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Lab for Determinants and Conic Section Curves
Exercise 10. Determine the conic that
passes through the five points
.
Solution 10. The points are entered into Mathematica with the command:
![[Graphics:../Images/cof_gr_117.gif]](../Images/cof_gr_117.gif)
Then a row vector corresponding to equation (11) is defined:
![[Graphics:../Images/cof_gr_118.gif]](../Images/cof_gr_118.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_119.gif]](../Images/cof_gr_119.gif)
For the given five points, the homogeneous system AC = 0 is:
![[Graphics:../Images/cof_gr_120.gif]](../Images/cof_gr_120.gif)
The determinant of this matrix is computed by typing:
![[Graphics:../Images/cof_gr_121.gif]](../Images/cof_gr_121.gif)
This quantity is multiplied
by
to
get the desired equation:
![[Graphics:../Images/cof_gr_123.gif]](../Images/cof_gr_123.gif)
The conic is the hyperbola shown in Figure 10. It is plotted using the commands:
![[Graphics:../Images/cof_gr_124.gif]](../Images/cof_gr_124.gif)
![[Graphics:../Images/cof_gr_125.gif]](../Images/cof_gr_125.gif)
(c) John H. Mathews