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

1 Answer
May 12, 2018

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

Explanation:

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#