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Canonical form/ 2y^2+x-y-1=0

2y^2+x-y-1=0 canonical form

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The solution

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                2    
-1 + x - y + 2*y  = 0
x+2y2−y−1=0x + 2 y^{2} - y - 1 = 0 
x + 2*y^2 - y - 1 = 0
Detail solution
Given line equation of 2-order:
x+2y2−y−1=0x + 2 y^{2} - y - 1 = 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=0a_{11} = 0
a12=0a_{12} = 0
a13=12a_{13} = \frac{1}{2}
a22=2a_{22} = 2
a23=−12a_{23} = - \frac{1}{2}
a33=−1a_{33} = -1
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣0002∣\Delta = \left|\begin{matrix}0 & 0\\0 & 2\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:
x+2y2−y−1=0x + 2 y^{2} - y - 1 = 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=0a_{11} = 0
a12=0a_{12} = 0
a13=12a_{13} = \frac{1}{2}
a22=2a_{22} = 2
a23=−12a_{23} = - \frac{1}{2}
a33=−1a_{33} = -1
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=2I_{1} = 2
     |0  0|
I2 = |    |
     |0  2|

I3=∣001202−1212−12−1∣I_{3} = \left|\begin{matrix}0 & 0 & \frac{1}{2}\\0 & 2 & - \frac{1}{2}\\\frac{1}{2} & - \frac{1}{2} & -1\end{matrix}\right|
I(λ)=∣−λ002−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & 2 - \lambda\end{matrix}\right|
     | 0   1/2|   | 2    -1/2|
K2 = |        | + |          |
     |1/2  -1 |   |-1/2   -1 |

I1=2I_{1} = 2
I2=0I_{2} = 0
I3=−12I_{3} = - \frac{1}{2}
I(λ)=λ2−2λI{\left(\lambda \right)} = \lambda^{2} - 2 \lambda
K2=−52K_{2} = - \frac{5}{2}
Because
I2=0∧I3≠0I_{2} = 0 \wedge I_{3} \neq 0
then by line type:
this equation is of type : parabola
I1y~2+2x~−I3I1=0I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0
or
x~+2y~2=0\tilde x + 2 \tilde y^{2} = 0
y~2=x~2\tilde y^{2} = \frac{\tilde x}{2}
- reduced to canonical form
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