How do you convert 3y=2x22xyx into a polar equation?

1 Answer
Oct 5, 2016

3sin(θ)+cos(θ)2cos2(θ)2cos(θ)sin(θ)=r

Explanation:

x=rcos(θ)
y=rsin(θ)

r2=x2+y2

Make the necessary substitutions

3rsin(θ)=2(rcos(θ))22rcos(θ)rsin(θ)rcos(θ)

Simplify

3rsin(θ)=2r2cos2(θ)2r2cos(θ)sin(θ)rcos(θ)

Add rcos(θ) to both sides

3rsin(θ)+rcos(θ)=2r2cos2(θ)2r2cos(θ)sin(θ)

Factor out r and r2

r(3sin(θ)+cos(θ))=r2(2cos2(θ)2cos(θ)sin(θ))

Isolated r2

r(3sin(θ)+cos(θ))2cos2(θ)2cos(θ)sin(θ)=r22cos2(θ)2cos(θ)sin(θ)2cos2(θ)2cos(θ)sin(θ)

r(3sin(θ)+cos(θ))2cos2(θ)2cos(θ)sin(θ)=r2

Gather r to the right hand side

r(3sin(θ)+cos(θ))2cos2(θ)2cos(θ)sin(θ)r=r2r

Simplify

3sin(θ)+cos(θ)2cos2(θ)2cos(θ)sin(θ)=r

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