Lab for Determinants and Conic Section Curves

 

 

Exercise 8.  Determine the conic that passes through the five points  [Graphics:Images/cof_gr_94.gif].  

Solution 8.  The points are entered into Mathematica with the command:

[Graphics:../Images/cof_gr_95.gif]

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

[Graphics:../Images/cof_gr_96.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/cof_gr_97.gif]

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

[Graphics:../Images/cof_gr_98.gif]

The determinant of this matrix is computed by typing:

[Graphics:../Images/cof_gr_99.gif]

This quantity is multiplied by  [Graphics:../Images/cof_gr_100.gif]to get the desired equation:

[Graphics:../Images/cof_gr_101.gif]

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

[Graphics:../Images/cof_gr_102.gif]

[Graphics:../Images/cof_gr_103.gif]

[Graphics:../Images/cof_gr_104.gif]

 

 

 

 

 

 

(c) John H. Mathews