Precalculus: Polar to x,y representation

Question

1. R=sin theta-cos theta
2. R=3/(3-cos theta)

Answer 1

r = sin(θ) - cos(θ)

Using the polar conversion equations sin(θ) = y/r and cos(θ) = x/r substitute:

r = (y/r)-(x/r)

r = (y - x)/r

r² = y - x

Using the polar conversion equation r² = x² + y², substitute to get:

x² + y² = y - x

x² + x + y² - y = 0

x² + x + 1/4 + y² - y + 1/4 = 0 + 1/2

(x + 1/2)² + (y - 1/2)² = 1/2

r = 3/(3 - cos(θ))

Using the polar conversion equation cos(θ) = x/r substitute:

r = 3/(3 - x/r)

r = 3/(3r/r - x/r)

r = 3/[(3r - x)/r]

r = 3r/(3r - x)

1 = 3/(3r - x)

3r - x = 3

3r = x + 3

Square both sides:

9r² = x² + 6x + 9

Using the polar conversion equation r² = x² + y², substitute to get:

9(x² + y²) = x² + 6x + 9

9x² + 9y² = x² + 6x + 9

8x² - 6x + 9y² = 9

8(x² - 3/4x + ___) + 9y² = 9

8(x² - 3/4x + 9/64) + 9y² = 9 + 72/64

8(x - 3/8)² + 9y² = 9(64/64) + 72/64

8(x - 3/8)² + 9y² = 576/64 + 72/64

8(x - 3/8)² + 9y² = 648/64

8(x - 3/8)² + 9y² = 81/8

Multiply both sides by 8/81