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:
$r^2=x^2+y^2$
$x=rcostheta$
$y=rcostheta$

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

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

$2r-rcostheta=9$

$2r= rcostheta+9$

$2r= x+9$

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

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

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

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

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

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

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

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

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