How do you convert #r = 1/(4 - costheta)# into cartesian form?

Hey, Socratic: Is it really necessary to tell us this was asked 9 minutes ago? I don't like being lied to. Tell us it was asked two years ago and no one's been able to do it yet. Also what's up with the suspiciously identically phrased questions asked from multiple places? Not to mention Santa Cruz, United States? There's almost certainly more than one, though I hear the one in California in nice. Credibility and reputation are important, especially in a homework site. Don't mislead people. End rant.

When converting equations from polar to rectangular coordinates the brute force rectangular to polar substitution

#r = \sqrt{x^2 + y^2}#

#theta = text{arctan2}(y "/," x) quad #

is seldom the best approach. (I'm intentionally indicating the four quadrant inverse tangent here, but let's not get diverted.)

Ideally we want to use the polar to rectangular substitutions,

#x = r cos theta #

#y= r sin theta #

# x^2 + y^2 = r^2 cos^2 theta + r^2 sin ^2 theta = r^2 #

OK let's look at the question.

# r = 1/{4 - cos theta }#

These polar equations generally allow for negative #r#, but here we're sure #r# is always positive.

#r (4 - cos theta) = 1#

These I think are ellipses, which doesn't really matter, but does give us some idea what we hope the rectangular form to look like. We want to aim for something without square roots or arctangents #r=\sqrt{x^2+y^2}# has square roots, but #rcos theta =x # doesn't, so we expand.

#4r - rcos theta = 1#

Now we just substitute; we'll do it in steps.

# 4r -x = 1#

# 4r = x+1#

Let's square now. We know #r>0.#

#16 r^2 = (x+1)^2 #

#16 (x^2 + y^2) = (x+1)^2 = x^2 + 2x + 1#

# 15 x^2 - 2 x + 16 y^2 = 1 #

This is a pretty circular looking ellipse. (A smaller constant than #4# in the original would give a more eccentric ellipse.) We could complete the square to put it in standard form, but let's leave it here.