C. Characteristics of the Conic Sections
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Section 8.1 A Brief Introduction to Analytical Geometry
EXAMPLE 3
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711
Finding an Equation for All Points That Form a Certain Parabola
With Example 2 as a pattern, use the analytic definition to find a formula
(equation) for the set of all points that form the parabola.
Solution
Y1 ϭ
ᮣ
1 2
X , Y2 ϭ Ϫ2
8
Use the ordered pair (x, y) to represent an arbitrary point on the parabola. Since
any point on the line y ϭ Ϫ2 has coordinates 1x, Ϫ22 , we set the distance from
1x, Ϫ22 to (x, y) equal to the distance from (0, 2) to (x, y). The result is
9
Ϫ12
12
Ϫ6
21x Ϫ x2 2 ϩ 3y Ϫ 1Ϫ22 4 2 ϭ 21x Ϫ 02 2 ϩ 1y Ϫ 22 2 distances are equal
simplify
21y ϩ 22 2 ϭ 2x2 ϩ 1y Ϫ 22 2
power property
1y ϩ 22 2 ϭ x2 ϩ 1y Ϫ 22 2
2
2
2
expand binomials
y ϩ 4y ϩ 4 ϭ x ϩ y Ϫ 4y ϩ 4
simplify
8y ϭ x2
1
result
y ϭ x2
8
All points satisfying these conditions are on the parabola defined by y ϭ 18x2.
See the figure.
Now try Exercises 27 and 28
ᮣ
At this point, it seems reasonable to ask what happens when the distance from the
focus to (x, y) is less than the distance from the directrix to (x, y). For example, what if the
distance is only two-thirds as long? As you might guess, the result is one of the other conic
sections, in this case an ellipse. If the distance from the focus to a point (x, y) is greater
than the distance from the directrix to (x, y), one branch of a hyperbola is formed. While
we will defer a development of their general equations until later in the chapter, the following diagrams serve to illustrate this relationship for the ellipse, and show why we refer
to the conic sections as a family of curves. In Figure 8.8, the line segment from the focus
to each point on the graph (shown in blue), is exactly two-thirds the length of the line segment from the directrix to the same point (shown in red). Note the graph of these points
forms the right half of an ellipse. In Figure 8.9, the lines and points forming the first half
are removed to more clearly show the remaining points that form the complete graph.
Figure 8.8
EXAMPLE 4
ᮣ
Figure 8.9
Finding an Equation for All Points That Form a Certain Ellipse
Suppose we arbitrarily select the point (1, 0) as a focus and the (vertical) line x ϭ 4 as
the directrix. Use these to find an equation for the set of all points where the distance
from the focus to a point (x, y) is 12 the distance from the directrix to (x, y).
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CHAPTER 8 Analytic Geometry and the Conic Sections
Solution
ᮣ
Since any point on the line x ϭ 4 has coordinates (4, y), we have:
1
Distance from 11, 02 to 1x, y2 ϭ 3 distance from 14, y2 to 1x, y2 4 in words
2
1
21x Ϫ 12 2 ϩ 3y Ϫ 102 4 2 ϭ 21x Ϫ 42 2 ϩ 1y Ϫ y2 2 resulting equation
2
1
21x Ϫ 12 2 ϩ y2 ϭ 21x Ϫ 42 2
simplify
2
1
1x Ϫ 12 2 ϩ y2 ϭ 1x Ϫ 42 2
power property
4
1
x2 Ϫ 2x ϩ 1 ϩ y2 ϭ 1x2 Ϫ 8x ϩ 162
expand binomials
4
1
x2 Ϫ 2x ϩ 1 ϩ y2 ϭ x2 Ϫ 2x ϩ 4
distribute
4
3 2
1
3
x ϩ y2 ϭ 3
simplify: 1x 2 Ϫ x 2 ϭ x 2
4
4
4
3x2 ϩ 4y2 ϭ 12
polynomial form
All points satisfying these conditions are on the ellipse defined by 3x2 ϩ 4y2 ϭ 12.
Now try Exercises 29 and 30
Figure 8.10
f1
f2
C. You’ve just seen how
we can use the defining
characteristics of a conic
section to find its equation
ᮣ
Actually, any given ellipse has two foci (see Figure 8.10) and the equation from
Example 4 could also have been developed using the left focus (with the directrix also
on the left). This symmetrical relationship leads us to an alternative definition for the
ellipse, which we will explore further in Section 8.2:
For foci f1 and f2, an ellipse is the set of all points
Figure 8.11
(x, y) where the sum of the distances from f1 to (x, y)
and f2 to (x, y) is constant.
d1
d2
See Figure 8.11 and Exercises 31 and 32. Both the
f1
focus/directrix definition and the two foci definition have
d3
f2
d4
merit, and simply tend to call out different characteris- (x, y)
tics and applications of the ellipse. The hyperbola also
d1 ϩ d2 ϭ d3 ϩ d4
has a focus/directrix definition and a two foci definition.
See Exercises 33 and 34.
8.1 EXERCISES
ᮣ
CONCEPTS AND VOCABULARY
Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.
1. Analytical geometry is a study of
the tools of
.
using
2. The distance formula is d ϭ
the midpoint formula is M ϭ
;
.
3. The distance between a point and a line always
refers to the
distance.
4. The conic sections are formed by the intersection
of a
and a
.
5. If a plane intersects a cone at its vertex, the result is
a
, a line, or a pair of
lines.
6. A circle is defined relative to an equal distance
between two
. A parabola is defined relative
to an equal distance between a
and a
.
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College Algebra G&M—
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ᮣ
DEVELOPING YOUR SKILLS
The three points given form a right triangle. Find the
midpoint of the hypotenuse and verify that the midpoint
is an equal distance from all three vertices.
7. P1 ϭ 1Ϫ5, 22
P2 ϭ 11, 22
P3 ϭ 1Ϫ5, Ϫ62
9. P1 ϭ 1Ϫ2, 12
P2 ϭ 16, Ϫ52
P3 ϭ 12, Ϫ72
11. P1 ϭ 110, Ϫ212
P2 ϭ 1Ϫ6, Ϫ92
P3 ϭ 13, 32
8. P1 ϭ 13, 22
P2 ϭ 13, 142
P3 ϭ 18, 22
16. Find an equation of the circle that circumscribes
the triangle in Exercise 10.
17. Find an equation of the circle that circumscribes
the triangle in Exercise 11.
18. Find an equation of the circle that circumscribes
the triangle in Exercise 12.
10. P1 ϭ 10, Ϫ52
P2 ϭ 1Ϫ6, 42
P3 ϭ 16, Ϫ12
19. Of the following six points, four are an equal
distance from the point A(2, 3) and two are not.
(a) Identify which four, and (b) find any two
additional points that are this same (nonvertical,
nonhorizontal) distance from (2, 3):
B(7, 15)
D(9, 14)
C1Ϫ10, 82
E1Ϫ3, Ϫ92
12. P1 ϭ 16, Ϫ62
P2 ϭ 1Ϫ12, 182
P3 ϭ 120, 422
F15, 4 ϩ 3 1102
13. Find an equation of the circle that circumscribes
the triangle in Exercise 7.
G12 Ϫ 2 130, 102
20. Of the following six points, four are an equal
distance from the point P1Ϫ1, 42 and two are not.
(a) Identify which four, and (b) find any two
additional points that are the same (nonvertical,
nonhorizontal) distance from (Ϫ1, 4).
Q1Ϫ9, 102 R(5, 12) S1Ϫ7, 112 T14, 4 ϩ 5 132
14. Find an equation of the circle that circumscribes
the triangle in Exercise 8.
15. Find an equation of the circle that circumscribes
the triangle in Exercise 9.
ᮣ
U1Ϫ1 ϩ 416, 62
V1Ϫ7, 4 ϩ 1512
WORKING WITH FORMULAS
The Perpendicular Distance from a Point to a Line: d ؍
ͦAx1 ؉ By1 ؉ Cͦ
(x1, y1)
. The perpendicular
2A2 ؉ B2
distance from a point (x1, y1) to a given line can be found using the formula shown, where
Ax ؉ By ؉ C ؍0 is the equation of the line in standard form (A, B, and C are integers).
21. Use the formula to verify that P1Ϫ6, 22 and Q(6, 4)
are an equal distance from the line y ϭ Ϫ12x ϩ 3.
ᮣ
713
Section 8.1 A Brief Introduction to Analytical Geometry
d
Ax ϩ By ϩ C ϭ 0
22. Find the value(s) for y that ensure
(1, y) is this same distance from
y ϭ Ϫ12x ϩ 3.
APPLICATIONS
23. Of the following four points, three are an equal
distance from the point A(0, 1) and the line
y ϭ Ϫ1. (a) Identify which three, and (b) find any
two additional points that satisfy these conditions.
B1Ϫ6, 92
C14, 42
D1Ϫ2 12, 62
E14 12, 82
24. Of the following four points, three are an equal
distance from the point P(2, 4) and the line
y ϭ Ϫ4. (a) Identify which three, and (b) find any
two additional points that satisfy these conditions.
Q1Ϫ10, 92
R12 ϩ 4 12, 32
S110, 42
T12 Ϫ 4 15, 52
25. Consider the fixed point 10, Ϫ42 and the fixed line
y ϭ 4. Verify that the distance from each point
given to 10, Ϫ42 , is equal to the distance from the
point to the line y ϭ 4.
25
C1412, Ϫ22
A14, Ϫ12
B a10, Ϫ b
4
D18 15, Ϫ202
26. Consider the fixed point 10, Ϫ22 and the fixed line
y ϭ 2. Verify that the distance from each point
given to 10, Ϫ22 , is equal to the distance from the
point to the line y ϭ 2.
9
R1415, Ϫ102
P112, Ϫ182
Qa6, Ϫ b
2
S14 16, Ϫ122
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27. The points from Exercise 25 are on the graph of a
parabola. Find an equation of the parabola.
28. The points from Exercise 26 are on the graph of a
parabola. Find an equation of the parabola.
29. Using 10, Ϫ22 as the focus and the horizontal line
y ϭ Ϫ8 as the directrix, find an equation for the set
of all points (x, y) where the distance from the
focus to (x, y) is one-half the distance from the
directrix to (x, y).
30. Using (4, 0) as the focus and the vertical line x ϭ 9 as
the directrix, find an equation for the set of all points
(x, y) where the distance from the focus to (x, y) is
two-thirds the distance from the directrix to (x, y).
31. From Exercise 29, verify the points 1Ϫ3, 22 and
1 112, 02 are on the ellipse defined
by 4x2 ϩ 3y2 ϭ 48. Then verify that
d1 ϩ d2 ϭ d3 ϩ d4.
y
5
f2 (0, 2)
(Ϫ3, 2)
d1
Ϫ5
d2
(0, Ϫ2)
d3
( 12, 0)
5 x
f1
d4
Ϫ5
Exercise 32
y
5
4, j
d1
(Ϫ4, 0)
Ϫ6 f1
f2
(4, 0) 6 x
d3
(Ϫ3, Ϫ 15)
d2
d4
Ϫ5
33. From the focus/directrix
definition of a hyperbola: If the distance from the
focus to a point (x, y) is greater than the distance
from the directrix to (x, y), one branch of a
hyperbola is formed. Using (2, 0) as the focus and
the vertical line x ϭ 12 as the directrix, find an
equation for the set of all points (x, y) where the
distance from the focus to (x, y), is twice the
distance from the directrix to (x, y).
5 x
EXTENDING THE CONCEPT
35. Properties of a circle: A
theorem from elementary
geometry states: If a
radius is perpendicular
to a chord, it bisects the
chord. Verify this is true
for the circle, radii, and
chords shown.
ᮣ
32. From Exercise 30, verify
the points 14, 10
3 2 and
1Ϫ3, Ϫ1152 are on the
ellipse defined by
5x2 ϩ 9y2 ϭ 180.
Then verify that
d1 ϩ d2 ϭ d3 ϩ d4.
y
34. From the two foci definition
5
of a hyperbola: For foci f1
(2, 3)
and f2, a hyperbola is the set
(Ϫ2, 0) d1
d2
f2
of all points (x, y) where the
(2, 0)
f1
Ϫ5
difference of the distances
d4
from f1 to (x, y) and f2 to (x, y)
d
is constant. Verify the points (Ϫ3, Ϫ2 6) Ϫ53
(2, 3) and 1Ϫ3, Ϫ2162 are
on the graph of the hyperbola from Exercise 33.
Then verify d1 Ϫ d2 ϭ d3 Ϫ d4.
Exercise 31
ᮣ
8–8
CHAPTER 8 Analytic Geometry and the Conic Sections
y
5
(Ϫ3, 4)
Q
(Ϫ4, 2)
T
C
P
Ϫ5
5 x
S
(2, Ϫ4) U
Ϫ5
(4, Ϫ3)
R
36. Verify that points C1Ϫ2, 32 and D12 12, 162 are
points on the ellipse with foci at A1Ϫ2, 02 and B(2, 0),
by verifying d1AC2 ϩ d1BC2 ϭ d1AD2 ϩ d1BD2.
The expression that results has the form
1U ϩ V ϩ 1U Ϫ V, which prior to the common
use of technology, had to be simplified using the
formula 1U ϩ V ϩ 1U Ϫ V ϭ 2a ϩ 1b,
where a ϭ 2U and b ϭ 41U2 Ϫ V2 2 . Use this
relationship to simplify the equation above.
MAINTAINING YOUR SKILLS
37. (5.6) $5000 is deposited at 4% compounded
continuously. How many years will it take for the
account to exceed $8000?
39. (4.3) Use the rational zeroes theorem and other
tools to factor f (x) and sketch its graph:
f 1x2 ϭ x4 Ϫ 3x3 Ϫ 3x2 ϩ 11x Ϫ 6.
38. (5.5) Solve for x in both exact and approximate form:
10
a. 5 ϭ
b. 345 ϭ 5e0.4x ϩ 75
1 ϩ 9eϪ0.5x
40. (4.4) Sketch a complete graph of h1x2 ϭ
x2 Ϫ 9
.
x2 Ϫ 4
Clearly label all intercepts and asymptotes.
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College Algebra G&M—
8.2
The Circle and the Ellipse
LEARNING OBJECTIVES
In Section 8.2 you will see how
we can:
A. Use the characteristics of
a circle and its graph to
understand the equation
of an ellipse
B. Use the equation of an
ellipse to graph central
and noncentral ellipses
C. Locate the foci of an
ellipse and use the foci
and other features to
write the equation
D. Solve applications
involving the foci
EXAMPLE 1
In Section 8.1, we introduced the equation of an ellipse using analytical geometry and
the focus-directrix definition. Here we’ll take a different approach, and use the equation of a circle to demonstrate that a circle is simply a special ellipse. In doing so, we’ll
establish a relationship between the foci and vertices of the ellipse, that enables us to
apply these characteristics in context.
A. The Equation and Graph of a Circle
Recall that the equation of a circle with radius r and center at (h, k) is
1x Ϫ h2 2 ϩ 1y Ϫ k2 2 ϭ r2.
As in Section 1.1, the standard form can be used to construct the equation of the
circle given the center and radius as in Example 1, or to graph the circle as in Example 2.
ᮣ
Determining the Equation of a Circle Given Its Center and Radius
Find an equation of the circle with radius 5 and center at (2, Ϫ1), then graph the
relation on a calculator.
Solution
ᮣ
With a center of (2, Ϫ1), we have h ϭ 2, k ϭ Ϫ1, and r ϭ 5. Making the
corresponding substitutions into the standard form we obtain
1x Ϫ h2 2 ϩ 1y Ϫ k2 2 ϭ r2
1x Ϫ 22 2 ϩ 3y Ϫ 1Ϫ12 4 2 ϭ 52
1x Ϫ 22 2 ϩ 1y ϩ 12 2 ϭ 25
standard form
substitute 2 for h, Ϫ1 for k, and 5 for r
simplify
The equation of this circle is 1x Ϫ 22 ϩ 1y ϩ 12 2 ϭ 25.
Recall from Section 1.1 that circles (and other relations) can be graphed by
solving for y, then graphing the upper and lower halves of the circle.
2
1x Ϫ 22 2 ϩ 1y ϩ 12 2 ϭ 25
1y ϩ 12 2 ϭ 25 Ϫ 1x Ϫ 22 2
y ϩ 1 ϭ Ϯ 225 Ϫ 1x Ϫ 22 2
y ϭ Ϯ 225 Ϫ 1x Ϫ 22 2 Ϫ 1
original equation
isolate term containing y
take square roots
subtract 1
Y1 ϭ ϩ 225 Ϫ 1X Ϫ 22 Ϫ 1, Y2 ϭ Ϫ 225 Ϫ 1X Ϫ 22 2 Ϫ 1
2
The graph is shown in the figure using a
square window. Note the point (5, 3)
satisfies the original equation and is a
point on the graph and that 15, Ϫ52,
1Ϫ1, Ϫ52 , and 1Ϫ1, 32 must also be on the
graph due to symmetry.
6.2
Ϫ9.4
9.4
Ϫ6.2
Now try Exercises 7 through 12
ᮣ
If the equation is given in polynomial form, recall that we first complete the square
in x and y to identify the center and radius.
8–9
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CHAPTER 8 Analytic Geometry and the Conic Sections
EXAMPLE 2
ᮣ
Completing the Square to Graph a Circle
Find the center and radius of the circle whose equation is given, then sketch its
graph: x2 ϩ y2 Ϫ 6x ϩ 4y Ϫ 3 ϭ 0.
Solution
ᮣ
Begin by completing the square in both x and y.
1x2 Ϫ 6x ϩ __ 2 ϩ 1y2 ϩ 4y ϩ __ 2 ϭ 3
1x2 Ϫ 6x ϩ 92 ϩ 1y2 ϩ 4y ϩ 42 ϭ 3 ϩ 9 ϩ 4
adds 9 to left side
adds 4 to left side
1x Ϫ 32 2 ϩ 1y ϩ 22 2 ϭ
group x- and y-terms; add 3
complete the square
add 9 ϩ 4 to right side
16
factor and simplify
The center is at (3, Ϫ2), with radius r ϭ 116 ϭ 4.
y
3
(3, 2)
Circle
Center at (3, Ϫ2)
Ϫ2
8
rϭ4
(3, Ϫ2)
(Ϫ1, Ϫ2)
Ϫ7
Radius: r ϭ 4
x
Diameter: 2r ϭ 8
(7, Ϫ2)
Endpoints of horizontal diameter
(Ϫ1, Ϫ2) and (7, Ϫ2)
Endpoints of vertical diameter
(3, 2) and (3, Ϫ6)
(3, Ϫ6)
Now try Exercises 13 through 18
ᮣ
The equation of a circle in standard form provides a useful link to some of the
other conic sections, and is obtained by setting the equation equal to 1. In the case of
a circle, this means we simply divide by r2.
1x Ϫ h2 2 ϩ 1y Ϫ k2 2 ϭ r2
1x Ϫ h2
r
A. You’ve just seen how
we can use the characteristics
of a circle and its graph to
understand the equation of an
ellipse
2
2
ϩ
1y Ϫ k2
r2
standard form
2
ϭ1
divide by r 2
In this form, the value of r in each denominator gives the horizontal and vertical
distances, respectively, from the center to the graph. This is not so important in the case
of a circle, since this distance is the same in any direction. But for other conics, these
horizontal and vertical distances are not the same, making the new form a valuable tool
for graphing. To distinguish the horizontal from the vertical distance, r2 is replaced by
a2 in the “x-term” (horizontal distance), and by b2 in the “y-term” (vertical distance).
This distinction leads us directly into our study of the ellipse.
B. The Equation of an Ellipse
It then seems reasonable to ask, “What happens to the graph when a b?” To answer,
1x Ϫ 32 2
1y ϩ 22 2
consider the equation from Example 2. We have
ϩ
ϭ 1 (after
42
42
1x Ϫ 32 2
1y ϩ 22 2
dividing by 16), which we now compare to
ϩ
ϭ 1, where a ϭ 4
42
32
and b ϭ 3. The center of the graph is still at (3, Ϫ2), since h ϭ 3 and k ϭ Ϫ2 remain
unchanged. Substituting y ϭ Ϫ2 to find additional points, eliminates the y-term and
gives two values for x: