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Question: If \(2y\cos\theta = x\sin\theta\text{ and }2x\sec\theta - y\text{cosec}\theta = 3,\) then \[x^{2} +...

If 2ycos⁡θ=xsin⁡θ and 2xsec⁡θ−ycosecθ=3,2y\cos\theta = x\sin\theta\text{ and }2x\sec\theta - y\text{cosec}\theta = 3, then

x2+4y2=x^{2} + 4y^{2} =

A

4

B

– 4

C

± 4

D

None of these

Answer

4

Explanation

Solution

Given that 2ycos⁡θ=xsin⁡θ2y\cos\theta = x\sin\theta …..(i)

and 2xsec⁡θ−ycosecθ=32x\sec\theta - y\text{cosec}\theta = 3 …..(ii)

⇒2xcos⁡θ−ysin⁡θ=3\Rightarrow \frac{2x}{\cos\theta} - \frac{y}{\sin\theta} = 3

⇒2xsin⁡θ−ycos⁡θ−3sin⁡θcos⁡θ=0\Rightarrow 2x\sin\theta - y\cos\theta - 3\sin\theta\cos\theta = 0 …..(iii)

Solving (i) and (iii), we get y=sin⁡θy = \sin\theta and x=2cos⁡θx = 2\cos\theta

Now, x2+4y2=4cos⁡2θ+4sin⁡2θx^{2} + 4y^{2} = 4\cos^{2}\theta + 4\sin^{2}\theta

=4(cos⁡2θ+sin⁡2θ)=4= 4(\cos^{2}\theta + \sin^{2}\theta) = 4.