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Chapter 7: Problem 50

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?

Short Answer

Expert verified
The bulb should be placed 4 inches from the vertex.

Step by step solution

01

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.
02

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 \(y=0.0625x^2\).
03

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.

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Most popular questions from this chapter

Use a graphing utility to graph the parabolas.Write the given equation as a quadratic equation in \(y\) and use the quadratic formula to solve for \(y .\) Enter each of the equations to produce the complete graph. $$y^{2}+10 y-x+25=0$$

Find the standard form of the equation of each ellipse satisfying the given conditions. Major axis horizontal with length \(12 ;\) length of minor axis \(=6 ;\) center: \((0,0)\)

What is a hyperbola?

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$$

In Exercises \(1-18,\) graph each ellipse and locate the foci. $$\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$$
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