Problem 50 - College Algebra

Question

The reflector of a flashlight is in the shape of a parabolic surface. The casting has a diameter of 8 inches and a depth of 1 inch. How far from the vertex should the light bulb be placed?

Step-by-Step Solution

Answer

The bulb should be placed 4 inches from the vertex.

Step 1: Determine the values of x and y

In the problem, the diameter of the parabola is given as 8 inches and the depth is 1 inch. As the vertex is at the centre, half of the diameter (4 inches) is the value of x while the depth represents the value of y. Thus, x=4 and y=1.

Step 2: Calculate the constant a

The constants in the equation \(y=ax^2\) can be calculated using the values of x and y by rearranging the equation to \(a=y/x^2\). Substitute x=4 and y=1 into the formula to calculate a: \(a=1/(4^2)=0.0625\). The equation of the parabola is \(y=0.0625x^2\).

Step 3: Find the vertex position of the lightbulb

In a parabola, the focus is one fourth of 1/a from the vertex along the line of symmetry (y-axis). So, the distance between the vertex and the light bulb is calculated as 1/(4*a). Substituting the value of a, we get the distance as 1/(4*0.0625)=4 inches.