Problem 61 - College Algebra

Question

The reflector of a flashlight is in the shape of a parabolic surface. The casting has a diameter of 4 inches and a depth of 1 inch. How far from the vertex should the light bulb be placed?

Step-by-Step Solution

Answer

The light bulb should be placed 1 inch above the vertex of the parabola.

Step 1: Generate the parabola equation

First, determine the equation for this parabola. We know that when \(x\) is 2 inches (half the diameter), \(y\) = 1 inch (the depth). Substituting these values into the equation \(y = ax^2\), we have \[1 = a(2^2)\]. Solve the resulting equation for \(a\).

Step 2: Solve for variable a

Solving for \(a\) gives us \(a = 1/4\). This is the value of \(a\) that shapes our parabola.

Step 3: Find the bulb's position

The light bulb should be placed at the focus of the parabola, such that the light rays it emits are parallel to the axis of the parabola when they reach the reflector, and are hence reflected outwards in a narrow beam. The focus of a parabola \(y = ax^2\) lies at \(y = 1/(4a)\). Substituting our determined value of \(a\) into this yields the position of the lightbulb above the vertex of the parabola.

Step 4: Calculate the position of the bulb

Substituting \(a = 1/4\) into \(y = 1/(4a)\), we get \(y = 1\). This means the bulb should be 1 inch above the vertex of the parabola.