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Canonical form/ -x^2-2x+3

-x^2-2x+3 canonical form

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     2          
3 - x  - 2*x = 0
−x2−2x+3=0- x^{2} - 2 x + 3 = 0 
-x^2 - 2*x + 3 = 0
Detail solution
Given line equation of 2-order:
−x2−2x+3=0- x^{2} - 2 x + 3 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=−1a_{11} = -1
a12=0a_{12} = 0
a13=−1a_{13} = -1
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=3a_{33} = 3
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣−1000∣\Delta = \left|\begin{matrix}-1 & 0\\0 & 0\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
x0=x~cos⁡(ϕ)−y~sin⁡(ϕ)x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y0=x~sin⁡(ϕ)+y~cos⁡(ϕ)y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}
x0=0⋅0x_{0} = 0 \cdot 0
y0=0⋅0y_{0} = 0 \cdot 0
x0=0x_{0} = 0
y0=0y_{0} = 0
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e⃗1=(1, 0)\vec e_1 = \left( 1, \ 0\right)
e⃗2=(0, 1)\vec e_2 = \left( 0, \ 1\right)
Invariants method
Given line equation of 2-order:
−x2−2x+3=0- x^{2} - 2 x + 3 = 0
This equation looks like:
a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x + a_{22} y^{2} + 2 a_{23} y + a_{33} = 0
where
a11=−1a_{11} = -1
a12=0a_{12} = 0
a13=−1a_{13} = -1
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=3a_{33} = 3
The invariants of the equation when converting coordinates are determinants:
I1=a11+a22I_{1} = a_{11} + a_{22}
     |a11  a12|
I2 = |        |
     |a12  a22|

I3=∣a11a12a13a12a22a23a13a23a33∣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(λ)=∣a11−λa12a12a22−λ∣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
I1=−1I_{1} = -1
     |-1  0|
I2 = |     |
     |0   0|

I3=∣−10−1000−103∣I_{3} = \left|\begin{matrix}-1 & 0 & -1\\0 & 0 & 0\\-1 & 0 & 3\end{matrix}\right|
I(λ)=∣−λ−100−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 1 & 0\\0 & - \lambda\end{matrix}\right|
     |-1  -1|   |0  0|
K2 = |      | + |    |
     |-1  3 |   |0  3|

I1=−1I_{1} = -1
I2=0I_{2} = 0
I3=0I_{3} = 0
I(λ)=λ2+λI{\left(\lambda \right)} = \lambda^{2} + \lambda
K2=−4K_{2} = -4
Because
I2=0∧I3=0∧K2<0∧I1≠0I_{2} = 0 \wedge I_{3} = 0 \wedge K_{2} < 0 \wedge I_{1} \neq 0
then by line type:
this equation is of type : two parallel lines
I1y~2+K2I1=0I_{1} \tilde y^{2} + \frac{K_{2}}{I_{1}} = 0
or
4−y~2=04 - \tilde y^{2} = 0
None

- reduced to canonical form
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