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Question

If xacosθ+ybsinθ=1andxasinθybcosθ=1, prove that x2a2+y2b2=2

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Solution

We have,
xacosθ+ybsinθ=1

On squaring both sides, we get
x2a2cos2θ+y2b2sin2θ+2xyabsinθcosθ=1 .........(1)

Simliarly,
xasinθybcosθ=1

x2a2sin2θ+y2b2cos2θ2xyabsinθcosθ=1 .........(2)

On adding both equations, we get
x2a2(cos2θ+sin2θ)+y2b2(sin2θ+cos2θ)=1+1

Since,
cos2θ+sin2θ=1

Therefore,
x2a2+y2b2=2

Hence, this is the answer.

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