Example 10.  Determine the conic that passes through the five points  [Graphics:Images/ConicFitMod_gr_128.gif] .  

Solution 10.

The points are entered into Mathematica with the command:

 

[Graphics:../Images/ConicFitMod_gr_129.gif]

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

 

[Graphics:../Images/ConicFitMod_gr_130.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_131.gif]


[Graphics:../Images/ConicFitMod_gr_132.gif]

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

 

[Graphics:../Images/ConicFitMod_gr_133.gif]

 

The determinant of this matrix is computed by typing:

[Graphics:../Images/ConicFitMod_gr_134.gif]


[Graphics:../Images/ConicFitMod_gr_135.gif]

This quantity is multiplied by  [Graphics:../Images/ConicFitMod_gr_136.gif]  to get the desired equation:

[Graphics:../Images/ConicFitMod_gr_137.gif]


[Graphics:../Images/ConicFitMod_gr_138.gif]

The conic is the hyperbola shown in Figure 10.  It is plotted using the commands:

[Graphics:../Images/ConicFitMod_gr_139.gif]


[Graphics:../Images/ConicFitMod_gr_140.gif]

[Graphics:../Images/ConicFitMod_gr_141.gif]

[Graphics:../Images/ConicFitMod_gr_142.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004