Example 7.  Use the determinant method to find the standard ellipse through the points (6,1), (2,2), (1,4), (9,2).

Solution 7.

The points are entered into Mathematica with the command:

 

[Graphics:../Images/ConicFitMod_gr_82.gif]

Then a row vector corresponding to equation (9) is defined:

 

[Graphics:../Images/ConicFitMod_gr_83.gif]

The matrix A for the linear system in (10) 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_84.gif]


[Graphics:../Images/ConicFitMod_gr_85.gif]

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

 

[Graphics:../Images/ConicFitMod_gr_86.gif]

 

The determinant of this matrix is computed by typing:

[Graphics:../Images/ConicFitMod_gr_87.gif]


[Graphics:../Images/ConicFitMod_gr_88.gif]

This quantity is multiplied by  [Graphics:../Images/ConicFitMod_gr_89.gif] to get the desired equation:

[Graphics:../Images/ConicFitMod_gr_90.gif]


[Graphics:../Images/ConicFitMod_gr_91.gif]

The conic is the circle shown in Figure 7.  It is plotted using the commands:

[Graphics:../Images/ConicFitMod_gr_92.gif]


[Graphics:../Images/ConicFitMod_gr_93.gif]

[Graphics:../Images/ConicFitMod_gr_94.gif]

[Graphics:../Images/ConicFitMod_gr_95.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004