Example 12.  Determine the conic that passes through the five points [Graphics:Images/ConicFitMod_gr_158.gif].  

Solution 12.

The points are entered into Mathematica with the command:

 

[Graphics:../Images/ConicFitMod_gr_159.gif]

Then a row vector corresponding to equation (11) is defined:

 

[Graphics:../Images/ConicFitMod_gr_160.gif]

The matrix A for the linear system in (12) and the determinant is now created.  The vector R is stored in the first row by issuing the command  A = {R}.  Then the remaining five rows of A are generated with the loop command:

[Graphics:../Images/ConicFitMod_gr_161.gif]


[Graphics:../Images/ConicFitMod_gr_162.gif]

For the given five points, the homogeneous system AC = 0 is:

 

[Graphics:../Images/ConicFitMod_gr_163.gif]

 

The determinant of this matrix is computed by typing:

[Graphics:../Images/ConicFitMod_gr_164.gif]


[Graphics:../Images/ConicFitMod_gr_165.gif]

This quantity is multiplied by [Graphics:../Images/ConicFitMod_gr_166.gif] to get the desired equation:

[Graphics:../Images/ConicFitMod_gr_167.gif]


[Graphics:../Images/ConicFitMod_gr_168.gif]

The conic is the ellipse shown in Figure 12.  It is plotted using the commands:

[Graphics:../Images/ConicFitMod_gr_169.gif]


[Graphics:../Images/ConicFitMod_gr_170.gif]

[Graphics:../Images/ConicFitMod_gr_171.gif]

[Graphics:../Images/ConicFitMod_gr_172.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004