Quadratic/Conic Ponderings
I was thinking about the problem we were working on last night, and I came up with a construction that has fewer steps
than what we were doing.
D
A
A
F
I
C
G
J
G
K
B
Figure 1
H
E
L
Figure 2
Given a parabola (Figure 1), construct the midpoint C of an arbitrary chord AB. Construct the line through arbitrary point
D parallel to AB and find the intersection point E, then construct the midpoint F of chord DE. Construct the point of
intersection G of the parabola with line FC. As noted by Todd, the line through G parallel to AB is tangent to the
parabola at G, and the line FC is parallel to the axis of symmetry.
Now, instead of constructing the perpendicular to the tangent line through G, which is a normal to the parabola at G,
and reflecting the line FC across this normal, we can accomplish the same thing by just reflecting line FC across the
tangent line. This reflected line (dotted in the figure) must go through the focus based on the reflective properties of
parabolas (all incoming lines parallel to the axis reflect towards the focus).
Now, instead of repeating this whole process again like we did last night, we can construct a perpendicular to line FC
through any arbitrary point (why not A) and find the intersection point H of this line with the parabola (Figure 2). The
perpendicular bisector through midpoint I of segment AH is the axis of the parabola. Point J is the focus (the
intersection of the axis and the image of the parallel to the axis reflected over the tangent line), point K is the vertex.
The perpendicular to the axis through point L, which is the symmetric of J through K, is the directrix.