Example
10. Determine
the conic that passes through the five
points
.
Solution 10.
The points are entered into Mathematica with the command:
Then a row vector corresponding to equation (11) is defined:
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:
For the given five points, the homogeneous system AC = 0 is:
![[Graphics:../Images/ConicFitMod_gr_133.gif]](../Images/ConicFitMod_gr_133.gif)
The determinant of this matrix is computed by typing:
This quantity is multiplied
by
to
get the desired equation:
The conic is the hyperbola shown in Figure 10. It is plotted using the commands:
![[Graphics:../Images/ConicFitMod_gr_140.gif]](../Images/ConicFitMod_gr_140.gif)
(c) John H. Mathews 2004