Example 9.  Determine the conic that passes through the five points [Graphics:Images/ConicFitMod_gr_113.gif].  

Solution 9.

The points are entered into Mathematica with the command:

 

[Graphics:../Images/ConicFitMod_gr_114.gif]

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

 

[Graphics:../Images/ConicFitMod_gr_115.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_116.gif]


[Graphics:../Images/ConicFitMod_gr_117.gif]

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

 

[Graphics:../Images/ConicFitMod_gr_118.gif]

 

The determinant of this matrix is computed by typing:

[Graphics:../Images/ConicFitMod_gr_119.gif]


[Graphics:../Images/ConicFitMod_gr_120.gif]

This quantity is multiplied by [Graphics:../Images/ConicFitMod_gr_121.gif] to get the desired equation:

[Graphics:../Images/ConicFitMod_gr_122.gif]


[Graphics:../Images/ConicFitMod_gr_123.gif]

The conic is the ellipse shown in Figure 9.  It is plotted using the commands:

[Graphics:../Images/ConicFitMod_gr_124.gif]


[Graphics:../Images/ConicFitMod_gr_125.gif]

[Graphics:../Images/ConicFitMod_gr_126.gif]

[Graphics:../Images/ConicFitMod_gr_127.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004