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Canonical form/ x^2+5x-y-60=0

x^2+5x-y-60=0 canonical form

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-60 + x  - y + 5*x = 0
x2+5x−y−60=0x^{2} + 5 x - y - 60 = 0 
x^2 + 5*x - y - 60 = 0
Detail solution
Given line equation of 2-order:
x2+5x−y−60=0x^{2} + 5 x - y - 60 = 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=0a_{12} = 0
a13=52a_{13} = \frac{5}{2}
a22=0a_{22} = 0
a23=−12a_{23} = - \frac{1}{2}
a33=−60a_{33} = -60
To calculate the determinant
Δ=∣a11a12a12a22∣\Delta = \left|\begin{matrix}a_{11} & a_{12}\\a_{12} & a_{22}\end{matrix}\right|
or, substitute
Δ=∣1000∣\Delta = \left|\begin{matrix}1 & 0\\0 & 0\end{matrix}\right|
Δ=0\Delta = 0
Because
Δ\Delta
is equal to 0, then
(x~+52)2=y~+2654\left(\tilde x + \frac{5}{2}\right)^{2} = \tilde y + \frac{265}{4}
x~′2=y~′\tilde x'^{2} = \tilde y'
Given equation is by parabola
- reduced to canonical form
The center of the canonical coordinate system in OXY
x0=x~cos⁡(ϕ)−y~sin⁡(ϕ)x_{0} = \tilde x \cos{\left(\phi \right)} - \tilde y \sin{\left(\phi \right)}
y0=x~sin⁡(ϕ)+y~cos⁡(ϕ)y_{0} = \tilde x \sin{\left(\phi \right)} + \tilde y \cos{\left(\phi \right)}
x0=0⋅0x_{0} = 0 \cdot 0
y0=0⋅0y_{0} = 0 \cdot 0
x0=0x_{0} = 0
y0=0y_{0} = 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:
x2+5x−y−60=0x^{2} + 5 x - y - 60 = 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=0a_{12} = 0
a13=52a_{13} = \frac{5}{2}
a22=0a_{22} = 0
a23=−12a_{23} = - \frac{1}{2}
a33=−60a_{33} = -60
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
     |1  0|
I2 = |    |
     |0  0|

I3=∣105200−1252−12−60∣I_{3} = \left|\begin{matrix}1 & 0 & \frac{5}{2}\\0 & 0 & - \frac{1}{2}\\\frac{5}{2} & - \frac{1}{2} & -60\end{matrix}\right|
I(λ)=∣1−λ00−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & 0\\0 & - \lambda\end{matrix}\right|
     | 1   5/2|   | 0    -1/2|
K2 = |        | + |          |
     |5/2  -60|   |-1/2  -60 |

I1=1I_{1} = 1
I2=0I_{2} = 0
I3=−14I_{3} = - \frac{1}{4}
I(λ)=λ2−λI{\left(\lambda \right)} = \lambda^{2} - \lambda
K2=−1332K_{2} = - \frac{133}{2}
Because
I2=0∧I3≠0I_{2} = 0 \wedge I_{3} \neq 0
then by line type:
this equation is of type : parabola
I1y~2+2x~−I3I1=0I_{1} \tilde y^{2} + 2 \tilde x \sqrt{- \frac{I_{3}}{I_{1}}} = 0
or
x~+y~2=0\tilde x + \tilde y^{2} = 0
y~2=x~\tilde y^{2} = \tilde x
- reduced to canonical form
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