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Canonical form/ 0.25(x-4)(x+5)

0.25(x-4)(x+5) canonical form

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The solution

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(-4 + x)*(5 + x)    
---------------- = 0
       4
(x−4)(x+5)4=0\frac{\left(x - 4\right) \left(x + 5\right)}{4} = 0 
(x - 4)*(x + 5)/4 = 0
Detail solution
Given line equation of 2-order:
(x−4)(x+5)4=0\frac{\left(x - 4\right) \left(x + 5\right)}{4} = 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=14a_{11} = \frac{1}{4}
a12=0a_{12} = 0
a13=18a_{13} = \frac{1}{8}
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=−5a_{33} = -5
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣14000∣\Delta = \left|\begin{matrix}\frac{1}{4} & 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⋅1+0⋅0x_{0} = 0 \cdot 1 + 0 \cdot 0
y0=0⋅0+0⋅1y_{0} = 0 \cdot 0 + 0 \cdot 1
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:
(x−4)(x+5)4=0\frac{\left(x - 4\right) \left(x + 5\right)}{4} = 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=14a_{11} = \frac{1}{4}
a12=0a_{12} = 0
a13=18a_{13} = \frac{1}{8}
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=−5a_{33} = -5
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=14I_{1} = \frac{1}{4}
     |1/4  0|
I2 = |      |
     | 0   0|

I3=∣14018000180−5∣I_{3} = \left|\begin{matrix}\frac{1}{4} & 0 & \frac{1}{8}\\0 & 0 & 0\\\frac{1}{8} & 0 & -5\end{matrix}\right|
I(λ)=∣14−λ00−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}\frac{1}{4} - \lambda & 0\\0 & - \lambda\end{matrix}\right|
     |1/4  1/8|   |0  0 |
K2 = |        | + |     |
     |1/8  -5 |   |0  -5|

I1=14I_{1} = \frac{1}{4}
I2=0I_{2} = 0
I3=0I_{3} = 0
I(λ)=λ2−λ4I{\left(\lambda \right)} = \lambda^{2} - \frac{\lambda}{4}
K2=−8164K_{2} = - \frac{81}{64}
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
y~24−8116=0\frac{\tilde y^{2}}{4} - \frac{81}{16} = 0
None

- reduced to canonical form
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