Question

1. Convert r= 2/1+sinθ to cartesian

2. Convert y=3 to polar

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

Answer 1

just use the parametric formulas...

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

y=3
rsinθ = 3
r = 3cscθ

3x^2+3y^2-6x=0
3r^2 = 6rcosθ
r=2cosθ