Mister Exam

f=2x^2+x^2-4xx canonical form

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The solution

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     2    
f + x  = 0
$$f + x^{2} = 0$$
f + x^2 = 0
Detail solution
Given line equation of 2-order:
$$f + x^{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
To calculate the determinant
$$\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|$$
or, substitute
$$\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|$$
$$\Delta = 0$$
Because
$$\Delta$$
is equal to 0, then
$$\tilde x^{2} = - \tilde f$$
$$\tilde x'^{2} = - \tilde f'$$
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
$$x_{0} = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}$$
$$f_{0} = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}$$
$$x_{0} = 0 \cdot 0$$
$$f_{0} = 0 \cdot 0$$
$$x_{0} = 0$$
$$f_{0} = 0$$
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
$$\vec e_1 = \left( 1, \ 0\right)$$
$$\vec e_2 = \left( 0, \ 1\right)$$
Invariants method
Given line equation of 2-order:
$$f + x^{2} = 0$$
This equation looks like:
$$a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0$$
where
$$a_{11} = 1$$
$$a_{12} = 0$$
$$a_{13} = 0$$
$$a_{22} = 0$$
$$a_{23} = \frac{1}{2}$$
$$a_{33} = 0$$
The invariants of the equation when converting coordinates are determinants:
$$I_{1} = a_{11} + a_{22}$$
     |a11  a12|
I2 = |        |
     |a12  a22|

$$I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|$$
$$I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12}\\a_{12} & a_{22} - \lambda\end{matrix}\right|$$
     |a11  a13|   |a22  a23|
K2 = |        | + |        |
     |a13  a33|   |a23  a33|

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

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

$$I_{1} = 1$$
$$I_{2} = 0$$
$$I_{3} = - \frac{1}{4}$$
$$I{\left(\lambda \right)} = \lambda^{2} - \lambda$$
$$K_{2} = - \frac{1}{4}$$
Because
$$I_{2} = 0 \wedge I_{3} \neq 0$$
then by line type:
this equation is of type : parabola
$$I_{1} \tilde f^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0$$
or
$$\tilde f^{2} + \tilde x = 0$$
$$\tilde f^{2} = \tilde x$$
- reduced to canonical form