Mister Exam

# 4x^2+y^2+4z^2+4y+24z+36=0 canonical form

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

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

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### The solution

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      2            2      2
36 + y  + 4*y + 4*x  + 4*z  + 24*z = 0
$$4 x^{2} + y^{2} + 4 y + 4 z^{2} + 24 z + 36 = 0$$
4*x^2 + y^2 + 4*y + 4*z^2 + 24*z + 36 = 0
Invariants method
Given equation of the surface of 2-order:
$$4 x^{2} + y^{2} + 4 y + 4 z^{2} + 24 z + 36 = 0$$
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
$$a_{11} = 4$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{14} = 0$$
$$a_{22} = 1$$
$$a_{23} = 0$$
$$a_{24} = 2$$
$$a_{33} = 4$$
$$a_{34} = 12$$
$$a_{44} = 36$$
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
$$I_{1} = 9$$
     |4  0|   |1  0|   |4  0|
I2 = |    | + |    | + |    |
|0  1|   |0  4|   |0  4|

$$I_{3} = \left|\begin{matrix}4 & 0 & 0\\0 & 1 & 0\\0 & 0 & 4\end{matrix}\right|$$
$$I_{4} = \left|\begin{matrix}4 & 0 & 0 & 0\\0 & 1 & 0 & 2\\0 & 0 & 4 & 12\\0 & 2 & 12 & 36\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}4 - \lambda & 0 & 0\\0 & 1 - \lambda & 0\\0 & 0 & 4 - \lambda\end{matrix}\right|$$
     |4  0 |   |1  2 |   |4   12|
K2 = |     | + |     | + |      |
|0  36|   |2  36|   |12  36|

     |4  0  0 |   |1  0   2 |   |4  0   0 |
|        |   |         |   |         |
K3 = |0  1  2 | + |0  4   12| + |0  4   12|
|        |   |         |   |         |
|0  2  36|   |2  12  36|   |0  12  36|

$$I_{1} = 9$$
$$I_{2} = 24$$
$$I_{3} = 16$$
$$I_{4} = -64$$
$$I{\left(\lambda \right)} = - \lambda^{3} + 9 \lambda^{2} - 24 \lambda + 16$$
$$K_{2} = 176$$
$$K_{3} = 112$$
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
$$\lambda^{3} - 9 \lambda^{2} + 24 \lambda - 16 = 0$$
Solve this equation
$$\lambda_{1} = 1$$
$$\lambda_{2} = 4$$
$$\lambda_{3} = 4$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$\tilde x^{2} + 4 \tilde y^{2} + 4 \tilde z^{2} - 4 = 0$$
$$\frac{\tilde x^{2}}{4} + \tilde y^{2} + \tilde z^{2} = 1$$
this equation is fora type ellipsoid
- reduced to canonical form