How do you convert (x^2 + y^2)^2 = 4(x^2- y^2) into polar form?

1 Answer
Jun 16, 2016

r = 2 cos(2theta)/abs(cos(2theta))sqrt(abs(cos(2theta)))

Explanation:

Using the pass equations

{ (x=r cos(theta)), (y=r sin(theta)) :}

in

f(x,y)=(x^2 + y^2)^2 - 4(x^2- y^2)=0

we get at

g(r,theta) = r^4-4r^2(cos(theta)^2-sin(theta)^2) = 0

but

cos(2theta)=cos(theta)^2-sin(theta)^2

so

g(r,theta)=r^2-4cos(2theta)=0

Finally we get at

r = 2 cos(2theta)/abs(cos(2theta))sqrt(abs(cos(2theta)))

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