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Canonical form/ x^2-8xy+7y^2

x^2-8xy+7y^2 canonical form

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The solution

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 2      2            
x  + 7*y  - 8*x*y = 0
x2−8xy+7y2=0x^{2} - 8 x y + 7 y^{2} = 0 
x^2 - 8*x*y + 7*y^2 = 0
Detail solution
Given line equation of 2-order:
x2−8xy+7y2=0x^{2} - 8 x y + 7 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=1a_{11} = 1
a12=−4a_{12} = -4
a13=0a_{13} = 0
a22=7a_{22} = 7
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=∣1−4−47∣\Delta = \left|\begin{matrix}1 & -4\\-4 & 7\end{matrix}\right|
Δ=−9\Delta = -9
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
x0−4y0=0x_{0} - 4 y_{0} = 0
−4x0+7y0=0- 4 x_{0} + 7 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
x′2−8x′y′+7y′2=0x'^{2} - 8 x' y' + 7 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)=34\cot{\left(2 \phi \right)} = \frac{3}{4}
then
ϕ=acot⁡(34)2\phi = \frac{\operatorname{acot}{\left(\frac{3}{4} \right)}}{2}
sin⁡(2ϕ)=45\sin{\left(2 \phi \right)} = \frac{4}{5}
cos⁡(2ϕ)=35\cos{\left(2 \phi \right)} = \frac{3}{5}
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)=255\cos{\left(\phi \right)} = \frac{2 \sqrt{5}}{5}
sin⁡(ϕ)=55\sin{\left(\phi \right)} = \frac{\sqrt{5}}{5}
substitute coefficients
x′=25x~5−5y~5x' = \frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}
y′=5x~5+25y~5y' = \frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}
then the equation turns from
x′2−8x′y′+7y′2=0x'^{2} - 8 x' y' + 7 y'^{2} = 0
to
7(5x~5+25y~5)2−8(5x~5+25y~5)(25x~5−5y~5)+(25x~5−5y~5)2=07 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right)^{2} - 8 \left(\frac{\sqrt{5} \tilde x}{5} + \frac{2 \sqrt{5} \tilde y}{5}\right) \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right) + \left(\frac{2 \sqrt{5} \tilde x}{5} - \frac{\sqrt{5} \tilde y}{5}\right)^{2} = 0
simplify
−x~2+9y~2=0- \tilde x^{2} + 9 \tilde y^{2} = 0
x~2−9y~2=0\tilde x^{2} - 9 \tilde y^{2} = 0
Given equation is degenerate hyperbole
x~212−y~2(13)2=0\frac{\tilde x^{2}}{1^{2}} - \frac{\tilde y^{2}}{\left(\frac{1}{3}\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=(255, 55)\vec e_1 = \left( \frac{2 \sqrt{5}}{5}, \ \frac{\sqrt{5}}{5}\right)
e⃗2=(−55, 255)\vec e_2 = \left( - \frac{\sqrt{5}}{5}, \ \frac{2 \sqrt{5}}{5}\right)
Invariants method
Given line equation of 2-order:
x2−8xy+7y2=0x^{2} - 8 x y + 7 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=1a_{11} = 1
a12=−4a_{12} = -4
a13=0a_{13} = 0
a22=7a_{22} = 7
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=8I_{1} = 8
     |1   -4|
I2 = |      |
     |-4  7 |

I3=∣1−40−470000∣I_{3} = \left|\begin{matrix}1 & -4 & 0\\-4 & 7 & 0\\0 & 0 & 0\end{matrix}\right|
I(λ)=∣1−λ−4−47−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -4\\-4 & 7 - \lambda\end{matrix}\right|
     |1  0|   |7  0|
K2 = |    | + |    |
     |0  0|   |0  0|

I1=8I_{1} = 8
I2=−9I_{2} = -9
I3=0I_{3} = 0
I(λ)=λ2−8λ−9I{\left(\lambda \right)} = \lambda^{2} - 8 \lambda - 9
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
λ2−8λ−9=0\lambda^{2} - 8 \lambda - 9 = 0
λ1=9\lambda_{1} = 9
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
9x~2−y~2=09 \tilde x^{2} - \tilde y^{2} = 0
x~2(13)2−y~212=0\frac{\tilde x^{2}}{\left(\frac{1}{3}\right)^{2}} - \frac{\tilde y^{2}}{1^{2}} = 0
- reduced to canonical form
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