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

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Answer 1 · 2016-06-16T17:02:42

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

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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