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Lab for Determinants and Conic Section Curves
Exercise 12. Determine the conic that
passes through the five points
.
Solution 12. The points are entered into Mathematica with the command:
![[Graphics:../Images/cof_gr_139.gif]](../Images/cof_gr_139.gif)
Then a row vector corresponding to equation (11) is defined:
![[Graphics:../Images/cof_gr_140.gif]](../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]](../Images/cof_gr_141.gif)
For the given five points, the homogeneous system AC = 0 is:
![[Graphics:../Images/cof_gr_142.gif]](../Images/cof_gr_142.gif)
The determinant of this matrix is computed by typing:
![[Graphics:../Images/cof_gr_143.gif]](../Images/cof_gr_143.gif)
This quantity is multiplied by
to get
the desired equation:
![[Graphics:../Images/cof_gr_145.gif]](../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]](../Images/cof_gr_146.gif)
![[Graphics:../Images/cof_gr_147.gif]](../Images/cof_gr_147.gif)
(c) John H. Mathews