PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
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Expert Solution & Answer
Chapter 8.3, Problem 69E
Solution
a.
To graph: The coordinate axes in the construction window.
Given information:
Use a geometry software package, such as Cabri Geometry IITM The Geometer’s Sketchpad®.
Graph:
Interpretation:
The coordinated axes have been plotted using graphing utility.
b.
To graph: To construct the two points on the x-axis at (±5,0) as the foci.
Given information:
The coordinates (±5,0) .
Graph:
Interpretation:
The two points on the x-axis at (±5,0) as the foci has been constructed in the above graph.
c.
To graph: To construct concentric circles of radii r=1,2,3,.....,12 centered at these two foci.
Given information:
Concentric circles of radii r=1,2,3,.....,12 centered at these two foci.
Graph:
Interpretation:
The concentric circles of radii r=1,2,3,.....,12 centered at these two foci have been constructed in the above graph.
d.
To graph: To construct the points where these concentric circles meet and have a
diference of radii of 2a=6 .
Given information:
The distance between the vertices is 2a .
Graph:
Interpretation:
Start with the circle's radius of 12 to get a sense of where the points of intersection are where the difference in radii is at six.
Thus to locate the circle a with radius 6 their difference to be 6 .
11/5,10/4,9/3 and 8/2 are the other combinations.
Therefore, the points where these concentric circles meet and have a difference of radii of 2a=6 has been constructed in the above graph.
e.
To find: The equation whose graph includes all of these points.
The equation whose graph includes all of these points is x29−y216=1 .
Given information:
The distance between the vertices is 2a .
Calculation:
The standard form of the hyperbola is given by,
x2a2−y2b2=1
The distance between the vertices is 2a .
2a=6a=3
Let’s consider the distance between the foci is 2c .
2c=10c=5
By using the Pythagorean Theorem,
c2=a2+b2b2=c2−a2b2=52−32b2=16
Substituting the values in the above equation to get,
x232−y216=1x29−y216=1
Therefore, the equation whose graph includes all of these points is x29−y216=1 .
Chapter 8 Solutions
PRECALCULUS:GRAPHICAL,...-NASTA ED.
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