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If I'm given a curve as an implicit function, is there some algorithm for obtaining a parametrization of the curve?

For example, how would I parameterize

$\frac{2x}{y} - \frac{3y}{x} = 8$

I know I might use $\cosh$ and $\sinh$, but I'm wondering if there's a surefire way to parameterize a curve, when you have a formula for it like this.

I can't isolate y, clearly, but you can re-arrange it to

$ 2x^2 - 3y^2 - 8xy = 0 $

I'm hoping there's a way to find a trig rep for an arbitrary equation that I don't know about.

Where do you go from here?

Is it just trial and error / recognizing the form and choosing trig functions?

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@bobobobo: Again, the curve $\frac{2x}{y}-\frac{3y}{x}=8$ and the curve $2x^2-3y^2 -8xy = 0$ are not the same curve; the former is undefined at $x=y=0$, but the latter is defined. They are closely related, but they are not the same, and that is important if you are trying to parametrize the former. –  Arturo Magidin Feb 4 '11 at 20:46
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@bobobobo: As it happens, you can isolate $y$: you just need to use the quadratic formula, thinking of the equation $3y^2 +(8x)y - 2x^2 = 0$ as a quadratic in $y$ (and then throwing away the solution $y=x=0$. You'll get one formula for $x\lt 0$ and another for $x\gt 0$, though. –  Arturo Magidin Feb 4 '11 at 20:47
    
I did that, but this question is inquiring about a method for finding a trigonometric identity for this equation, if possible. –  bobobobo Feb 4 '11 at 21:44
    
@bobobobo: So really, you are not asking how to parametrize this curve, despite what you state in the question, but rather your real question is "If I'm given a curve as an implicit function, is there some algorithm for obtaining a parametrization of the curve?" You should state the question you are asking clearly and explicitly, not buried in the middle of a different question. –  Arturo Magidin Feb 5 '11 at 3:25
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@bobobobo: Now, to answer your real question, as far as I know there is no general algorithm that will always yield a good parametrization of any curve. Relatively simple trigonometric parametrizations cannot work in general, because they always yield closed curves (being periodic). Parametrizing curves generally takes ingenuity, not unthinking application of general rules, though practice and familiarity with a variety of tricks will help (much as with integration). –  Arturo Magidin Feb 5 '11 at 3:28

3 Answers 3

Solve a quadratic equation for $y$.

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Ok, you get y = 0.23t. But that's not the answer I was looking for, I was hoping for a trigonometric function –  bobobobo Feb 4 '11 at 19:01
    
Where did t come from? 0.23 is approximately one of the two allowable values for $y/x$, but there is another. –  Ross Millikan Feb 5 '11 at 3:40
    
@Ross: he forgot to write $x=t$ (and to take the other possible values, when $t\lt 0$; and to give exact answers and not relatively bad approximations...) –  Arturo Magidin Feb 5 '11 at 3:48
    
@Arturo Magidin: it looks to me like the solution is two curves through the origin(one a straight line), with the origin deleted. Your second comment on the original post scared me a bit as I have the same two formulas for y/x regardless of the sign of x. Am I missing something? –  Ross Millikan Feb 5 '11 at 4:07
    
@Ross: Not really; if you work it out, you get $$y = \frac{-8x\pm\sqrt{88x^2}}{6} = \frac{-4x \pm \sqrt{22}|x|}{3}.$$ If $x$ is positive, you get two formulas, if $x$ is negative, you get two other formulas... which are the same ones you got for positive $x$, only "switched". So in the end, you only get the two curves you mention. If you keep $x$ inside the radical, of course, it's just the two formulas yielding the two curves. But I had simplified and got the absolute value. –  Arturo Magidin Feb 5 '11 at 4:28

You can define $z=\frac{y}{x}$ and get $2z^{-1}-3z=8$, $3z^2+8z-2=0$, $z=\frac{-8 \pm \sqrt{88}}{6}$. Plugging them in, both solutions check. So you can have $x=t, y=\frac{-8 \pm \sqrt{88}}{6}t$ for $t \ne 0$. Why did you expect trigonometry?

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up vote -1 down vote accepted

Quoting Arturo Magidin:

There is no general algorithm that will always yield a good parametrization of any curve. Relatively simple trigonometric parametrizations cannot work in general, because they always yield closed curves (being periodic). Parametrizing curves generally takes ingenuity, not unthinking application of general rules, though practice and familiarity with a variety of tricks will help (much as with integration)
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-1 Much as I appreciate Arturo's wisdom(and this is a good example), what does it achieve to quote this statement that is already available as a comment to your question? –  Ross Millikan Feb 6 '11 at 5:54
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