So far what is solved is x squared + y squared = 4/(1 + 2y). The book has an answer showing a completing the square step and multiplying the conjugate of 1 + 2y on the right side, but there is a cubed y value that is not in the book's answer. If anyone can help with this, it is awesome.
I need to see detailed to steps if possible to know how exactly to move the 1 + 2y over to the left without foiling which usually leads to a cubed root.
Thank you.
I need to see detailed to steps if possible to know how exactly to move the 1 + 2y over to the left without foiling which usually leads to a cubed root.
Thank you.
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r = 4/[1 + 2cosθ]
r = √(x² + y²)
x = r∙cosθ
cosθ = x/r = x/√(x² + y²)
√(x² + y²) = 4/[1 + 2x/√(x² + y²)]
√(x² + y²) [1 + 2x/√(x² + y²)] = 4
√(x² + y²) + 2x = 4
√(x² + y²) = 4 - 2x
x² + y² = 16 - 16x + 4x²
y² = 16 - 16x + 3x²
y = ±√(16 - 16x + 3x²)
r = √(x² + y²)
x = r∙cosθ
cosθ = x/r = x/√(x² + y²)
√(x² + y²) = 4/[1 + 2x/√(x² + y²)]
√(x² + y²) [1 + 2x/√(x² + y²)] = 4
√(x² + y²) + 2x = 4
√(x² + y²) = 4 - 2x
x² + y² = 16 - 16x + 4x²
y² = 16 - 16x + 3x²
y = ±√(16 - 16x + 3x²)
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r + 2rcos t = 4
(x^2+y)^2^(1/2) + 2x = 4
or x^2+y^2 = (4-2x)^2 , expand and simplify
(x^2+y)^2^(1/2) + 2x = 4
or x^2+y^2 = (4-2x)^2 , expand and simplify