How do I convert #r=3+3sec(theta)# to a Cartesian equation?

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Answer 1 · 2018-05-12T22:37:56

#x^2+y^2=(9x^2)/(x-3)^2#

Multily all terms by #rcostheta#, since #costheta*sectheta=1#

#r^2costheta=3rcostheta+3r#

#rcostheta=x#
#r=sqrt(x^2+y^2)#

#xsqrt(x^2+y^2)=3x+3sqrt(x^2+y^2)#

#sqrt(x^2+y^2)(x-3)=3x#

#sqrt(x^2+y^2)=(3x)/(x-3)#

#x^2+y^2=(9x^2)/(x-3)^2#