Answer:
For x(θ) = 3cosθ + 2
y² = [9x²/(x - 2)²] - x²
For x(θ) = 3cosθ + 2
x² = [4y²/(y - 1)²] - y²
Step-by-step explanation:
Given the following equivalence:
x² + y² = r²
r = √(x² + y²)
x = rcosθ
cosθ = x/r
y = rsinθ
sinθ = y/r
Applying these to the given equations,
x(θ) = 3cosθ + 2
x = 3(x/r) + 2
xr = 3x + 2r
(x - 2)r = 3x
r = 3x/(x - 2)
Square both sides
r² = 9x²/(x - 2)²
(x - 2)²r² = 9x²
(x - 2)²(x² + y²) = 9x²
(x² + y²) = 9x²/(x - 2)²
y² = [9x²/(x - 2)²] - x²
y(θ) = 2sinθ - 1
y = 2y/r - 1
yr = 2y - r
(y - 1)r = 2y
r = 2y/(y - 1)
Square both sides
r² = 4y²/(y - 1)²
x² + y² = 4y²/(y - 1)²
x² = [4y²/(y - 1)²] - y²