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Canonical form/ 2x^2-y^2-2y=0

2x^2-y^2-2y=0 canonical form

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The solution

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   2            2    
- y  - 2*y + 2*x  = 0
2x2−y2−2y=02 x^{2} - y^{2} - 2 y = 0 
2*x^2 - y^2 - 2*y = 0
Detail solution
Given line equation of 2-order:
2x2−y2−2y=02 x^{2} - y^{2} - 2 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=2a_{11} = 2
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=−1a_{22} = -1
a23=−1a_{23} = -1
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=∣200−1∣\Delta = \left|\begin{matrix}2 & 0\\0 & -1\end{matrix}\right|
Δ=−2\Delta = -2
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=02 x_{0} = 0
−y0−1=0- y_{0} - 1 = 0
then
x0=0x_{0} = 0
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′=−y0a'_{33} = - y_{0}
a33′=1a'_{33} = 1
then equation turns into
2x′2−y′2+1=02 x'^{2} - y'^{2} + 1 = 0
Given equation is hyperbole
        2           2     
\tilde x    \tilde y      
--------- - --------- = -1
   1/2          1

- reduced to canonical form
The center of canonical coordinate system at point O
(0, -1)

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:
2x2−y2−2y=02 x^{2} - y^{2} - 2 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=2a_{11} = 2
a12=0a_{12} = 0
a13=0a_{13} = 0
a22=−1a_{22} = -1
a23=−1a_{23} = -1
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
     |2  0 |
I2 = |     |
     |0  -1|

I3=∣2000−1−10−10∣I_{3} = \left|\begin{matrix}2 & 0 & 0\\0 & -1 & -1\\0 & -1 & 0\end{matrix}\right|
I(λ)=∣2−λ00−λ−1∣I{\left(\lambda \right)} = \left|\begin{matrix}2 - \lambda & 0\\0 & - \lambda - 1\end{matrix}\right|
     |2  0|   |-1  -1|
K2 = |    | + |      |
     |0  0|   |-1  0 |

I1=1I_{1} = 1
I2=−2I_{2} = -2
I3=−2I_{3} = -2
I(λ)=λ2−λ−2I{\left(\lambda \right)} = \lambda^{2} - \lambda - 2
K2=−1K_{2} = -1
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−λ−2=0\lambda^{2} - \lambda - 2 = 0
λ1=2\lambda_{1} = 2
λ2=−1\lambda_{2} = -1
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−y~2+1=02 \tilde x^{2} - \tilde y^{2} + 1 = 0
        2           2     
\tilde x    \tilde y      
--------- - --------- = -1
   1/2          1

- reduced to canonical form
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