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    Hopefully this is the right place for this

    Hey I would like some help with these questions.



    1.) x=t/(2t-1) y=t/(t+3)
    2.) x=sin2theta y=costheta.3
    .)x=sintheta +2costheta y=2sintheta +costheta


    For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer
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    (Original post by Sylarus)
    Hopefully this is the right place for this

    Hey I would like some help with these questions.



    1.) x=t/(2t-1) y=t/(t+3)
    2.) x=sin2theta y=costheta.3
    .)x=sintheta +2costheta y=2sintheta +costheta


    For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer
    1) and 2) are correct.

    For 3), try squaring x and y and also consider their product.
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    (Original post by Sylarus)
    Hopefully this is the right place for this

    Hey I would like some help with these questions.



    1.) x=t/(2t-1) y=t/(t+3)
    2.) x=sin2theta y=costheta.3
    .)x=sintheta +2costheta y=2sintheta +costheta


    For all I need to find the Cartesian form. For the first one I got it down to y=x/(7x-3) and for the second one I got it to be x^2=y^2(4-4y^2). As for the third one I have no answer
    Agree with your first two.

    For the third one, obtain expressions for \cos\theta and \sin\theta, in terms of x,y, and use a well known identity to eliminate theta.

    Edit: Nothing for 50 minutes and then two at once!
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    If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

    Costheta= y-2sintheta
    Sintheta= x-2costheta

    Then squaring I got this big messy expansion which I can't seem to get the Thetas out of
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    (Original post by Sylarus)
    If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

    Costheta= y-2sintheta
    Sintheta= x-2costheta

    Then squaring I got this big messy expansion which I can't seem to get the Thetas out of
    Ghostwalker may have to advise you here because I'm not sure.

    An alternative method (as mentioned in my last post) : Compare x^2+y^2 and xy.
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    (Original post by Sylarus)
    If you are saying use cos squared theta add sign squared theta = 1 I've tried using it like this.

    Costheta= y-2sintheta
    Sintheta= x-2costheta

    Then squaring I got this big messy expansion which I can't seem to get the Thetas out of
    You haven't got cos/sin in terms of x,y.

    Using the two equations, you can eliminate sin, and get cos as a function of x,y.
    Similarly you can eliminate cos, and get sin as a function of x,y.

    Then apply the identity.
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    (Original post by ghostwalker)
    You haven't got cos/sin in terms of x,y.

    Using the two equations, you can eliminate sin, and get cos as a function of x,y.
    Similarly you can eliminate cos, and get sin as a function of x,y.

    Then apply the identity.
    Yes that's a much better way. I should have spotted it.
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    K so after some rearranging got it to (2x-y)^2+(2y-x)^2=1
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    (Original post by Sylarus)
    K so after some rearranging got it to (2x-y)^2+(2y-x)^2=1
    Close. But it doesn't equal 1.

    As a check you can choose a value for the parameter, theta = 0 say. Sub into your original equations, and check if the x,y values that come out satisfy your cartesian equation.
 
 
 
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