Find semi-major/semi-minor axis of ellipse from parametric equations with different phasex=x^⋅cos⁡(Ωt−θ)y=y^⋅sin⁡(Ωt−φ)

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Sabrina Campos · Accepted Answer

Position vector&amp;nbsp;r=(x,y)=(cos⁡Ωt−θ,sin⁡Ωt−φ)Define the variable&amp;nbsp;t′=Ωt−θ, then&amp;nbsp;Ωt−φ=t′+θ−φ=t′+ψ&amp;nbsp;where&amp;nbsp;ψ=θ−φhence,&amp;nbsp;r=(x,y)=(cos⁡t′,sin⁡(t′+ψ))Expanding the sine function,&amp;nbsp;r=(x,y)=(cos⁡t′,sin⁡t′cos⁡ψ+cos⁡t′sin⁡ψ)And this can be written as&amp;nbsp;r=(x,y)=cos⁡t′(1,sin⁡ψ)+sin⁡t′(0,cos⁡ψ)In matrix-vector form use a&amp;nbsp;2×1&amp;nbsp;vector for r, i.e.&amp;nbsp;r=xy, and letu=cost&#039;sint&#039;and let&amp;nbsp;V=1&amp;nbsp;0sinψ&amp;nbsp;cosψThen&amp;nbsp;r=VuFrom which&amp;nbsp;u=V−1rSince&amp;nbsp;uTu=cos2t′+sin2t′=1, then it follows that the algebraic equation of the ellipse isrTV−TV−1r=1Now all we have to do is find the eigenvalues of the matrix&amp;nbsp;V−TV−1.We have thatV-1=1cosψcosψ&amp;nbsp;0-sinψ&amp;nbsp;1So,&amp;nbsp;G=V-TV-1=1cos2ψ1&amp;nbsp;-sinψ-sinψ&amp;nbsp;1We want to put G in the form:&amp;nbsp;G=RDRTUsing standard diagonalization of a&amp;nbsp;2×2&amp;nbsp;matrix, and since&amp;nbsp;G11=G22=1, then the angle between the axes of the ellipse and the coordinate axes is&amp;nbsp;ϕ=π4.The diagonal matrix D entries are computed as followsD11=G11cos2ϕ+G22sin2ϕ+G12sin⁡2ϕD22=G11sin2ϕ+G22cos2ϕ−G12sin⁡2ϕThese evaluate to,D11=1−sin⁡ψcos2ψ=11+sin⁡ψD22=1+sin⁡ψcos2ψ=11−sin⁡ψTherefore, the semi-major axis of the ellipse has length&amp;nbsp;1+sin⁡ψ&amp;nbsp;and the semi-minor axis has length&amp;nbsp;1−sin⁡ψ.

posciad48o · Accepted Answer

Express the squared distance to the origin as a function of&amp;nbsp;2Ωt. Then find the extrema.2d2=(x^2cos⁡(2θ)−y^2cos⁡(2ϕ))cos⁡(2Ωt)+(x^2sin⁡(2θ)−y^2sin⁡(2ϕ))sin⁡(2Ωt)+x^2−y^2acos⁡(u)+bsin⁡(u)+c∈[c−a2+b2,c+a2+b2].

Find semi-major/semi-minor axis of ellipse from parametric equations

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