(a) Given that $2y\,\,\cos \theta = x\,\sin \theta $…..$(i)$

and $2x\,\sec \theta – y\,\,{\rm{cosec}}\,\theta = 3$…..$(ii)$

$ \Rightarrow \,\,\frac{{2x}}{{\cos \theta }} – \frac{y}{{\sin \theta }} = 3$

$ \Rightarrow \,\,2x\,\sin \theta – y\,\cos \theta – 3\,\sin \theta \cos \theta = 0$…..$(iii)$

Solving $(i)$ and $(iii)$, 

we get $y = \sin \theta $ and $x = 2\,\,\cos \theta $

Now, ${x^2} + 4{y^2} = 4\,\,{\cos ^2}\theta + 4\,\,{\sin ^2}\theta $ 

$ = 4\,({\cos ^2}\theta + {\sin ^2}\theta ) = 4$.