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Answer 1 · 2017-04-27T02:24:02

#x^2+y^2=((x^2-y^2)/(x^2+y^2))^2#

Here is the graph of the original equation #r = cos(2theta)#:

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Use the identity #cos(2theta) = cos^2(theta)-sin^2(theta)#

#r = cos^2(theta)-sin^2(theta)#:

Substitute the #sqrt(x^2+y^2)# for r, #x^2/(x^2+y^2)# for #cos^2(theta)#, and #y^2/(x^2+y^2)# for #sin^2(theta)#:

#sqrt(x^2+y^2)=x^2/(x^2+y^2)-y^2/(x^2+y^2)#

#sqrt(x^2+y^2)=(x^2-y^2)/(x^2+y^2)#

Here is the graph of that equation:

graph{sqrt(x^2+y^2)=x^2/(x^2+y^2)-y^2/(x^2+y^2) [-10, 10, -5, 5]}

It is only half of the loops

Square both sides:

#x^2+y^2=((x^2-y^2)/(x^2+y^2))^2#

Here is the graph of that equation:

graph{x^2+y^2=((x^2-y^2)/(x^2+y^2))^2 [-10, 10, -5, 5]}

This looks like the polar equation.