A Method For Constructing A Conic Using The Geometer's Sketchpad

One of the beautiful theorems of Projective Geometry is Pascal's Theorem
on conics. It can be stated as follows:

Pascal's Theorem  Let  C be a conic section in the projective plane. Let 1, 2, 3, 4, 5, 6 denote six points of  C such that points  1, ..., 5 are ordinary points of the projective plane. Next, let <a, b> denote the line through the points  a and b. Then the intersection point  A  of lines  <1,2> and  <4,5>, B of  <2,3> and < 5,6>, and C  of <3,4> and < 6,1> are collinear. We dub the line through
A, B and C the  Pascal line of the ordered sextuplet of points (1,2,3,4,5,6).
 

Throughout the activity, we will use the notation established in the statement of Pascal's Theorem. The goal of the activity is to construct the unique conic C through the given five points. The idea is to use Pascal's Theorem to express the desired conic  C  as the ``locus of sixth points". In other words, we want a way to construct all possible Pascal lines where the first five
points have been given as above.  Click here to see a sketch of Pascal's Theorem.

Step 1. Construct five ordinary points of the projective plane such that no three of them are collinear. Label these points 1, 2, 3, 4, 5.
 
Step 2. In a remote corner of the sketchpad screen, construct a small circle and an arbitrary point P of the circle. Animate the point P along the circle. Next, construct the radius r determined by P followed by the line L through point 5 that is parallel to  segment  r.

Comment: The ``locus of sixth points", namely  C, is in 1-1 correspondence with the set of lines through the given fifth point in such a way that each line L through point 5 determines a unique point of  C and conversely. Step 1 gives a method of dynamically sweeping out the set of
lines through point 5. Note that classically, a set of lines through a point is called a  pencil of lines.

Step 3. Construct the intersection point of the lines L and  < 2,3>.  Label this point B. Next, construct the line  <A, B>. Because point 6 lies somewhere on line L, this line is also described as the line  <5,6> so that the intersection point B of L and line <2,3> gives the Pascal line corresponding to L, namely, <A,B>.

Step 4.  Line <3,4> intersects the Pascal line from step 2. Construct the intersection point
of <3,4> and  <A, B>.  Label it C.

Step 5. The desired point 6 lies on lines <5,B> and  <1, C>. Why? Construct the intersection point of
< 5,B> and  <1, C>. Label this point 6.

Comment: Point 6 depends upon line  L  which in turn depends upon the circle in step 1. This says that as  P moves around the circle on which it is defined, we obtain the ``locus of all possible sixth points".

Step 6. Click on point 6 and point P. Go to the Construct menu and click on Locus. Voila, you have the desired conic.

Comment: Let P denote the pencil of lines through point 5 and as earlier let  C be the desired conic. The above method describes a 1-1 function from  P to  C indicated in the diagram below. This function is not described by a formula. Rather, it is described by a sequence of geometric operations:                                P -----> C