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Troubles with Ellipses (Cartesian -> Polar)

  1. Dec 5, 2011 #1
    Okay, so I have just broken into the polar coordinate system, and I like to derive things on my own to strengthen my intuition. I decided to try and derive the equation of an ellipse swiftly on my own, and had the high ambitions of eventually deriving the area of an ellipse with polar integration. When I think of a radius as a function of theta, especially in regards to circles/ellipses, I think of distance with regards to a point, so when I think of finding the equation for the radius of an ellipse as a function of theta, I immediately go to Pythagoreas.

    My exact line of reasoning is this:

    Pythagorean theorem holds for any point on an ellipse.
    c^2 = a^2 + b^2
    Except c changes, of course.

    If I think of a as x and b as y, I can easily find them independently in terms of theta.

    X would be some constant "a" (max x value) times cos(theta): [itex]a*cos(\theta)[/itex].

    Y would be some other constant "b" (max y value) times sin(theta): [itex]b*cos(\theta)[/itex].

    R would therefore be c, making it: [itex]r^{2} = (a*cos(\theta))^{2} + (b*sin(\theta))^{2}[/itex]
    (Or r = sqrt of that)

    This makes perfect sense to me, but then I read that r is officially defined as [itex]\frac{ab}{\sqrt{(b*cos(\theta))^{2} + (a*sin(\theta))^{2}}}[/itex]

    Plugging in a value for theta shows that these are not equivalent, and I have no doubt that I went wrong somewhere, but how? If I were to pick any point on an ellipse, and were given theta/a/b, that is how I would solve for r. I have also established that my definition of a and b matches the official equation's definition. It goes without saying that my integration attempts later did not turn up sensible results. I am utterly confused.
    Last edited: Dec 5, 2011
  2. jcsd
  3. Dec 5, 2011 #2


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    Right at the beginning is where your trouble lies, I think. Do you just want to derive the equation of an ellipse in polar form? Then take the equation in rectangular form (with the assumption that the center is (0, 0)):
    [tex]\frac{x^2}{a^2} + {y^2}{b^2} = 1[/tex]
    ... substitute x = r cos θ and y = r sin θ, and then solve for r.
  4. Dec 5, 2011 #3


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    The "Edit" button is disabled at the moment. I made a mistake above -- the rectangular form of the equation of the ellipse is
    [tex]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/tex]
  5. Dec 5, 2011 #4
    That makes a kind of sense in its own right, and I can easily derive the official equation for r from that. However, intuitively I am still at a loss. What exactly did I create, if not an equation for finding the length r, given any angle? Perhaps this will make more sense after I practice some problems.

    Edit: Okay, my approach relies entirely on [itex]x^{2} + y^{2} = r^{2}[/itex] applying for circles AND ellipses, the difference being that it is constant for a circle and varying for an ellipse. This makes sense to me, and all the logic that follows hinges on it. However, I can see that it is not consistent with the definition of an ellipse. Yet still, I can see how [itex]x^{2} + y^{2} = r^{2}[/itex] logically applies, it's obvious when you compare the radius at an angle on a circle to that on an ellipse. I feel a headache coming on.

    Edit2: Okay, here is what gets me. I am using [itex] r = \sqrt{ x^{2} + y^{2}}[/itex], just a "distance" formula, and f(x) and f(y) are both functions of theta. I must have a fundamental misunderstanding of polar equations then.
    Last edited: Dec 5, 2011
  6. Dec 7, 2011 #5
    It has been resolved, and I figured out precisely where I was wrong. My understanding of an ellipse was derived from the parametric representation of a circle, where x = acos(t) and y = bsin(t). In other words I just imagined taking a circle and stretching it in one dimension or another. This does produce an ellipse perfectly fine, but with respect to time, not angle. That distinction is important because when you take a circle and distort it, the angles become distorted as well. (2cos(pi/4), sin(pi/4)) does not refer to a point that is at an angle of pi/4 from the center, but rather one that is arctan(1/2), around pi/6. So of course I cant refer to that point to find the length of the radius at pi/4 about the center of the ellipse.

    So in short, I am an idiot because I assumed equivalency between parametric t and polar theta. Hopefully if someone else has this same problem, they will find this thread and save themselves sleepless nights ^^
  7. Nov 7, 2012 #6
    Thank you so much for this.
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