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Lab for Determinants and Conic Section Curves
Exercise 8. Determine the conic that
passes through the five points
.
Solution 8. The points are entered into Mathematica with the command:
![[Graphics:../Images/cof_gr_95.gif]](../Images/cof_gr_95.gif)
Then a row vector corresponding to equation (11) is defined:
![[Graphics:../Images/cof_gr_96.gif]](../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]](../Images/cof_gr_97.gif)
For the given five points, the homogeneous system AC = 0 is:
![[Graphics:../Images/cof_gr_98.gif]](../Images/cof_gr_98.gif)
The determinant of this matrix is computed by typing:
![[Graphics:../Images/cof_gr_99.gif]](../Images/cof_gr_99.gif)
This quantity is multiplied
by
to get the
desired equation:
![[Graphics:../Images/cof_gr_101.gif]](../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]](../Images/cof_gr_102.gif)
![[Graphics:../Images/cof_gr_103.gif]](../Images/cof_gr_103.gif)
(c) John H. Mathews