please help with parabola algebra?

Question

A spotlight is made by placing a strong bulb inside a reflective paraboloid formed by rotating the parabola x^2=4y around its axis of symmetry (assume that x and y are in units of inches). In order to have the brightest, most concentrated light beam, how far from the vertex should the bulb be placed?

pleasea help, i dont know even know where to begin.

Answer 1

You'll want to put the bulb at the focus of the parabola (see the Wikipedia article below). For a parabola with equation

y = ax^2,

the focus is at the point (0, 1/4a). In your equation, a = 1/4, so 1/4a = 1, and the focus is at (0,1).

Answer 2

The bulb should be placed at the FOCUS*** of the parabola! When placed there, a perfect parallel beam will emerge from the reflective parabaloid, vertically upwards, i.e. in the +ve y direction in your graph.

The focus has a mathematical definition, but in practical terms you can find it as follows. You need to know the value of y when dy/dx =1. (Why that?: because then a horizontal ray from the bulb will be reflected precisely vertically.) Well, dy/dx = x/2. If this = 1 then x = 2.

For x =2, the parabola's equation gives y = 1.

So the answer is: the bulb must be placed 1 inch from the vertex.

*** NOTE that it is no coincidence that the mathematically defined place, the FOCUS, is the same as that used to describe a "concentrated point of light." This is nicely illustrated by ellipses, parabolae being a limiting case. Ellipses have TWO FOCI, symmetrically placed along the major axis. Light emitted from one passes through the other on reflection; after TWO reflections, it's back where it started! So each focus can be viewed as a "concentrated point of light," both at emission and at reception.

An analogous thing works for sound, too. In San Francisco's Exploratorium, you can stand within a few feet of an aural focus, and hear people whispering or speaking quietly some 50 or more feet away. In fact, if they don't know that they're at that spot, or that you're listening in at the other one, you can surprise people by whispering to them as though you were right beside them! Provided that a sufficient part of a curved ceiling approximates an appropriate ellipsoidal shape, you can get the same effect. There are political meeting chambers where members discovered this empirically, and have used it to communicate with others across the chamber.

So how is this property for ellipses related to that for parabolae? Both the ellipse and parabola are CONIC SECTIONS, known clasically to, and called that, by Greek mathematicians. In other words, they're part of the same mathematical family. A parabola can be viewed as a "degenerate Ellipse," in which the second focus has "moved off to infinity." That means that the light now going to IT has become a parallel beam. That's why the question mentions "the brightest, most concentrated light beam." Put the bulb anywhere else, and the rays will either cross at some other point and then diverge, or simply diverge without ever crossing. Only if the bulb is at the focus willl you get this "brightest, most concentrated light beam."

This also works in reverse. Parallel light, received "from infinity" will itself focus at the (mathematical) focus of a paraboloidal mirror. (!That verbal connection again.) For practical purposes, that's not only why most reflectors in car headlights are paraboloidal, or nearly so, but also why the primary reflecting mirror in all reflecting telescopes has a paraboloidal shape.

Live long and prosper.

Answer 3

x^2 = 4y
The standard equation of a parabola with focus on the y-axis is x^2=2py, where p is the distance between the directrix and the focus. The focus is located at (0,p/2).

In the equation given, p=2, so the focus is at (0,1)

To get the most concentrated light beam, the bulb should be placed at the focus or at (0,1).

Hope this shines a light on things. :)