Converting Polar Equations into Rectangular Form

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convert r=2/(2-3cos(theta)) into rectangular form.
this is how i attempted it:
r(2-3cos(theta))=2
2r-3rcos(theta)=2
2r-3x=2
2r=2+3x
r=1+3x/2
x^2+y^2=(1+3x/2)^2
y^2=(1+3x/2)^2-x^2
y=sqrt((1+3x/2)^2-x^2)
but apparently thats not correct?

 

[DrPhil]

convert r=2/(2-3cos(theta)) into rectangular form.
this is how i attempted it:
r(2-3cos(theta))=2
2r-3rcos(theta)=2
2r-3x=2
2r=2+3x
r=1+3x/2
x^2+y^2=(1+3x/2)^2 <-- you have to expand the square to combine x^2 terms
y=sqrt((1+3x/2)^2-x^2)
but apparently thats not correct?

Not completely simplified.
Once the right side is a quadratic in x, you could write in in terms of the roots of the quadratic, but that is probably not necessary - especially since the roots are irrational. You might also note that to express the complete function, you would have to include both signs of the square root. might be just as well to leave it as y^2 = . . .

 

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Thanks!