![]()
![]()
Lab for Determinants and Conic Section Curves
Exercise 9. Determine the conic that
passes through the five points
.
Solution 9. The points are entered into Mathematica with the command:
![[Graphics:../Images/cof_gr_106.gif]](../Images/cof_gr_106.gif)
Then a row vector corresponding to equation (11) is defined:
![[Graphics:../Images/cof_gr_107.gif]](../Images/cof_gr_107.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_108.gif]](../Images/cof_gr_108.gif)
For the given five points, the homogeneous system AC = 0 is:
![[Graphics:../Images/cof_gr_109.gif]](../Images/cof_gr_109.gif)
The determinant of this matrix is computed by typing:
![[Graphics:../Images/cof_gr_110.gif]](../Images/cof_gr_110.gif)
This quantity is multiplied by
to get
the desired equation:
![[Graphics:../Images/cof_gr_112.gif]](../Images/cof_gr_112.gif)
The conic is the ellipse shown in Figure 9. It is plotted using the commands:
![[Graphics:../Images/cof_gr_113.gif]](../Images/cof_gr_113.gif)
![[Graphics:../Images/cof_gr_114.gif]](../Images/cof_gr_114.gif)
(c) John H. Mathews