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Question

x = a sec θ + b tan θ

y = a tan θ + b sec θ


A

x2 + y2 = a2 + b2

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B

x2 + y2 = a2 - b2

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C

x2 - y2 = a2 + b2

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D

x2 - y2 = a2 - b2

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Solution

The correct option is D

x2 - y2 = a2 - b2


x2 = a2sec2θ + b2tan2θ + 2ab secθ tanθ ... (i)

y2 = a2tan2θ + b2sec2θ + 2ab tanθ secθ ... (ii)

Subtracting (ii) from (i), we get

x2 - y2 = a2 - b2 [ sec2θ - tan2θ = 1 ]


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