Converting from Cartesian to Polar form

I want to show that $y = \sqrt{2x+1}$ is $r= \frac{1}{1+cos\theta}$ in Polar form.

I used $x=r \cos\theta$ and $y = r \sin\theta$ to eliminate $x$ and $y$ from the cartesian equation and end up with a quadratic in $r$

Then using the quadratic equation yields:

$r = \frac{cos\theta \pm 1}{1-cos^2\theta}$

Thus $r = \frac{1}{1-cos\theta}$ or $r = \frac{-1}{1+cos\theta}$

I am not sure how to justify taking one root - and also what significance the other root has.

i think its because, for the range of values that theta can be, the second root would give values of r<0 which can't happen..cant have a negative magnitude..at least we dont for a-level i think.

I want to show that $y = \sqrt{2x+1}$ is $r= \frac{1}{1+cos\theta}$ in Polar form.

It isn't.

I want to show that $y = \sqrt{2x+1}$ is $r= \frac{1}{1-cos\theta}$ in Polar form.

I used $x=r \cos\theta$ and $y = r \sin\theta$ to eliminate $x$ and $y$ from the cartesian equation and end up with a quadratic in $r$

Then using the quadratic equation yields:

$r = \frac{cos\theta \pm 1}{1-cos^2\theta}$

Thus $r = \frac{1}{1- \cos \theta}$ or $r = \frac{-1}{1+cos\theta}$

I am not sure how to justify taking one root - and also what significance the other root has.

What's wrong with negative r? Some of the Mei fp2 papers use to ask you to sketch the graph with any parts r<0 as a dotted line. (Not sure if they still do now) but it doesn't cause any problems does it?