Coburn J.W., Herdlick J.D. College Algebra: Graphs and Models
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710 CHAPTER 8 Analytic Geometry and the Conic Sections 8–4
College Algebra G&M—
C. Characteristics of the Conic Sections
Examples 1 and 2 bring us one step closer to the wider application of these ideas in a
study of the conic sections. But before the connection is clearly made, we’ll introduce
some background on this family of curves. In common use, a cone might bring to mind
the conical paper cups found at a water cooler. The point of the cone is called the
vertex and the sheet of paper forming the sides is called a nappe. In mathematical
terms, a cone has two nappes, formed by rotating a nonvertical line (called the genera-
tor), about a vertical line (called the axis), at their point of intersection—the vertex
(see Figure 8.5). The conic sections are so named because all curves in the family can
be formed by a section of the cone, or more precisely the intersection of a plane and a
cone. Figure 8.6 shows that if the plane does not go through the vertex, the intersection
will produce a circle, ellipse, parabola, or hyperbola.
If the plane does go through the vertex, the result is a single point, a single line (if
the plane contains the generator), or a pair of intersecting lines (if the plane contains
the axis).
The connection we seek to make is that each conic section can be defined in terms of
the distance between points in the plane, as in Example 1, or the distance between a given
point and a line, as in Example 2. In Example 1, we noted the points ,
and (4, 4) were all on a circle of radius 5 with center (0, 1), in line with the analytic defini-
tion of a circle: A circle is the set of all points that are an equal distance (called the radius)
from a given point (called the center).
In Example 2, you may have no-
ticed that the points seemed to form
the right branch of a parabola (see Fig-
ure 8.7), and in fact, this example illus-
trates the analytic definition of a
parabola: A parabola is the set of all
points that are an equal distance from
a given point (called the focus), and a
given line (called the directrix).
The focus and directrix are not ac-
tually part of the graph, they are sim-
ply used to locate points on the graph.
For this reason all foci (plural of focus)
will be represented by a “” symbol
rather than a point.
14, 2214, 22,
Ellipse HyperbolaCircle Parabola
Figure 8.5
Vertex
Generato
r
Axis
Nappe
Nappe
Figure 8.6
55
5
y
x
(0, 0)
(4, 2)
(8, 8)
Focus
(0, 2)
directrix
d
1
d
2
d
1
d
2
y 2
Figure 8.7
Circle
Ellipse
Hyperbola
Parabola
cob19545_ch08_707-715.qxd 10/25/10 2:43 PM Page 710
EXAMPLE 3
䊳
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
䊳
Use the ordered pair (x, y) to represent an arbitrary point on the parabola. Since
any point on the line has coordinates , we set the distance from
to (x, y) equal to the distance from (0, 2) to (x, y). The result is
distances are equal
simplify
power property
expand binomials
simplify
result
All points satisfying these conditions are on the parabola defined by .
See the figure.
Now try Exercises 27 and 28
䊳
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 fol-
lowing 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 seg-
ment 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.
y
1
8
x
2
y
1
8
x
2
8 y x
2
y
2
4y 4 x
2
y
2
4y 4
1y 22
2
x
2
1y 22
2
21y 22
2
2x
2
1y 22
2
21x x2
2
3y 1224
2
21x 02
2
1y 22
2
1x, 22
1x, 22y 2
8–5 Section 8.1 A Brief Introduction to Analytical Geometry 711
College Algebra G&M—
12 12
6
9
Y
1
X
2
,
1
8
Y
2
2
Figure 8.9
Figure 8.8
EXAMPLE 4
䊳
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 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 the distance from the directrix to (x, y).
1
2
x 4
cob19545_ch08_707-715.qxd 12/15/10 10:41 AM Page 711
Solution
䊳
Since any point on the line has coordinates (4, y), we have:
in words
resulting equation
simplify
power property
expand binomials
distribute
simplify:
polynomial form
All points satisfying these conditions are on the ellipse defined by
Now try Exercises 29 and 30
䊳
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 f
1
and f
2
, an ellipse is the set of all points
(x, y) where the sum of the distances from f
1
to (x, y)
and f
2
to (x, y) is constant.
See Figure 8.11 and Exercises 31 and 32. Both the
focus/directrix definition and the two foci definition have
merit, and simply tend to call out different characteris-
tics and applications of the ellipse. The hyperbola also
has a focus/directrix definition and a two foci definition.
See Exercises 33 and 34.
3x
2
4y
2
12.
3 x
2
4y
2
12
1x
2
1
4
x
2
3
4
x
2
3
4
x
2
y
2
3
x
2
2x 1 y
2
1
4
x
2
2x 4
x
2
2x 1 y
2
1
4
1x
2
8x 162
1x 12
2
y
2
1
4
1x 42
2
21x 12
2
y
2
1
2
21x 42
2
21x 12
2
3y 1024
2
1
2
21x 42
2
1y y2
2
Distance from 11, 02 to 1x, y2
1
2
3distance from 14, y2 to 1x, y24
x 4
712 CHAPTER 8 Analytic Geometry and the Conic Sections 8–6
College Algebra G&M—
f
1
f
2
d
3
d
2
f
1
f
2
d
1
d
4
(x, y)
d
1
d
2
d
3
d
4
Figure 8.11
Figure 8.10
C. You’ve just seen how
we can use the defining
characteristics of a conic
section to find its equation
䊳
CONCEPTS AND VOCABULARY
Fill in the blank with the appropriate word or phrase. Carefully reread the section if needed.
8.1 EXERCISES
2. The distance formula is ;
the midpoint formula is .
4. The conic sections are formed by the intersection
of a and a .
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 .
M
d 1. Analytical geometry is a study of using
the tools of .
3. The distance between a point and a line always
refers to the distance.
5. If a plane intersects a cone at its vertex, the result is
a , a line, or a pair of lines.
cob19545_ch08_707-715.qxd 10/25/10 2:43 PM Page 712
8–7 Section 8.1 A Brief Introduction to Analytical Geometry 713
College Algebra G&M—
䊳
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. 8.
9. 10.
11. 12.
13. Find an equation of the circle that circumscribes
the triangle in Exercise 7.
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.
P
3
120, 422P
3
13, 32
P
2
112, 182P
2
16, 92
P
1
16, 62P
1
110, 212
P
3
16, 12P
3
12, 72
P
2
16, 42P
2
16, 52
P
1
10, 52P
1
12, 12
P
3
18, 22P
3
15, 62
P
2
13, 142P
2
11, 22
P
1
13, 22P
1
15, 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.
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)
20. Of the following six points, four are an equal
distance from the point and two are not.
(a) Identify which four, and (b) find any two
additional points that are the same (nonvertical,
nonhorizontal) distance from ( , 4).
R(5, 12)
V17, 4 151
2U11 416, 62
T14, 4 513
2S17, 112Q19, 102
1
P11, 42
G12 2130
, 102F15, 4 31102
E13, 92C110, 82
䊳
WORKING WITH FORMULAS
The Perpendicular Distance from a Point to a Line: . The perpendicular
distance from a point (x
1
, y
1
) to a given line can be found using the formula shown, where
is the equation of the line in standard form (A, B, and C are integers).Ax ⴙ By ⴙ C ⴝ 0
d ⴝ
円Ax
1
ⴙ By
1
ⴙ C円
2A
2
ⴙ B
2
21. Use the formula to verify that and Q(6, 4)
are an equal distance from the line y
1
2
x 3.
P16, 22
22. Find the value(s) for y that ensure
(1, y) is this same distance from
y
1
2
x 3.
(x
1
, y
1
)
d
Ax By C 0
䊳
APPLICATIONS
23. Of the following four points, three are an equal
distance from the point A(0, 1) and the line
(a) Identify which three, and (b) find any
two additional points that satisfy these conditions.
24. Of the following four points, three are an equal
distance from the point P(2, 4) and the line
(a) Identify which three, and (b) find any
two additional points that satisfy these conditions.
T12 415
, 52
S110, 42R12 412
, 32Q110, 92
y 4.
E1412
, 82D1212, 62C14, 42B16, 92
y 1.
25. Consider the fixed point and the fixed line
. Verify that the distance from each point
given to , is equal to the distance from the
point to the line
26. Consider the fixed point and the fixed line
. Verify that the distance from each point
given to , is equal to the distance from the
point to the line
S1416
, 122
R1415
, 102Qa6,
9
2
bP112, 182
y 2.
10, 22
y 2
10, 22
D1815
, 202
C1412
, 22B a10,
25
4
bA14, 12
y 4.
10, 42
y 4
10, 42
cob19545_ch08_707-715.qxd 12/15/10 10:41 AM Page 713
714 CHAPTER 8 Analytic Geometry and the Conic Sections 8–8
College Algebra G&M—
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 as the focus and the horizontal line
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 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 and
are on the ellipse defined
by Then verify that
.
d
1
d
2
d
3
d
4
4x
2
3y
2
48.
1112
, 02
13, 22
x 9
y 8
10, 22
32. From Exercise 30, verify
the points and
are on the
ellipse defined by
Then verify that
.
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 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).
34. From the two foci definition
of a hyperbola: For foci f
1
and f
2
, a hyperbola is the set
of all points (x, y) where the
difference of the distances
from f
1
to (x, y) and f
2
to (x, y)
is constant. Verify the points
(2, 3) and are
on the graph of the hyperbola from Exercise 33.
Then verify d
1
d
2
d
3
d
4
.
13, 216
2
x
1
2
d
1
d
2
d
3
d
4
5x
2
9y
2
180.
13, 115
2
14,
10
3
2
55
5
5
x
y
(3, 2)
(0, 2)
(0, 2)
d
1
d
3
d
4
d
2
f
1
f
2
( 12, 0)
Exercise 32
6
5
5
x
6
y
(3, 15)
d
3
d
4
d
1
d
2
冢4, j冣
f
2
f
1
(4, 0)
(4, 0)
Exercise 31
55
5
5
x
y
(3, 2 6)
d
1
d
2
d
3
d
4
f
1
f
2
(2, 0)
(2, 3)
(2, 0)
䊳
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.
36. Verify that points and are
points on the ellipse with foci at and B(2, 0),
by verifying
The expression that results has the form
which prior to the common
use of technology, had to be simplified using the
formula ,
where and . Use this
relationship to simplify the equation above.
b 41U
2
V
2
2a 2U
1U V
1U V 2a 1b
1U V 1U V,
d1AC2 d1BC2 d1AD2 d1BD2.
A12, 02
D1212
, 162C12, 32
5
5
5
x
5
y
(4, 3)
(2, 4)
(4, 2)
(3, 4)
R
P
T
Q
C
S
U
䊳
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?
38. (5.5) Solve for x in both exact and approximate form:
a. b. 345 5e
0.4x
755
10
1 9e
0.5x
39. (4.3) Use the rational zeroes theorem and other
tools to factor f(x) and sketch its graph:
40. (4.4) Sketch a complete graph of
Clearly label all intercepts and asymptotes.
h1x2
x
2
9
x
2
4
.
f
1x2 x
4
3x
3
3x
2
11x 6.
cob19545_ch08_707-715.qxd 12/15/10 10:41 AM Page 714
College Algebra G&M—
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 equa-
tion 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
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.
EXAMPLE 1
䊳
Determining the Equation of a Circle Given Its Center and Radius
Find an equation of the circle with radius 5 and center at (2, ), then graph the
relation on a calculator.
Solution
䊳
With a center of (2, ), we have , and . Making the
corresponding substitutions into the standard form we obtain
standard form
substitute 2 for
h
, 1 for
k
, and 5 for
r
simplify
The equation of this circle is .
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.
original equation
isolate term containing
y
take square roots
subtract 1
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
, and must also be on the
graph due to symmetry.
Now try Exercises 7 through 12
䊳
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.
11, 3211, 52
15, 52,
Y
1
225 1X 22
2
1, Y
2
225 1X 22
2
1
y 225 1x 22
2
1
y 1 225 1x 22
2
1y 12
2
25 1x 22
2
1x 22
2
1y 12
2
25
1x 22
2
1y 12
2
25
1x 22
2
1y 12
2
25
1x 22
2
3y 1124
2
5
2
1x h2
2
1y k2
2
r
2
r 5h 2, k 11
1
1x h2
2
1y k2
2
r
2
.
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
9.4 9.4
6.2
6.2
8–9 715
cob19545_ch08_707-715.qxd 10/25/10 2:43 PM Page 715
The center is at (3, ), with radius .
Now try Exercises 13 through 18
䊳
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 r
2
.
standard form
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, r
2
is replaced by
a
2
in the “x-term” (horizontal distance), and by b
2
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 ” To answer,
consider the equation from Example 2. We have (after
dividing by 16), which we now compare to where
and The center of the graph is still at (3, ), since and remain
unchanged. Substituting to find additional points, eliminates the y-term and
gives two values for x:
y 2
k 2h 32b 3.
a 4
1x 32
2
4
2
1y 22
2
3
2
1,
1x 32
2
4
2
1y 22
2
4
2
1
a b?
1x h2
2
r
2
1y k2
2
r
2
1
1x h2
2
1y k2
2
r
2
82
7
3
Circle
Center at (3, 2)
Radius: r 4
Diameter: 2r 8
Endpoints of horizontal diameter
(1, 2) and (7, 2)
Endpoints of vertical diameter
(3, 2) and (3, 6)
x
y
(1, 2)
(3, 2)
r 4
(3, 2)
(3, 6)
(7, 2)
r 116 42
716 CHAPTER 8 Analytic Geometry and the Conic Sections 8–10
College Algebra G&M—
EXAMPLE 2
䊳
Completing the Square to Graph a Circle
Find the center and radius of the circle whose equation is given, then sketch its
graph: .
Solution
䊳
Begin by completing the square in both x and y.
group
x
- and
y
-terms; add 3
complete the square
adds 9 to left side adds 4 to left side add to right side
factor and simplify 1x 32
2
1y 22
2
16
9 4
1x
2
6x 92 1y
2
4y 42 3 9 4
1x
2
6x
__
2 1y
2
4y
__
2 3
x
2
y
2
6x 4y 3 0
A. You’ve just seen how
we can use the characteristics
of a circle and its graph to
understand the equation of an
ellipse
cob19545_ch08_716-730.qxd 10/25/10 2:44 PM Page 716
College Algebra G&M—
8–11 Section 8.2 The Circle and the Ellipse 717
substitute 2 for
y
simplify
multiply by
property of square roots
add 3
This shows the horizontal distance from the center to the graph is still , and
the points ( and (7, ) are on the graph (see Figure 8.12). Similarly, for
we have giving and , and showing the vertical dis-
tance from the center to the graph is now with points (3, 1) and (3, ) on the
graph. Using this information to sketch the curve reveals the “circle” is elongated and
has become a horizontal ellipse.
For this ellipse, the line segment through the center, parallel the x-axis, and with
endpoints on the ellipse is called the major axis, with the endpoints of the major axis
called vertices. The segment perpendicular to and bisecting the major axis (with its
endpoints on the ellipse) is called the minor axis, as shown in Figure 8.13.
• If , the major axis is horizontal (parallel to the x-axis) with length 2a, and
the minor axis is vertical with length 2b (see Example 3).
• If the major axis is vertical (parallel to the y-axis) with length 2b, and the
minor axis is horizontal with length 2a (see Example 4).
Generalizing this observation we obtain the equation of an ellipse in standard form.
The Equation of an Ellipse in Standard Form
Given
If the equation represents the graph of an ellipse with center at (h, k).
• gives the horizontal distance from center to graph.
• gives the vertical distance from center to graph.
Finally, note the line segment from center to vertex is called the semimajor axis, with
the perpendicular line segment from center to graph called the semiminor axis.
EXAMPLE 3
䊳
Graphing a Horizontal Ellipse
Sketch the graph of the ellipse defined by
1x 22
2
25
1y 12
2
9
1.
冟
b
冟
冟
a
冟
a b
1x h2
2
a
2
1y k2
2
b
2
1.
a
2
6 b
2
a
2
7 b
2
82
5
3
x
y
(1, 2)
(3, 2)
(3, 1)
(3, 5)
(7, 2)
Ellipse
b 3
a 4
a
Vertex
The case where
a > b
Vertex
Minor axis
Major axis
b
Figure 8.12
Figure 8.13
5b 3,
y 1y 51y 22
2
9, x 3
21, 22
a 4
x 7 and x 1
x 3 4
x 3 4
4
2
16 1x 32
2
16
1x 32
2
4
2
0 1
1x 32
2
4
2
12 22
2
3
2
1
WORTHY OF NOTE
If you point a flashlight at the floor
keeping it perpendicular to the
ground, a circle is formed with the
bulb pointing directly at the center
and every point along the outer
edge of the beam an equal
distance from this center. If you
hold the flashlight at an angle, the
circle is elongated and becomes an
ellipse, with the bulb pointing
directly at one focus.
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718 CHAPTER 8 Analytic Geometry and the Conic Sections 8–12
College Algebra G&M—
Solution
䊳
Noting we have an ellipse with
center The horizontal
distance from the center to the graph is
and the vertical distance from the
center to the graph is After plotting
the corresponding points and connecting
them with a smooth curve, we obtain the
graph shown.
Now try Exercises 19 through 24
䊳
As with the circle, the equation of an ellipse can be given in polynomial form, and
here our knowledge of circles is helpful. For the equation , we
know the graph cannot be a circle since the coefficients are unequal, and the center of
the graph must be at the origin since . To actually draw the graph, we con-
vert the equation to standard form. Note that a circle whose center is at (0, 0) is called
a central circle, and an ellipse with center at (0, 0) is called a central ellipse.
EXAMPLE 4
䊳
Graphing a Vertical Ellipse
For
a. Write the equation in standard form and identify the center and the values of a
and b.
b. Identify the major and minor axes and name the vertices.
c. Sketch the graph.
d. Graph the relation on a graphing calculator using a “friendly” window, then
use the feature to find four additional points on the graph whose
coordinates are rational.
Solution
䊳
The coefficients of x
2
and y
2
are unequal, and 25, 4, and 100 have like signs. The
equation represents an ellipse with center at (0, 0). To obtain standard form:
a.
given equation
divide by 100
standard form
write denominators in squared form;
b. The result shows and indicating the major axis will be vertical and
the minor axis will be horizontal. With the center at the origin, the x-intercepts will
be and (2, 0),
with the vertices (and
y-intercepts) at
and (0, 5).
c. Plotting these
intercepts and
sketching the ellipse
results in the graph
shown in
Figure 8.14.
10, 52
12, 02
b 5,a 2
a 2, b 5
x
2
2
2
y
2
5
2
1
x
2
4
y
2
25
1
25x
2
100
4y
2
100
1
25x
2
4y
2
100
TRACE
25x
2
4y
2
100,
h k 0
25x
2
4y
2
100
b 3.
a 5,
1h, k2 12, 12.
a b,
(2, 1)
(2, 4)
(3, 1)
(2, 2)
Ellipse
(7, 1)
y
a 5
b 3
x
WORTHY OF NOTE
In general, for the equation
( ), the
equation represents a circle if
, and an ellipse if .A BA B
A, B, F 7 0Ax
2
By
2
F
Figure 8.14
x
y
(0, 5)
(2, 0)
(0, 5)
(2, 0)
Vertical ellipse
Center at (0, 0)
Endpoints of major axis (vertices)
(0, 5) and (0, 5)
Endpoints of minor axis
(2, 0) and (2, 0)
Length of major axis 2b: 2(5) 10
Length of minor axis 2a: 2(2) 4
a 2
b 5
cob19545_ch08_716-730.qxd 10/25/10 2:44 PM Page 718
8–13 Section 8.2 The Circle and the Ellipse 719
College Algebra G&M—
d. As with the circle, we begin by solving for y.
original equation
isolate term containing
y
divide by 4
take square roots
The graph is shown in Figure 8.15, where
we note that (1.6, 3) is a point on the graph.
Due to the symmetry of the ellipse,
, , and are
also on the graph.
Now try Exercises 25 through 36
䊳
If the center of the ellipse is not
at the origin, the polynomial form
has additional linear terms and we
must first complete the square in x
and y, then write the equation in stan-
dard form to sketch the graph (see the
Reinforcing Basic Concepts feature
for more on completing the square).
Figure 8.16 illustrates how the cen-
tral ellipse and the shifted ellipse are
related.
EXAMPLE 5
䊳
Completing the Square to Graph an Ellipse
Sketch the graph of , then state the domain and range of
the relation.
Solution
䊳
The coefficients of x
2
and y
2
are unequal and have like signs, and we assume the equation
represents an ellipse but wait until we have the factored form to be certain (it could be a
degenerate ellipse).
given equation (polynomial form)
group like terms; subtract 141
factor out leading coefficient from each group
complete the square
adds adds
add to right
225 16
4142 1625192 225
251x
2
6x 92 41y
2
4y 42141 225 16
251x
2
6x
__
2 41y
2
4y
__
2141
25x
2
150x 4y
2
16y 141
25x
2
4y
2
150x 16y 141 0
25x
2
4y
2
150x 16y 141 0
11.6, 3211.6, 3211.6, 32
Y
1
B
100 25X
2
4
, Y
2
B
100 25X
2
4
y
B
100 25x
2
4
y
2
100 25x
2
4
4 y
2
100 25x
2
25x
2
4y
2
100
9.4 9.4
6.2
6.2
WORTHY OF NOTE
After writing the equation in
standard form, it is possible to end
up with a constant that is zero or
negative. In the first case, the graph
is a single point. In the second
case, no graph is possible since
roots of the equation will be
complex numbers. These are called
degenerate cases. See Exercise 84.
Ellipse with center
at (h, k)
All points shift
h units horizontally,
k units vertically,
opposite the sign
x
2
a
2
a b
a
2
b
2
(0, b)
(a, 0)
(0, b)
(a, 0)
y
2
b
2
1
a
2
(x h)
2
1
b
2
(y k)
2
h
k
(h, k)
(0, 0)
x
y
Central
ellipse
Figure 8.16
Figure 8.15
c
c
cc
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