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Homework Help: Question concerning defining the general equation of a conic from 5-points

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    I understand that the general equation of a conic is:
    ca5d856e1fbe0fdf05d99def75d82005.png

    And I know that I can use 5 points to determine the unique conic that passes through these points.

    What I don't understand, however, is this site's explanation of the process:
    http://home.att.net/~srschmitt/zenosamples/zs_conic_eqn_5points.html" [Broken]

    The reason being is that, under the header of "Fitting a Conic Section Through Five Points,"
    they declare that there is one and only one conic that may fit through these five points. However, they have to define the F variable before they solve the matrix. This seems contradictory to me. Why do they have to define F in order to pick a unique conic? Wouldn't F be defined already?

    My overarching problem, however, is merely to find an ellipse based on a number of points. It is for a program that will predict the orbit of a controllable-planet by using past positions. Is there an easier way to find the length, width, position, and orientation of an ellipse?

    Wolfram-Alpha's page on ellipses is very helpful in giving me a way to find those values if I have all the variables in the general equation, but is there an easier way?

    2. Relevant equations

    Ax2 + Bxy + Cy2 + Dx + Ey + F = 0




    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 24, 2010 #2

    HallsofIvy

    User Avatar
    Science Advisor

    No. Note that there are 6 constants in the general formula:A, B, C, D, E, and F. But we only need 5 points because they are not independent. In particular, [itex]F= -(Ax^2+ Bxy+ Cy^2+ Dx+ Ey)[/itex]. Or, more simply, you can divide through by any number and get a formula for the same conic section with different coefficients. In particular, if you divide through by F you get [itex]A'x^3+ B'xy+ C'y^2+ D'x+ E'y+ 1= 0[/itex] where A', B', C', D', and E' are A, B, C, D, and E, each divided by F. "Choosing" F just determines which of many formulations for the same conic section we get.

    I won't say it is "easier" but all those things are determined by the eigenvalues and eigenvectors of the matrix giving the quadratic part of that:
    [tex]Ax^2+ Bxy+ Cy^2= \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}A & \frac{1}{2}B \\ \frac{1}{2}B & C\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}[/tex]

     
    Last edited by a moderator: May 4, 2017
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