Respuesta :

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

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