PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Concept explainers

Maximization
Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.
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Question
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Chapter 9.1, Problem 99E

(a)

To determine

To find:The equation for the cross section of the reflector.

(a)

Expert Solution
Check Mark

Answer to Problem 99E

The equation for the cross section of the reflector is y2=6x .

Explanation of Solution

Given information:

The filament of an automobile headlight is at the focus of a parabolic reflector, which sends light out in a straight beam.(See figure). The filament of the headlight is 1.5 inches from the vertex.

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 9.1, Problem 99E , additional homework tip 1

Calculation:

As clear from the figure that the cross section is in form of a rightwards parabola.

As the filament is placed at focus of the parabola and it is 1.5 inches from the vertex.

So, the value of p is 1 1.5 .

The general equation of the parabola is 3 y2=4px .

Substitute 1.5 for p in the equation.

  y2=4pxy2=4(1.5)xy2=6px

Therefore, the equation for the cross section of the reflector is y2=6x .

(b)

To determine

To calculate:The depth of the reflector.

(b)

Expert Solution
Check Mark

Answer to Problem 99E

The depth of the reflector is 2.67 inches.

Explanation of Solution

Given information:

The filament of an automobile headlight is at the focus of a parabolic reflector, which sends light out in a straight beam.(See figure). The reflector is 8 inches wide.

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 9.1, Problem 99E , additional homework tip 2

Calculation:

As calculated in part(a) the equation of the cross section is y2=6x .

As the depth of the reflector is 8 in. , therefore the value of y=4 .

Substitute 4 for y in the equation.

  y2=6x42=6xx=166x≈2.67

Therefore, the depth of the reflector is 2.67 inches.

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Chapter 9.1, Problem 98E
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Chapter 9.1, Problem 100E

Chapter 9 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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Prob. 24CTCh. 9 - Prob. 25CTCh. 9 - Prob. 1STPCh. 9 - Prob. 2STPCh. 9 - Prob. 3STPCh. 9 - Prob. 4STPCh. 9 - Prob. 5STPCh. 9 - Prob. 6STPCh. 9 - Prob. 7STPCh. 9 - Prob. 8STPCh. 9 - Prob. 9STPCh. 9 - Prob. 10STPCh. 9 - Prob. 11STPCh. 9 - Prob. 12STPCh. 9 - Prob. 13STPCh. 9 - Prob. 14STPCh. 9 - Prob. 15STPCh. 9 - Prob. 16STPCh. 9 - Prob. 17STPCh. 9 - Prob. 18STPCh. 9 - Prob. 19STPCh. 9 - Prob. 20STPCh. 9 - Prob. 21STPCh. 9 - Prob. 22STPCh. 9 - Prob. 23STPCh. 9 - Prob. 24STPCh. 9 - Prob. 25STPCh. 9 - Prob. 26STPCh. 9 - Prob. 27STP
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Finding The Focus and Directrix of a Parabola - Conic Sections; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=KYgmOTLbuqE;License: Standard YouTube License, CC-BY