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Question

If xacosθ+ybsinθ=1,xasinθybcosθ=1, then eliminate θ

A
x2a2+y2b2=2
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B
x2a2y2b2=2
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C
x2b2+y2a2=2
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D
x2b2y2a2=2
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Solution

The correct option is A x2a2+y2b2=2
Given, xacosθ+ybsinθ=1,xasinθybcosθ=1
Squaring and adding them
x2a2cos2θ+y2b2sin2θ+2xyabsinθcosθ+
x2a2sin2θ+y2b2cos2θ2xyabsinθcosθ=1+1
x2a2+y2b2=2

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