$
then the canonical form of the equation will be
$$\tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
and
$$- \tilde z 2 \sqrt{\frac{\left(-1\right) I_{4}}{I_{2}}} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) = 0$$
$$\tilde x^{2} + \tilde y^{2} + 4 \tilde z = 0$$
and
$$\tilde x^{2} + \tilde y^{2} - 4 \tilde z = 0$$
$$2 \tilde z + \left(\frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2}\right) = 0$$
and
$$- 2 \tilde z + \left(\frac{\tilde x^{2}}{2} + \frac{\tilde y^{2}}{2}\right) = 0$$
this equation is fora type elliptical paraboloid
- reduced to canonical form
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