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Slide 8- 1
Chapter 8
Analytic Geometry in Two and Three Dimensions
8.1
Conic Sections and Parabolas
Slide 8- 4
Quick Review
2
2 2
1. Find the distance between ( 1,2) and (3, 4).
2. Solve for in terms of . 2 6
3. Complete the square to rewrite the equation in vertex form.
2 5
4. Find
52
3
( 1
the ver
4
tex
)
y x y x
y xy
y
x x
x
2
2
and axis of the graph of ( ) 2( 1) 3.
Describe how the graph of can be obtained from the graph
vertex:( 1,3); axis: 1; translation left 1 unit,
vertical stretch by a factor of
of ( ) .
2,
f x x
f
g xx x
2
5. Write an equation for the quadratic function whose graph
contains the vertex (2, 3) and
translation up 3 u
the point (0,3).
nits.
32 3
2y x
Slide 8- 5
What you’ll learn about
Conic Sections Geometry of a Parabola Translations of Parabolas Reflective Property of a Parabola
… and whyConic sections are the paths of nature: Any free-moving object in a gravitational field follows the path of a conic section.
Slide 8- 6
Parabola
A parabola is the set of all points in a plane equidistant from a particular line (the directrix) and a particular point (the focus) in the plane.
Slide 8- 7
A Right Circular Cone (of two nappes)
Slide 8- 8
Conic Sections and Degenerate Conic Sections
Slide 8- 9
Conic Sections and Degenerate Conic Sections (cont’d)
Animation
Slide 8- 10
Second-Degree (Quadratic) Equations in Two Variables
2 2 0, where , , and , are not all zero.Ax Bxy Cy Dx Ey F A B C
Slide 8- 11
Structure of a Parabola
Slide 8- 12
Graphs of x2=4py
Slide 8- 13
Parabolas with Vertex (0,0)
Standard equation x2 = 4py y2 = 4px Opens Upward or To the right or to the
downward left
Focus (0,p) (p,0) Directrix y = -p x = -p Axis y-axis x-axis Focal length p p Focal width |4p| |4p|
Slide 8- 14
Graphs of y2 = 4px
Slide 8- 15
Example Finding an Equation of a Parabola
Find an equation in standard form for the parabola whose directrix
is the line 3 and whose focus is the point ( 3,0).x
2
2
Because the directrix is 3 and the focus is ( 3,0), the focal
length is 3 and the parabola opens to the left. The equation of
the parabola in standard from is:
4
12
x
y px
y x
Slide 8- 16
Parabolas with Vertex (h,k)
Standard equation (x-h)2 = 4p(y-k) (y-k)2 = 4p(x-h)
Opens Upward or To the right or to the downward left
Focus (h,k+p) (h+p,k) Directrix y = k-p x = h-p Axis x = h y = k Focal length p p Focal width |4p| |4p|
Slide 8- 17
Example Finding an Equation of a Parabola
Find the standard form of the equation for the parabola with
vertex at (1,2) and focus at (1, 2).
2
2
The parabola is opening downward so the equation has the form
( ) 4 ( ).
( , ) (1,2) and the distance between the vertex and the focus is
4. Thus, the equation is ( 1) 16( 2).
x h p y k
h k
p x y
8.2
Ellipses
Slide 8- 19
Quick Review
2 2
2
2
2
2
2
1 2
3
1. Find the distance between ( , ) and (1,2).
2. Solve for in terms of . 1 9 4
Solve for algebraically.
3. 3 8 3 1
6 9
2
82 10
4. 6 1 6 12 11
5. Find the exact solut
2
a b a b
xy
x
x
y xy x
x
x x
x x
2
ion by completing the square.
2 8 21 0 29
22
xx x
Slide 8- 20
What you’ll learn about
Geometry of an Ellipse Translations of Ellipses Orbits and Eccentricity Reflective Property of an Ellipse
… and whyEllipses are the paths of planets and comets around the Sun, or of moons around planets.
Slide 8- 21
Ellipse
An ellipse is the set of all points in a plane whose distance from two fixed points in the plane have a constant sum. The fixed points are the foci (plural of focus) of the ellipse. The line through the foci is the focal axis. The point on the focal axis midway between the foci is the center. The points where the ellipse intersects its axis are the vertices of the ellipse.
Slide 8- 22
Key Points on the Focal Axis of an Ellipse
Slide 8- 23
Ellipse with Center (0,0)
2 2 2 2
2 2 2 2 Standard equation 1 1
Focal axis -axis -axis
Foci ( ,0)
x y y x
a b a bx y
c
(0, )
Vertices ( ,0) (0, )
Semimajor axis
Semiminor axis
c
a a
a a
b b
2 2 2 2 2 2 Pythagorean relation a b c a b c
Slide 8- 24
Pythagorean Relation
Slide 8- 25
Example Finding the Vertices and Foci of an Ellipse
2 2Find the vertices and the foci of the ellipse 9 4 36.x y
2 2
2
2 2 2 2 2
Divide both sides by 36 to put the equation in standard form.
14 9
Since the larger number is the denominator of the , the focal
axis is the -axis. So 9, 4, and 5.
Thus the vertice
x y
y
y a b c a b
s are (0, 3), and the foci are (0, 5).
Slide 8- 26
Example Finding an Equation of an Ellipse
Find an equation of the ellipse with foci ( 2,0) and (2,0) whose minor
axis has length 2.
2 2 2
2 2 2
2 2
The center is (0,0). The foci are on the -axis with 2.
The semiminor axis is 2 / 2 1. Using , find
1 2 5. Thus the equation of the ellipse is
1.5 1
x c
b a b c
a
x y
Slide 8- 27
Ellipse with Center (h,k)
2 2 2 2
2 2 2 2 Standard equation 1 1
Focal axis
Foci ( , )
x h y k y k x h
a b a by k x h
h c k
( , )
Vertices ( , ) ( , )
Semimajor axis
Semiminor ax
h k c
h a k h k a
a a
2 2 2 2 2 2
is
Pythagorean relation
b b
a b c a b c
Slide 8- 28
Ellipse with Center (h,k)
Slide 8- 29
Example Locating Key Points of an Ellipse
2 2
Find the center, vertices, and foci of the ellipse
1 11
4 9
x y
2 2
2 2
The standard form of the equations is
1 11.
9 4
The center is at ( 1,1). Because the semimajor axis 9 3,
the vertices are at ( , ) ( 1,1 3) which are ( 1,4) and ( 1, 2).
Because 9 4 5, th
y x
a
h k a
c a b
e foci ( , ) are ( 1,1 5) or
approximately ( 1,3.24) and ( 1, 1.24).
h k c
Slide 8- 30
Elliptical Orbits Around the Sun
Slide 8- 31
Eccentricity of an Ellipse
2 2
The of an ellipse is ,
where is the semimajor axis, is the semiminor
axis, and is the distance from the center of the
ellipse to either focus.
c a be
a aa b
c
eccentricity
8.3
Hyperbolas
Slide 8- 33
Quick Review
2 2
2
2
2
2 2
1. Find the distance between the points ( , ) and ( , 4).
2. Solve for in terms of . 1 16 2
Solve for algebraically.
3.
4
3 12 3 8 10
4. 6 1
8 16
no solution
222
62 6 1 1
5.
a b c
y x
a c b
y x
x
y x
x
x x
x x
2 2
Solve the system of equations:
2
16 / no soluti n o
c a
c a a c
Slide 8- 34
What you’ll learn about
Geometry of a Hyperbola Translations of Hyperbolas Eccentricity and Orbits Reflective Property of a Hyperbola Long-Range Navigation
… and whyThe hyperbola is the least known conic section, yet it is used astronomy, optics, and navigation.
Slide 8- 35
Hyperbola
A hyperbola is the set of all points in a plane whose distances from two fixed points in the plane have a constant difference. The fixed points are the foci of the hyperbola. The line through the foci is the focal axis. The point on the focal axis midway between the foci is the center. The points where the hyperbola intersects its focal axis are the vertices of the hyperbola.
Slide 8- 36
Hyperbola
Slide 8- 37
Hyperbola
Slide 8- 38
Hyperbola with Center (0,0)
2 2 2 2
2 2 2 2 Standard equation 1 1
Focal axis -axis -axis
Foci ( ,0)
x y y x
a b a bx y
c
(0, )
Vertices ( ,0) (0, )
Semitransverse axis
Semiconjugate axis
P
c
a a
a a
b b
2 2 2 2 2 2ythagorean relation
Asymptotes
c a b c a b
b ay x y x
a b
Slide 8- 39
Hyperbola Centered at (0,0)
Slide 8- 40
Example Finding the Vertices and Foci of a Hyperbola
2 2Find the vertices and the foci of the hyperbola 9 4 36.x y
2 2
2 2 2 2 2
Divide both sides of the equation by 36 to find the standard form
36. So 4, 9, and 13. Thus the4 9
vertices are ( 2,0) and the foci are ( 13, 0).
x ya b c a b
1
Slide 8- 41
Example Finding an Equation of a Hyperbola
Find an equation of the hyperbola with foci (0,4) and (0, 4)
whose conjugate axis has length 2.
2 2 2
2 2
The center is at (0,0). The foci are on the -axis with 4.
The semiconjugate axis is 2 / 2 1. Thus 16 1 15.
The standard form of the hyperbola is 1.15 1
y c
b a c b
y x
Slide 8- 42
Hyperbola with Center (h,k)
2 2 2 2
2 2 2 2 Standard equation 1 1
Focal axis
Foci ( , )
x h y k y k x h
a b a by k x h
h c k
( , )
Vertices ( , ) ( , )
Semimajor axis
Semiminor ax
h k c
h a k h k a
a a
2 2 2 2 2 2
is
Pythagorean relation
Asymptotes ( )
b b
c a b c a b
by x h k y
a
( )a
x h kb
Slide 8- 43
Hyperbola with Center (h,k)
Slide 8- 44
Example Locating Key Points of a Hyperbola
221
Find the center, vertices, and foci of the hyperbola 1.4 9
x y
2 2
The center ( , ) ( 1,0). Because the semitransverse axis 4 2,
the vertices are at ( , ) 1 2,0 or ( 3,0) and (1,0).
Because 4 9 13, the foci are at
( , ) ( 1 13,0) or approximately (2.61,0)
h k a
h a k
c a b
h c k
and ( 4.61,0).
Slide 8- 45
Eccentricity of a Hyperbola
2 2
The of a hyperbola is ,
where is the semitransverse axis, is the semiconjugate
axis, and is the distance from the center to either focus.
c a be
a aa b
c
eccentricity
8.4
Translations and Rotations of Axes
Slide 8- 47
Quick Review
Assume 0 < /2.
1. Given cot 2 3/ 4, find cos 2 .
2. Given cot 2 1/ 3, find cos 2 .
3. Given cot 2 1, find .
4. Given cot 2 1/ 3, find .
5. Given cot 2 3/ 4, find cos .
3/5
1/2
/8
/6
2/ 5
Slide 8- 48
What you’ll learn about
Second-Degree Equations in Two Variables Translating Axes versus Translating Graphs Rotation of Axes Discriminant Test
… and whyYou will see ellipses, hyperbolas, and parabolas as members of the family of conic sections rather than as separate types of curves.
Slide 8- 49
Translation-of-Axes Formulas
The coordinates ( , ) and ( ', ') based on parallel sets of axes are
related by either of the following :
' and ' or ' and ' .
x y x y
x x h y y k x x h y y k translations formulas
Slide 8- 50
Example Translation Formula
2 2Prove that 9 4 18 16 11 0 is the equation of an ellipse.
Translate the coordinate axes so that the origin is at the center of this ellipse.
x y x y
2 2
2 2
2 2
2 2
Complete the square for both the and .
9 18 4 16 11
9( 2 1) 4( 4 4) 11 9 16
9( 1) 4( 2) 36
( 1) ( 2)1
4 9This is a standard equation of an ellipse. If we let ' 1
and ' 2, then
x y
x x y y
x x y y
x y
x y
x x
y y
22
the equation of the ellipse becomes
' ( ')1.
4 9
x y
Slide 8- 51
Rotation-of-Axes Formulas
The coordinates ( , ) and ( ', ') based on rotated sets of axes are
related by either of the following :
' cos sin and ' sin cos , or
'cos 'sin and 'sin 'cos ,
where
x y x y
x x y y x y
x x y y x y
rotation formulas
, 0 / 2, is the . angle of rotation
Slide 8- 52
Rotation of Cartesian Coordinate Axes
Slide 8- 53
Example Rotation of Axes
Prove that 2 25 0 is the equation of a hyperbola by rotating the
coordinate axes through an angle / 4.
xy
Slide 8- 54
Example Rotation of Axes
Prove that 2 25 0 is the equation of a hyperbola by rotating the
coordinate axes through an angle / 4.
xy
2 2
2 2
The rotation equations are
'cos / 4 'sin / 4 and 'sin / 4 'cos / 4
' ' ' ' and .
2 2The equation 2 25 0 becomes
' ' ' '2 25 0
2 2
' ' 25
' '1
25 25
x x y y x y
x y x yx y
xy
x y x y
x y
x y
Slide 8- 55
Coefficients for a Conic in a Rotated System
2 2
If we apply the rotation formulas to the general second-degree equation in
and , we obtain a second-degree equation in ' and ' of the form
' ' ' ' ' ' ' ' ' ' ' ' 0, where the coefficients
x y x y
A x B x y C y D x E y F 2 2
2 2
are
' cos cos sin sin
' cos 2 ( )sin 2
' cos cos sin sin
' cos sin
' cos sin
'
A A B C
B B C A
C C B A
D D E
E E D
F F
Slide 8- 56
Angle of Rotation to Eliminate the Cross-Product Term
If 0, an angle of rotation such that cot 2 and 0 / 2
will eliminate the term ' ' ' from the second degree equation in the
rotated ' ' coordinate system.
A CB
BB x y
x y
Slide 8- 57
Discriminant Test
2 2
2
2
2
The second-degree equation 0 graphs as
a hyperbola if 4 0,
a parabola if 4 0,
an ellipse if 4 0,
except for degenerate cases.
Ax Bxy Cy Dx Ey F
B AC
B AC
B AC
Slide 8- 58
Conics and the Equation Ax2+Bxy+Cy2+Dx+Ey+F=0
8.5
Polar Equations of Conics
Slide 8- 60
Quick Review
2
1. Solve for . (4, ) ( , )
2. Solve for . (3, 5 /3)=( 3, ), 2 2
3. Find the focus and the directrix of the parabola.
12
4
4 / 3
(0,3
Find the focus and the vertices of the conic.
)
4
; 3
r r
x y y
2 2
2 2
. 1 16 9
5.
( 5,0); ( 4,0)
(0, 7); 1 (0, 4) 9 16
x y
x y
Slide 8- 61
What you’ll learn about
Eccentricity Revisited Writing Polar Equations for Conics Analyzing Polar Equations of Conics Orbits Revisited
… and whyYou will learn the approach to conics used by astronomers.
Slide 8- 62
Focus-Directrix Definition Conic Section
A conic section is the set of all points in a plane whose distances from a particular point (the focus) and a particular line (the directrix) in the plane have a constant ratio. (We assume that the focus does not lie on the directrix.)
Slide 8- 63
Focus-Directrix Eccentricity Relationship
If is a point of a conic section, is the conic's focus, and is the
point of the directrix closest to , then and ,
where is a constant and the eccentricity of the conic.
Moreo
P F D
PFP e PF e PD
PDe
ver, the conic is
a hyperbola if 1,
a parabola if 1,
an ellipse if 1.
e
e
e
Slide 8- 64
The Geometric Structure of a Conic Section
Slide 8- 65
A Conic Section in the Polar Plane
Slide 8- 66
Three Types of Conics for r = ke/(1+ecosθ)
Slide 8- 67
Polar Equations for Conics
Slide 8- 68
Example Writing Polar Equations of Conics
Given that the focus is at the pole, write a polar equation for the conic
with eccentricity 4/5 and directrix 3.x
Setting 4 / 5 and 3 in yields1 cos
3 4 / 5
1 4 / 5 cos
12
5 cos
kee k r
e
r
Slide 8- 69
Example Identifying Conics from Their Polar Equations
Determine the eccentricity, the type of conic, and the directrix.
6
3 2cosr
Divide the numerator and the denominator by 3.
2 The eccentricity is 2/3 which means
1 (2 / 3)cos
the conic is an ellipse.
The numerator 2 (2 / 3) , so 3 and the directrix is 3.
r
ke k k y
Slide 8- 70
Semimajor Axes and Eccentricities of the Planets
Slide 8- 71
Ellipse with Eccentricity e and Semimajor Axis a
21
1 cos
a er
e
8.6
Three-Dimensional Cartesian Coordinate System
Slide 8- 73
Quick Review
2 2
Let ( , ) and (3,2) be points in the -plane.
1. Compute the distance between and .
2. Find the midpoint of the line segment .
3. If P is 5 units from ,
3
describ
2
3
e
2,
2 2
the
x
P x y Q xy
P Q
P
y
x yQ
Q
2 2
position of .
Let 4,5 be a vector in the - plane.
4. Find the maginitude of .
5. Find a unit vector in the directio
3 2 25
41
4 5,
41 41n of .
xP
xy
y
v
v
v
Slide 8- 74
What you’ll learn about
Three-Dimensional Cartesian Coordinates Distances and Midpoint Formula Equation of a Sphere Planes and Other Surfaces Vectors in Space Lines in Space
… and whyThis is the analytic geometry of our physical world.
Slide 8- 75
The Point P(x,y,z) in Cartesian Space
Slide 8- 76
The Coordinate Planes Divide Space into Eight Octants
Slide 8- 77
Distance Formula (Cartesian Space)
1 1 1 2 2 2
2 2 2
1 2 1 2 1 2
The distance ( , ) between the points ( , , ) and Q( , , )
in space is ( , ) .
d P Q P x y z x y z
d P Q x x y y z z
Slide 8- 78
Midpoint Formula (Cartesian Space)
1 1 1
1 2 1 2 1 2
2 2 2
The midpoint of the line segment with endpoints ( , , )
and ( , , ) is , , .2 2 2
M PQ P x y z
x x y y z zQ x y z M
Slide 8- 79
Example Calculating a Distance and Finding a Midpoint
Find the distance between the points (1, 2,3) and (4,5,6), and
find the midpoint of the line segment .
P Q
PQ
2 2 2
( , ) 1 4 2 5 3 6
3 3
1 4 2 5 3 6 5 7 9The midpoint is , , , , .
2 2 2 2 2 2
d P Q
M
Slide 8- 80
Standard Equation of a Sphere
2 2 22
A point ( , , ) is on the sphere with center ( , , ) and radius
if and only if .
P x y z h k l r
x h y k z l r
Slide 8- 81
Drawing Lesson
Slide 8- 82
Drawing Lesson (cont’d)
Slide 8- 83
Example Finding the Standard Equation of a Sphere
Find the standard equation of the sphere with center (1,2,3)
and radius 4.
2 2 2
1 2 3 16x y z
Slide 8- 84
Equation for a Plane in Cartesian Space
Every plane can be written as 0, where , , and
are not all zero. Conversely, every first-degree equation in three
variables represents a plane in Cartesian space.
Ax By Cz D A B C
Slide 8- 85
The Vector v = <v1,v2,v3>
Slide 8- 86
Vector Relationships in Space
1 2 3 1 2 3
1 1 2 2 3 3
1 1 2 2 3 3
1 1 2 2 3 3
2 2 2
1 2 3
For vectors , , and , , ,
= if and only if , ,
+ = , ,
= , ,
v v v w w w
v w v w v w
v w v w v w
v w v w v w
v v v
Equality :
Addition :
Subtraction :
Magnitude :
v w
v w
v w
v w
v
1 1 2 2 3 3 = +
/ , 0, is the unit vector in the direction of .
v w v w v w
Dot Product :
Unit Vector :
v w
u v v v v
Slide 8- 87
Equations for a Line in Space
O O
If is a line through the point ( , , ) in the direction of a
nonzero vector , , , then a point ( , , ) is on if and only if
Vector form: = , where , , and , , ; or
Param
O O O O
O O O
P x y z
a b c P x y z
t x y z x y z
v
r r v r r
etric form: , , and , where is
a real number.O O O
x x at y y bt z z ct t
Slide 8- 88
Example Finding Equations for a Line
Using the standard unit vector , , and , write a vector equation for
the line containing the points ( 2,0,3) and (4, 1,3).A B i j k
O
O
The line is in the direction of 4 2, 1 0,3 3 6, 1,0 .
Using , the vector equation is
=
= 2,0,3 6, 1,0
AB
OA
t
t
v
r
r r v
r
Slide 8- 89
Chapter Test
2
2 2
2
1. Find the vertex, focus, directrix, and focal width of the parabola
12 .
2 12. Given 1. Identify the type of conic, find
16 7the center, vertices, and foci.
3. Given 6 3 0. Identify the
y x
x y
x x y
2 2
conic and complete the
square to write it in standard form.
4. Given 2 3 12 24 60 0. Identify the conic and
complete the square to write it in standard form.
5. Find the equation in standard form
x y x y
for the ellipse with center (0,2),
semimajor axis = 3, and one focus at (2,2).
Slide 8- 90
Chapter Test
6. Find the equation for the conic in standard form.
5 3cos , 3 3sin , 2 2 .
Use the vectors 3,1 2 and 3, 4,0 .
7. Compute .
8. Write the unit vector in the direction of
9. Write paramet
x t y t t
v w
v w
w.
ric equations for the line through
P( 1,0,3) and Q(3, 2, 4).
10. B-Ball Network uses a parabolic microphone to capture all the
sounds from the basketball players and coaches during each regular
season g
2
ame. If one of its microphones has a parabolic surface
generated by the parabola 18 , locate the focus (the electronic
receiver) of the parabola.
y x
Slide 8- 91
Chapter Test Solutions
2
2 2
vertex (0,0), focus (3,0), directrix 3, foca
1. Find the vertex, focus, directrix, and focal width of the parabola
12 .
2 12. Given 1. Identify the type of conic, find
1
l w
6 7the ce
idth:12
n
y
xy x
x
2
ter, vertices, and foci.
3. Given 6 3 0. Identify the
Ellipse, center (2, 1), vert
conic and complete the
squa
ices (6, 1)
( 2, 1), foci
re to writ
(5, 1
e it
) ( 1, 1)
parabola ( in standard form 3.
x x y
x
2
2
2 2
24. Given 2 3 12 24 60 0. Identify the conic and
complete the square to write it in standard form.
5. Find the equation in sta
) 12
hyperbola
ndard form for the el
4 31
30 4lipse with c n
5e te
y
x
x
x
y
y y
22
r (0,2),
semimajor axis = 3, and one focus at (2,22
1.9 5
) yx
Slide 8- 92
Chapter Test Solutions
2 2
6. Find the equation for the conic in standard form.
5 3cos , 3 3sin , 2 2 .
Use the vectors 3,1 2 and 3, 4,0 .
7. Compute .
8. Write the unit vector i
5 31
9 9
6,5,
n the dir
2
ectio
x t y ty
tx
v w
v w
n of
9. Write parametric equations for the line through
P( 1,0,3) and Q(3, 2, 4).
10. B-Ball Network uses a parabolic microphone to capture all the
sounds
3/ 5, 4 / 5,0
from the
1 4 , 2
ba
,
3 7x t y t z t
w.
2
sketball players and coaches during each regular
season game. If one of its microphones has a parabolic surface
generated by the parabola 18 , locate the focus (the electronic
receiver) of the parab
y x(0,ola. 4.5)
