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Canonical form/ 2x^2-5xy+3y^2=0

2x^2-5xy+3y^2=0 canonical form

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   2      2            
2*x  + 3*y  - 5*x*y = 0
2x2−5xy+3y2=02 x^{2} - 5 x y + 3 y^{2} = 0 
2*x^2 - 5*x*y + 3*y^2 = 0
Detail solution
Given line equation of 2-order:
2x2−5xy+3y2=02 x^{2} - 5 x y + 3 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=2a_{11} = 2
a12=−52a_{12} = - \frac{5}{2}
a13=0a_{13} = 0
a22=3a_{22} = 3
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=∣2−52−523∣\Delta = \left|\begin{matrix}2 & - \frac{5}{2}\\- \frac{5}{2} & 3\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
2x0−5y02=02 x_{0} - \frac{5 y_{0}}{2} = 0
−5x02+3y0=0- \frac{5 x_{0}}{2} + 3 y_{0} = 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
2x′2−5x′y′+3y′2=02 x'^{2} - 5 x' y' + 3 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)=15\cot{\left(2 \phi \right)} = \frac{1}{5}
then
ϕ=acot⁡(15)2\phi = \frac{\operatorname{acot}{\left(\frac{1}{5} \right)}}{2}
sin⁡(2ϕ)=52626\sin{\left(2 \phi \right)} = \frac{5 \sqrt{26}}{26}
cos⁡(2ϕ)=2626\cos{\left(2 \phi \right)} = \frac{\sqrt{26}}{26}
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)=2652+12\cos{\left(\phi \right)} = \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}
sin⁡(ϕ)=12−2652\sin{\left(\phi \right)} = \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}
substitute coefficients
x′=x~2652+12−y~12−2652x' = \tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}
y′=x~12−2652+y~2652+12y' = \tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}
then the equation turns from
2x′2−5x′y′+3y′2=02 x'^{2} - 5 x' y' + 3 y'^{2} = 0
to
3(x~12−2652+y~2652+12)2−5(x~12−2652+y~2652+12)(x~2652+12−y~12−2652)+2(x~2652+12−y~12−2652)2=03 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right)^{2} - 5 \left(\tilde x \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} + \tilde y \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right) \left(\tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right) + 2 \left(\tilde x \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right)^{2} = 0
simplify
−5x~212−26522652+12−26x~252+5x~22−526x~y~26+2x~y~12−26522652+12+26y~252+5y~212−26522652+12+5y~22=0- 5 \tilde x^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} - \frac{\sqrt{26} \tilde x^{2}}{52} + \frac{5 \tilde x^{2}}{2} - \frac{5 \sqrt{26} \tilde x \tilde y}{26} + 2 \tilde x \tilde y \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} + \frac{\sqrt{26} \tilde y^{2}}{52} + 5 \tilde y^{2} \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}} \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}} + \frac{5 \tilde y^{2}}{2} = 0
−26x~22+5x~22+5y~22+26y~22=0- \frac{\sqrt{26} \tilde x^{2}}{2} + \frac{5 \tilde x^{2}}{2} + \frac{5 \tilde y^{2}}{2} + \frac{\sqrt{26} \tilde y^{2}}{2} = 0
Given equation is degenerate hyperbole
x~2(1−52+262)2−y~2(152+262)2=0\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{26}}{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=(2652+12, 12−2652)\vec e_1 = \left( \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}, \ \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}\right)
e⃗2=(−12−2652, 2652+12)\vec e_2 = \left( - \sqrt{\frac{1}{2} - \frac{\sqrt{26}}{52}}, \ \sqrt{\frac{\sqrt{26}}{52} + \frac{1}{2}}\right)
Invariants method
Given line equation of 2-order:
2x2−5xy+3y2=02 x^{2} - 5 x y + 3 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=2a_{11} = 2
a12=−52a_{12} = - \frac{5}{2}
a13=0a_{13} = 0
a22=3a_{22} = 3
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=5I_{1} = 5
     | 2    -5/2|
I2 = |          |
     |-5/2   3  |

I3=∣2−520−5230000∣I_{3} = \left|\begin{matrix}2 & - \frac{5}{2} & 0\\- \frac{5}{2} & 3 & 0\\0 & 0 & 0\end{matrix}\right|
I(λ)=∣2−λ−52−523−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & - \frac{5}{2}\\- \frac{5}{2} & 3 - \lambda\end{matrix}\right|
     |2  0|   |3  0|
K2 = |    | + |    |
     |0  0|   |0  0|

I1=5I_{1} = 5
I2=−14I_{2} = - \frac{1}{4}
I3=0I_{3} = 0
I(λ)=λ2−5λ−14I{\left(\lambda \right)} = \lambda^{2} - 5 \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
λ2−5λ−14=0\lambda^{2} - 5 \lambda - \frac{1}{4} = 0
λ1=52−262\lambda_{1} = \frac{5}{2} - \frac{\sqrt{26}}{2}
λ2=52+262\lambda_{2} = \frac{5}{2} + \frac{\sqrt{26}}{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(52−262)+y~2(52+262)=0\tilde x^{2} \left(\frac{5}{2} - \frac{\sqrt{26}}{2}\right) + \tilde y^{2} \left(\frac{5}{2} + \frac{\sqrt{26}}{2}\right) = 0
x~2(1−52+262)2−y~2(152+262)2=0\frac{\tilde x^{2}}{\left(\frac{1}{\sqrt{- \frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{\sqrt{\frac{5}{2} + \frac{\sqrt{26}}{2}}}\right)^{2}} = 0
- reduced to canonical form
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