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    Obtain a cartesian equation for this pair of parametric equations: x=cot(theta), y=2cosec(theta). I have no idea how to go about this as I`m used to x and y being in terms of cos and sin, respectively. What, at least, would be the first step?
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    (Original post by itlstallion69)
    Obtain a parametric equation for this pair of parametric equations: x=cot(theta), y=cosec(theta). I have no idea how to go about this as I`m used to x and y being in terms of cos and sin, respectively. What, at least, would be the first step?
    You wish to obtain the equation of the curve entirely in terms of x,y. Since you got trig in there, a good start is to have a think what identities link \cot \theta and \mathrm{cosec \ } \theta
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    (Original post by RDKGames)
    Do you mean obtain the Cartesian equation...?

    In which case, you wish to obtain the equation of the curve entirely in terms of x,y. Since you got trig in there, a good start is to have a think what identities link \cot \theta and \mathrm{cosec \ } \theta
    I know that cot^(2)x =cosec^(2)x -1, but I can`t see how that would apply in getting the cartesian equations
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    (Original post by itlstallion69)
    I know that cot^(2)x =cosec^(2)x -1, but I can`t see how that would apply in getting the cartesian equations
    That's right. So then you want to get \cot^2 \theta and \mathrm{cosec^2 \ } \theta from what you got.

    Note that if x= \cot \theta then x^2 = \cot^2 \theta. So in the identity you can replace \cot^2 \theta by x^2. Repeat similarly for y
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    (Original post by RDKGames)
    That's right. So then you want to get \cot^2 \theta and \mathrm{cosec^2 \ } \theta from what you got.

    Note that if x= \cot \theta then x^2 = \cot^2 \theta. So in the identity you can replace \cot^2 \theta by x^2. Repeat similarly for y
    Following that, my answer was 4x^2 +y^2-4=0, but the book gives 4x^2 -y^2 +4=0
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    (Original post by itlstallion69)
    Following that, my answer was 4x^2 +y^2-4=0, but the book gives 4x^2 -y^2 +4=0
    Didnt notice you changes cosec to 2cosec.

    So then, anyway, x=\cot \theta and \frac{y}{2} = \csc \theta

    Hence x^2 = \cot^2 \theta and \frac{y^2}{4} = \csc^2 \theta

    Subbing these into the identity \cot^2 \theta = \csc^2 \theta -1 yields x^2 = \frac{y^2}{4}-1 which arranges to their given answer.

    Not sure where you're error lies without you showing your working out.
 
 
 
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