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This discussion is about the relationship between parabolas and quadratic functions. Quadratic functions are here.

Definition: Let d be a line (the directrix) and F be a point (the focus) not on d.
A parabola is all the points C such that the length of \overline{CF} equals the distance from C to d.

If the directrix is a horizontal line, the parabola is a quadratic function.

Screencast Construction

                 

GeoGebra InterActivity

Click and drag the point Q along d to trace the parabola. Move F along the y-axis to change the focus. (Here, the focus F is restricted to a point on the y-axis and the directrix d is set so that the vertex of the parabola is always (0,0).)

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Why this construction?

A friend of mine wrote me and asked "Why didn't you find C_q using a perpendicular bisector of \overline{\color{teal}{F}\color{#999900}{Q}} ?"
(Using a perpendicular bisector would have made the construction a proper compass & straightedge construction.)
Answer: This construction started as "A Personal GeoGebra Exploration". That is, I was using GeoGebra to think myself. At the time, I did not plan to turn it into a webpage or mathcast or anything.
  1. I was trying to understand how the locus tool in GeoGebra works.
  2. I was looking at the new parabola tool in GeoGebra. I did not even know what a directrix was. (Without really thinking about it, I guess I assumed that a parabola was the graph of a quadratic function and not vice-versa.)
Result: I would explore the connection between locus, the definition of parabola and graphs of quadratics.
Problem: I did not know how to construct a parabola from definition.
Result: I just started "constructing" - that is, using GeoGebra to understand. The explanation in the video is exactly the way I “realized the construction”.
From quadratic to parabola ... (not an interactivity)
\color{green}{ y=x^2}
Here f(x)=x^2 . We want to find F and d.
Since V=(0,0) is the vertex, the focus F is on the y-axis. Let F=(0,b) .
Since f(x) is a quadratic, the directrix is a horizontal line.
The vertex V=(0,0) is halfway between the focus and the directrix so the directrix is d: \,\, y=-b
We know the point C(1,1) lies on the graph of this function.
The point Q on the directrix corresponding to C must be perpendicular to (1,1) so Q=(1,-b).
We calculate the distances |\overline{CF}| and |\overline{CQ}| and set them equal.   Calculation
d(C,F) = \sqrt {(1 - 0)^2 + (1 - b)^2 } = 2 - 2b + b^2
d(C,Q) = \sqrt {(1 - 1)^2 + (1 + b)^2 } = 1 + 2b + b^2
\displaylines{ 2 - 2b + b^2 = 1 + 2b + b^2 \cr 4b = 1 \cr b = 0.25 \cr}
Solving, we get b=0.25 . So \color{#009999}{F=(0,0.25)} and \color{#999900}{d: \,\, y=-0.25} .

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