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Canonical form/ 9x2+121y2=1089

9x2+121y2=1089 canonical form

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The solution

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-1089 + 9*x2 + 121*y2 = 0
9x2+121y2−1089=09 x_{2} + 121 y_{2} - 1089 = 0 
9*x2 + 121*y2 - 1089 = 0
Detail solution
Given line equation of 2-order:
9x2+121y2−1089=09 x_{2} + 121 y_{2} - 1089 = 0
This equation looks like:
a11y22+2a12x2y2+2a13y2+a22x22+2a23x2+a33=0a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0
where
a11=0a_{11} = 0
a12=0a_{12} = 0
a13=1212a_{13} = \frac{121}{2}
a22=0a_{22} = 0
a23=92a_{23} = \frac{9}{2}
a33=−1089a_{33} = -1089
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣0000∣\Delta = \left|\begin{matrix}0 & 0\\0 & 0\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
Given equation is straight line
- reduced to canonical form
The center of the canonical coordinate system in OXY
y20=−x~2sin⁡(ϕ)+y~2cos⁡(ϕ)y_{20} = - \tilde x2 \sin{\left(\phi \right)} + \tilde y2 \cos{\left(\phi \right)}
x20=x~2cos⁡(ϕ)+y~2sin⁡(ϕ)x_{20} = \tilde x2 \cos{\left(\phi \right)} + \tilde y2 \sin{\left(\phi \right)}
y20=0⋅0y_{20} = 0 \cdot 0
x20=0⋅0x_{20} = 0 \cdot 0
y20=0y_{20} = 0
x20=0x_{20} = 0
The center of canonical coordinate system at point O
(0, 0)

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:
9x2+121y2−1089=09 x_{2} + 121 y_{2} - 1089 = 0
This equation looks like:
a11y22+2a12x2y2+2a13y2+a22x22+2a23x2+a33=0a_{11} y_{2}^{2} + 2 a_{12} x_{2} y_{2} + 2 a_{13} y_{2} + a_{22} x_{2}^{2} + 2 a_{23} x_{2} + a_{33} = 0
where
a11=0a_{11} = 0
a12=0a_{12} = 0
a13=1212a_{13} = \frac{121}{2}
a22=0a_{22} = 0
a23=92a_{23} = \frac{9}{2}
a33=−1089a_{33} = -1089
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  0|
I2 = |    |
     |0  0|

I3=∣0012120092121292−1089∣I_{3} = \left|\begin{matrix}0 & 0 & \frac{121}{2}\\0 & 0 & \frac{9}{2}\\\frac{121}{2} & \frac{9}{2} & -1089\end{matrix}\right|
I(λ)=∣−λ00−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}- \lambda & 0\\0 & - \lambda\end{matrix}\right|
     |  0    121/2|   | 0    9/2 |
K2 = |            | + |          |
     |121/2  -1089|   |9/2  -1089|

I1=0I_{1} = 0
I2=0I_{2} = 0
I3=0I_{3} = 0
I(λ)=λ2I{\left(\lambda \right)} = \lambda^{2}
K2=−73612K_{2} = - \frac{7361}{2}
Because
I2=0∧I3=0∧(I1=0∨K2=0)I_{2} = 0 \wedge I_{3} = 0 \wedge \left(I_{1} = 0 \vee K_{2} = 0\right)
then by line type:
this equation is of type : two coincident straight lines
I1x~22+K2I1=0I_{1} \tilde x2^{2} + \frac{K_{2}}{I_{1}} = 0
or
False

None

- reduced to canonical form
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