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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!

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  • $\begingroup$ Good question, how doubling up in this case does not show, right? $\endgroup$
    – Narasimham
    Commented Mar 8, 2017 at 23:01

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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.

enter image description here

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.

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  • $\begingroup$ I agree that is what I think would happen. But that doesn't happen for my particular example. Why not? $\endgroup$ Commented Mar 8, 2017 at 16:24

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