Example 5.  Use the determinant method to find the standard equation of a parabola through the points (6,1),  (2,2)  and  (1,4).
Remark. In Exercises 4 and 6 the same points are used to find the circle and alternate parabola.

Solution 5.

The points are entered into Mathematica with the command:

 

[Graphics:../Images/ConicFitMod_gr_53.gif]

Then a row vector corresponding to equation (5) is defined:

 

[Graphics:../Images/ConicFitMod_gr_54.gif]

The matrix A for the linear system in (6) and the determinant is now created.  The vector R is stored in the first row by issuing the command  A = {R}.  Then the remaining three rows of A are generated with the loop command:

[Graphics:../Images/ConicFitMod_gr_55.gif]


[Graphics:../Images/ConicFitMod_gr_56.gif]

For the given three points, the homogeneous system AC = 0 is:

 

[Graphics:../Images/ConicFitMod_gr_57.gif]

 

The determinant of this matrix is computed by typing:

[Graphics:../Images/ConicFitMod_gr_58.gif]


[Graphics:../Images/ConicFitMod_gr_59.gif]

The desired equation is:

[Graphics:../Images/ConicFitMod_gr_60.gif]


[Graphics:../Images/ConicFitMod_gr_61.gif]

The conic is the standard parabola shown in Figure 5.  It is plotted using the commands:

[Graphics:../Images/ConicFitMod_gr_62.gif]


[Graphics:../Images/ConicFitMod_gr_63.gif]

[Graphics:../Images/ConicFitMod_gr_64.gif]

[Graphics:../Images/ConicFitMod_gr_65.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) John H. Mathews 2004