changing $r=1+2r\cos \theta$ to its cartesian equivalent
Andrew Henderson 
My textbook says the polar equation, $r=1+2r\cos \theta$, its cartesian equivalent is $y^2-3x^2-4x-1=0.$ I understand that I get this if I square $r$; $r^2=x^2+y^2=(1+2x)^2.$ But don't I need to square root this back or something?? I thought the answer is supposed to be $ \sqrt{y^2-3x-4x-1}=0 $... I'm confused..
$\endgroup$ 12 Answers
$\begingroup$From $r=1+2r\cos\theta$, we get $$r(1-2\cos\theta)=1$$ or $$r(1-2\frac xr)=1$$ $$r-2x=1$$ $$\sqrt{x^2+y^2}=2x+1$$ and finally square to get $$x^2+y^2=4x^2+4x+1$$
$\endgroup$ $\begingroup$You are right that the polar equation $r = (1+2r \cos \theta)$ and the cartesian equation $y^2 - 3x^2 - 4x -1 = 0$ are not equivalent. The solutions of the first one are a subset of the solutions of the second.
$y^2 - 3x^2 - 4x -1 = 0$ is equivalent to $r^2=(1+2r \cos \theta)^2$ and that is equivalent to
$$ r = (1+2r \cos \theta) \quad\textbf{ or }\quad r = -(1+2r \cos \theta) \quad . $$
So your polar equation $r = (1+2r \cos \theta)$ describes one branch of the hyperbola $y^2 - 3x^2 - 4x -1 = 0$ and $r = -(1+2r \cos \theta) $ is the other branch.
Here is an image from Wolfram Alpha:
Addendum: I did assume that $r$ must be positive (see below comments), and it is easy to see that then $x \ge -1/3$ for the first equation and $x \le -1$ for the second, so each polar equation produces one branch of the hyperbola.
Without this restriction, either equation produces the entire hyperbola, because the transformation $r \to -r, \theta \to \theta + \pi$ transforms one equation into the other.
In other words: If $x = r \cos \theta, y = r \sin \theta $ is a solution of the cartesian equation $y^2 - 3x^2 - 4x -1 = 0$ then either $(r, \theta)$ or $(-r, \theta + \pi)$ is a solution of the polar equation $r = 1 + 2r \cos \theta$.
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