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Canonical form/ 𝒇(𝒙)=𝟏𝟐(𝒙−𝟏)^𝟐−𝟐

𝒇(𝒙)=𝟏𝟐(𝒙−𝟏)^𝟐−𝟐 canonical form

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               2          
2 - 12*(-1 + x)  + f*x = 0
fx−12(x−1)2+2=0f x - 12 \left(x - 1\right)^{2} + 2 = 0 
f*x - 12*(x - 1)^2 + 2 = 0
Detail solution
Given line equation of 2-order:
fx−12(x−1)2+2=0f x - 12 \left(x - 1\right)^{2} + 2 = 0
This equation looks like:
a11x2+2a12fx+2a13x+a22f2+2a23f+a33=0a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0
where
a11=−12a_{11} = -12
a12=12a_{12} = \frac{1}{2}
a13=12a_{13} = 12
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=−10a_{33} = -10
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣−1212120∣\Delta = \left|\begin{matrix}-12 & \frac{1}{2}\\\frac{1}{2} & 0\end{matrix}\right|
Δ=−14\Delta = - \frac{1}{4}
Because
Δ\Delta
is not equal to 0, then
find the center of the canonical coordinate system. To do it, solve the system of equations
a11x0+a12f0+a13=0a_{11} x_{0} + a_{12} f_{0} + a_{13} = 0
a12x0+a22f0+a23=0a_{12} x_{0} + a_{22} f_{0} + a_{23} = 0
substitute coefficients
f02−12x0+12=0\frac{f_{0}}{2} - 12 x_{0} + 12 = 0
x02=0\frac{x_{0}}{2} = 0
then
x0=0x_{0} = 0
f0=−24f_{0} = -24
Thus, we have the equation in the coordinate system O'x'y'
a33′+a11x′2+2a12f′x′+a22f′2=0a'_{33} + a_{11} x'^{2} + 2 a_{12} f' x' + a_{22} f'^{2} = 0
where
a33′=a13x0+a23f0+a33a'_{33} = a_{13} x_{0} + a_{23} f_{0} + a_{33}
or
a33′=12x0−10a'_{33} = 12 x_{0} - 10
a33′=−10a'_{33} = -10
then equation turns into
f′x′−12x′2−10=0f' x' - 12 x'^{2} - 10 = 0
Rotate the resulting coordinate system by an angle φ
x′=−f~sin⁡(ϕ)+x~cos⁡(ϕ)x' = - \tilde f \sin{\left(\phi \right)} + \tilde x \cos{\left(\phi \right)}
f′=f~cos⁡(ϕ)+x~sin⁡(ϕ)f' = \tilde f \cos{\left(\phi \right)} + \tilde x \sin{\left(\phi \right)}
φ - determined from the formula
cot⁡(2ϕ)=a11−a222a12\cot{\left(2 \phi \right)} = \frac{a_{11} - a_{22}}{2 a_{12}}
substitute coefficients
cot⁡(2ϕ)=−12\cot{\left(2 \phi \right)} = -12
then
ϕ=−acot⁡(12)2\phi = - \frac{\operatorname{acot}{\left(12 \right)}}{2}
sin⁡(2ϕ)=−145145\sin{\left(2 \phi \right)} = - \frac{\sqrt{145}}{145}
cos⁡(2ϕ)=12145145\cos{\left(2 \phi \right)} = \frac{12 \sqrt{145}}{145}
cos⁡(ϕ)=cos⁡(2ϕ)2+12\cos{\left(\phi \right)} = \sqrt{\frac{\cos{\left(2 \phi \right)}}{2} + \frac{1}{2}}
sin⁡(ϕ)=1−cos⁡2(ϕ)\sin{\left(\phi \right)} = \sqrt{1 - \cos^{2}{\left(\phi \right)}}
cos⁡(ϕ)=6145145+12\cos{\left(\phi \right)} = \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}
sin⁡(ϕ)=−12−6145145\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}
substitute coefficients
x′=f~12−6145145+x~6145145+12x' = \tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}
f′=f~6145145+12−x~12−6145145f' = \tilde f \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}
then the equation turns from
f′x′−12x′2−10=0f' x' - 12 x'^{2} - 10 = 0
to
−12(f~12−6145145+x~6145145+12)2+(f~12−6145145+x~6145145+12)(f~6145145+12−x~12−6145145)−10=0- 12 \left(\tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right)^{2} + \left(\tilde f \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} + \tilde x \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right) \left(\tilde f \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}\right) - 10 = 0
simplify
−6f~2+f~212−61451456145145+12+72145f~2145−24f~x~12−61451456145145+12+12145f~x~145−6x~2−72145x~2145−x~212−61451456145145+12−10=0- 6 \tilde f^{2} + \tilde f^{2} \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} + \frac{72 \sqrt{145} \tilde f^{2}}{145} - 24 \tilde f \tilde x \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} + \frac{12 \sqrt{145} \tilde f \tilde x}{145} - 6 \tilde x^{2} - \frac{72 \sqrt{145} \tilde x^{2}}{145} - \tilde x^{2} \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}} \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}} - 10 = 0
−145f~22+6f~2+6x~2+145x~22+10=0- \frac{\sqrt{145} \tilde f^{2}}{2} + 6 \tilde f^{2} + 6 \tilde x^{2} + \frac{\sqrt{145} \tilde x^{2}}{2} + 10 = 0
Given equation is hyperbole
−f~2101−6+1452+x~21016+1452=−1- \frac{\tilde f^{2}}{10 \frac{1}{-6 + \frac{\sqrt{145}}{2}}} + \frac{\tilde x^{2}}{10 \frac{1}{6 + \frac{\sqrt{145}}{2}}} = -1
- reduced to canonical form
The center of canonical coordinate system at point O
(0, -24)

Basis of the canonical coordinate system
e⃗1=(6145145+12, −12−6145145)\vec e_1 = \left( \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}\right)
e⃗2=(12−6145145, 6145145+12)\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{6 \sqrt{145}}{145}}, \ \sqrt{\frac{6 \sqrt{145}}{145} + \frac{1}{2}}\right)
Invariants method
Given line equation of 2-order:
fx−12(x−1)2+2=0f x - 12 \left(x - 1\right)^{2} + 2 = 0
This equation looks like:
a11x2+2a12fx+2a13x+a22f2+2a23f+a33=0a_{11} x^{2} + 2 a_{12} f x + 2 a_{13} x + a_{22} f^{2} + 2 a_{23} f + a_{33} = 0
where
a11=−12a_{11} = -12
a12=12a_{12} = \frac{1}{2}
a13=12a_{13} = 12
a22=0a_{22} = 0
a23=0a_{23} = 0
a33=−10a_{33} = -10
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=−12I_{1} = -12
     |-12  1/2|
I2 = |        |
     |1/2   0 |

I3=∣−1212121200120−10∣I_{3} = \left|\begin{matrix}-12 & \frac{1}{2} & 12\\\frac{1}{2} & 0 & 0\\12 & 0 & -10\end{matrix}\right|
I(λ)=∣−λ−121212−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda - 12 & \frac{1}{2}\\\frac{1}{2} & - \lambda\end{matrix}\right|
     |-12  12 |   |0   0 |
K2 = |        | + |      |
     |12   -10|   |0  -10|

I1=−12I_{1} = -12
I2=−14I_{2} = - \frac{1}{4}
I3=52I_{3} = \frac{5}{2}
I(λ)=λ2+12λ−14I{\left(\lambda \right)} = \lambda^{2} + 12 \lambda - \frac{1}{4}
K2=−24K_{2} = -24
Because
I2<0∧I3≠0I_{2} < 0 \wedge I_{3} \neq 0
then by line type:
this equation is of type : hyperbola
Make the characteristic equation for the line:
−I1λ+I2+λ2=0- I_{1} \lambda + I_{2} + \lambda^{2} = 0
or
λ2+12λ−14=0\lambda^{2} + 12 \lambda - \frac{1}{4} = 0
λ1=−1452−6\lambda_{1} = - \frac{\sqrt{145}}{2} - 6
λ2=−6+1452\lambda_{2} = -6 + \frac{\sqrt{145}}{2}
then the canonical form of the equation will be
f~2λ2+x~2λ1+I3I2=0\tilde f^{2} \lambda_{2} + \tilde x^{2} \lambda_{1} + \frac{I_{3}}{I_{2}} = 0
or
f~2(−6+1452)+x~2(−1452−6)−10=0\tilde f^{2} \left(-6 + \frac{\sqrt{145}}{2}\right) + \tilde x^{2} \left(- \frac{\sqrt{145}}{2} - 6\right) - 10 = 0
−f~2101−6+1452+x~21016+1452=−1- \frac{\tilde f^{2}}{10 \frac{1}{-6 + \frac{\sqrt{145}}{2}}} + \frac{\tilde x^{2}}{10 \frac{1}{6 + \frac{\sqrt{145}}{2}}} = -1
- reduced to canonical form
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