Determine whether (x,y)=(mcos⁡(ωt),ncos⁡(ωt−ϕ)) describes an ellipse for all m, n, ω,ϕ

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Nathanial Carey · Accepted Answer

Step 1x=mcos⁡ωty=ncos⁡(ωt−ϕ)=n(cos⁡ωtcos⁡ϕ+sin⁡ωtsin⁡ϕ)This is a linear system in&amp;nbsp;cos⁡ωt&amp;nbsp;and&amp;nbsp;sin⁡ωt, the system is given bym&amp;nbsp;0ncosϕ&amp;nbsp;nsinϕcosωtsinωt=xyThe solution (assuming&amp;nbsp;sin⁡ϕ≠0&amp;nbsp;and&amp;nbsp;n≠0) iscosωtsinωt=1mnsinϕnsinϕ&amp;nbsp;0-ncosϕ&amp;nbsp;mxyStep 2Now, since&amp;nbsp;cos2ωt+sin2ωt=1, then the equation governing x,y isx&amp;nbsp;yn2&amp;nbsp;-nmcosϕ-nmcosϕ&amp;nbsp;m2xy=m2n2sin2ϕwritten concisely,rTQr=cThe principal determinants are&amp;nbsp;n2&amp;nbsp;and&amp;nbsp;n2m2(1−cos2ϕ)=n2m2sin2ϕSo to have a positive definite matrix Q, we must also have that&amp;nbsp;m≠0If these three conditions are met, then the above equation represents an ellipse, because then the positive definite matrix Q can be diagonalized intoQ=RDRTwhere D is the diagonal matrix of Q&#039;s eigenvalues, and R is the rotation matrix whose columns are the corresponding (orthogonal) unit eigenvectors. Now the equation above can be re-written asrTRDRTr=cDivide through by c,rTRERTr=1with&amp;nbsp;E=d1cDNow let&amp;nbsp;v=RTr&amp;nbsp;so that&amp;nbsp;r=Rv, thenvTEv=1and this is clearly an equation of an ellipse because the diagonal entries of E are both positive. Finally the vector r is just a rotation about the origin of vector v, thus it describes a rotated ellipse whose angle of rotation is determined by the angle between the first column of R and the positive x axis.

Determine whether \(\displaystyle{\left({x},{y}\right)}={\left({m}{\cos{{\left(\omega{t}\right)}}},{n}{\cos{{\left(\omega{t}-\phi\right)}}}\right)}\) describes an ellipse for

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