Lab for Determinants and Conic Section Curves

 

 

Exercise 12.  Determine the conic that passes through the five points [Graphics:Images/cof_gr_138.gif].  

Solution 12.  The points are entered into Mathematica with the command:

[Graphics:../Images/cof_gr_139.gif]

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

[Graphics:../Images/cof_gr_140.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_141.gif]

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

[Graphics:../Images/cof_gr_142.gif]

The determinant of this matrix is computed by typing:

[Graphics:../Images/cof_gr_143.gif]

This quantity is multiplied by [Graphics:../Images/cof_gr_144.gif] to get the desired equation:

[Graphics:../Images/cof_gr_145.gif]

The conic is the ellipse shown in Figure 12.  It is plotted using the commands:

[Graphics:../Images/cof_gr_146.gif]

[Graphics:../Images/cof_gr_147.gif]

[Graphics:../Images/cof_gr_148.gif]

 

 

 

 

 

 

(c) John H. Mathews