How do you convert 3xy=-x^2+4y^2 3xy=−x2+4y2 into a polar equation?

1 Answer
Cesareo R.
May 13, 2016

cos(theta)sin(theta)+cos(theta)^2+4sin(theta)^2=0cos(θ)sin(θ)+cos(θ)2+4sin(θ)2=0

Explanation:

We must be carefull with those equations. Here 4y^2-x^2-3x y = 04y2−x2−3xy=0 can be written as (4y-x)(x+y) = 0(4y−x)(x+y)=0 so, the function graphic is build with two straigths: 4y-x=04y−x=0 and x + y=0x+y=0. This phenomenon will appear as well in polar coordinates. Making x = r cos(theta), y = r sin(theta)x=rcos(θ),y=rsin(θ) and substituting, we get:
r^2 cos(theta)sin(theta) = -r^2 cos(theta)^2+4 r^2 sin(theta)^2r2cos(θ)sin(θ)=−r2cos(θ)2+4r2sin(θ)2 and after simplification, we get the condition: cos(theta)sin(theta)+cos(theta)^2+4sin(theta)^2=0cos(θ)sin(θ)+cos(θ)2+4sin(θ)2=0. Solving for thetaθ we will obtain the straigths directions.

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