Problem 69 - Algebra and Trigonometry

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 right at the base of the reflector which corresponds to the vertex of the parabolic surface, hence the height is 0 inch.

Step 1: Convert the given dimensions to the same units and determine the equation of the parabola

The diameter of the reflector is 4 inches, so the radius is half of the diameter which is 2 inches. The depth of the reflector is 1 inch. So, given the points (2,1) and (0,0), we can create the equation of the parabola. We start with the formula \(y=ax^2\), and substitute the point (2,1) into the equation which gives us \(1=a*2^2\). Solve for a to get the equation of the parabola.

Step 2: Solve for a

Substitute \(x=2, y=1\) into the equation \(y=ax^2\) and solve for a which results in \(a =1/4\). This gives us the equation for the parabola \(y=1/4*x^2\).

Step 3: Find the vertex of the parabola

The vertex of a parabola \(y=ax^2\) is always at (0, 0), so in this context, that's the base of the reflector of the flashlight. In the other words, the light bulb should be placed right on the base of the reflector, which is the vertex of the parabola, hence the height from the vertex to the flashlight bulb is 0 inch.