Precalculus: Mathematics for Calculus - 6th Edition
Precalculus: Mathematics for Calculus - 6th Edition
6th Edition
ISBN: 9780840068071
Author: Stewart, James, Redlin, Lothar, Watson, Saleem
Publisher: Cengage Learning
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Vertices Of An Ellipse
In the study of the conic section of the geometry, An ellipse is a plane curve surrounding two focal points, or foci where the set of all locus of points in the plane has their sum of the distances to the two foci is constant. On the ellipse curve, it is…
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Chapter 10.8, Problem 47E

Global Positioning System (GPS) The Global Positioning System determines the location of an object from its distances to satellites in orbit around the earth. In the simplified, two-dimensional situation shown in the following figure, determine the coordinates of P from the fact that P is 26 units from satellite A and 20 units from satellite B.

Chapter 10.8, Problem 47E, Global Positioning System (GPS) The Global Positioning System determines the location of an object

Expert Solution & Answer
Check Mark
To determine

The coordinates of P from given global positioning system.

Answer to Problem 47E

The coordinates of point P is (12,8) .

Explanation of Solution

Given:

Point P is 26 units far from satellite A and 20 units far from satellite B, and the coordinates of satellite A and B are shows in Figure (1).

The Figure of global positioning system is,

Precalculus: Mathematics for Calculus - 6th Edition, Chapter 10.8, Problem 47E

Figure (1)

Calculation:

The formula to find distance between two points (x1,y1) and (x2,y2) is,

d=(x2−x1)2+(y2−y1)2 (1)

Where, d is distance between two points.

In Figure (1) the coordinates of P is (x,y) and coordinates of satellite A is (22,32) and distance between P and A is 26 units.

Substitute 22 for x2 , x for x1 , 32 for y2 , y for y1 and 26 for d in equation (1) to get the equation,

26=(22−x)2+(32−y)2(26)2=(22)2+x2−2⋅22⋅x+(32)2+y2−2⋅32⋅y676=484+x2−44x+1024+y2−64y−832=x2+y2−44x−64y

Further, solve the above equation to get the equation of circle,

x2+y2−44x−64y+832=0 (2)

In Figure (1) the coordinates of P is (x,y) and coordinates of satellite B is (28,20) and distance between P and B is 20 units.

Substitute 28 for x2 , x for x1 , 20 for y2 , y for y1 and 20 for d  in equation (1) to get the equation,

20=(28−x)2+(20−y)2(20)2=(28)2+x2−2⋅28⋅x+(20)2+y2−2⋅20⋅y400=784+x2−56x+400+y2−40y−784=x2+y2−56x−40y

Further, solve the above equation to get the equation of circle,

x2+y2−56x−40y+784=0 (3)

In the Figure (1), if the radius of planet centered at the corresponding satellite the equation (2) and equation (3) of two circles intersect at a point, thus the equation (2) is equal to the equation (3), solve it for x in y term,

x2+y2−44x−64y+832=x2+y2−56x−40y+784−44x−64y+832=−56x−40y+78412x=24y−48x=2y−4

Substitute 2y−4 for x in equation (3) to get the value of y,

(2y−4)2+y2−56(2y−4)−40y+784=04y2+16−16y+y2−112y+224−40y+784=05y2−168y+1024=05y2−(40+128)y+1024=0

Further, solve the above equation to get the value of y,

5y2−40y−128y+1024=05y(y−8)−128(y−8)=0(y−8)(5y−128)=0y=8,25.6

Two possible value of y give two corresponding values of x.

Substitute 8 for y in equation x=2y−4 to get the value of x,

x=16−4x=12

Substitute 25.6 for y in equation x=2y−4 to get the value of x,

x=51.2−4x=47.2

Avoid the fractional values of coordinates of point P.

Thus, the coordinates of point P is (12,8) .

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Chapter 10.8, Problem 46E
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Chapter 10.8, Problem 48E

Chapter 10 Solutions

Precalculus: Mathematics for Calculus - 6th Edition

Ch. 10.1 - The system of equations {2x+3y=75xy=9 is a system...Ch. 10.1 - Prob. 2ECh. 10.1 - A system of two linear equations in two variables...Ch. 10.1 - The following is a system of two linear equations...Ch. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.1 - Prob. 8ECh. 10.1 - Prob. 9ECh. 10.1 - Prob. 10E
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