1. Convert r= 2/1+sinθ to cartesian
2. Convert y=3 to polar
3. Convert 3x^2+3y^2-6x=0 to polar
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
1. To convert the polar equation r = 2/(1 + sinθ) to a Cartesian equation, we need to use the relations between polar and Cartesian coordinates. The key relations are:
x = r * cosθ
y = r * sinθ
Substituting these equations into the given polar equation, we get:
x = (2/(1 + sinθ)) * cosθ
y = (2/(1 + sinθ)) * sinθ
Therefore, the Cartesian equation is:
x = (2cosθ) / (1 + sinθ)
y = (2sinθ) / (1 + sinθ)
2. To convert the Cartesian equation y = 3 to a polar equation, we can use the relation:
r = √(x^2 + y^2)
Substituting y = 3 into the equation, we get:
r = √(x^2 + 3^2)
r = √(x^2 + 9)
Therefore, the polar equation is:
r = √(x^2 + 9)
3. To convert the Cartesian equation 3x^2 + 3y^2 - 6x = 0 to a polar equation, we can again use the relation:
r = √(x^2 + y^2)
First, let's rearrange the given Cartesian equation:
3x^2 + 3y^2 - 6x = 0
3x^2 - 6x + 3y^2 = 0
3(x^2 - 2x) + 3y^2 = 0
3(x^2 - 2x + 1) + 3y^2 = 3
3(x - 1)^2 + 3y^2 = 3
(x - 1)^2 + y^2 = 1
Now, we can convert this equation to polar form by substituting x = r * cosθ and y = r * sinθ:
(r * cosθ - 1)^2 + (r * sinθ)^2 = 1
r^2 * cos^2θ - 2r * cosθ + 1 + r^2 * sin^2θ = 1
r^2 * (cos^2θ + sin^2θ) - 2r * cosθ = 0
r^2 - 2r * cosθ = 0
Therefore, the polar equation is:
r = 2 * cosθ