How do you convert #r=2(cos(theta))^2# into rectangular form?

1 Answer
Mar 6, 2016

#=>(x^2+y^2)^3=4x^4#

Explanation:

we know,#x=rcostheta#,,#y=rsintheta#,#r=sqrt(x^2+y^2)#
Given relation is
#r=2(costheta)^2#
multiplying both sides by #r^2# we have
#r^3=2(rcostheta)^2#
putting #rcostheta=x# and,#r=sqrt(x^2+y^2)#, we get
#(x^2+y^2)^(3/2)=2x^2#
#=>(x^2+y^2)^3=(2x^2)^2#
#=>(x^2+y^2)^3=4x^4#

Is it OK?