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How can I plot this implicit function: $$y^2=\dfrac{x^3}{a-x}$$

Should I use a polar coordinate transformation?

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I've copyedited your question; please check that I didn't change the intended meaning. –  Ilmari Karonen Oct 17 '12 at 11:29
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Why don't you use a Math software? –  Nancy Rutkowskie Oct 17 '12 at 11:30
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You should specify where you want to plot this function, as in Mathematica, or Maple, or on your TI, or just as a rough sketch - that will make the answers more relevant. –  jlv Oct 17 '12 at 11:34
    
sorry. I am not good at speaking English. –  wlantakumi Oct 17 '12 at 13:04

2 Answers 2

up vote 1 down vote accepted

This is another command via Maple for doing that plot for any certain value of $a$:

$[>$ with(plots):

$[>$ implicitplot({seq($x^3/(a-x)-y^2 = 0$, $a = 1 .. 14$)}, $x = 0 .. 10, y = -20 .. 20$, numpoints $= 20000$);

Above I chose $a=14$, you can choose it $40$. :)

enter image description here

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That's an incredibly beautiful graph! –  amWhy Mar 29 '13 at 1:56

You can just rewrite it as $$y = \pm \sqrt{\dfrac{x^3}{a-x}}$$ and plot the positive and negative branches of that (which will be just mirror images of each other). Here's a quick example I did in Maple, with $a = 1$:

Plot of y^2=x^3/(1-x) made using Maple 13

Note that, as the plot shows, your equation only has real solutions in the range $0 \le x < a$ (or $a < x \le 0$ if $a < 0$). Outside that range, $\frac{x^3}{a-x}$ will be negative (or undefined, for $x = a$), so $y$ would need to be imaginary.

In principle, you could also solve the original equation for $x$ as a function of $y$ and plot that. However, that entails solving the cubic polynomial $x^3+y^2x-y^2a=0$, and while there are straightforward methods for doing that, the result will not be particularly pretty.

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I really appreciate for your prompt arrengiment. –  wlantakumi Oct 17 '12 at 13:00

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