How do you convert #4x^2y=1# into polar form?

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Answer 1 · 2016-01-26T00:18:54

Use these two identities ...

#x=rcostheta#
#y=rsintheta#

Now use those identities by substitution into the original equation ...

#4x^2y=1#

#4(rcostheta)^2(rsintheta)=1#

Simplify ...

#r=(4cos^2thetasintheta)^(-1/3)#

hope that helped