Mister Exam

x²+y²+z²=z canonical form

The teacher will be very surprised to see your correct solution 😉

v

The graph:

x: [, ]
y: [, ]
z: [, ]

Quality:

 (Number of points on the axis)

Plot type:

The solution

You have entered [src]
 2    2    2        
x  + y  + z  - z = 0
x2+y2+z2z=0x^{2} + y^{2} + z^{2} - z = 0
x^2 + y^2 + z^2 - z = 0
Invariants method
Given equation of the surface of 2-order:
x2+y2+z2z=0x^{2} + y^{2} + z^{2} - z = 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=0a_{14} = 0
a22=1a_{22} = 1
a23=0a_{23} = 0
a24=0a_{24} = 0
a33=1a_{33} = 1
a34=12a_{34} = - \frac{1}{2}
a44=0a_{44} = 0
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=a11a12a13a12a22a23a13a23a33I_{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=a11a12a13a14a12a22a23a24a13a23a33a34a14a24a34a44I_{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)=a11λa12a13a12a22λa23a13a23a33λI{\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=3I_{1} = 3
     |1  0|   |1  0|   |1  0|
I2 = |    | + |    | + |    |
     |0  1|   |0  1|   |0  1|

I3=100010001I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right|
I4=100001000011200120I_{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(λ)=1λ0001λ0001λI{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 1 - \lambda\end{matrix}\right|
     |1  0|   |1  0|   | 1    -1/2|
K2 = |    | + |    | + |          |
     |0  0|   |0  0|   |-1/2   0  |

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

I1=3I_{1} = 3
I2=3I_{2} = 3
I3=1I_{3} = 1
I4=14I_{4} = - \frac{1}{4}
I(λ)=λ3+3λ23λ+1I{\left(\lambda \right)} = - \lambda^{3} + 3 \lambda^{2} - 3 \lambda + 1
K2=14K_{2} = - \frac{1}{4}
K3=12K_{3} = - \frac{1}{2}
Because
I3 != 0

then by type of surface:
you need to
Make the characteristic equation for the surface:
I1λ2+I2λI3+λ3=0- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0
or
λ33λ2+3λ1=0\lambda^{3} - 3 \lambda^{2} + 3 \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~214=0\tilde x^{2} + \tilde y^{2} + \tilde z^{2} - \frac{1}{4} = 0
z~2(12)2+(x~2(12)2+y~2(12)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
this equation is fora type ellipsoid
- reduced to canonical form