Mister Exam

# 3x^2+2y^2-4z^2=0 canonical form

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

#### Quality:

(Number of points on the axis)

### The solution

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

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

     |3  0  0|   |2  0   0|   |3  0   0|
|       |   |        |   |        |
K3 = |0  2  0| + |0  -4  0| + |0  -4  0|
|       |   |        |   |        |
|0  0  0|   |0  0   0|   |0  0   0|

$$I_{1} = 1$$
$$I_{2} = -14$$
$$I_{3} = -24$$
$$I_{4} = 0$$
$$I{\left(\lambda \right)} = - \lambda^{3} + \lambda^{2} + 14 \lambda - 24$$
$$K_{2} = 0$$
$$K_{3} = 0$$
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
$$\lambda^{3} - \lambda^{2} - 14 \lambda + 24 = 0$$
$$\lambda_{1} = 3$$
$$\lambda_{2} = 2$$
$$\lambda_{3} = -4$$
then the canonical form of the equation will be
$$\left(\tilde z^{2} \lambda_{3} + \left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right)\right) + \frac{I_{4}}{I_{3}} = 0$$
$$3 \tilde x^{2} + 2 \tilde y^{2} - 4 \tilde z^{2} = 0$$
$$- \frac{\tilde z^{2}}{\left(\frac{1}{2}\right)^{2}} + \left(\frac{\tilde x^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{2}}{2}\right)^{2}}\right) = 0$$
this equation is fora type cone
- reduced to canonical form
To see a detailed solution - share to all your student friends
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share to all your student friends: