The Ellipse and It's Properties

Discussion in 'Math' started by MrAl, Aug 27, 2015.

  1. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello there,

    I have been trying to remember some of the properties of the ellipse as we started talking about it in another thread and i realized that it had been so long since i studied this stuff that i had forgotten some of it. The category would be under geometry, elemental two dimensional shapes, conics.

    There are a host of properties and formulas, such as that the distance from one focus to the curve and from the curve back to the second focus is a constant and this is the idea behind drawing the Ellipse with a length of cord and a pencil, where the two ends of the cord are tacked down and the pencil point is then constrained by the cord so that it can only reach so far away from either focus, and that, if done right, creates an ellipse.

    There are other properties too, but what i am after is a little different than the usually handful. Since i dont remember much about how this works, i'll ask some very general questions and maybe that will jar my memory and then i can remember some more specific things. I can assure you though that what i am after here is much more interesting than most of the more common properties. I might also mention that there have been several writings on the ellipse that contain several pages, just for this one amazing 2d shape. Compare that to the circle, which although interesting for many reasons in itself still pales in comparison.
    So without any further delay, let me get down to it.

    At the heart of this exploration we have the equation for the Ellipse in cartesian coordinates:
    [EQU 1] y^2/b^2+x^2/a^2=1

    and in Latex (latex code not working perhaps):
    <br />
EQU 1:  \,\, \frac{{y}^{2}}{{b}^{2}}+\frac{{x}^{2}}{{a}^{2}}=1<br />

    The first two questions are simple and straightforward, and replies may lead to other questions.

    Question 1:
    Eliminate either 'a' or 'b' from EQU 1.

    Question 2:
    Convert EQU1 to an equation with only one variable instead of both x and y.
    (Please note that solving for x or y in terms of the other variables and constants is not the same thing).

    Notes:
    'a' and 'b' are considered to be constants, while 'x' and 'y' are considered to be variables.
    The attachment is a picture file of the EQU 1 formula for easy viewing.
     
    Last edited: Aug 28, 2015
  2. WBahn

    Moderator

    Mar 31, 2012
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    I'm not following what you are asking for.

    Since the parameters 'a' and 'b' define the specific ellipse, how can you eliminate either of them?

    If solving for y in terms of x and the other parameters isn't what you are looking for, then what ARE you looking for? This is a two-dimensional shape and so equation is going to have to relate 'x' and 'y' in order to define it.

    Note that the equation for an ellipse is not a function (or at least not in the normal sense of being single-valued).
     
  3. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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  4. BR-549

    Well-Known Member

    Sep 22, 2013
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    That’s fine for 2D.

    But an ellipse is always tilted. 3D.

    That’s because in reality, an ellipse always travels on the surface of a toroid.

    The proper formula for an ellipse is the toroid formula.

    It also has two origins, and they are related in the proper manner.
     
  5. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hello again,

    First, thanks for the replies.

    BR-549:
    Can you show an equation you are talking about there? Sounds interesting.

    Kermit:
    That link is interesting, but doesnt have what i am looking for. It does have other good information however which should be useful, so thanks.

    WBahn:
    If anyone can figure this out it will probably be you :)
    If you feel more comfortable we can look at it as a function by just considering the top side, the positive y values, and ignore the bottom side negative y values for now. Note that this helps a little because y is constrained by x in that if x is a certain value then y must be a certain value.
    I am going by memory now from a long time ago so please bear with me.
    Variables a and b or x and y can sometimes be combined to form a single variable. This helps to generalize sometimes or even find solutions that are otherwise very hard to find. Let me give a quick example that is simpler than with the problem about the Ellipse.
    Take for example the equation:
    z=(y^3+x*y^2+x^2*y+x^3)/x^3
    If we examine the top and bottom, we find that combined powers come out to 3 for every term. So let's divide top and bottom by x^3, simplify, and see what happens. We get:
    z=(y^3/x^3)+(y^2/x^2)+(y/x)+1
    which simplified a bit more gives us:
    z=(y/x)^3+(y/x)^2+(y/x)+1
    And now we can make the substitution u=y/x and get:
    z=u^3+u^2+u+1
    and this is what we want; the two variables x and y have been combined into one variable.
    This worked so easy because the top and bottom simplified, and doing this for the Ellipse may require a transformation to another type of coordinate system. For example, petal coordinates, but i am having trouble remembering how to do this, possibly with a focus as the center point (dont quote me on that though).
    To reduce the a and b however may not be as simple, but i cant remember much about this either. This may require transformation first too, then reduction to a form of either (some guesses):
    a/b, a^2/b, b^2/a, a^2/b^2, a/c, c/a, a^2/c, c^2/a, a^2/c^2, etc., where c is one of the foci.
    Sorry i cant remember more about this, but it is interesting once the solution(s) is/are found.
    I looked around on the web and there is little to be found.
    If i remember right too, solving these problems helps to simplify a special type of elliptic integral for certain values which helps with a certain magnetic field calculation. Cant remember much about that either though except maybe it is the mag field surrounding an elliptical shaped ring. Im going back 30 years here. To give an idea what this means, i can calculate the field around an elliptical shaped ring but the equation comes out in a form that requires numerical integration to integrate the dB. Knowing this simplification leads to a closed form solution because the integration can be performed symbolically. There is another form that requires common elliptic integrals, but those integrals have to be evaluated numerically also, so solving this problem at hand helps to solve those kinds of problems without requiring numerical integration. I wish i could remember more about this more specifically :)
     
    Last edited: Aug 30, 2015
  6. BR-549

    Well-Known Member

    Sep 22, 2013
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    From Wikipedia, the free encyclopedia.

    A torus can be defined parametrically by:[2]
    [​IMG]
    where
    θ, φ are angles which make a full circle, so that their values start and end at the same point,
    R is the distance from the center of the tube to the center of the torus,
    r is the radius of the tube.
    R is known as the "major radius" and r is known as the "minor radius".[3] The ratio R divided by r is known as the "aspect ratio". A doughnut has an aspect ratio of about 2 to 3.

    An implicit equation in Cartesian coordinates for a torus radially symmetric about the z-axis is [​IMG]
    or the solution of f(x, y, z) = 0, where

    [​IMG]
    Algebraically eliminating the square root gives a quartic equation, [​IMG]

    The three different classes of standard tori correspond to the three possible aspect ratios between R and r:
    • When R > r, the surface will be the familiar ring torus.
    • R = r corresponds to the horn torus, which in effect is a torus with no "hole".
    • R < r describes the self-intersecting spindle torus.
    • When R = 0, the torus degenerates to the sphere.
    When Rr, the interior [​IMG] of this torus is diffeomorphic (and, hence, homeomorphic) to a product of an Euclidean open disc and a circle. The surface area and interior volume of this torus are easily computed using Pappus's centroid theorem giving[4]
    [​IMG]
    These formulas are the same as for a cylinder of length 2πR and radius r, created by cutting the tube and unrolling it by straightening out the line running around the center of the tube. The losses in surface area and volume on the inner side of the tube exactly cancel out the gains on the outer side.
    As a torus is the product of two circles, a modified version of the spherical coordinate system is sometimes used. In traditional spherical coordinates there are three measures, R, the distance from the center of the coordinate system, and θ and φ, angles measured from the center point.
    As a torus has, effectively, two center points, the centerpoints of the angles are moved; φ measures the same angle as it does in the spherical system, but is known as the "toroidal" direction. The center point of θ is moved to the center of r, and is known as the "poloidal" direction. These terms were first used in a discussion of the Earth's magnetic field, where "poloidal" was used to denote "the direction toward the poles".[5]
    In modern use these terms are more commonly used to discuss magnetic confinement fusion devices.

    Unquote. If the rotation of r is equal to the rotation of R, you have an ellipse.
     
    Last edited: Aug 30, 2015
  7. MrAl

    Thread Starter Well-Known Member

    Jun 17, 2014
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    Hi again,

    That looks interesting and i will probably look more into that regardless of how it affects this problem at hand. So what you are saying is that you wish to view the Ellipse as a 3d space curve rather than just a 2d curve as we had been doing so far. That's really good to know and study.

    I am sure you have a good reason for showing this, so maybe you can state your reason for pointing this out. BTW i am a firm believer in the rough rule that going to an extra dimension can sometimes help solve a problem in a lower dimension. For a simpler example, the conic sections.
     
    nsaspook likes this.
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