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

GeoGebraInterActivity

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

Why this construction?

A friend of mine wrote me and asked "Why didn't you findC_qusing 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.

I was trying to understand how the locus tool in GeoGebra works.

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

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.

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).)

Why this construction?

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.)Related topics: