6965_CH08_pp579-640.qxd
1/14/10
2:01 PM
Page 629
SECTION 8.6
Three-Dimensional Cartesian Coordinate System
629
8.6 Three-Dimensional Cartesian
Coordinate System
What you’ll learn about
• Three-Dimensional Cartesian
Coordinates
• Distance and Midpoint Formulas
• Equation of a Sphere
• Planes and Other Surfaces
• Vectors in Space
Three-Dimensional Cartesian Coordinates
In Sections P.2 and P.4, we studied Cartesian coordinates and the associated basic
formulas and equations for the two-dimensional plane; we now extend these ideas to
three-dimensional space. In the plane, we used two axes and ordered pairs to name
points; in space, we use three mutually perpendicular axes and ordered triples of real
numbers to name points (Figure 8.46).
• Lines in Space
z
... and why
This is the analytic geometry of
our physical world.
z ⫽ constant
(0, 0, z)
(0, y, z)
(x, 0, z)
P(x, y, z)
(0, y, 0)
y
(x, 0, 0)
x
x ⫽ constant
y ⫽ constant
(x, y, 0)
FIGURE 8.46 The point P1x, y, z2 in Cartesian space.
Notice that Figure 8.46 exhibits several important features of the three-dimensional
Cartesian coordinate system:
z
y⫽0
x⫽0
• The axes are labeled x, y, and z, and these three coordinate axes form a righthanded coordinate frame: When you hold your right hand with fingers curving
from the positive x-axis toward the positive y-axis, your thumb points in the direction of the positive z-axis.
• A point P in space uniquely corresponds to an ordered triple 1x, y, z2 of real numbers. The numbers x, y, and z are the Cartesian coordinates of P.
• Points on the axes have the form 1x, 0, 02, 10, y, 02, or 10, 0, z2, with 1x, 0, 02 on
the x-axis, 10, y, 02 on the y-axis, and 10, 0, z2 on the z-axis.
In Figure 8.47, the axes are paired to determine the coordinate planes:
z⫽0
Origin
(0, 0, 0)
x
FIGURE 8.47 The coordinate planes
divide space into eight octants.
y
• The coordinate planes are the xy-plane, the xz-plane, and the yz-plane, and have
equations z = 0, y = 0, and x = 0, respectively.
• Points on the coordinate planes have the form 1x, y, 02, 1x, 0, z2, or 10, y, z2, with
1x, y, 02 on the xy-plane, 1x, 0, z2 on the xz-plane, and 10, y, z2 on the yz-plane.
• The coordinate planes meet at the origin, (0, 0, 0).
• The coordinate planes divide space into eight regions called octants, with the first
octant containing all points in space with three positive coordinates.
6965_CH08_pp579-640.qxd
630
1/14/10
2:01 PM
Page 630
CHAPTER 8 Analytic Geometry in Two and Three Dimensions
EXAMPLE 1 Locating a Point in Cartesian Space
Draw a sketch that shows the point 12, 3, 52.
SOLUTION To locate the point 12, 3, 52, we first sketch a right-handed three-di-
mensional coordinate frame. We then draw the planes x = 2, y = 3, and z = 5,
which parallel the coordinate planes x = 0, y = 0, and z = 0, respectively. The
point 12, 3, 52 lies at the intersection of the planes x = 2, y = 3, and z = 5, as
shown in Figure 8.48.
Now try Exercise 1.
z
(0, 0, 5)
(2, 3, 5)
Line y ⫽ 3, z ⫽ 5
Plane z ⫽ 5
Line x ⫽ 2, z ⫽ 5
Plane y ⫽ 3
Plane x ⫽ 2
0
(0, 3, 0)
(2, 0, 0)
y
x
Line x ⫽ 2, y ⫽ 3
FIGURE 8.48 The planes x = 2, y = 3, and z = 5 determine the point 12, 3, 52. (Example 1)
Distance and Midpoint Formulas
The distance and midpoint formulas for space are natural extensions of the corresponding formulas for the plane.
Distance Formula (Cartesian Space)
The distance d1P, Q2 between the points P1x 1, y1, z 12 and Q1x 2, y2, z 22 in
space is
d1P, Q2 = 21x 1 - x 222 + 1y1 - y222 + 1z 1 - z 222.
Just as in the plane, the coordinates of the midpoint of a line segment are the averages
for the coordinates of the endpoints of the segment.
Midpoint Formula (Cartesian Space)
The midpoint M of the line segment PQ with endpoints P1x 1, y1, z 12 and
Q1x 2, y2, z 22 is
M = a
x 1 + x 2 y1 + y2 z 1 + z 2
,
,
b.
2
2
2
6965_CH08_pp579-640.qxd
1/14/10
2:01 PM
Page 631
SECTION 8.6
Three-Dimensional Cartesian Coordinate System
631
EXAMPLE 2 Calculating a Distance and Finding a Midpoint
Find the distance between the points P1- 2, 3, 12 and Q14, -1, 52, and find the midpoint of line segment PQ.
SOLUTION The distance is given by
d1P, Q2 = 21- 2 - 422 + 13 + 122 + 11 - 522
= 136 + 16 + 16
= 168 L 8.25
The midpoint is
M = a
-2 + 4 3 - 1 1 + 5
,
,
b = 11, 1, 32.
2
2
2
Now try Exercises 5 and 9.
Drawing Lesson
How to Draw Three-Dimensional Objects to Look Three-Dimensional
z
1. Make the angle between the
positive x-axis and the
positive y-axis large enough.
z
y
y
x
x
This
2. Break lines. When one line
passes behind another, break
it to show that it doesn’t
touch and that part of it is
hidden.
A
A
D
C
B
Intersecting
4. Spheres: Draw the sphere
first (outline and equator);
draw axes, if any, later.
Use line breaks and dashed
lines.
A
D
D
C
3. Dash or omit hidden portions
of lines. Don’t let the line
touch the boundary of the
parallelogram that represents
the plane, unless the line lies
in the plane.
Not this
Line below plane
C
B
B
CD behind AB
AB behind CD
Line above plane
Line in plane
z
Hidden part
dashed
Break
A contact dot
sometimes helps
y
Break
x
Sphere first
Axes later
6965_CH08_pp579-640.qxd
632
1/14/10
2:01 PM
Page 632
CHAPTER 8 Analytic Geometry in Two and Three Dimensions
Equation of a Sphere
A sphere is the three-dimensional analogue of a circle: In space, the set of points that
lie a fixed distance from a fixed point is a sphere. The fixed distance is the radius, and
the fixed point is the center of the sphere. The point P1x, y, z2 is a point of the sphere
with center (h, k, l) and radius r if and only if
21x - h22 + 1y - k22 + 1z - l22 = r.
Squaring both sides gives the standard equation shown below.
Standard Equation of a Sphere
A point P1x, y, z2 is on the sphere with center 1h, k, l2 and radius r if and only if
1x - h22 + 1y - k22 + (z - l22 = r 2.
EXAMPLE 3 Finding the Standard Equation of a Sphere
The standard equation of the sphere with center 12, 0, - 32 and radius 7 is
1x - 222 + y 2 + 1z + 322 = 49.
Now try Exercise 13.
Planes and Other Surfaces
In Section P.4, we learned that every line in the Cartesian plane can be written as a firstdegree (linear) equation in two variables; that is, every line can be written as
Ax + By + C = 0,
where A and B are not both zero. Conversely, every first-degree equation in two variables represents a line in the Cartesian plane.
In an analogous way, every plane in Cartesian space can be written as a first-degree
equation in three variables:
Equation for a Plane in Cartesian Space
Every plane can be written as
Ax + By + Cz + D = 0,
where A, B, and C are not all zero. Conversely, every first-degree equation in
three variables represents a plane in Cartesian space.
z
(0, 0, 3)
EXAMPLE 4 Sketching a Plane in Space
Sketch the graph of 12x + 15y + 20z = 60.
(0, 4, 0)
y
(5, 0, 0)
12x + 15y + 20z = 60
x
FIGURE 8.49 The intercepts 15, 0, 02,
10, 4, 02, and 10, 0, 32 determine the plane
12x + 15y + 20z = 60. (Example 4)
SOLUTION Because this is a first-degree equation, its graph is a plane. Three
points determine a plane. To find three points, we first divide both sides of
12x + 15y + 20z = 60 by 60:
y
z
x
+ + = 1
5
4
3
In this form, it is easy to see that the points 15, 0, 02, 10, 4, 02, and 10, 0, 32 satisfy
the equation. These are the points where the graph crosses the coordinate axes.
Now try Exercise 17.
Figure 8.49 shows the completed sketch.
6965_CH08_pp579-640.qxd
1/14/10
2:01 PM
Page 633
SECTION 8.6
Three-Dimensional Cartesian Coordinate System
633
Equations in the three variables x, y, and z generally graph as surfaces in three-dimensional
space. Just as in the plane, second-degree equations are of particular interest. Recall
that second-degree equations in two variables yield conic sections in the Cartesian
plane. In space, second-degree equations in three variables yield quadric surfaces:
The paraboloids, ellipsoids, and hyperboloids of revolution that have special reflective
properties are all quadric surfaces, as are such exotic-sounding surfaces as hyperbolic
paraboloids and elliptic hyperboloids.
Other surfaces of interest include graphs of functions of two variables, whose equations have the form z = ƒ1x, y2. Some examples are z = x ln y, z = sin 1xy2, and
z = 21 - x 2 - y 2. The last equation graphs as a hemisphere (see Exercise 63).
Equations of the form z = ƒ1x, y2 can be graphed using some graphing calculators and
most computer algebra software. Quadric surfaces and functions of two variables are
studied in most university-level calculus course sequences.
Vectors in Space
z
v3
0, 0, 1
1, 0, 0
v1
i
v
k
0, 1, 0
j
v1, v2, v3
v2
y
x
FIGURE 8.50 The vector v = 8v1, v2, v39.
In space, just as in the plane, the sets of equivalent directed line segments (or arrows)
are vectors. They are used to represent forces, displacements, and velocities in three dimensions. In space, we use ordered triples to denote vectors:
v = 8v1, v2, v39
The zero vector is 0 = 80, 0, 09, and the standard unit vectors are i = 81, 0, 09,
j = 80, 1, 09, and k = 80, 0, 19. As shown in Figure 8.50, the vector v can be expressed in terms of these standard unit vectors:
v = 8v1, v2, v39 = v1i + v2 j + v3k
The vector v that is represented by the arrow from P1a, b, c2 to Q1x, y, z2 is
!
v = PQ = 8x - a, y - b, z - c9 = 1x - a2i + 1y - b2j + 1z - c2k.
A vector v = 8v1, v2, v39 can be multiplied by a scalar (real number) c as follows:
cv = c8v1, v2, v39 = 8cv1, cv2, cv39
Many other properties of vectors extend in a natural way when we move from two to
three dimensions:
Vector Relationships in Space
For vectors v = 8v1, v2, v39 and w = 8w1, w2, w39,
• Equality:
• Addition:
• Subtraction:
v = w if and only if v1 = w1, v2 = w2, and v3 = w3
v + w = 8v1 + w1, v2 + w2, v3 + w39
v - w = 8v1 - w1, v2 - w2, v3 - w39
• Magnitude:
ƒ v ƒ = 2v12 + v22 + v32
• Dot product:
v # w = v1w1 + v2w2 + v3w3
• Unit vector:
u = v/ ƒ v ƒ , v Z 0, is the unit vector in the direction of v.
EXAMPLE 5 Computing with Vectors
(a) 38- 2, 1, 49 = 83 # -2, 3 # 1, 3 # 49 = 8-6, 3, 129
(b) 80, 6, -79 + 8- 5, 5, 89 = 80 - 5, 6 + 5, -7 + 89 = 8-5, 11, 19
(c) 81, -3, 49 - 8- 2, -4, 59 = 81 + 2, -3 + 4, 4 - 59 = 83, 1, -19
(d) ƒ 82, 0, - 69 ƒ = 222 + 0 2 + 62 = 140 L 6.32
(e) 85, 3, -19 # 8-6, 2, 39 = 5 # 1-62 + 3 # 2 + 1- 12 # 3
= - 30 + 6 - 3 = - 27 Now try Exercises 23–26.
6965_CH08_pp579-640.qxd
634
1/14/10
2:01 PM
Page 634
CHAPTER 8 Analytic Geometry in Two and Three Dimensions
EXAMPLE 6 Using Vectors in Space
A jet airplane just after takeoff is pointed due east. Its air velocity vector makes an
angle of 30° with flat ground with an airspeed of 250 mph. If the wind is out of the
southeast at 32 mph, calculate a vector that represents the plane’s velocity relative to
the point of takeoff.
SOLUTION Let i point east, j point north, and k point up. The plane’s air velocity is
a = 8250 cos 30°, 0, 250 sin 30°9 L 8216.506, 0, 1259,
and the wind velocity, which is pointing northwest, is
w = 832 cos 135°, 32 sin 135°, 09 L 8- 22.627, 22.627, 09.
The velocity relative to the ground is v = a + w, so
v L 8216.506, 0, 1259 + 8- 22.627, 22.627, 09
L 8193.88, 22.63, 1259
= 193.88i + 22.63j + 125k
Now try Exercise 33.
In Exercise 64, you will be asked to interpret the meaning of the velocity vector obtained in Example 6.
z
Lines in Space
v = a, b, c
y
P(x, y, z)
P0(x0, y0, z0)
x
ᐉ
FIGURE 8.51 The line / is parallel to the
direction vector v = 8a, b, c9.
We have seen that first-degree equations in three variables graph as planes in space. So
how do we get lines? There are several ways. First notice that to specify the x-axis,
which is a line, we could use the pair of first-degree equations y = 0 and z = 0. As
alternatives to using a pair of Cartesian equations, we can specify any line in space using
• one vector equation, or
• a set of three parametric equations.
Suppose / is a line through the point P01x 0, y0, z 02 and in the direction of a nonzero
vector v = 8a, b, c9 (Figure 8.51). Then for any point P1x, y, z9 on /,
!
P0P = tv
!
for some real number! t. The vector v is a direction vector for line /. If r = OP =
8x, y, z9 and r0 = OP0 = 8x 0, y0, z 09, then r - r0 = tv. So an equation of the line / is
r = r0 + tv.
Equations for a Line in Space
If / is a line through the point P01x 0, y0, z 02 in the direction of a nonzero vector
v = 8a, b, c9, then a point P1x, y, z2 is on / if and only if
r = r0 + tv, where r = 8x, y, z9 and r0 = 8x 0, y0, z 09; or
• Vector form:
• Parametric form: x = x 0 + at, y = y0 + bt, and z = z 0 + ct,
where t is a real number.
EXAMPLE 7 Finding Equations for a Line
The line through P014, 3, -12 with direction vector v = 8-2, 2, 79 can be written
r = 84, 3, -19 + t8- 2, 2, 79; or
• in vector form as
• in parametric form as x = 4 - 2t, y = 3 + 2t, and z = - 1 + 7t.
Now try Exercise 35.
6965_CH08_pp579-640.qxd
1/14/10
2:01 PM
Page 635
SECTION 8.6
Three-Dimensional Cartesian Coordinate System
635
EXAMPLE 8 Finding Equations for a Line
Using the standard unit vectors i, j, and k, write a vector equation for the line containing the points A13, 0, -22 and B1-1, 2, -52, and compare it to the parametric
equations for the line.
SOLUTION The line is in the direction of
!
v = AB = 8-1 - 3, 2 - 0, -5 + 29 = 8-4, 2, - 39.
!
Using r0 = OA, the vector equation of the line becomes:
r
8x, y, z9
8x, y, z9
xi + yj + zk
=
=
=
=
r0 + tv
83, 0, -29 + t8-4, 2, -39
83 - 4t, 2t, - 2 - 3t9
13 - 4t2i + 2tj + 1-2 - 3t2k
The parametric equations are the three component equations
x = 3 - 4t, y = 2t, and z = -2 - 3t.
Now try Exercise 41.
QUICK REVIEW 8.6
(For help, go to Sections 6.1 and 6.3.)
Exercise numbers with a gray background indicate problems
that the authors have designed to be solved without a calculator.
In Exercises 1–3, let P1x, y2 and Q12, -32 be points in the xy-plane.
1. Compute the distance between P and Q.
2. Find the midpoint of the line segment PQ.
3. If P is 5 units from Q, describe the position of P.
In Exercises 4–6, let v = 8-4, 59 = - 4i + 5j be a vector in the
xy-plane.
4. Find the magnitude of v.
5. Find a unit vector in the direction of v.
6. Find a vector 7 units long in the direction of - v.
7. Give a geometric description of the graph of
1x + 122 + 1y - 522 = 25 in the xy-plane.
8. Give a geometric description of the graph of
x = 2 - t, y = -4 + 2t in the xy-plane.
9. Find the center and radius of the circle
x 2 + y 2 + 2x - 6y + 6 = 0 in the xy-plane.
10. Find a vector from P12, 52 to Q1 -1, -42 in the xy-plane.
SECTION 8.6 EXERCISES
In Exercises 1–4, draw a sketch that shows the point.
1. 13, 4, 22
3. 11, - 2, - 42
2. 12, - 3, 62
4. 1 - 2, 3, -52
In Exercises 5–8, compute the distance between the points.
5. 1 -1, 2, 52, 13, - 4, 62
In Exercises 13–16, write an equation for the sphere with the given
point as its center and the given number as its radius.
13. 15, - 1, -22, 8
15. 11, -3, 22, 1a, a 7 0
14. 1- 1, 5, 82, 15
16. 1p, q, r2, 6
In Exercises 17–22, sketch a graph of the equation. Label all intercepts.
6. 12, -1, - 82, 16, - 3, 42
17. x + y + 3z = 9
18. x + y - 2z = 8
19. x + z = 3
20. 2y + z = 6
8. 1x, y, z2, 1p, q, r2
21. x - 3y = 6
22. x = 3
9. P1 - 1, 2, 52, Q13, - 4, 62
23. r + v
24. r - w
10. P12, - 1, - 82, Q16, -3, 42
25. v w
#
#
27. r 1v + w2
26. ƒ w ƒ
29. w/ ƒ w ƒ
30. i r
7. 1a, b, c2, 11, - 3, 22
In Exercises 9–12, find the midpoint of the segment PQ.
11. P12x, 2y, 2z2, Q1- 2, 8, 62
12. P1 - a, - b, - c2, Q13a, 3b, 3c2
In Exercises 23–32, evaluate the expression using r = 81, 0, - 39,
v = 8 -3, 4, - 59, and w = 84, -3, 129.
28. 1r v2 + 1r w2
#
#
31. 8i v, j v, k v9
# #
#
32. 1r v2w
#
#
6965_CH08_pp579-640.qxd
636
1/14/10
2:01 PM
Page 636
CHAPTER 8 Analytic Geometry in Two and Three Dimensions
In Exercises 33 and 34, let i point east, j point north, and k point up.
33. Three-Dimensional Velocity An airplane just after
takeoff is headed west and is climbing at a 20° angle relative to
flat ground with an airspeed of 200 mph. If the wind is out of
the northeast at 10 mph, calculate a vector v that represents the
plane’s velocity relative to the point of takeoff.
34. Three-Dimensional Velocity A rocket soon after
takeoff is headed east and is climbing at an 80° angle relative
to flat ground with an airspeed of 12,000 mph. If the wind is
out of the southwest at 8 mph, calculate a vector v that represents the rocket’s velocity relative to the point of takeoff.
In Exercises 35–38, write the vector and parametric forms of the line
through the point P0 in the direction of v.
35. P012, -1, 52, v = 83, 2, - 79
56. Generalizing a Property of the Dot Product
Prove u # u = ƒ u ƒ 2 where u is a vector in three-dimensional
space.
Standardized Test Questions
57. True or False x 2 + 4y 2 = 1 represents a surface in
space. Justify your answer.
58. True or False The parametric equations x = 1 + 0t,
y = 2 - 0t, z = -5 + 0t represent a line in space. Justify
your answer.
In Exercises 59–62, you may use a graphing calculator to solve the
problem.
59. Multiple Choice A first-degree equation in three variables graphs as
36. P01 -3, 8, - 12, v = 8- 3, 5, 29
(A) a line.
38. P010, -1, 42, v = 80, 0, 19
(C) a sphere.
37. P016, -9, 02, v = 81, 0, - 49
In Exercises 39–48, use the points A1 - 1, 2, 42, B10, 6, -32, and
C12, -4, 12.
39. Find the distance from A to the midpoint of BC.
40. Find the vector from A to the midpoint of BC.
(B) a plane.
(D) a paraboloid.
(E) an ellipsoid.
60. Multiple Choice Which of the following is not a
quadric surface?
41. Write a vector equation of the line through A and B.
(A) A plane
42. Write a vector equation of the line through A and the midpoint
of BC.
(B) A sphere
43. Write parametric equations for the line through A and C.
(D) An elliptic paraboloid
44. Write parametric equations for the line through B and C.
(E) A hyperbolic paraboloid
45. Write parametric equations for the line through B and the midpoint of AC.
46. Write parametric equations for the line through C and the midpoint of AB.
47. Is ¢ ABC equilateral, isosceles, or scalene?
48. If M is the midpoint of BC, what is the midpoint of AM?
In Exercises 49–52, (a) sketch the line defined by the pair of equations,
and (b) Writing to Learn give a geometric description of the line,
including its direction and its position relative to the coordinate frame.
49. x = 0, y = 0
50. x = 0, z = 2
51. x = -3, y = 0
52. y = 1, z = 3
53. Write a vector equation for the line through the distinct points
P1x 1, y1, z 12 and Q1x 2, y2, z 22.
54. Write parametric equations for the line through the distinct
points P1x 1, y1, z 12 and Q1x 2, y2, z 22.
55. Generalizing the Distance Formula
Prove that the distance d1P, Q2 between the points
P1x 1, y1, z 12 and Q1x 2, y2, z 22 in space is
21x 1 - x 222 + 1y1 - y222 + 1z 1 - z 222 by using the point
R1x 2, y2, z 12, the two-dimensional distance formula within the
plane z = z 1, the one-dimensional distance formula within the
line r = 8x 2, y2, t9, and the Pythagorean Theorem. [Hint: A
sketch may help you visualize the situation.]
(C) An ellipsoid
61. Multiple Choice If v and w are vectors and c is a scalar,
which of these is a scalar?
(A) v + w
(B) v - w
(C) v # w
(D) cv
(E) ƒ v ƒ w
62. Multiple Choice The parametric form of the line
r = 82, -3, 09 + t81, 0, - 19 is
(A) x = 2 - 3t, y = 0 + 1t, z = 0 - 1t.
(B) x = 2t, y = -3 + 0t, z = 0 - 1t.
(C) x = 1 + 2t, y = 0 - 3t, z = - 1 + 0t.
(D) x = 1 + 2t, y = - 3, z = - 1t.
(E) x = 2 + t, y = -3, z = - t.
Explorations
63. Group Activity Writing to Learn The figure
shows a graph of the ellipsoid x 2/9 + y 2/4 + z 2/16 = 1 drawn
in a box using Mathematica computer software.
(a) Describe its cross sections in each of the three coordinate
planes, that is, for z = 0, y = 0, and x = 0. In your description, include the name of each cross section and its
position relative to the coordinate frame.
6965_CH08_pp579-640.qxd
1/25/10
11:05 AM
Page 637
CHAPTER 8
(b) Explain algebraically why the graph of z = 21 - x 2 - y 2
is half of a sphere. What is the equation of the related
whole sphere?
Key Ideas
637
64. Revisiting Example 6 Read Example 6. Then using
v = 193.88i + 22.63j + 125k, establish the following:
(a) The plane’s compass bearing is 83.34°.
(c) By hand, sketch the graph of the hemisphere
z = 21 - x 2 - y 2. Check your sketch using a 3D
grapher if you have access to one.
(b) Its speed downrange (that is, ignoring the vertical component) is 195.2 mph.
(d) Explain how the graph of an ellipsoid is related to the
graph of a sphere and why a sphere is a degenerate
ellipsoid.
(d) The plane’s overall speed is 231.8 mph.
(c) The plane is climbing at an angle of 32.63°.
Extending the Ideas
–2
2
–1
0
1
2
0
–2
The cross product u : v of the vectors u = u 1i + u 2 j + u 3k and
v = v1i + v2 j + v3k is
i
u * v = 3 u1
v1
j
u2
v2
k
u3 3
v3
= 1u 2v3 - u 3v22i + 1u 3v1 - u 1v32j + 1u 1v2 - u 2v12k.
4
Use this definition in Exercises 65–68.
65. 81, -2, 39 * 8-2, 1, -19
66. 84, -1, 29 * 81, -3, 29
2
67. Prove that i * j = k.
0
68. Assuming the theorem about angles between vectors (Section
6.2) holds for three-dimensional vectors, prove that u * v is
perpendicular to both u and v if they are nonzero.
–2
–4
CHAPTER 8 Key Ideas
Properties, Theorems, and Formulas
Parabolas with Vertex (0, 0) 583
Parabolas with Vertex (h, k) 584
Ellipses with Center (0, 0) 592
Ellipses with Center (h, k) 594
Hyperbolas with Center (0, 0) 604
Hyperbolas with Center (h, k) 605
Translation Formulas 614
Rotation Formulas 614
Coefficients for a Conic in a Rotated System 615
Angle of Rotation to Eliminate the Cross-Product Term 616
Discriminant Test 617
Focus–Directrix–Eccentricity Relationship 620
Polar Equations for Conics 622
Ellipse with Eccentricity e and Semimajor Axis a 625
Distance Formula (Cartesian Space) 630
Midpoint Formula (Cartesian Space) 630
Standard Equation of a Sphere 632
Equation for a Plane in Cartesian Space 632
Vector Relationships in Space 633
Equations for a Line in Space 634
Procedures
How to Sketch the Ellipse
x2
a
2
y2
+
How to Sketch the Hyperbola
2
b
x2
a
2
= 1
y2
-
2
b
593
= 1
603
Translation of Axes 613–614
Rotation of Axes 614–616
How to Draw Three-Dimensional Objects to Look
Three-Dimensional 631