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Just for fun, I decided to try one method of converting a parabola to polar coordinates. Without loss of generality and assuming $a \neq 0,$ let's suppose our quadratic $y = ax^2 + bx + c$ can be factored as

$$y(x) = a(x-r_1)(x-r_2)$$

Now I notice that if our expression is of the form $u(u + c),$ it's easier to solve because we can then divide both sides. So, we make the substitution $u = x - r_1$ we end up with

$y(u + r_1) = au(u+r_1 - r_2),$ then if we convert to polar coordinates $y = r\sin(\theta),$ $u = r\cos(\theta)$ we can then divide both sides by $r \cos(\theta)$ to obtain

$\tan(\theta) = a(r \cos(\theta) + r_1 - r_2)$ which is then explicitly solvable for $r$ in terms of $\theta.$

But in going about this process, I'm confused as to how I "back-substitute" $u$ in terms of $x$ to complete the conversion process. Can $u$ substitution of this nature help simplify conversions to polar coordinates? Or is this concept flawed?

askquestions4
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  • A useful parameterization would take the turning point as the origin, but $y=ax^2$ is equivalent to $r=a^{-1}\sec\theta\tan\theta$. – J.G. Oct 19 '22 at 19:36
  • What's wrong with saying $x - r_1 = r \cos(\theta)$? Is there a way to salvage this? – askquestions4 Oct 19 '22 at 19:42
  • You can use that in a parametrization, partnered with say $y-r_2=r\sin\theta$, but it means you're basically taking $X:=x-r_1,,Y:=y-r_2$ as a new system of Cartesian coordinates, which is the point of my origin-shifting proposal. – J.G. Oct 19 '22 at 19:45
  • So how do we "un"shift once we're done solving for $r$ with the Cartesian to Polar conversion process and finish the solving? – askquestions4 Oct 19 '22 at 19:47
  • That would mean using the original Cartesian coordinates with $x=r\cos\theta,,y=r\sin\theta$, so the result is nowhere near as neat. If you're genuinely interested in that polar form, the one I've been talking about isn't worth obtaining, as it can't be converted except via a Cartesian intermediate. – J.G. Oct 19 '22 at 19:49
  • So in general, u-substitution can't help convert from Cartesian to Polar? – askquestions4 Oct 19 '22 at 19:50

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