Example 11.  Determine the conic that passes through the five points  [Graphics:Images/ConicFitMod_gr_143.gif].  

Solution 11.

The points are entered into Mathematica with the command:

 

[Graphics:../Images/ConicFitMod_gr_144.gif]

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

 

[Graphics:../Images/ConicFitMod_gr_145.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_146.gif]


[Graphics:../Images/ConicFitMod_gr_147.gif]

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

 

[Graphics:../Images/ConicFitMod_gr_148.gif]

 

The determinant of this matrix is computed by typing:

[Graphics:../Images/ConicFitMod_gr_149.gif]


[Graphics:../Images/ConicFitMod_gr_150.gif]

This quantity is multiplied by  [Graphics:../Images/ConicFitMod_gr_151.gif]  to get the desired equation:

[Graphics:../Images/ConicFitMod_gr_152.gif]


[Graphics:../Images/ConicFitMod_gr_153.gif]

The conic is the pair of intersecting lines shown in Figure 11.  It is plotted using the commands:

[Graphics:../Images/ConicFitMod_gr_154.gif]


[Graphics:../Images/ConicFitMod_gr_155.gif]

[Graphics:../Images/ConicFitMod_gr_156.gif]

[Graphics:../Images/ConicFitMod_gr_157.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004