1. Convert r= 2/1+sinθ to cartesian

2. Convert y=3 to polar

3. Convert 3x^2+3y^2-6x=0 to polar

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  1. just use the parametric formulas...

    r = 2/(1+sinθ)
    r+rsinθ = 2
    r = 2-y
    r^2 = (2-y)^2
    x^2+y^2 = y^2-4y+4
    x^2 = 4-4y
    check the polar form for conics and you will see that the curve is a parabola, since e=1

    rsinθ = 3
    r = 3cscθ

    3r^2 = 6rcosθ

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