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Canonical form/ x^2-xy

x^2-xy canonical form

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The solution

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x  - x*y = 0
x2−xy=0x^{2} - x y = 0 
x^2 - x*y = 0
Detail solution
Given line equation of 2-order:
x2−xy=0x^{2} - 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=1a_{11} = 1
a12=−12a_{12} = - \frac{1}{2}
a13=0a_{13} = 0
a22=0a_{22} = 0
a23=0a_{23} = 0
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=∣1−12−120∣\Delta = \left|\begin{matrix}1 & - \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+a12y0+a13=0a_{11} x_{0} + a_{12} y_{0} + a_{13} = 0
a12x0+a22y0+a23=0a_{12} x_{0} + a_{22} y_{0} + a_{23} = 0
substitute coefficients
x0−y02=0x_{0} - \frac{y_{0}}{2} = 0
−x02=0- \frac{x_{0}}{2} = 0
then
x0=0x_{0} = 0
y0=0y_{0} = 0
Thus, we have the equation in the coordinate system O'x'y'
a33′+a11x′2+2a12x′y′+a22y′2=0a'_{33} + a_{11} x'^{2} + 2 a_{12} x' y' + a_{22} y'^{2} = 0
where
a33′=a13x0+a23y0+a33a'_{33} = a_{13} x_{0} + a_{23} y_{0} + a_{33}
or
a33′=0a'_{33} = 0
a33′=0a'_{33} = 0
then equation turns into
x′2−x′y′=0x'^{2} - x' y' = 0
Rotate the resulting coordinate system by an angle φ
x′=x~cos⁡(ϕ)−y~sin⁡(ϕ)x' = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y′=x~sin⁡(ϕ)+y~cos⁡(ϕ)y' = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\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)=−1\cot{\left(2 \phi \right)} = -1
then
ϕ=−π8\phi = - \frac{\pi}{8}
sin⁡(2ϕ)=−22\sin{\left(2 \phi \right)} = - \frac{\sqrt{2}}{2}
cos⁡(2ϕ)=22\cos{\left(2 \phi \right)} = \frac{\sqrt{2}}{2}
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)=24+12\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}
sin⁡(ϕ)=−12−24\sin{\left(\phi \right)} = - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}
substitute coefficients
x′=x~24+12+y~12−24x' = \tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}
y′=−x~12−24+y~24+12y' = - \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}
then the equation turns from
x′2−x′y′=0x'^{2} - x' y' = 0
to
−(−x~12−24+y~24+12)(x~24+12+y~12−24)+(x~24+12+y~12−24)2=0- \left(- \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} + \tilde y \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right) + \left(\tilde x \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)^{2} = 0
simplify
2x~24+x~212−2424+12+x~22−2x~y~2+2x~y~12−2424+12−2y~24−y~212−2424+12+y~22=0\frac{\sqrt{2} \tilde x^{2}}{4} + \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{\tilde x^{2}}{2} - \frac{\sqrt{2} \tilde x \tilde y}{2} + 2 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} - \frac{\sqrt{2} \tilde y^{2}}{4} - \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}} \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}} + \frac{\tilde y^{2}}{2} = 0
x~22+2x~22−2y~22+y~22=0\frac{\tilde x^{2}}{2} + \frac{\sqrt{2} \tilde x^{2}}{2} - \frac{\sqrt{2} \tilde y^{2}}{2} + \frac{\tilde y^{2}}{2} = 0
Given equation is degenerate hyperbole
x~2(112+22)2−y~2(1−12+22)2=0\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{\frac{1}{2} + \frac{\sqrt{2}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{- \frac{1}{2} + \frac{\sqrt{2}}{2}}}\right)^{2}} = 0
- reduced to canonical form
The center of canonical coordinate system at point O
(0, 0)

Basis of the canonical coordinate system
e⃗1=(24+12, −12−24)\vec e_1 = \left( \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}, \ - \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}\right)
e⃗2=(12−24, 24+12)\vec e_2 = \left( \sqrt{\frac{1}{2} - \frac{\sqrt{2}}{4}}, \ \sqrt{\frac{\sqrt{2}}{4} + \frac{1}{2}}\right)
Invariants method
Given line equation of 2-order:
x2−xy=0x^{2} - 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=1a_{11} = 1
a12=−12a_{12} = - \frac{1}{2}
a13=0a_{13} = 0
a22=0a_{22} = 0
a23=0a_{23} = 0
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    -1/2|
I2 = |          |
     |-1/2   0  |

I3=∣1−120−1200000∣I_{3} = \left|\begin{matrix}1 & - \frac{1}{2} & 0\\- \frac{1}{2} & 0 & 0\\0 & 0 & 0\end{matrix}\right|
I(λ)=∣1−λ−12−12−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & - \frac{1}{2}\\- \frac{1}{2} & - \lambda\end{matrix}\right|
     |1  0|   |0  0|
K2 = |    | + |    |
     |0  0|   |0  0|

I1=1I_{1} = 1
I2=−14I_{2} = - \frac{1}{4}
I3=0I_{3} = 0
I(λ)=λ2−λ−14I{\left(\lambda \right)} = \lambda^{2} - \lambda - \frac{1}{4}
K2=0K_{2} = 0
Because
I3=0∧I2<0I_{3} = 0 \wedge I_{2} < 0
then by line type:
this equation is of type : degenerate hyperbole
Make the characteristic equation for the line:
−I1λ+I2+λ2=0- I_{1} \lambda + I_{2} + \lambda^{2} = 0
or
λ2−λ−14=0\lambda^{2} - \lambda - \frac{1}{4} = 0
λ1=12−22\lambda_{1} = \frac{1}{2} - \frac{\sqrt{2}}{2}
λ2=12+22\lambda_{2} = \frac{1}{2} + \frac{\sqrt{2}}{2}
then the canonical form of the equation will be
x~2λ1+y~2λ2+I3I2=0\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2} + \frac{I_{3}}{I_{2}} = 0
or
x~2(12−22)+y~2(12+22)=0\tilde x^{2} \left(\frac{1}{2} - \frac{\sqrt{2}}{2}\right) + \tilde y^{2} \left(\frac{1}{2} + \frac{\sqrt{2}}{2}\right) = 0
x~2(1−12+22)2−y~2(112+22)2=0\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{1}{2} + \frac{\sqrt{2}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{1}{2} + \frac{\sqrt{2}}{2}}}\right)^{2}} = 0
- reduced to canonical form
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