How do you convert #r = 9 / (2 - cos(theta))# to rectangular form?

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Answer 1 · 2018-05-14T18:01:20

#(x-3)^2/36+y^2/27=1#

This is the standard ellipse form

Just a few things:
#r^2=x^2+y^2#
#x=rcostheta#
#y=rcostheta#

#r = 9 / (2 - cos(theta))#

#1=9/(2r-rcostheta)#

#2r-rcostheta=9#

#2r= rcostheta+9#

#2r= x+9#

#(2r)^2= (x+9)^2#

#4r^2= x^2+18x+81#

#4x^2+4y^2= x^2+18x+81#

#3x^2-18x+4y^2=81#

#3(x^2-6x+9)+4y^2= 108#

#3(x-3)^2+4y^2= 108#

#(x-3)^2/(1/3)+y^2/(1/4)= 108#

#(x-3)^2/36+y^2/27=1#