B.1 Conic Sections

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Appendix B.1
B
■
Conic Sections
B1
Conic Sections
B.1 Conic Sections
■ Recognize the four basic conics: circles, parabolas, ellipses, and hyperbolas.
■ Recognize, graph, and write equations of parabolas (vertex at origin).
■ Recognize, graph, and write equations of ellipses (center at origin).
■ Recognize, graph, and write equations of hyperbolas (center at origin).
Introduction to Conic Sections
Conic sections were discovered during the classical Greek period, which lasted from
600 to 300 B.C. By the beginning of the Alexandrian period, enough was known of
conics for Apollonius (262–190 B.C.) to produce an eight-volume work on the subject.
This early Greek study was largely concerned with the geometric properties of conics.
It was not until the early seventeenth century that the broad applicability of conics
became apparent.
A conic section (or simply conic) can be described as the intersection of a plane
and a double-napped cone. Notice from Figure B.1 that in the formation of the four
basic conics, the intersecting plane does not pass through the vertex of the cone. When
the plane does pass through the vertex, the resulting figure is a degenerate conic, as
shown in Figure B.2.
Conic Sections
FIGURE B.1
Degenerate Conics
FIGURE B.2
There are several ways to approach the study of conics. You could begin by defining
conics in terms of the intersections of planes and cones, as the Greeks did, or you could
define them algebraically, in terms of the general second-degree equation
Ax 2 ⫹ Bxy ⫹ Cy 2 ⫹ Dx ⫹ Ey ⫹ F ⫽ 0.
However, you will study a third approach in which each of the conics is defined as a
locus, or collection, of points satisfying a certain geometric property. For example, in
Section 2.1 you saw how the definition of a circle as the collection of all points 共x, y兲 that
are equidistant from a fixed point 共h, k兲 led easily to the standard equation of a circle,
共x ⫺ h兲2 ⫹ 共 y ⫺ k兲2 ⫽ r 2.
You will restrict your study of conics in Appendix B.1 to parabolas with vertices at
the origin, and ellipses and hyperbolas with centers at the origin. In Appendix B.2, you
will look at the general cases.
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Conic Sections
Parabolas
In Section 3.1, you determined that the graph of the quadratic function given by
Note that the term parabola
is a technical term used in
mathematics and does not
simply refer to any U-shaped
curve.
f 共x兲 ⫽ ax2 ⫹ bx ⫹ c
is a parabola that opens upward or downward. The definition of a parabola given below
is more general in the sense that it is independent of the orientation of the parabola.
Definition of a Parabola
A parabola is the set of all points
共x, y兲 in a plane that are equidistant
from a fixed line called the directrix
and a fixed point called the focus
(not on the line). The midpoint
between the focus and the directrix
is called the vertex, and the line
passing through the focus and the
vertex is called the axis of the parabola.
y
d2
Focus
(x, y)
d1
d1
Vertex
d2
x
Directrix
Axis
Using this definition, you can derive the following standard form of the equation
of a parabola.
Standard Equation of a Parabola (Vertex at Origin)
The standard form of the equation of a parabola with vertex at 共0, 0兲 and
directrix y ⫽ ⫺p is given by
x2 ⫽ 4py,
p ⫽ 0.
Vertical axis
For directrix x ⫽ ⫺p, the equation is given by
y2 ⫽ 4px,
p ⫽ 0.
Horizontal axis
The focus is on the axis p units (directed distance) from the vertex. See Figure B.3.
y
y
x 2 = 4py, p ≠ 0
Focus (0, p)
Vertex (0, 0)
p
(x, y)
Vertex
(0, 0)
(x, y)
Focus (p, 0)
Axis
x
p
Directrix: y = − p
Axis
Parabola with Vertical Axis
FIGURE B.3
Directrix: x = −p
STUDY TIP
p
p
y 2 = 4px, p ≠ 0
Parabola with Horizontal Axis
x
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Appendix B.1
Example 1
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Conic Sections
B3
Finding the Focus of a Parabola
Find the focus of the parabola whose equation is y ⫽ ⫺2x2.
Because the squared term in the equation involves x, you know that the
axis is vertical, and the equation is of the form
SOLUTION
y
(
Focus 0, − 18
x2 ⫽ 4py.
(
x
−1
1
Standard form, vertical axis
You can write the original equation in this form, as shown.
⫺2x2 ⫽ y
y = −2x2
Write original equation.
1
x2 ⫽ ⫺ y
2
−1
冢 81冣y.
x2 ⫽ 4 ⫺
−2
Divide each side by ⫺2.
Write in standard form.
So, p ⫽ ⫺ 18. Because p is negative, the parabola opens downward and the focus of the
parabola is 共0, p兲 ⫽ 共0, ⫺ 18 兲, as shown in Figure B.4.
FIGURE B.4
Checkpoint 1
Find the focus of the parabola whose equation is y2 ⫽ 2x.
■
y
Example 2
Finding the Standard Equation of a Parabola
2
Write the standard form of the equation of the parabola with vertex at the origin and
focus at 共2, 0兲.
y2 = 8x
1
Vertex
1
−1
(0, 0)
Focus
(2, 0)
2
The axis of the parabola is horizontal, passing through 共0, 0兲 and 共2, 0兲, as
shown in Figure B.5. So, the standard form is
SOLUTION
x
3
4
y2 ⫽ 4px.
Standard form, horizontal axis
Because the focus is p ⫽ 2 units from the vertex, the equation is
−2
FIGURE B.5
y2 ⫽ 4共2兲x
Standard form
y2 ⫽ 8x.
Checkpoint 2
Write the standard form of the equation of the parabola with vertex at the origin and
focus at 共0, 2兲.
■
Parabolas occur in a wide variety of
applications. For instance, a parabolic
reflector can be formed by revolving a
parabola about its axis. The resulting
surface has the property that all incoming
rays parallel to the axis are reflected through
the focus of the parabola. This is the principle
behind the construction of the parabolic
mirrors used in reflecting telescopes.
Conversely, the light rays emanating from
the focus of the parabolic reflector used
in a flashlight are all reflected parallel to
one another, as shown in Figure B.6.
Light source
at focus
Focus
Axis
Parabolic reflector:
Light is reflected
in parallel rays.
FIGURE B.6
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Conic Sections
Ellipses
Another basic type of conic is called an ellipse.
Definition of an Ellipse
An ellipse is the set of all points 共x, y兲 in a plane the sum of whose distances from
two distinct fixed points, called foci, is constant.
(x, y)
d2
d1
Focus
Major axis
Center
Vertex
Focus
Vertex
Minor
axis
d 1 + d 2 is constant.
The line through the foci intersects the ellipse at two points, called the vertices.
The chord joining the vertices is called the major axis, and its midpoint is called the
center of the ellipse. The chord perpendicular to the major axis at the center is called
the minor axis of the ellipse. The endpoints of the minor axis of an ellipse are commonly
referred to as the co-vertices.
You can visualize the definition of an ellipse by imagining two thumbtacks placed
at the foci, as shown in Figure B.7. If the ends of a fixed length of string are fastened
to the thumbtacks and the string is drawn taut with a pencil, the path traced by the pencil
will be an ellipse.
The standard form of the equation of an ellipse takes one of two forms, depending
on whether the major axis is horizontal or vertical.
Standard Equation of an Ellipse (Center at Origin)
FIGURE B.7
The standard form of the equation of an ellipse with the center at the origin and
major and minor axes of lengths 2a and 2b, respectively 共where 0 < b < a兲, is
x2
y2
⫹ 2⫽1
2
a
b
or
y2
x2
⫹ 2 ⫽ 1.
2
b
a
y
x2
a2
+
y2
=1
b2
x2
y2
+ 2 =1
2
b
a
(0, b)
x
(c, 0)
(a, 0)
Vertex
(− b, 0)
(b, 0)
(0, − c)
(0, − b )
Vertex
Major axis is horizontal.
Minor axis is vertical.
(0, a)
(0, c)
Vertex
(− a, 0)
Vertex (− c, 0)
y
x
(0, −a)
Major axis is vertical.
Minor axis is horizontal.
The vertices and foci lie on the major axis, a and c units, respectively, from the
center. Moreover, a, b, and c are related by the equation c2 ⫽ a2 ⫺ b2.
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Appendix B.1
Example 3
y
Conic Sections
B5
Finding the Standard Equation of an Ellipse
Find the standard form of the equation of the ellipse that has a major axis of length 6
and foci at 共⫺2, 0兲 and 共2, 0兲, as shown in Figure B.8.
x2
y2
+ 2 =1
2
a
b
3
■
Because the foci occur at 共⫺2, 0兲 and 共2, 0兲, the center of the ellipse is
共0, 0兲, and the major axis is horizontal. So, the ellipse has an equation of the form
SOLUTION
1
(− 2, 0)
(2, 0)
x
−2
−1
1
2
−1
x2
y2
⫹ 2 ⫽ 1.
2
a
b
Standard form, horizontal major axis
Because the length of the major axis is 6, you have 2a ⫽ 6, which implies that a ⫽ 3.
Moreover, the distance from the center to either focus is c ⫽ 2. Finally, you have
b2 ⫽ a2 ⫺ c2 ⫽ 32 ⫺ 22 ⫽ 9 ⫺ 4 ⫽ 5.
−3
Substituting a2 ⫽ 32 and b 2 ⫽ 共冪5兲 yields the following equation in standard form.
2
FIGURE B.8
x2
y2
⫹
⫽1
32 共冪5兲2
Standard form
This equation simplifies to
x2 y2
⫹ ⫽ 1.
9
5
Checkpoint 3
Find the standard form of the equation of the ellipse that has a major axis of length 8
and foci at 共0, ⫺3兲 and 共0, 3兲.
■
Example 4
Sketching an Ellipse
Sketch the ellipse given by
4x2 ⫹ y2 ⫽ 36
and identify the vertices.
SOLUTION
y
x2
y2
+ 2 =1
32
6
(0, 6)
4
2
(− 3, 0)
(3, 0)
Begin by writing the equation in standard form.
4x2 ⫹ y2 ⫽ 36
Write original equation.
4x2
y2
36
⫹
⫽
36
36 36
Divide each side by 36.
x2
y2
⫹
⫽1
9
36
Simplify.
x2
y2
⫹ 2⫽1
2
3
6
Write in standard form.
x
−6
−4
−2
2
−2
−4
(0, − 6)
FIGURE B.9
4
6
Because the denominator of the y2-term is greater than the denominator of the x2-term,
you can conclude that the major axis is vertical. Also, because a2 ⫽ 62, the endpoints
of the major axis lie six units up and down from the center 共0, 0兲. So, the vertices of the
ellipse are 共0, 6兲 and 共0, ⫺6兲. Similarly, because the denominator of the x2-term is
b2 ⫽ 32, the endpoints of the minor axis (or co-vertices) lie three units to the right and
left of the center at 共3, 0兲 and 共⫺3, 0兲. The ellipse is shown in Figure B.9.
Checkpoint 4
Sketch the ellipse given by x2 ⫹ 4y2 ⫽ 64, and identify the vertices.
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Conic Sections
Hyperbolas
The definition of a hyperbola is similar to that of an ellipse. The distinction is that, for an
ellipse, the sum of the distances between the foci and a point on the ellipse is constant,
whereas for a hyperbola, the difference of the distances between the foci and a point on
the hyperbola is constant.
Definition of a Hyperbola
A hyperbola is the set of all points 共x, y兲 in a plane the difference of whose
distances from two distinct fixed points, called foci, is constant.
(x, y)
d2
d1
Focus
Branch
Branch
Transverse axis
Vertex
Focus
Vertex
Center
a
d 2 − d 1 is constant.
c
The graph of a hyperbola has two disconnected parts, called branches. The line
through the two foci intersects the hyperbola at two points, called vertices. The line
segment connecting the vertices is called the transverse axis, and the midpoint of the
transverse axis is called the center of the hyperbola.
Standard Equation of a Hyperbola (Center at Origin)
The standard form of the equation of a hyperbola with the center at the origin
(where a ⫽ 0 and b ⫽ 0) is
x2
y2
⫺ 2⫽1
2
a
b
or
y2
x2
⫺ 2 ⫽ 1.
2
a
b
x2 y2
−
=1 y
a2 b2
y
Vertex:
(0, a)
Transverse
axis
Vertex:
(− a, 0)
Focus:
(− c, 0)
Vertex:
(a, 0)
Focus:
(0, c)
y2 x2
−
=1
a2 b2
x
Transverse axis
x
Focus:
(c, 0)
Vertex:
(0, − a)
Focus:
(0, −c)
The vertices and foci are, respectively, a and c units from the center. Moreover, a,
b, and c are related by the equation
b2 ⫽ c2 ⫺ a2.
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Appendix B.1
Example 5
3
From the graph, c ⫽ 3 because the foci are three units from the center.
Also, a ⫽ 2 because the vertices are two units from the center. So,
SOLUTION
2
1
−3
−1
(2, 0)
1
−1
−2
−3
FIGURE B.10
Finding the Standard Equation of a Hyperbola
Find the standard form of the equation of the hyperbola with foci at 共⫺3, 0兲 and 共3, 0兲
and vertices at 共⫺2, 0兲 and 共2, 0兲, as shown in Figure B.10.
y
(− 3, 0) (− 2, 0)
B7
Conic Sections
■
(3, 0)
3
b2 ⫽ c2 ⫺ a2 ⫽ 32 ⫺ 22 ⫽ 9 ⫺ 4 ⫽ 5.
x
Because the transverse axis is horizontal, the standard form of the equation is
x2
y2
⫺ 2 ⫽ 1.
2
a
b
Standard form, horizontal transverse axis
Finally, substituting a2 ⫽ 22 and b2 ⫽ 共冪5兲 , you have
2
x2
y2
⫺
⫽ 1.
2
2
共冪5兲2
Write in standard form.
Checkpoint 5
Find the standard form of the equation of the hyperbola with foci at 共0, ⫺4兲 and 共0, 4兲
and vertices at 共0, ⫺3兲 and 共0, 3兲.
■
An important aid in sketching the graph of a hyperbola is the determination of its
asymptotes, as shown in Figure B.11. Each hyperbola has two asymptotes that intersect
at the center of the hyperbola. Furthermore, the asymptotes pass through the corners of
a rectangle of dimensions 2a by 2b. The line segment of length 2b, joining 共0, b兲 and
共0, ⫺b兲 关or 共⫺b, 0兲 and 共b, 0兲兴, is referred to as the conjugate axis of the hyperbola.
y
x2
a2
−
y2
b2
=1
y2
x2
− 2 =1
2
y
a
b
Asymptote:
b
y= a x
(0, b)
(− a, 0)
(0, a)
(a, 0)
Asymptote:
a
y= x
b
x
(− b, 0)
(b, 0)
(0, − b)
Asymptote:
a
y=− x
b
Asymptote:
b
y=− a x
Transverse axis is horizontal.
FIGURE B.11
x
(0, −a)
Transverse axis is vertical.
Asymptotes of a Hyperbola (Center at Origin)
b
y ⫽ x and
a
b
y⫽⫺ x
a
Transverse axis is horizontal.
a
y ⫽ x and
b
a
y⫽⫺ x
b
Transverse axis is vertical.
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Conic Sections
Example 6
Sketching a Hyperbola
Sketch the hyperbola whose equation is 4x2 ⫺ y2 ⫽ 16.
SOLUTION
4x2 ⫺ y2 ⫽ 16
4x2
16
⫺
y2
16
⫽
Write original equation.
16
16
Divide each side by 16.
x2
y2
⫺ 2⫽1
2
2
4
Write in standard form.
Because the x2-term is positive, you can conclude that the transverse axis is horizontal
and the vertices occur at 共⫺2, 0兲 and 共2, 0兲. Moreover, the endpoints of the conjugate
axis occur at 共0, ⫺4兲 and 共0, 4兲, and you can sketch the rectangle shown in Figure B.12.
Finally, by drawing the asymptotes through the corners of this rectangle, you can
complete the sketch shown in Figure B.13.
y
y
6
6
(0, 4)
x2 y2
−
=1
22 42
(−2, 0)
(2, 0)
x
−6
−4
4
6
−6
x
−4
4
6
(0, − 4)
−6
−6
FIGURE B.12
FIGURE B.13
Checkpoint 6
Sketch the hyperbola whose equation is 9y2 ⫺ x2 ⫽ 9.
Example 7
y
2
SOLUTION
(0, 3)
x
−4
y = 2x
−2
2
−2
−4
FIGURE B.14
Finding the Standard Equation of a Hyperbola
Find the standard form of the equation of the hyperbola that has vertices at 共0, ⫺3兲 and
共0, 3兲 and asymptotes y ⫽ ⫺2x and y ⫽ 2x, as shown in Figure B.14.
4
y = −2x
■
(0, − 3)
4
Because the transverse axis is vertical, the asymptotes are of the form
a
y ⫽ x and
b
a
y ⫽ ⫺ x.
b
Transverse axis is vertical.
Using the fact that y ⫽ 2x and y ⫽ ⫺2x, you can determine that a兾b ⫽ 2. Because
a ⫽ 3, you can determine that b ⫽ 32. Finally, you can conclude that the hyperbola has
the equation
y2
x2
⫺
⫽ 1.
32 共3兾2兲2
Write in standard form.
Checkpoint 7
Find the standard form of the equation of the hyperbola that has vertices at 共⫺5, 0兲
and 共5, 0兲 and asymptotes y ⫽ ⫺x and y ⫽ x.
■
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Appendix B.1
■
Conic Sections
B9
The following warm-up exercises involve skills that were covered in earlier sections. You will use
these skills in the exercise set for this section. For additional help, review Sections 0.7, 1.3, and 2.1.
SKILLS WARM UP B.1
In Exercises 1– 4, rewrite the equation so that it has no fractions.
1.
x2
y2
⫹ ⫽1
16
9
2.
x2
4y2 32
⫹
⫽
32
32
32
3.
x2
y2
⫺ ⫽1
1兾4
4
4.
3x2
4y2
⫹
⫽1
1兾9
9
In Exercises 5–8, solve for c. (Assume c > 0.)
5. c2 ⫽ 32 ⫺ 12
6. c2 ⫽ 22 ⫹ 32
7. c2 ⫹ 22 ⫽ 42
8. c2 ⫺ 12 ⫽ 22
In Exercises 9 and 10, find the distance between the point and the origin.
9. 共0, ⫺4兲
10. 共⫺2, 0兲
Exercises B.1
Matching In Exercises 1–8, match the equation with
its graph. [The graphs are labeled (a), (b), (c), (d), (e), (f),
(g), and (h).]
y
(a)
y
(b)
4
6
4
x
2
−4
x
− 4 −2
−2
2
y
(d)
4
x
2
4
−2
x
−4
−2
−2
−6
−4
y
(e)
4
−4
2
−4
2
4
y
(c)
−2
2
4
6
4
2
2
x
−4
−2
2
1
−2
−4
y
(g)
x
−1
4
y
(h)
4
1
x
x
−2
2
−12
−8
−4
−4
2. x2 ⫽ ⫺4y
4. y2 ⫽ ⫺4x
Finding the Vertex and Focus of a Parabola In
Exercises 9–16, find the vertex and focus of the parabola
and sketch its graph. See Example 1.
9.
10.
11.
12.
13.
14.
15.
16.
y
(f)
1. x2 ⫽ 4y
3. y2 ⫽ 4x
x2 y2
5.
⫹ ⫽1
1
4
2
x
y2
6.
⫹ ⫽1
4
1
2
x
y2
7.
⫺ ⫽1
1
4
2
y
x2
8.
⫺ ⫽1
4
1
y ⫽ 4x2
1
y ⫽ 2x2
y2 ⫽ ⫺6x
y2 ⫽ 3x
x2 ⫹ 8y ⫽ 0
x2 ⫹ 12y ⫽ 0
x ⫹ y2 ⫽ 0
y2 ⫺ 8x ⫽ 0
Finding the Standard Equation of a Parabola In
Exercises 17–26, find the standard form of the equation
of the parabola with the given characteristic(s) and vertex
at the origin. See Example 2.
17. Focus: 共0, ⫺ 32 兲
18. Focus: 共0, ⫺2兲
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21.
23.
25.
26.
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Conic Sections
Focus: 共⫺2, 0兲
20. Focus: 共52, 0兲
Directrix: y ⫽ ⫺1
22. Directrix: y ⫽ 2
Directrix: x ⫽ 3
24. Directrix: x ⫽ ⫺2
Passes through the point 共4, 6兲; horizontal axis
Passes through the point 共⫺2, ⫺2兲; vertical axis
Finding the Standard Equation of an Ellipse In
Exercises 27– 34, find the standard form of the equation
of the ellipse with the given characteristics and center at
the origin. See Example 3.
27.
28.
29.
30.
31.
32.
33.
34.
Vertices: 共0, ± 2兲; minor axis of length 2
Vertices: 共± 2, 0兲; minor axis of length 3
Vertices: 共± 5, 0兲; foci: 共± 2, 0兲
Vertices: 共0, ± 10兲; foci: 共0, ± 4兲
Foci: 共± 5, 0兲; major axis of length 12
Foci: 共± 2, 0兲; major axis of length 8
Vertices: 共0, ± 5兲; passes through the point 共4, 2兲
Major axis vertical; passes through the points 共0, 4兲 and
共2, 0兲
Sketching an Ellipse In Exercises 35–42, find the
center and vertices of the ellipse and sketch its graph.
See Example 4.
x2
y2
⫹
⫽1
25 16
x2 y2
37. 25 ⫹ 16 ⫽ 1
35.
9
9
x2 y2
39.
⫹ ⫽1
9
5
2
41. 5x ⫹ 3y2 ⫽ 15
x2
y2
⫹
⫽1
144 169
x2 y2
38.
⫹ 1 ⫽1
4
4
x2
y2
40.
⫹
⫽1
28 64
42. x2 ⫹ 4y2 ⫽ 4
Vertices:
Vertices:
Vertices:
Vertices:
共0, ± 2兲; foci:
共0, ± 5兲; foci:
共± 6, 0兲; foci:
共± 3, 0兲; foci:
共0, ± 4兲
共0, ± 8兲
共± 9, 0兲
共± 5, 0兲
Sketching a Hyperbola In Exercises 47–54, find the
center and vertices of the hyperbola and sketch its
graph. See Example 6.
47. x2 ⫺ y2 ⫽ 1
y2 x2
⫺ ⫽1
1
4
2
y
x2
51.
⫺
⫽1
25 144
49.
54. 3y2 ⫺ 5x2 ⫽ 15
Finding the Standard Equation of a Hyperbola In
Exercises 55– 60, find the standard form of the equation
of the hyperbola with the given characteristics and
center at the origin. See Example 7.
55.
56.
57.
58.
59.
60.
Vertices: 共± 1, 0兲; asymptotes: y ⫽ ± 3x
Vertices: 共0, ± 3兲; asymptotes: y ⫽ ± 3x
1
Foci: 共0, ± 4兲; asymptotes: y ⫽ ± 2x
3
Foci: 共± 10, 0兲; asymptotes: y ⫽ ± 4 x
Vertices: 共0, ± 3兲; passes through the point 共⫺2, 5兲
Vertices: 共± 2, 0兲; passes through the point 共3, 冪3兲
61. Satellite Antenna The receiver in a parabolic
television dish antenna is 3 feet from the vertex and is
located at the focus (see figure). Write an equation for a
cross section of the reflector. (Assume that the dish is
directed upward and the vertex is at the origin.)
y
Receiver
3 ft
x
36.
Finding the Standard Equation of a Hyperbola In
Exercises 43–46, find the standard form of the equation
of the hyperbola with the given characteristics and center
at the origin. See Example 5.
43.
44.
45.
46.
53. 2x2 ⫺ 3y2 ⫽ 6
x2
y2
⫺
⫽1
9
16
y2 x2
50.
⫺ ⫽1
9
1
2
y2
x
52.
⫺ ⫽1
36
4
62. Suspension Bridge Each cable of the Golden Gate
Bridge is suspended (in the shape of a parabola) between
two towers that are 1280 meters apart. The top of each
tower is 152 meters above the roadway (see figure). The
cables touch the roadway at the midpoint between
the towers. Write an equation for the parabolic shape of
each cable.
y
(− 640, 152)
(640, 152)
Cable
x
Roadway
63. Architecture A fireplace arch is to be constructed in
the shape of a semiellipse. The opening is to have a
height of 2 feet at the center and a width of 5 feet along
the base (see figure). The contractor draws the outline
of the ellipse by the method shown in Figure B.7.
Where should the tacks be placed and what should be
the length of the piece of string?
48.
y
3
1
−3 −2 −1
x
1
2
3
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Appendix B.1
64. Mountain Tunnel A semielliptical arch over a tunnel
for a road through a mountain has a major axis of 100 feet,
and its height at the center is 30 feet (see figure).
Determine the height of the arch 5 feet from the edge of
the tunnel.
y
x
45 ft
100 ft
65. Sketching an Ellipse Sketch a graph of the ellipse
that consists of all points 共x, y兲 such that the sum of the
distances between 共x, y兲 and two fixed points is 15 units
and the foci are located at the centers of the two sets of
concentric circles, as shown in the figure.
B11
Conic Sections
Using Latera Recta In Exercises 67–70, sketch the
ellipse using the latera recta (see Exercise 66).
x2 y2
⫹ ⫽1
4
1
69. 9x2 ⫹ 4y2 ⫽ 36
70. 5x2 ⫹ 3y2 ⫽ 15
67.
71. Think About It
30 ft
■
68.
x2
y2
⫹
⫽1
9
16
Consider the ellipse
x2
y2
⫹
⫽ 1.
328 327
Is this ellipse better described as elongated or nearly
circular? Explain your reasoning.
72. Navigation Long-range navigation for aircraft and
ships is accomplished by synchronized pulses transmitted
by widely separated transmitting stations. These pulses
travel at the speed of light (186,000 miles per second).
The difference in the times of arrival of these pulses at
an aircraft or ship is constant on a hyperbola having the
transmitting stations as foci. Assume that two stations
300 miles apart are positioned on a rectangular coordinate
system at points with coordinates 共⫺150, 0兲 and 共150, 0兲
and that a ship is traveling on a path with coordinates
共x, 75兲 (see figure). Find the x-coordinate of the position
of the ship if the time difference between the pulses
from the transmitting stations is 1000 microseconds
(0.001 second).
y
150
66. Think About It A line segment through a focus of an
ellipse with endpoints on the ellipse and perpendicular
to its major axis is called a latus rectum of the ellipse.
An ellipse has two latera recta. Knowing the length of
the latera recta is helpful in sketching an ellipse because
this information yields other points on the curve (see
figure). Show that the length of each latus rectum is
2b2
.
a
y
Latera recta
x
−150
75
150
73. Hyperbolic Mirror A hyperbolic mirror (used in some
telescopes) has the property that a light ray directed at
one focus will be reflected to the other focus (see figure).
The focus of the hyperbolic mirror has coordinates
共12, 0兲. Find the vertex of the mirror if its mount at the
top edge of the mirror has coordinates 共12, 12兲.
y
(12, 12)
x
F1
F2
x
(−12, 0)
(12, 0)
74. Writing Explain how the central rectangle of a
hyperbola can be used to sketch its asymptotes.
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Appendix B
■
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Page B12
Conic Sections
B.2 Conic Sections and Translations
■ Recognize equations of conics that have been shifted vertically or horizontally
in the plane.
■ Write and graph equations of conics that have been shifted vertically or
horizontally in the plane.
Vertical and Horizontal Shifts of Conics
In Appendix B.1, you studied conic sections whose graphs were in standard position.
In this section, you will study the equations of conic sections that have been shifted
vertically or horizontally in the plane. The following summary lists the standard forms
of the equations of the four basic conics.
Standard Forms of Equations of Conics
y
Circle: Center ⫽ 共h, k兲; Radius ⫽ r; See Figure B.15.
(x − h) 2 + (y − k) 2 = r 2
Parabola: Vertex ⫽ 共h, k兲; Directed distance from vertex to focus ⫽ p
y
y
(h, k)
p>0
(x − h) 2 = 4 p (y − k)
r
(y − k) 2 = 4p (x − h)
x
Vertex:
(h, k)
p>0
Vertex:
(h, k)
FIGURE B.15
Focus:
(h, k + p)
Focus:
(h + p, k)
x
x
Ellipse: Center ⫽ 共h, k兲; Major axis length ⫽ 2a; Minor axis length ⫽ 2b
y
2
y (x − h)
(x − h) 2 (y − k) 2
+
=1
a2
b2
b2
(h, k)
+
2b
(y − k) 2
=1
a2
(h, k)
2a
2a
x
2b
x
Hyperbola: Center ⫽ 共h, k兲; Transverse axis length ⫽ 2a;
Conjugate axis length ⫽ 2b
y
(x − h) 2 (y − k) 2
−
=1
a2
b2
y
(y − k) 2 (x − h) 2
−
=1
a2
b2
(h, k)
(h, k)
2a
2b
2a
x
2b
x
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Appendix B.2
Example 1
B13
Conic Sections and Translations
■
Equations of Conic Sections
Describe the translation of the graph of each conic.
a. 共x ⫺ 1兲2 ⫹ 共 y ⫹ 2兲2 ⫽ 32
c.
y
2
x
−1
−2
−3
共x ⫺ 3兲2 共 y ⫺ 2兲2
⫺
⫽1
12
32
1
2
3
3
4
5
(1, − 2)
−5
−6
a. The graph of 共x ⫺ 1兲2 ⫹ 共 y ⫹ 2兲2 ⫽ 32 is a circle whose center is the point 共1, ⫺2兲
and whose radius is 3, as shown in Figure B.16. The graph of the circle has been
shifted one unit to the right and two units downward from standard position.
c. The graph of
FIGURE B.16
共x ⫺ 3兲2 共 y ⫺ 2兲2
⫺
⫽1
12
32
(x − 2) 2 = 4(−1)(y − 3)
4
(2, 3)
3
p = −1
2
(2, 2)
is a hyperbola whose center is the point 共3, 2兲. The transverse axis is horizontal with
a length of 2共1兲 ⫽ 2. The conjugate axis is vertical with a length of 2共3兲 ⫽ 6, as
shown in Figure B.18. The graph of the hyperbola has been shifted three units to the
right and two units upward from standard position.
d. The graph of
1
−1
共x ⫺ 2兲2 共 y ⫺ 1兲2
⫹
⫽1
32
22
b. The graph of 共x ⫺ 2兲2 ⫽ 4共⫺1兲共 y ⫺ 3兲 is a parabola whose vertex is the point
共2, 3兲. The axis of the parabola is vertical. Moreover, because p ⫽ ⫺1, it follows
that the focus lies below the vertex, as shown in Figure B.17. The graph of the
parabola has been reflected in the x-axis, and shifted two units to the right and three
units upward from standard position.
−4
y
d.
SOLUTION
(x − 1) 2 + (y + 2) 2 = 3 2
1
− 3 −2
b. 共x ⫺ 2兲2 ⫽ 4共⫺1兲共 y ⫺ 3兲
共x ⫺ 2兲2 共 y ⫺ 1兲2
⫹
⫽1
32
22
x
1
−1
−2
FIGURE B.17
3
4
5
is an ellipse whose center is the point 共2, 1). The major axis of the ellipse is horizontal
with a length of 2共3兲 ⫽ 6. The minor axis of the ellipse is vertical with a length of
2共2兲 ⫽ 4, as shown in Figure B.19. The graph of the ellipse has been shifted two
units to the right and one unit upward from standard position.
y
(x − 3) 2 (y − 2 ) 2
=1
−
32
12
y
4
8
6
3
(x − 2) 2 (y − 1) 2
=1
+
22
32
1
2
4
(2, 1)
3
(3, 2)
2
x
x
6
8
10
−1
1
3
5
−1
−2
FIGURE B.18
FIGURE B.19
Checkpoint 1
Describe the translation of the graph of
共x ⫹ 4兲2 ⫹ 共 y ⫺ 3兲2 ⫽ 52.
■
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Conic Sections
■
Writing Equations of Conics in Standard Form
Example 2
Finding the Standard Form of a Parabola
Find the vertex and focus of the parabola given by x2 ⫺ 2x ⫹ 4y ⫺ 3 ⫽ 0.
SOLUTION
y
Complete the square to write the equation in standard form.
x2 ⫺ 2x ⫹ 4y ⫺ 3 ⫽ 0
2
x ⫺ 2x ⫽ ⫺4y ⫹ 3
(1, 1)
1
1
−1
2
3
4
(1, 0)
−2
(x − 1) 2 = 4(−1)(y − 1)
−3
Group terms.
x2 ⫺ 2x ⫹ 1 ⫽ ⫺4y ⫹ 3 ⫹ 1
x
−2
Write original equation.
2
Complete the square.
共x ⫺ 1兲2 ⫽ ⫺4y ⫹ 4
Write in completed
square form.
共x ⫺ 1兲2 ⫽ 4共⫺1兲共 y ⫺ 1兲
Standard form,
共x ⫺ h兲2 ⫽ 4p共 y ⫺ k兲
From this standard form, it follows that h ⫽ 1, k ⫽ 1, and p ⫽ ⫺1. Because the axis is
vertical and p is negative, the parabola opens downward. The vertex is 共h, k兲 ⫽ 共1, 1兲
and the focus is 共h, k ⫹ p兲 ⫽ 共1, 0兲. (See Figure B.20.)
−4
FIGURE B.20
Checkpoint 2
Find the vertex and focus of the parabola given by x2 ⫹ 2x ⫺ 4y ⫹ 5 ⫽ 0.
■
In Examples 1(b) and 2, p is the directed distance from the vertex to the focus.
Because the axis of the parabola is vertical and p ⫽ ⫺1, the focus is one unit below the
vertex, and the parabola opens downward.
Example 3
Sketching an Ellipse
Sketch the ellipse given by x2 ⫹ 4y2 ⫹ 6x ⫺ 8y ⫹ 9 ⫽ 0.
SOLUTION
Complete the square to write the equation in standard form.
x2 ⫹ 4y2 ⫹ 6x ⫺ 8y ⫹ 9 ⫽ 0
共x2 ⫹ 6x ⫹ 䊏兲 ⫹ 共4y2 ⫺ 8y ⫹ 䊏兲 ⫽ ⫺9
共x2 ⫹ 6x ⫹ 䊏兲 ⫹ 4共 y2 ⫺ 2y ⫹ 䊏兲 ⫽ ⫺9
共x2 ⫹ 6x ⫹ 9兲 ⫹ 4共 y2 ⫺ 2y ⫹ 1兲 ⫽ ⫺9 ⫹ 9 ⫹ 4共1兲
y
3) 2
(x +
22
+
(y −
12
1)2
=1
共x ⫹ 3兲 ⫹ 4共 y ⫺ 1兲 ⫽ 4
2
4
共x ⫹ 3兲2 共 y ⫺ 1兲2
⫹
⫽1
4
1
共x ⫹ 3兲2 共 y ⫺ 1兲2
⫹
⫽1
22
12
3
(− 5, 1)
(− 3, 2)
(− 1, 1) 2
(−3, 1)
1
(− 3, 0)
x
−5
−4
−3
−2
−1
−1
FIGURE B.21
2
Write original equation.
Group terms.
Factor 4 out of y-terms.
Complete the squares.
Write in completed
square form.
Divide each
side by 4.
共x ⫺ h兲2 共 y ⫺ k兲2
⫹
⫽1
a2
b2
From this standard form, it follows that the center is 共h, k兲 ⫽ 共⫺3, 1兲. Because the
denominator of the x-term is a2 ⫽ 22, the endpoints of the major axis lie two units to
the right and left of the center. Similarly, because the denominator of the y-term is
b2 ⫽ 12, the endpoints of the minor axis lie one unit up and down from the center. The
ellipse is shown in Figure B.21.
Checkpoint 3
Sketch the ellipse given by x2 ⫹ 9y2 ⫺ 4x ⫹ 18y ⫹ 4 ⫽ 0.
■
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Appendix B.2
Example 4
B15
Conic Sections and Translations
■
Sketching a Hyperbola
Sketch the hyperbola given by
y2 ⫺ 4x2 ⫹ 4y ⫹ 24x ⫺ 41 ⫽ 0.
SOLUTION
Complete the square to write the equation in standard form.
y2 ⫺ 4x2 ⫹ 4y ⫹ 24x ⫺ 41 ⫽ 0
Write original equation.
共 y2 ⫹ 4y ⫹ 䊏兲 ⫺ 共4x 2 ⫺ 24x ⫹ 䊏兲 ⫽ 41
共y2 ⫹ 4y ⫹ 䊏兲 ⫺ 4共x 2 ⫺ 6x ⫹ 䊏兲 ⫽ 41
共 y2 ⫹ 4y ⫹ 4兲 ⫺ 4共x2 ⫺ 6x ⫹ 9兲 ⫽ 41 ⫹ 4 ⫺ 4共9兲
2
(3, 1)
x
−2
2
4
6
−2
(3, − 2)
−4
−6
FIGURE B.22
(3, − 5)
Factor 4 out of x-terms.
Complete the squares.
共 y ⫹ 2兲 ⫺ 4共x ⫺ 3兲 ⫽ 9
Write in completed
square form.
共 y ⫹ 2兲2 4共x ⫺ 3兲2
⫺
⫽1
9
9
Divide each side by 9.
2
(y + 2 ) 2
(x − 3 ) 2
−
=1
2
3
(3/2) 2
y
Group terms.
2
共 y ⫹ 2兲2 共x ⫺ 3兲2
⫺
⫽1
9
9兾4
Change 4 to
共 y ⫹ 2兲2 共x ⫺ 3兲2
⫺
⫽1
32
共3兾2兲2
共 y ⫺ k兲2 共x ⫺ h兲2
⫺
⫽1
a2
b2
1
.
1兾4
From the standard form, it follows that the transverse axis is vertical and the center lies
at 共h, k兲 ⫽ 共3, ⫺2兲. Because the denominator of the y-term is a2 ⫽ 32, you know that
the vertices lie three units above and below the center.
Vertices: 共3, 1兲 and
共3, ⫺5兲
To sketch the hyperbola, draw a rectangle whose top and bottom pass through the vertices.
2
Because the denominator of the x-term is b2 ⫽ 共32 兲 , locate the sides of the rectangle
3
2 units to the right and left of the center, as shown in Figure B.22. Finally, sketch the
asymptotes by drawing lines through the opposite corners of the rectangle. Using these
asymptotes, you can complete the graph of the hyperbola, as shown in Figure B.22.
Checkpoint 4
Sketch the hyperbola given by
4x2 ⫺ y2 ⫹ 16x ⫺ 2y ⫺ 1 ⫽ 0.
■
To find the foci in Example 4, first find c. Recall from Appendix B.1 that
b2 ⫽ c2 ⫺ a2. So,
c2 ⫽ a2 ⫹ b2
c2 ⫽ 9 ⫹
c2 ⫽
c⫽
9
4
45
4
3冪5
.
2
Because the transverse axis is vertical, the foci lie c units above and below the center.
Foci:
冢3, ⫺2 ⫹ 3 2 5 冣
冪
and
冢3, ⫺2 ⫺ 3 2 5 冣
冪
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Page B16
Conic Sections
Example 5
Write the standard form of the equation of the ellipse whose vertices are 共2, ⫺2兲 and
共2, 4兲. The length of the minor axis of the ellipse is 4, as shown in Figure B.23.
y
4
(2, 4)
SOLUTION
3
4
共h, k兲 ⫽
1
x
1
2
3
4
5
−1
−2
The center of the ellipse lies at the midpoint of its vertices. So, the
center is
2
−1
Writing the Equation of an Ellipse
(2, − 2)
FIGURE B.23
冢2 ⫹2 2, 4 ⫹ 2共⫺2兲冣
⫽ 共2, 1兲.
Center
Because the vertices lie on a vertical line and are six units apart, it follows that the major
axis is vertical and has a length of 2a ⫽ 6. So, a ⫽ 3. Moreover, because the minor axis
has a length of 4, it follows that 2b ⫽ 4, which implies that b ⫽ 2. So, you can conclude
that the standard form of the equation of the ellipse is
共x ⫺ h兲2 共 y ⫺ k兲2
⫹
⫽1
b2
a2
Major axis is vertical.
共x ⫺ 2兲2 共 y ⫺ 1兲2
⫹
⫽ 1.
22
32
Write in standard form.
Checkpoint 5
Write the standard form of the equation of the ellipse whose vertices are 共⫺1, 4兲 and
共4, 4兲. The length of the minor axis of the ellipse is 3.
■
Hyperbolic orbit
Vertex
Elliptical orbit
p
Sun (Focus)
Parabolic orbit
FIGURE B.24
An interesting application of conic sections involves the orbits of comets in our
solar system. Of the 610 comets identified prior to 1970, 245 have elliptical orbits, 295
have parabolic orbits, and 70 have hyperbolic orbits. For example, Halley’s comet has
an elliptical orbit, and reappearance of this comet can be predicted every 76 years. The
center of the sun is a focus of each of these orbits, and each orbit has a vertex at the
point where the comet is closest to the sun, as shown in Figure B.24.
If p is the distance between the vertex and the focus (in meters), and v is the speed
of the comet at the vertex (in meters per second), then the type of orbit is determined
as follows.
冪2GM
p
2GM
2. Parabola: v ⫽ 冪
p
2GM
3. Hyperbola: v > 冪
p
1. Ellipse: v <
In these expressions, M ⫽ 1.989 ⫻ 1030 kilograms (the mass of the sun) and G ⬇ 6.67 ⫻ 10⫺11
cubic meter per kilogram-second squared (the universal gravitational constant).
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Appendix B.2
■
Conic Sections and Translations
B17
The following warm-up exercises involve skills that were covered in earlier sections. You will use
these skills in the exercise set for this section. For additional help, review Section B.1.
SKILLS WARM UP B.2
In Exercises 1–10, identify the conic represented by the equation.
1.
x2 y2
⫺ ⫽1
4
4
x2 y2
⫹ ⫽1
9
1
2
x
y2
4.
⫹
⫽1
4
16
2.
3. 2x ⫹ y2 ⫽ 0
y2 x2
⫺ ⫽1
4
2
5. 4x2 ⫹ 4y2 ⫽ 25
6.
7. x2 ⫺ 6y ⫽ 0
8. 3x ⫺ y2 ⫽ 0
2
9.
y2
9x
⫹ ⫽1
16
4
10. 3y2 ⫺ 3x2 ⫽ 48
Exercises B.2
Equations of Conic Sections In Exercises 1– 6,
describe the translation of the graph of the conic from
the standard position. See Example 1.
y
1.
y
2.
5
8
4
3
2
−4
1
x
−4
(x +
+ (y −
1) 2
y
4.
4
3
2
1
x
−2
−2
2
4
6
x
−4
1 2 3 4
−6
6
−3
−4
(x − 2 ) 2
(y + 3) 2
− (x − 1) 2 = 1
4
9
y
5.
12
(y − 1) 2 = 4 (2)(x + 2)
=4
y
3.
−4
8
−8
1
−1
2) 2
x
4
+
(y + 1 ) 2
=1
4
y
6.
2
2
x
−1
−2
−3
1 2 3
5
x
−6
−2
−2
2
−4
−6
−8
−6
(x − 1 ) 2
9
+
(y + 2 ) 2
16
=1
(x + 2 ) 2
4
−
(y + 3 ) 2
9
=1
Finding the Standard Form of a Parabola In
Exercises 7–16, find the vertex, focus, and directrix of the
parabola. Then sketch its graph. See Example 2.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
共x ⫺ 1兲2 ⫹ 8共 y ⫹ 2兲 ⫽ 0
共x ⫹ 3兲 ⫹ 共 y ⫺ 2兲2 ⫽ 0
共y ⫹ 12 兲2 ⫽ 2共x ⫺ 5兲
共x ⫹ 12 兲2 ⫽ 4共 y ⫺ 3兲
y ⫽ 14 共x2 ⫺ 2x ⫹ 5兲
y ⫽ ⫺ 16 共x2 ⫹ 4x ⫺ 2兲
4x ⫺ y2 ⫺ 2y ⫺ 33 ⫽ 0
y2 ⫹ x ⫹ y ⫽ 0
y2 ⫹ 6y ⫹ 8x ⫹ 25 ⫽ 0
x2 ⫺ 2x ⫹ 8y ⫹ 9 ⫽ 0
Finding the Standard Form of a Parabola In
Exercises 17–24, find the standard form of the equation
of the parabola with the given characteristics.
17. Vertex: 共3, 2兲; focus: 共1, 2兲
18. Vertex: 共⫺1, 2兲; focus: 共⫺1, 0兲
19. Vertex: 共0, 4兲;
directrix: y ⫽ 2
20. Vertex: 共⫺2, 1兲;
directrix: x ⫽ 1
21. Focus: 共2, 2兲;
directrix: x ⫽ ⫺2
22. Focus: 共0, 0兲;
directrix: y ⫽ 4
23. Vertex: 共0, 4兲;
passes through 共⫺2, 0兲 and 共2, 0兲
24. Vertex: 共2, 4兲;
passes through 共0, 0兲 and 共4, 0兲
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Conic Sections
Sketching an Ellipse In Exercises 25–32, find the
center, foci, and vertices of the ellipse. Then sketch its
graph. See Example 3.
25.
26.
27.
28.
29.
30.
31.
32.
共x ⫺ 1兲2 共 y ⫺ 5兲2
⫹
⫽1
9
25
共 y ⫹ 4兲2
共x ⫹ 2兲2 ⫹
⫽1
1兾4
9x2 ⫹ 4y2 ⫹ 36x ⫺ 24y ⫹ 36 ⫽ 0
9x2 ⫹ 4y2 ⫺ 36x ⫹ 8y ⫹ 31 ⫽ 0
16x2 ⫹ 25y2 ⫺ 32x ⫹ 50y ⫹ 16 ⫽ 0
9x2 ⫹ 25y2 ⫺ 36x ⫺ 50y ⫹ 61 ⫽ 0
12x2 ⫹ 20y2 ⫺ 12x ⫹ 40y ⫺ 37 ⫽ 0
36x2 ⫹ 9y2 ⫹ 48x ⫺ 36y ⫹ 43 ⫽ 0
Sketching a Hyperbola In Exercises 33–42, find the
center, vertices, and foci of the hyperbola. Then sketch
its graph, using asymptotes as sketching aids. See
Example 4.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
共x ⫺ 1兲2 共 y ⫹ 2兲2
⫺
⫽1
4
1
共x ⫹ 1兲2 共 y ⫺ 4兲2
⫺
⫽1
144
25
共 y ⫹ 6兲2 ⫺ 共x ⫺ 2兲2 ⫽ 1
共 y ⫺ 1兲2 共x ⫹ 3兲2
⫺
⫽1
1兾4
1兾9
9x2 ⫺ y2 ⫺ 36x ⫺ 6y ⫹ 18 ⫽ 0
x2 ⫺ 9y2 ⫹ 36y ⫺ 72 ⫽ 0
9y2 ⫺ x2 ⫹ 2x ⫹ 54y ⫹ 62 ⫽ 0
16y2 ⫺ x2 ⫹ 2x ⫹ 64y ⫹ 63 ⫽ 0
x2 ⫺ 9y2 ⫹ 2x ⫺ 54y ⫺ 107 ⫽ 0
9x2 ⫺ y2 ⫹ 54x ⫹ 10y ⫹ 55 ⫽ 0
Finding the Standard Equation of an Ellipse In
Exercises 43–52, find the standard form of the equation of
the ellipse with the given characteristics. See Example 5.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
Vertices: 共0, 2兲, 共4, 2兲; minor axis of length 2
Vertices: 共3, 1兲, 共3, 9兲; minor axis of length 6
Foci: 共0, 0兲, 共4, 0兲; major axis of length 8
Foci: 共0, 0兲, 共0, 8兲; major axis of length 16
Center: 共2, ⫺1兲; vertex: 共2, 12 兲;
minor axis of length 2
Center: 共4, 0兲; vertex: 共4, ⫺4兲; minor axis of length 6
Center: 共3, 2兲; a ⫽ 3c; foci: 共1, 2兲, 共5, 2兲
Center: 共0, 4兲; a ⫽ 2c; vertices 共⫺4, 4兲, 共4, 4兲
Vertices: 共5, 0兲, 共5, 12兲; endpoints of minor axis:
共0, 6兲, 共10, 6兲
Vertices: 共⫺12, 0兲, 共0, 0兲; endpoints of minor axis:
共⫺6, 3兲, 共⫺6, ⫺3兲
Finding the Standard Equation of a Hyperbola In
Exercises 53–60, find the standard form of the equation
of the hyperbola with the given characteristics.
Vertices: 共2, 0兲, 共6, 0兲; foci: 共0, 0兲, 共8, 0兲
Vertices: 共2, 3兲, 共2, ⫺3兲; foci: 共2, 5兲, 共2, ⫺5兲
Vertices: 共4, 1兲, 共4, 9兲; foci: 共4, 0兲, 共4, 10兲
Vertices: 共⫺2, 1兲, 共2, 1兲; foci: 共⫺3, 1兲, 共3, 1兲
Vertices: 共2, 3兲, 共2, ⫺3兲;
passes through 共0, 5兲
58. Vertices: 共⫺2, 1兲, 共2, 1兲;
passes through 共4, 3兲
59. Vertices: 共0, 2兲, 共6, 2兲;
asymptotes: y ⫽ 23x, y ⫽ 4 ⫺ 23x
60. Vertices: 共3, 0兲, 共3, 4兲;
asymptotes: y ⫽ 23x, y ⫽ 4 ⫺ 23x
53.
54.
55.
56.
57.
Identifying Conics In Exercises 61– 68, identify the
conic by writing the equation in standard form. Then
sketch its graph.
61.
62.
63.
64.
65.
66.
67.
68.
x2 ⫹ y2 ⫺ 6x ⫹ 4y ⫹ 9 ⫽ 0
x2 ⫹ 4y2 ⫺ 6x ⫹ 16y ⫹ 21 ⫽ 0
4x2 ⫺ y2 ⫺ 4x ⫺ 3 ⫽ 0
y2 ⫺ 4y ⫺ 4x ⫽ 0
4x2 ⫹ 3y2 ⫹ 8x ⫺ 24y ⫹ 51 ⫽ 0
4y2 ⫺ 2x2 ⫺ 4y ⫹ 8x ⫺ 15 ⫽ 0
25x2 ⫺ 10x ⫺ 200y ⫺ 119 ⫽ 0
4x2 ⫹ 4y2 ⫺ 16y ⫹ 15 ⫽ 0
69. Satellite Orbit A satellite in a 100-mile-high circular
orbit around Earth has a velocity of approximately
17,500 miles per hour. When this velocity is multiplied
by 冪2, the satellite has the minimum velocity necessary
to escape Earth’s gravity, and it follows a parabolic path
with the center of Earth as the focus (see figure).
y
Circular
orbit
4100
Parabolic
path
x
Not drawn to scale
(a) Find the escape velocity of the satellite.
(b) Find an equation of its path (assume the radius of
Earth is 4000 miles).
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Appendix B.2
70. Fluid Flow Water is flowing from a horizontal pipe
48 feet above the ground. The falling stream of water
has the shape of a parabola whose vertex 共0, 48兲 is at the
end of the pipe (see figure). The stream of water strikes
the ground at the point 共10冪3, 0兲. Find the equation of
the path taken by the water.
y
40
30
48 ft
20
10
x
10 20 30 40
Eccentricity of an Ellipse In Exercises 71–78, the
flatness of an ellipse is measured by its eccentricity e,
defined by
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Conic Sections and Translations
B19
74. Planetary Motion The dwarf planet Pluto moves in an
elliptical orbit with the sun at one of the foci (see figure).
The length of half of the major axis is 3.670 ⫻ 109
miles and the eccentricity is 0.249. Find the shortest
distance and the greatest distance between Pluto and the
sun.
75. Planetary Motion Saturn moves in an elliptical
orbit with the sun at one of the foci (see figure). The
shortest distance and the greatest distance between
Saturn and the sun are 1.3495 ⫻ 109 kilometers
and 1.5040 ⫻ 109 kilometers, respectively. Find the
eccentricity of the orbit.
76. Satellite Orbit The first artificial satellite to orbit
Earth was Sputnik I (launched by the former Soviet
Union in 1957). Its highest point above Earth’s surface
was 588 miles, and its lowest point was 142 miles (see
figure). Assume that the center of Earth is one of the
foci of the elliptical orbit and that the radius of Earth is
4000 miles. Find the eccentricity of the orbit.
c
e ⫽ , where 0 < e < 1.
a
When an ellipse is nearly circular, e is close to 0. When an
ellipse is elongated, e is close to 1 (see figures).
y
y
Focus
c
c
x
c
e = a : close to 0
x
c
e = a : close to 1
a
142 miles
a
71. Find an equation of the ellipse with vertices 共± 5, 0兲 and
eccentricity e ⫽ 35.
72. Find an equation of the ellipse with vertices 共0, ± 8兲 and
eccentricity e ⫽ 12.
73. Planetary Motion Earth moves in an elliptical orbit
with the sun at one of the foci (see figure). The length
of half of the major axis is a ⫽ 9.2956 ⫻ 107 miles
and the eccentricity is 0.017. Find the shortest distance
(perihelion) and the greatest distance (aphelion) between
Earth and the sun.
y
x
Sun
a
Not drawn to scale
Figure for 73–75
Not drawn to scale
588 miles
77. Alternate Form of Equation of an Ellipse Show
that the equation of an ellipse can be written as
共x ⫺ h兲2
共 y ⫺ k兲2
⫹ 2
⫽ 1.
2
a
a 共1 ⫺ e2兲
78. Comet Orbit Halley’s comet has an elliptical orbit
with the sun at one focus. The eccentricity of the orbit
is approximately 0.97. The length of the major axis of
the orbit is approximately 35.88 astronomical units. (An
astronomical unit is about 93 million miles.) Find the
standard form of the equation of the orbit. Place the
center of the orbit at the origin and place the major axis
on the x-axis.
79. Australian Football In Australia, football by
Australian Rules (or rugby) is played on elliptical fields.
The fields can be a maximum of 170 yards wide and a
maximum of 200 yards long. Let the center of a field of
maximum size be represented by the point 共0, 85兲. Find
the standard form of the equation of the ellipse that
represents this field. (Source: Australian Football
League)