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Canonical form/ 4xy+4x-4y-2=0

4xy+4x-4y-2=0 canonical form

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The solution

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-2 - 4*y + 4*x + 4*x*y = 0
4xy+4x−4y−2=04 x y + 4 x - 4 y - 2 = 0 
4*x*y + 4*x - 4*y - 2 = 0
Detail solution
Given line equation of 2-order:
4xy+4x−4y−2=04 x y + 4 x - 4 y - 2 = 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=2a_{12} = 2
a13=2a_{13} = 2
a22=0a_{22} = 0
a23=−2a_{23} = -2
a33=−2a_{33} = -2
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣0220∣\Delta = \left|\begin{matrix}0 & 2\\2 & 0\end{matrix}\right|
Δ=−4\Delta = -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
2y0+2=02 y_{0} + 2 = 0
2x0−2=02 x_{0} - 2 = 0
then
x0=1x_{0} = 1
y0=−1y_{0} = -1
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′=2x0−2y0−2a'_{33} = 2 x_{0} - 2 y_{0} - 2
a33′=2a'_{33} = 2
then equation turns into
4x′y′+2=04 x' y' + 2 = 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)=0\cot{\left(2 \phi \right)} = 0
then
ϕ=π4\phi = \frac{\pi}{4}
sin⁡(2ϕ)=1\sin{\left(2 \phi \right)} = 1
cos⁡(2ϕ)=0\cos{\left(2 \phi \right)} = 0
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)=22\cos{\left(\phi \right)} = \frac{\sqrt{2}}{2}
sin⁡(ϕ)=22\sin{\left(\phi \right)} = \frac{\sqrt{2}}{2}
substitute coefficients
x′=2x~2−2y~2x' = \frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}
y′=2x~2+2y~2y' = \frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}
then the equation turns from
4x′y′+2=04 x' y' + 2 = 0
to
4(2x~2−2y~2)(2x~2+2y~2)+2=04 \left(\frac{\sqrt{2} \tilde x}{2} - \frac{\sqrt{2} \tilde y}{2}\right) \left(\frac{\sqrt{2} \tilde x}{2} + \frac{\sqrt{2} \tilde y}{2}\right) + 2 = 0
simplify
2x~2−2y~2+2=02 \tilde x^{2} - 2 \tilde y^{2} + 2 = 0
Given equation is hyperbole
x~21−y~21=−1\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{1} = -1
- reduced to canonical form
The center of canonical coordinate system at point O
(1, -1)

Basis of the canonical coordinate system
e⃗1=(22, 22)\vec e_1 = \left( \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)
e⃗2=(−22, 22)\vec e_2 = \left( - \frac{\sqrt{2}}{2}, \ \frac{\sqrt{2}}{2}\right)
Invariants method
Given line equation of 2-order:
4xy+4x−4y−2=04 x y + 4 x - 4 y - 2 = 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=2a_{12} = 2
a13=2a_{13} = 2
a22=0a_{22} = 0
a23=−2a_{23} = -2
a33=−2a_{33} = -2
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=0I_{1} = 0
     |0  2|
I2 = |    |
     |2  0|

I3=∣02220−22−2−2∣I_{3} = \left|\begin{matrix}0 & 2 & 2\\2 & 0 & -2\\2 & -2 & -2\end{matrix}\right|
I(λ)=∣−λ22−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 2\\2 & - \lambda\end{matrix}\right|
     |0  2 |   |0   -2|
K2 = |     | + |      |
     |2  -2|   |-2  -2|

I1=0I_{1} = 0
I2=−4I_{2} = -4
I3=−8I_{3} = -8
I(λ)=λ2−4I{\left(\lambda \right)} = \lambda^{2} - 4
K2=−8K_{2} = -8
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
λ2−4=0\lambda^{2} - 4 = 0
λ1=−2\lambda_{1} = -2
λ2=2\lambda_{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
−2x~2+2y~2+2=0- 2 \tilde x^{2} + 2 \tilde y^{2} + 2 = 0
x~21−y~21=1\frac{\tilde x^{2}}{1} - \frac{\tilde y^{2}}{1} = 1
- reduced to canonical form
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