If 2y cos theta = x sin theta and 2x sec thet | vedclass

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(A) Given equations are:$2y \cos 	heta = x \sin 	heta$  --- $(i)$$2x \sec 	heta - y \csc 	heta = 3$  --- $(ii)$From $(i)$,we have $\frac{x}{\cos \

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$4$

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(A) Given equations are:$2y \cos 	heta = x \sin 	heta$  --- $(i)$$2x \sec 	heta - y \csc 	heta = 3$  --- $(ii)$From $(i)$,we have $\frac{x}{\cos 	heta} = \frac{2y}{\sin 	heta} = k$ (let).So,$x = k \cos 	heta$ and $2y = k \sin 	heta$,which implies $y = \frac{k}{2} \sin 	heta$.Substitute these into $(ii)$:$2(k \cos 	heta) \sec 	heta - (\frac{k}{2} \sin 	heta) \csc 	heta = 3$$2k(1) - \frac{k}{2}(1) = 3$$\frac{3k}{2} = 3 \Rightarrow k = 2$.Thus,$x = 2 \cos 	heta$ and $y = \sin 	heta$.Now,$x^2 + 4y^2 = (2 \cos 	heta)^2 + 4(\sin 	heta)^2$$= 4 \cos^2 	heta + 4 \sin^2 	heta$$= 4(\cos^2 	heta + \sin^2 	heta) = 4(1) = 4$.

If $2y \cos \theta = x \sin \theta$ and $2x \sec \theta - y \csc \theta = 3$,then $x^2 + 4y^2 = $

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