Question

$$\begin{gathered} \mathrm{Eliminate}\mathrm{\theta },\mathrm{if} \\ \left(\mathrm{ii}\right)x=2\cos \theta -3\sin \theta ,y=\cos \theta +2\sin \theta \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{Given}:x=2\cos \theta -3\sin \theta \mathrm{and}y=\cos \theta +2\sin \theta \\ \therefore x-2y=2\cos \theta -3\sin \theta -2(\cos \theta +2\sin \theta ) \\ =2\cos \theta -3\sin \theta -2\cos \theta -4\sin \theta \\ =-7\sin \theta \\ \Rightarrow \sin \theta =-\frac{x-2y}{7} \\ \mathrm{Also},2x+3y=2\left(2\cos \theta -3\sin \theta \right)+3(\cos \theta +2\sin \theta ) \\ =4\cos \theta -6\sin \theta +3\cos \theta +6\sin \theta \\ =7\cos \theta \\ \Rightarrow \cos \theta =\frac{2x+3y}{7} \\ \mathrm{We}\mathrm{know}\mathrm{that}{\sin }^2\theta +{\cos }^2\theta =1 \\ \Rightarrow {\left(-\frac{x-2y}{7}\right)}^2+{\left(\frac{2x+3y}{7}\right)}^2=1 \\ \therefore (x-2y{)}^2+{\left(2x+3y\right)}^2=49 \end{gathered}$$