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Canonical form/ y=4x^2-16x

y=4x^2-16x canonical form

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y - 4*x  + 16*x = 0
−4x2+16x+y=0- 4 x^{2} + 16 x + y = 0 
-4*x^2 + 16*x + y = 0
Detail solution
Given line equation of 2-order:
−4x2+16x+y=0- 4 x^{2} + 16 x + y = 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=−4a_{11} = -4
a12=0a_{12} = 0
a13=8a_{13} = 8
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=∣−4000∣\Delta = \left|\begin{matrix}-4 & 0\\0 & 0\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
(2x~+4)2=y~+16\left(2 \tilde x + 4\right)^{2} = \tilde y + 16
(x~+2)2=y~4+4\left(\tilde x + 2\right)^{2} = \frac{\tilde y}{4} + 4
x~′2=y~4+4\tilde x'^{2} = \frac{\tilde y}{4} + 4
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⋅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:
−4x2+16x+y=0- 4 x^{2} + 16 x + y = 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=−4a_{11} = -4
a12=0a_{12} = 0
a13=8a_{13} = 8
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=−4I_{1} = -4
     |-4  0|
I2 = |     |
     |0   0|

I3=∣−40800128120∣I_{3} = \left|\begin{matrix}-4 & 0 & 8\\0 & 0 & \frac{1}{2}\\8 & \frac{1}{2} & 0\end{matrix}\right|
I(λ)=∣−λ−400−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 4 & 0\\0 & - \lambda\end{matrix}\right|
     |-4  8|   | 0   1/2|
K2 = |     | + |        |
     |8   0|   |1/2   0 |

I1=−4I_{1} = -4
I2=0I_{2} = 0
I3=1I_{3} = 1
I(λ)=λ2+4λI{\left(\lambda \right)} = \lambda^{2} + 4 \lambda
K2=−2574K_{2} = - \frac{257}{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
x~−4y~2=0\tilde x - 4 \tilde y^{2} = 0
y~2=−x~4\tilde y^{2} = - \frac{\tilde x}{4}
- reduced to canonical form
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