If $2 y \cos \theta=x \sin \theta$ and $2 x \sec \theta-y \operatorname{cosec} \theta=3$, then $x^2+4 y^2=$

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If $2 y \cos \theta=x \sin \theta$ and $2 x \sec \theta-y \operatorname{cosec} \theta=3$, then $x^2+4 y^2=$, 2, 4, 1, none

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We have $2 \mathrm{y} \cos \theta=\mathrm{x} \sin \theta$ or $\frac{\cos \theta}{\mathrm{x}}=\frac{\sin \theta}{2 \mathrm{y}}=\mathrm{k}$ (say) Then $\cos \theta=\mathrm{kx}$ and $\sin \theta=2 \mathrm{ky}$ ...(1) Again $2 \mathrm{x} \sec \theta-\mathrm{y} \operatorname{cosec} \theta=3$ or $\frac{2 \mathrm{x}}{\cos \theta}-\frac{\mathrm{y}}{\sin \theta}=3$ or $\frac{2 \mathrm{x}}{\mathrm{kx}}-\frac{\mathrm{y}}{2 \mathrm{ky}}=3 \quad$ [from(1)] or $\frac{2}{\mathrm{k}}-\frac{1}{2 \mathrm{k}}=3$, giving $\mathrm{k}=\frac{1}{2}$ We now get $\cos \theta=\frac{\mathrm{x}}{2}$ and $\sin \theta=\mathrm{y}$ Squaring and adding we get $\begin{aligned} & \cos ^2 \theta+\sin ^2 \theta=\frac{x^2}{4+y^2} \ & \Rightarrow \frac{x^2}{4}+y^2=1, \text { or } x^2+4 y^2=4 \end{aligned}$

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