Example 8.  Determine the conic that passes through the five points  .  

Solution 8.

The points are entered into Mathematica with the command:

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

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:

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

The determinant of this matrix is computed by typing:

This quantity is multiplied by  to get the desired equation:

The conic is the parabola shown in Figure 8.  It is plotted using the commands:

(c) John H. Mathews 2004