Squaring a polar equation, extraneous solutions? If I have the equation $y=x$, it provides a graph of a single line.  However, if I square both sides, I have $y^2=x^2$ whose graph is both $y=x$ and $y=-x$.  Should the same be true for polar equations and graphs instead of rectangular?  For instance, say I have $$r=\frac{4}{1+\sin \theta}$$ and want to write it as a rectangular equation.  If I first graph it by hand, I can see it looks like it is producing a downward opening parabola.  And, in fact, if I find its rectangular equation, it is $y=-\frac{x^2}{8}+2$.  However, in obtaining that equation, I had at one point $r=4-y$.  I then squared both sides to obtain $r^2$ on the left so I could substitute $r^2=x^2+y^2$.  Why didn't I create extraneous solutions?  I know I did not because the polar graph of $r=\frac{4}{1+\sin \theta}$ and rectangular graph of $y=-\frac{x^2}{8}+2$ match up.  Thanks!
 A: By squaring you get another curve in polar coordinates. By substituting  $ r\rightarrow -r$ we get $ r=-f(\theta) $  anti-symmetric curve to the original $ r = f(\theta)$  (and vice-versa) formed around the origin. 

EDIT1:
During the process of squaring you introduce a negative sign unwittingly.
When you decided to square you were agreeing to see another changed sign in front of the radical sign, $r$ can be either $+\sqrt{x^2+y^2}$ or$\,-\sqrt{x^2+y^2}.$ 
Although you took 
$ r=\dfrac{4}{1+\sin \theta} $
but the problem was you were not a priori prepared to take 
$ r=\dfrac{-4}{1+\sin \theta} $
which is the parabola hat is facing upwards.
Alternately if you square first time
$$ (4-r)^2 = y^2 $$
$$ 16- 8 r + r^2=16- 8 r + x^2+y^2 = y^2$$
$$ r = 2 + \dfrac{x^2}{8}$$
And again square for a second time
$$ x^2+y^2 = 4+ \dfrac{x^4}{64}+  \dfrac{x^2}{2}$$
and simplify
$$[ y- (2- \dfrac{x^2}{8})] \cdot [y+ (2+ \dfrac{x^2}{8})] =0 $$
which appears as a product of the original and "extraneous" parabolas.
It is interesting even a conic in canonical form ( focus at origin)
$ \dfrac {p}{r}= (1 - e \cos \theta )$ or $\quad p = r -  e \, x $
may not produce a contour plot for $  p = -r -  e \, x $ as a default option in some CAS.
