Lab for Determinants and Conic Section Curves

 

 

Exercise 11.  Determine the conic that passes through the five points  [Graphics:Images/cof_gr_127.gif].  

Solution 11.  The points are entered into Mathematica with the command:

[Graphics:../Images/cof_gr_128.gif]

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

[Graphics:../Images/cof_gr_129.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_130.gif]

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

[Graphics:../Images/cof_gr_131.gif]

The determinant of this matrix is computed by typing:

[Graphics:../Images/cof_gr_132.gif]

This quantity is multiplied by  [Graphics:../Images/cof_gr_133.gif]  to get the desired equation:

[Graphics:../Images/cof_gr_134.gif]

The conic is the pair of intersecting lines shown in Figure 11.  It is plotted using the commands:

[Graphics:../Images/cof_gr_135.gif]

[Graphics:../Images/cof_gr_136.gif]

[Graphics:../Images/cof_gr_137.gif]

 

 

 

 

 

 

(c) John H. Mathews