# Canonical to Parametric, Ellipse Equation

I've done some algebra tricks in this derivation and I'm not sure if it's okay to do those things.

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2\theta + \sin^2\theta$$

Can I really do this next step?
$$\frac{x^2}{a^2} = \cos^2\theta\quad\text{and}\quad\frac{y^2}{b^2} = \sin^2\theta$$

$$x^2 = a^2\cos^2\theta\quad\text{and}\quad y^2 = b^2\sin^2\theta$$ Ignoring the negative numbers: $$x = a\cos\theta\quad\text{and}\quad y = b\sin\theta$$

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Which kind of "okay" do you aim for here? It is usually "okay" to use dodgy steps to find an answer if only you can prove it's correct once you know what it is. In this case it's clear by inspection that the parametric curve you end up with lies within the solution to the original equation. Then you only need to prove that you actually generate the entire solution set. (Given an arbitrary solution to the equation, consider how you could find a $\theta$ that yields it). –  Henning Makholm Aug 31 '11 at 23:53
@Hennig Makholm. What's the correct way to do this, I mean, without dodgy steps, because I was only able to dos this. And what do you mean "by inspection"? And finnaly, an arbitrary solution would be a (x,y) point? Thank you, you are really helping me. –  Vandell Sep 1 '11 at 0:08
@Thijs Laarhoven, surely. It's easy to show that with a counter example. –  Vandell Sep 1 '11 at 0:10
@Vandell, that took a lot of text to reply to, so I have expanded my earlier comment into an answer instead. –  Henning Makholm Sep 1 '11 at 1:12

The idea behind your argument is absolutely fine. Any two non-negative numbers $u$ and $v$ such that $u+v=1$ can be expressed as $u=\cos^2\theta$, $v=\sin^2\theta$ for some $\theta$. This is so obvious that it probably does not require proof. Set $u=\cos^2\theta$. Then $v=1-\cos^2\theta=\sin^2\theta$.

The second displayed formula muddies things somewhat. You intended to say that if $x^2/a^2+y^2/b^2=1$, then there exists a $\theta$ such that $x^2/a^2=\cos^2\theta$ and $y^2/b^2=\sin^2\theta$. You did not mean that for any $\theta$, if $x^2/a^2+y^2/b^2=1$ then $x^2/a^2=\cos^2\theta$! But the transition from the second displayed equation to the third could be interpreted as asserting what you clearly did not intend to say.

It would be better to do exactly what you did, but to use more geometric language, as follows. $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad\text{iff}\quad \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2=1.$$

But the equation on the right holds iff the point $(x/a, y/b)$ lies on the unit circle. The points on the unit circle are parametrized by $(\cos \theta,\sin\theta)$, with $\theta$ ranging over $[0,2\pi)$, so the points on our ellipse are given by $x=a\cos\theta$, $y=a\sin\theta$.

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I think that people object to the OP deriving line 3 from line 2. Line 3 can be derived from line 1 as you say, for the right value of $\theta$. But if line 2 is established with "the wrong" value of $\theta$, then it's not logically valid to conclude line 3. –  alex.jordan Sep 1 '11 at 1:46
I was thinking in make some supositions to hold my equatios, I think that it's clear now, I can accept this fact now ^^. Thank you very much. –  Vandell Sep 1 '11 at 3:59
@Vandell: Short version of André's answer: you could let (yes, "let") $x=a\cos\,\theta$, substitute into your ellipse equation, and then derive an expression for $y$ solely in terms of $\theta$. Note that it will turn out that it doesn't matter which square root of $\sin^2$ are you going to take: one solution will have you traversing the ellipse anticlockwise, and the other, clockwise. –  Ｊ. Ｍ. Sep 1 '11 at 5:54

If you start with the line you wrote last and work your way backwards to the equation of the ellipse that you wrote at the top, then your logic is fine.

(But if your question gets edited after this answer appears, then this answer might be wrong after that. It's happened here before.)

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Which kind of "okay" do you aim for here -- it is usually "okay" to use dodgy steps to find an answer if only you can prove it's correct once you know what it is.

Doing mathematics has two parts: (1) you need to figure out what to prove, and (2) you need to actually prove it. The latter part is an exact science; it has clear and strict rules for what is allowed and what is not, and it's the one that usually gets all the press. However, the first one is no less important. It is sometimes easier and sometimes harder, but its rules are quite different, namely this: anything goes! Yes, really. No matter how you got the idea to prove such-and-such, the only thing that matters is that you can deliver in phase 2 (and that what you proved then turns out to be useful in the context of whichever problem you had originally, but that's a different matter).

Sometimes, the process by which you arrive at the-thing-to-prove is so straightforward that you can read a proof directly off it with essentially no effort. Teachers love these cases (and sometimes give the impression they are all there is), because they make things look nice and orderly, and they're easy to grade. But in real mathematics, there is no shame at all in using less direct methods to find the answer you prove correct later. It doesn't matter if you divided by zero in order to find it, or if an angel appeared in a dream and told you -- if you can deliver a proof that the answer is right at the end of the day, then that is "correct", no matter how you found it.

So it's not really meaningful to ask for a "correct" way to arrive at the parameterization. The important thing, once you have made the leap of faith to separate the two sums, is to prove that the result is right, i.e., that image of your parametric curve is exactly the set of solutions to the original equation.

One side of this is simple. If we substitute your expressions for $x$ and $y$ into the equation, we get $$\frac{(a\cos\theta)^2}{a^2} + \frac{(b\sin\theta)^2}{b^2} = 1$$ and it is then a simple matter of rigorous but uninspired rewriting to prove that this is indeed an identity. (This kind of verification is what is typically meant by "by inspection"). Now we have proved that the image of your curve is a subset of the set of solutions.

It then remains to prove that the set of solutions is a subset of the image of the curve. To do we assume that some given $x$ and $y$ satisfy the equation, and then aim to prove that there must exist a $\theta$ such that $x=a\cos\theta$ and $y=b\sin\theta$. How do we do this? Well, at this time of the night the best I can think of would be some horribly messy case analysis on the various combinations of signs for $x$ and $y$, with special cases if one of them is 0 and otherwise something like $\arctan(\frac{ay}{bx})$ turning up somewhere -- possibly a lemma proving that $(tx,ty)$ can only be a solution if $|t|=1$ will be necessary along the way. If would work out eventually, but it wouldn't be pretty. I'm not even going to try to get all of the details right just now.

Perhaps you can find a slicker argument. Perhaps there is none. Perhaps your audience will be happy with a more handwavy argument than the one I'm imagining.

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I'm doing this for myself, I'm revising some geometry and linear algebra to better understand some concepts in Computer Graphics. But I had pretty rigorous math teachers on my first semesters, so I got this "fear" of doing wrong things. Thank you for you enlightening explanations. –  Vandell Sep 1 '11 at 4:01