B. The Equation of an Ellipse
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Section 8.2 The Circle and the Ellipse
1x Ϫ 32 2
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.
42
ϩ
1Ϫ2 ϩ 22 2
32
1x Ϫ 32 2
ϭ1
ϩ0ϭ1
42
1x Ϫ 32 2 ϭ 16
x Ϫ 3 ϭ Ϯ4
xϭ3Ϯ4
x ϭ 7 and x ϭ Ϫ1
substitute Ϫ2 for y
simplify
multiply by 42 ϭ 16
property of square roots
add 3
This shows the horizontal distance from the center to the graph is still a ϭ 4, and
the points (Ϫ1, Ϫ22 and (7, Ϫ2) are on the graph (see Figure 8.12). Similarly, for
x ϭ 3 we have 1y ϩ 22 2 ϭ 9, giving y ϭ Ϫ5 and y ϭ 1, and showing the vertical distance from the center to the graph is now b ϭ 3, with points (3, 1) and (3, Ϫ5) 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.
Figure 8.12
y
3
Figure 8.13
(3, 1)
Major axis
bϭ3
Ϫ2
(Ϫ1, Ϫ2)
8
aϭ4
(3, Ϫ2)
Ϫ5
a
x
(7, Ϫ2)
b
Vertex
Vertex
Ellipse
(3, Ϫ5)
The case where
a>b
Minor axis
• If a2 7 b2, 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 a2 6 b2 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 a
1x Ϫ h2 2
ϩ
1y Ϫ k2 2
ϭ 1.
a2
b2
b the equation represents the graph of an ellipse with center at (h, k).
• ͿaͿ gives the horizontal distance from center to graph.
• ͿbͿ 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
ᮣ
Graphing a Horizontal Ellipse
Sketch the graph of the ellipse defined by
1y ϩ 12 2
1x Ϫ 22 2
ϩ
ϭ 1.
25
9
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Solution
ᮣ
Noting a b, we have an ellipse with
center 1h, k2 ϭ 12, Ϫ12. The horizontal
distance from the center to the graph is
a ϭ 5, and the vertical distance from the
center to the graph is b ϭ 3. After plotting
the corresponding points and connecting
them with a smooth curve, we obtain the
graph shown.
y
Ellipse
(2, 2)
(Ϫ3, Ϫ1)
bϭ3
aϭ5
(2, Ϫ1)
x
(7, Ϫ1)
(2, Ϫ4)
Now try Exercises 19 through 24
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 25x2 ϩ 4y2 ϭ 100, 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 h ϭ k ϭ 0. To actually draw the graph, we convert 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.
WORTHY OF NOTE
In general, for the equation
Ax2 ϩ By2 ϭ F (A, B, F 7 0), the
equation represents a circle if
A ϭ B, and an ellipse if A B.
EXAMPLE 4
ᮣ
ᮣ
Graphing a Vertical Ellipse
For 25x2 ϩ 4y2 ϭ 100,
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 TRACE feature to find four additional points on the graph whose
coordinates are rational.
Solution
ᮣ
The coefficients of x2 and y2 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. 25x2 ϩ 4y2 ϭ 100 given equation
4y2
25x2
ϩ
ϭ1
divide by 100
100
100
y2
x2
ϩ
ϭ1
standard form
4
25
y2
x2
ϩ
ϭ1
write denominators in squared form; a ϭ 2, b ϭ 5
22
52
b. The result shows a ϭ 2 and b ϭ 5, indicating the major axis will be vertical and
the minor axis will be horizontal. With the center at the origin, the x-intercepts will
Figure 8.14
be 1Ϫ2, 02 and (2, 0),
y
with the vertices (and
Vertical ellipse
(0, 5)
y-intercepts) at
Center at (0, 0)
10, Ϫ52 and (0, 5).
bϭ5
Endpoints of major axis (vertices)
c. Plotting these
(0, Ϫ5) and (0, 5)
intercepts and
(Ϫ2, 0)
(2, 0)
Endpoints of minor axis
x
sketching the ellipse
(Ϫ2, 0) and (2, 0)
a
ϭ
2
results in the graph
Length of major axis 2b: 2(5) ϭ 10
shown in
Length of minor axis 2a: 2(2) ϭ 4
Figure 8.14.
(0, Ϫ5)
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Section 8.2 The Circle and the Ellipse
d. As with the circle, we begin by solving for y.
25x2 ϩ 4y2 ϭ 100
4y2 ϭ 100 Ϫ 25x2
100 Ϫ 25x2
y2 ϭ
4
100 Ϫ 25x2
yϭϮ
B
4
Y1 ϭ ϩ
719
original equation
isolate term containing y
divide by 4
take square roots
100 Ϫ 25X2
100 Ϫ 25X2
, Y2 ϭ Ϫ
B
B
4
4
Figure 8.15
6.2
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,
1Ϫ1.6, 32 , 1Ϫ1.6, Ϫ32 , and 11.6, Ϫ32 are
also on the graph.
Ϫ9.4
9.4
Ϫ6.2
Now try Exercises 25 through 36
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.
EXAMPLE 5
ᮣ
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 standard form to sketch the graph (see the
Reinforcing Basic Concepts feature
for more on completing the square).
Figure 8.16 illustrates how the central ellipse and the shifted ellipse are
related.
ᮣ
Figure 8.16
y
Ellipse with center
at (h, k)
k
(h, k)
Central
ellipse (0, b)
(Ϫa, 0)
(a, 0)
(0, 0)
(0, Ϫb)
All points shift
h units horizontally,
k units vertically,
opposite the sign
(x Ϫ h)2 (y Ϫ k)2
ϭ1
ϩ
a2
b2
a2 Ͼ b2
x
h
x2
y2
ϩ 2 ϭ1
a2
b
aϾb
Completing the Square to Graph an Ellipse
Sketch the graph of 25x2 ϩ 4y2 ϩ 150x Ϫ 16y ϩ 141 ϭ 0, then state the domain and range of
the relation.
Solution
ᮣ
The coefficients of x2 and y2 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).
25x2 ϩ 4y2 ϩ 150x Ϫ 16y ϩ 141 ϭ 0
25x2 ϩ 150x ϩ 4y2 Ϫ 16y ϭ Ϫ141
2
251x ϩ 6x ϩ __ 2 ϩ 41y2 Ϫ 4y ϩ __ 2 ϭ Ϫ141
251x2 ϩ 6x ϩ 92 ϩ 41y2 Ϫ 4y ϩ 42 ϭ Ϫ141 ϩ 225 ϩ 16
c
c
adds 25192 ϭ 225
c
c
adds 4142 ϭ 16
add 225 ϩ 16 to right
given equation (polynomial form)
group like terms; subtract 141
factor out leading coefficient from each group
complete the square
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251x ϩ 32 2 ϩ 41y Ϫ 22 2 ϭ 100
41y Ϫ 22 2
251x ϩ 32 2
100
ϩ
ϭ
100
100
100
1y Ϫ 22 2
1x ϩ 32 2
ϩ
ϭ1
4
25
1x ϩ 32 2
1y Ϫ 22 2
ϩ
ϭ1
22
52
The result is a vertical ellipse with
center at 1Ϫ3, 22, with a ϭ 2 and
b ϭ 5. The vertices are a vertical
distance of 5 units from center,
and the endpoints of the minor
axis are a horizontal distance of
2 units from center. Note this is
the same ellipse as in Example 4,
but shifted 3 units left and 2 up.
The domain of this relation is
x ʦ 3 Ϫ5, Ϫ1 4 , and the range is
y ʦ 3 Ϫ3, 7 4.
factor
divide both sides by 100
simplify (standard form)
write denominators in squared form
(Ϫ3, 7)
y
Vertical ellipse
Center at (Ϫ3, 2)
(Ϫ5, 2)
B. You’ve just seen how
we can use the equation of an
ellipse to graph central and
noncentral ellipses
(Ϫ3, 2)
Endpoints of major axis (vertices)
(Ϫ3, Ϫ3) and (Ϫ3, 7)
(Ϫ1, 2)
Endpoints of minor axis
(Ϫ5, 2) and (Ϫ1, 2)
x Length of major axis 2b: 2(5) ϭ 10
Length of minor axis 2a: 2(2) ϭ 4
(Ϫ3, Ϫ3)
Now try Exercises 37 through 44
ᮣ
C. The Foci of an Ellipse
In Section 8.1, we noted that an ellipse could also be defined in terms of two special
points called the foci. The Museum of Science and Industry in Chicago, Illinois
(http://www.msichicago.org), has a permanent exhibit called the Whispering Gallery.
The construction of the room is based on some of the reflective properties of an ellipse.
If two people stand at designated points in the room and one of them whispers very
softly, the other person can hear the whisper quite clearly—even though they are over
40 ft apart! The point where each person stands is a focus of an ellipse. This reflective
property also applies to light and radiation, giving the ellipse some powerful applications in science, medicine, acoustics, and other areas. To understand and appreciate
these applications, we introduce the analytic definition of an ellipse.
WORTHY OF NOTE
You can easily draw an ellipse that
satisfies the definition. Press two
pushpins (these form the foci of the
ellipse) halfway down into a piece
of heavy cardboard about 6 in.
apart. Take an 8-in. piece of string
and loop each end around the pins.
Use a pencil to draw the string taut
and keep it taut as you move the
pencil in a circular motion—and
the result is an ellipse! A different
length of string or a different
distance between the foci will
produce a different ellipse.
Definition of an Ellipse
Given two fixed points f1 and f2 in a plane, an ellipse
is the set of all points (x, y) where the distance from
f1 to (x, y) added to the distance from f2 to (x, y)
remains constant.
y
P(x, y)
d1
d1 ϩ d2 ϭ k
The fixed points f1 and f2 are called the foci of the
ellipse, and the points P(x, y) are on the graph of the
ellipse.
f1
d2
f2
x
d1 ϩ d2 ϭ k
6 in.
3 in.
5 in.
To find the equation of an ellipse in terms of a and b we combine the definition
just given with the distance formula. Consider the ellipse shown in Figure 8.17 (for
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Section 8.2 The Circle and the Ellipse
Figure 8.17
y
(0, b)
P(x, y)
(a, 0)
x
(Ϫa, 0)
(Ϫc, 0)
(c, 0)
calculating ease we use a central ellipse). Note the vertices have coordinates 1Ϫa, 02 and
(a, 0), and the endpoints of the minor axis have coordinates 10, Ϫb2 and (0, b) as
before. It is customary to assign foci the coordinates f1 S 1Ϫc, 02 and f2 S 1c, 02. We
can calculate the distance between (c, 0) and any point P(x, y) on the ellipse using the
distance formula:
21x Ϫ c2 2 ϩ 1y Ϫ 02 2
Likewise the distance between 1Ϫc, 02 and any point (x, y) is
21x ϩ c2 2 ϩ 1y Ϫ 02 2
(0, Ϫb)
According to the definition, the sum must be constant:
21x Ϫ c2 2 ϩ y2 ϩ 21x ϩ c2 2 ϩ y2 ϭ k
EXAMPLE 6
ᮣ
Finding the Value of k from the Definition of an Ellipse
Use the definition of an ellipse and the diagram given to determine the constant k
used for this ellipse (also see the following Worthy of Note). Note that
a ϭ 5, b ϭ 3, and c ϭ 4.
y
(0, 3)
P(3, 2.4)
(Ϫ5, 0)
(Ϫ4, 0)
(4, 0)
(5, 0)
x
(0, Ϫ3)
Solution
ᮣ
21x Ϫ c2 2 ϩ 1y Ϫ 02 2 ϩ 21x ϩ c2 2 ϩ 1y Ϫ 02 2 ϭ k
213 Ϫ 42 ϩ 12.4 Ϫ 02 ϩ 213 ϩ 42 ϩ 12.4 Ϫ 02 ϭ k
2
WORTHY OF NOTE
Note that if the foci are coincident
(both at the origin) the “ellipse” will
k
actually be a circle with radius ;
2
2x2 ϩ y2 ϩ 2x2 ϩ y2 ϭ k leads to
k2
x2 ϩ y2 ϭ . In Example 6 we
4
10
ϭ 5, and if
found k ϭ 10, giving
2
we used the “string” to draw the
circle, the pencil would be 5 units
from the center, creating a circle of
radius 5.
2
2
2
21Ϫ12 ϩ 2.4 ϩ 27 ϩ 2.4 ϭ k
16.76 ϩ 154.76 ϭ k
2.6 ϩ 7.4 ϭ k
10 ϭ k
The constant value for this ellipse is 10 units.
2
2
2
2
given
substitute
add
simplify radicals
compute square roots
result
Now try Exercises 45 through 48
In Example 6, the sum of the distances
could also be found by moving the point (x, y)
to the location of a vertex (a, 0), then using
the symmetry of the ellipse. The sum is identical to the length of the major axis, since the
overlapping part of the string from (c, 0) to
(a, 0) is the same length as from (Ϫa, 0) to
(Ϫc, 0) (see Figure 8.18). This shows the
constant k is equal to 2a regardless of the distance between foci.
As we noted, the result is
ᮣ
Figure 8.18
y
d1 ϩ d2 ϭ 2a
d1
d2
(Ϫa, 0)
(Ϫc, 0)
21x Ϫ c2 2 ϩ y2 ϩ 21x ϩ c2 2 ϩ y2 ϭ 2a
(c, 0)
(a, 0)
x
These two segments
are equal
substitute 2a for k