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Lab for Determinants and Conic Section Curves
Exercise 11. Determine the conic that
passes through the five points
.
Solution 11. The points are entered into Mathematica with the command:
![[Graphics:../Images/cof_gr_128.gif]](../Images/cof_gr_128.gif)
Then a row vector corresponding to equation (11) is defined:
![[Graphics:../Images/cof_gr_129.gif]](../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]](../Images/cof_gr_130.gif)
For the given five points, the homogeneous system AC = 0 is:
![[Graphics:../Images/cof_gr_131.gif]](../Images/cof_gr_131.gif)
The determinant of this matrix is computed by typing:
![[Graphics:../Images/cof_gr_132.gif]](../Images/cof_gr_132.gif)
This quantity is multiplied
by
to
get the desired equation:
![[Graphics:../Images/cof_gr_134.gif]](../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]](../Images/cof_gr_135.gif)
![[Graphics:../Images/cof_gr_136.gif]](../Images/cof_gr_136.gif)
(c) John H. Mathews