Given line equation of 2-order: x2+4x+5=0 This equation looks like: a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0 where a11=1 a12=0 a13=2 a22=0 a23=0 a33=5 To calculate the determinant Δ=a11a12a12a22 or, substitute Δ=1000 Δ=0 Because Δ is equal to 0, then (x~+2)2=−1 x~′2=−1 Given equation is two parallel straight lines - reduced to canonical form where replacement made x~′=x~+2 y~′=y~ The center of the canonical coordinate system in OXY x0=x~cos(ϕ)−y~sin(ϕ) y0=x~sin(ϕ)+y~cos(ϕ) x0=−2+0⋅0 y0=−0⋅2 x0=−2 y0=0 The center of canonical coordinate system at point O
(-2, 0)
Basis of the canonical coordinate system e1=(1,0) e2=(0,1)
Invariants method
Given line equation of 2-order: x2+4x+5=0 This equation looks like: a11x2+2a12xy+2a13x+a22y2+2a23y+a33=0 where a11=1 a12=0 a13=2 a22=0 a23=0 a33=5 The invariants of the equation when converting coordinates are determinants: I1=a11+a22
I1=1 I2=0 I3=0 I(λ)=λ2−λ K2=1 Because I2=0∧I3=0∧K2>0∧I1=0 then by line type: this equation is of type : two imaginary parallel lines I1y~2+I1K2=0 or y~2+1=0