How do you convert #r=6/(2-cos(theta))# into cartesian form?

1 Answer
Narad T.
Nov 8, 2016

The cartesian form is #3x^2-12x+4y^2-36=0#

Explanation:

The formulae used are #r^2=x^2+y^2#
and #x=rcostheta# and #y=rsintheta#

Rewring the polar form as #r(2-costheta)=6#
Replacing #costheta=x/r#
We get #r(2-x/r)=6##=>##2r-x=6#
#=># #2r=x+6#
replacing #r#
#2sqrt(x^2+y^2)=x+6#
Squaring both sides
#4(x^2+y^2)=(x+6)^2#
#4x^2+4y^2=x^2+12x+36#
#3x^2-12x+4y^2-36=0#
graph{3x^2-12x+4y^2-36=0 [-7.9, 7.9, -3.95, 3.95]}

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