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x^2+y
Canonical form/ x^2+y

x^2+y canonical form

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y + x  = 0
x2+y=0x^{2} + y = 0 
x^2 + y = 0
Detail solution
Given line equation of 2-order:
x2+y=0x^{2} + y = 0
This equation looks like:
a11x2+2a12xy+a22y2+2a13x+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=0a_{22} = 0
a23=12a_{23} = \frac{1}{2}
a33=0a_{33} = 0
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
x~2=−y~\tilde x^{2} = - \tilde y
x~′2=−y~′\tilde x'^{2} = - \tilde y'
Given equation is by parabola
- 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:
x2+y=0x^{2} + y = 0
This equation looks like:
a11x2+2a12xy+a22y2+2a13x+2a23y+a33=0a_{11} x^{2} + 2 a_{12} x y + a_{22} y^{2} + 2 a_{13} x + 2 a_{23} y + a_{33} = 0
where
a11=1a_{11} = 1
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=0a_{22} = 0
a23=12a_{23} = \frac{1}{2}
a33=0a_{33} = 0
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=∣10000120120∣I_{3} = \left|\begin{matrix}1 & 0 & 0\\0 & 0 & \frac{1}{2}\\0 & \frac{1}{2} & 0\end{matrix}\right|
I(λ)=∣−λ+100−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda + 1 & 0\\0 & - \lambda\end{matrix}\right|
     |1  0|   | 0   1/2|
K2 = |    | + |        |
     |0  0|   |1/2   0 |

I1=1I_{1} = 1
I2=0I_{2} = 0
I3=−14I_{3} = - \frac{1}{4}
I(λ)=λ2−λI{\left(\lambda \right)} = \lambda^{2} - \lambda
K2=−14K_{2} = - \frac{1}{4}
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
y~2+x~=0\tilde y^{2} + \tilde x = 0
y~2=x~\tilde y^{2} = \tilde x
- reduced to canonical form
The graph
x^2+y canonical form
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