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

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The discussion centers on the process of defining a unique conic section using five points, specifically addressing the role of the variable F in the general equation of a conic. It is clarified that while there are six constants in the equation, only five points are needed because they are not independent, allowing for the definition of F to determine the specific formulation of the conic. The user seeks a simpler method to derive the properties of an ellipse for a program predicting planetary orbits, noting that while Wolfram Alpha is useful, they are looking for alternatives. The response suggests that the properties of the ellipse can be determined through eigenvalues and eigenvectors of a related matrix, although it may not be significantly easier. The conversation highlights the complexities involved in fitting conics to data points and the mathematical underpinnings of conic sections.
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Homework Statement



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"

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?

Homework Equations



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




The Attempt at a Solution



 
Last edited by a moderator:
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CyJackX said:

Homework Statement



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"

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?
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, F= -(Ax^2+ Bxy+ Cy^2+ Dx+ Ey). 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 A'x^3+ B'xy+ C'y^2+ D'x+ E'y+ 1= 0 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.

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?
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:
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}

Homework Equations



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




The Attempt at a Solution

 
Last edited by a moderator:
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