A parabolic reflector is a reflective device used to collect or project energy such as light, sound, or radio waves. Its shape is of a circular paraboloid and the surface is generated by a parabola revolving around it's axis. Parabolic reflectors are used in car headlights, telescopes, and dishes. They can project and collect energy in a parallel beam.

Parabolic reflectors started when the mathematician Diocles described them in his book On Burning Mirrors and proved that they focus a parallel beam to a point. Archimedes then later studied them and it is rumoured that he then used them to set a Roman fleet alight. Newton also used them for his telescope and they were also used in lighthouses until the late 19th century.
I traced my original graph on graph paper and went on Exel, then I graphed the known points which were: (9,7), (-9,7), and (0,0). The equation I got was y=.0864x^2 in standard form. Here is my graph:
I found it strange that my parabola's y-intercept, x-intercept,and vertex all equaled zero but I found out that because my parabola is used for things like car headlights, that it couldn't be anything else. An easy way to find the y-intercept or x-intercept without looking at the graph is to find is to find it in factored form equation. An easy way of finding the vertex would be looking at the vertex form equation. I decided to convert my equation to factored and vertex form as well. For factored I got y=.0864(x-0)(x-0), and for vertex I got y=.0864(x-0)^2
.

For the quadratic formula, I got x=-1 +- sqrt (1-4(.0864)(1)) over 2(1). Then I knew that my discriminant was .6544.

Problem:
When a headlight is 4 inches long, how wide is the light that is reflected off of it?
4=.0864x^2
0=.0864x^2 -4
x= -1 +- sqrt (1^2 - 4(.0864)(-4)) over 2(.0864)
x= 3.14536916 and -14.7193432

Parabolic reflectors started when the mathematician Diocles described them in his book

On Burning Mirrorsand proved that they focus a parallel beam to a point. Archimedes then later studied them and it is rumoured that he then used them to set a Roman fleet alight. Newton also used them for his telescope and they were also used in lighthouses until the late 19th century.I traced my original graph on graph paper and went on Exel, then I graphed the known points which were: (9,7), (-9,7), and (0,0). The equation I got was y=.0864x^2 in standard form. Here is my graph:

I found it strange that my parabola's y-intercept, x-intercept,and vertex all equaled zero but I found out that because my parabola is used for things like car headlights, that it couldn't be anything else. An easy way to find the y-intercept or x-intercept without looking at the graph is to find is to find it in factored form equation. An easy way of finding the vertex would be looking at the vertex form equation. I decided to convert my equation to factored and vertex form as well. For factored I got y=.0864(x-0)(x-0), and for vertex I got y=.0864(x-0)^2

.

For the quadratic formula, I got x=-1 +- sqrt (1-4(.0864)(1)) over 2(1). Then I knew that my discriminant was .6544.

Problem:

When a headlight is 4 inches long, how wide is the light that is reflected off of it?

4=.0864x^2

0=.0864x^2 -4

x= -1 +- sqrt (1^2 - 4(.0864)(-4)) over 2(.0864)

x= 3.14536916 and -14.7193432