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

1. Feb 23, 2010

### CyJackX

1. The problem statement, all variables and given/known data

I understand that the general equation of a conic is:

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. Feb 24, 2010

### HallsofIvy

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.

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:
$$Ax^2+ Bxy+ Cy^2= \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}A & \frac{1}{2}B \\ \frac{1}{2}B & C\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}$$

Last edited by a moderator: May 4, 2017