Picture can be found at Eric Weisstein's
World of Mathematics. (http://mathworld.wolfram.com/ConicSection.html)
We give credit to Apollonius for naming the conic sections (parabola,
hyperbola, and ellipse) around 220 B.C. Only a few short years later in
212 B.C., Archimedes studied the properties of the conic sections. In
this activity, we are going to explore the characteristics of one specific
conic section, the parabola.
We will investigate the locus of points in the plane that are equidistant
from a point and a line by using a paperfolding technique and by using
The Geometer's Sketchpad. Below you can find the definitions for three
geometric terms that we'll be using frequently in this activity.
directrix  a fixed line that is the same distance
as the focus from each point on the parabola.
focus  a fixed point that is the same distance as the directrix
from each point on the parabola.
locus  the set of all points that satisfy a given condition
or a set of given conditions. 
Part
1: Parabola  paper folding activity
Each student will need a 6inch
square piece of wax paper (or patty paper) and a straight edge.
 With a straight edge,
construct a line on your piece of wax paper. This line will serve as
the directrix.
 Construct a point anywhere
on the paper, except on the line, and label it F. This point will serve
as a focus. Construct another point anywhere on the line, and label
it G.
 Using your straightedge,
construct the line segment FG. Fold your paper so that the two points
are concurrent, and deliberately crease it so that you can easily see
the fold mark when the paper has been flattened out. Conjecture any
relationships you might see between the line segment FG and the folded
line.
 Fold your wax paper so
that point F falls anywhere along the directrix. Again, deliberately
crease your wax paper so that you can easily see the fold. Continue
this process approximately ten more times so that each time point F
falls on a different location of the directrix.
We are interested in the pattern
of the creases that are formed when point F is folded along the directrix.
Flatten out your paper and look for any geometric patterns. Describe the
boundary of the shape of the area containing the focus that is bounded
by all of the fold lines.
 Make a conjecture about
the relationship of the distance from the focus to the boundary, and
the distance from the boundary to the directrix. How could you test
your conjecture?
Part
2: Parabola as loci of lines
To see the pattern described
in Part 1 distinctly, we would have to fold the paper dozens of times.
Instead, we are going to simulate the activity using The Geometer's Sketchpad.
 With your neighbor, plan
a geometric construction we could use to simulate the folding process
as described in Part 1. Write down the key steps of the construction
and share your ideas with the class. Carry out your construction using
The Geometer's Sketchpad.
There are several ways to
simulate the activity in Part 1, using different features of The Geometer's
Sketchpad. Specific steps have been recorded below for three different
methods. (Instructor
Note: A
Sketchpad file illustrating the simulation using the animate
feature has been saved as parabolaanimate.gsp.)
All three
constructions start in a similar manner.
 Construct a line (d) that will serve as the directrix.
 Construct two points,
one as the focus (F) and one on the directrix (G).
 Construct a segment
from the focus to the point on the directrix (segment FG).
 Construct the perpendicular
bisector segment FG.

Simulation using the Trace
feature:
 Select the perpendicular
bisector, and choose the Trace Perpendicular Line command under the Display
menu.
 Drag the point G along the
directrix.
 To erase the
traces, choose the Erase Traces command under the Display
menu.

Simulation
using the Animate feature:

Select the perpendicular
bisector, and choose the Trace Perpendicular Line command under the Display
menu.

Select point G,
and choose Animate Point from the Display menu.

To cease
animation, you may click on the solid square button ( ■ ) on
the Motion Controller window, or choose Stop Animation
under the Display menu.

Simulation using the Locus feature:
 Select point G (the
driver point) and the perpendicular bisector. Select Locus
from the Construct menu.

Note: There is an advantage
to the construction using the Locus feature over the others. Once
you have constructed the sketch, you can easily manipulate the position of the focus and directrix
and investigate the connection between them. (Instructor
Note: A
Sketchpad file illustrating the simulation using the locus of
lines has been saved as parabolalocus.gsp.)
Part
3: Parabola as a loci of points
Why does the construction
tracing perpendicular bisectors of segment from the focus to the directrix
produce a parabola? To be able to answer this question we will need
to slightly modify our construction. In Part 2, we constructed a parabola
by loci of lines. Since each of the lines were tangent to the parabola,
we could not locate any specific points on the parabola. We would
like to be able to construct only those points on the parabola.
To
construct a parabola as a locus of points:

Construct a
dashed line through point G that is perpendicular to the
directrix.

Construct the
point of intersection of the dashed line and the perpendicular
bisector of segment FG. Call this point of intersection P.

Construct the locus
of point P as G moves along the directrix.


What
is the general shape formed by the loci of point P? Drag the focus
point to manipulate the loci. What do you observe about the shape
formed?

Confirm that point P satisfies
the definition of a parabola, that is, measure the distance from P to
G and from P to F. Formally prove that these distances are equal.

Algebraically, derive the
standard form of a quadratic function. (Hints: Use the labeled sketch
below to equate the distances from P to G and from P to F. Let the distance
(FV) = p.)
Algebraic
derivation of the standard form of a quadratic function:
distance (P to G)
= distance (P to F)
=
(xx)^{2}
+ (kpy)^{ 2 }=
(xh)^{2} + (ykp)^{ 2}
k^{2 }kpkykp+p^{2
}+pyky+py+y^{2} = (xh)^{2} + y^{2}ky
pyky+k^{2}+kppy+pk+ p^{2}
k^{2
}+p^{2 }+ y^{2 }–2kp2ky^{ }+2py
= (xh)^{2 }+ k^{2 }+p^{2 }+ y^{2 }+2kp
–2ky2py2kp^{
}+2py = (xh)^{2}+2kp 2py
4py = (xh)^{2}+4kp
y = 1/4p(xh)^{2}+k


Another common form of
a quadratic function is y = ax^{2} + bx + c. Further explore how the
constant a, b, and c affect the graph of the parabola. Compare and contrast
these two algebraic forms of quadratic functions.
Extensions:

In Part 2 of this activity,
instead of using the perpendicular bisector of FG, conjecture what would
happen when an arbitrary perpendicular line to FG is used.

Construct a line perpendicular
to FG through any point other than the midpoint. Trace this perpendicular
line. Is the shape formed a parabola? Does it meet the definition of
a parabola? If so, where are its focus and directrix? If not, how would
you classify its shape?

Further manipulate the
sketch to find when point F is the focus? When is line d the directrix?
What is special about the relationship of the perpendicular bisector
of segment FG to the parabola?

Refer back to Part 3 of
this activity. Since segment GP = segment FP, a circle centered at P
will pass through both F and G and be tangent to the directrix at G.
On the basis of these relationships, make up an alternative definition
of a parabola.