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Question
Eliminate θ from the following:
2x = 3 − 4 tan θ, 3y = 5 + 3 sec θ
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Solution
2x = 3 − 4 tan θ, 3y = 5 + 3 sec θ
∴ 2x − 3 = − 4 tan θ, 3y − 5 = 3 sec θ
∴ tanθ = `(2x - 3)/(− 4) and secθ = (3y - 5)/3`
We know that,
sec2θ = 1 + tan2θ
∴ sec2θ – tan2θ = 1
Therefore,
∴ `((3y - 5)/3)^2 - ((2x - 3)/(− 4))^2` = 1
∴ `(3y - 5)^2/9 - (2x - 3)^2/16` = 1
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