Slide 8- 1. Chapter 8 Analytic Geometry in Two and Three Dimensions.

Chapter 8

Analytic Geometry in Two and Three Dimensions

Conic Sections and Parabolas

Quick Review

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.

4. Find

the ver

y x y x

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 ( ) .

g xx x

5. Write an equation for the quadratic function whose graph

contains the vertex (2, 3) and

translation up 3 u

the point (0,3).

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.

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.

A Right Circular Cone (of two nappes)

Conic Sections and Degenerate Conic Sections

Conic Sections and Degenerate Conic Sections (cont’d)

Animation

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

Structure of a Parabola

Graphs of x2=4py

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|

Graphs of y2 = 4px

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

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:

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|

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).

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

Ellipses

Quick Review

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

4. 6 1 6 12 11

5. Find the exact solut

a b a b

y xy x

ion by completing the square.

2 8 21 0 29

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.

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.

Key Points on the Focal Axis of an Ellipse

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

Vertices ( ,0) (0, )

Semimajor axis

Semiminor axis

2 2 2 2 2 2 Pythagorean relation a b c a b c

Pythagorean Relation

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

Divide both sides by 36 to put the equation in standard form.

Since the larger number is the denominator of the , the focal

axis is the -axis. So 9, 4, and 5.

Thus the vertice

y a b c a b

s are (0, 3), and the foci are (0, 5).

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.

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

b a b c

Ellipse with Center (h,k)

2 2 2 2

Focal axis

Foci ( , )

x h y k y k x h

a b a by k x h

Vertices ( , ) ( , )

Semimajor axis

Semiminor ax

h a k h k a

2 2 2 2 2 2

Pythagorean relation

a b c a b c

Ellipse with Center (h,k)

Example Locating Key Points of an Ellipse

Find the center, vertices, and foci of the ellipse

The standard form of the equations is

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

e foci ( , ) are ( 1,1 5) or

approximately ( 1,3.24) and ( 1, 1.24).

Elliptical Orbits Around the Sun

Eccentricity of an Ellipse

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

eccentricity

Hyperbolas

Quick Review

1. Find the distance between the points ( , ) and ( , 4).

2. Solve for in terms of . 1 16 2

Solve for algebraically.

3 12 3 8 10

4. 6 1

no solution

62 6 1 1

Solve the system of equations:

16 / no soluti n o

c a a c

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.

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.

Hyperbola

Hyperbola with Center (0,0)

2 2 2 2

Focal axis -axis -axis

Foci ( ,0)

x y y x

a b a bx y

Vertices ( ,0) (0, )

Semitransverse axis

Semiconjugate axis

2 2 2 2 2 2ythagorean relation

Asymptotes

c a b c a b

b ay x y x

Hyperbola Centered at (0,0)

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

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

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.

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

b a c b

Hyperbola with Center (h,k)

2 2 2 2

Focal axis

Foci ( , )

x h y k y k x h

a b a by k x h

Vertices ( , ) ( , )

Semimajor axis

Semiminor ax

h a k h k a

2 2 2 2 2 2

Pythagorean relation

Asymptotes ( )

c a b c a b

by x h k y

x h kb

Hyperbola with Center (h,k)

Example Locating Key Points of a Hyperbola

Find the center, vertices, and foci of the hyperbola 1.4 9

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)

and ( 4.61,0).

Eccentricity of a Hyperbola

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

eccentricity

Translations and Rotations of Axes

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 .

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.

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

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

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 x y y

the equation of the ellipse becomes

' ( ')1.

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 ,

x y x y

x x y y x y

rotation formulas

, 0 / 2, is the . angle of rotation

Rotation of Cartesian Coordinate Axes

Example Rotation of Axes

Prove that 2 25 0 is the equation of a hyperbola by rotating the

coordinate axes through an angle / 4.

Example Rotation of Axes

Prove that 2 25 0 is the equation of a hyperbola by rotating the

coordinate axes through an angle / 4.

The rotation equations are

'cos / 4 'sin / 4 and 'sin / 4 'cos / 4

' ' ' ' and .

2 2The equation 2 25 0 becomes

' ' ' '2 25 0

' ' 25

x x y y x y

x y x yx y

x y x y

Coefficients for a Conic in a Rotated System

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

' cos cos sin sin

' cos 2 ( )sin 2

' cos cos sin sin

' cos sin

A A B C

B B C A

C C B A

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.

BB x y

Discriminant Test

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

Conics and the Equation Ax2+Bxy+Cy2+Dx+Ey+F=0

Polar Equations of Conics

Quick Review

1. Solve for . (4, ) ( , )

2. Solve for . (3, 5 /3)=( 3, ), 2 2

3. Find the focus and the directrix of the parabola.

Find the focus and the vertices of the conic.

. 1 16 9

( 5,0); ( 4,0)

(0, 7); 1 (0, 4) 9 16

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.

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.)

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.

PFP e PF e PD

ver, the conic is

a hyperbola if 1,

a parabola if 1,

an ellipse if 1.

The Geometric Structure of a Conic Section

A Conic Section in the Polar Plane

Three Types of Conics for r = ke/(1+ecosθ)

Polar Equations for Conics

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

kee k r

Example Identifying Conics from Their Polar Equations

Determine the eccentricity, the type of conic, and the directrix.

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.

ke k k y

Semimajor Axes and Eccentricities of the Planets

Ellipse with Eccentricity e and Semimajor Axis a

Three-Dimensional Cartesian Coordinate System

Quick Review

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 ,

describ

P x y Q xy

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 41n of .

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.

The Point P(x,y,z) in Cartesian Space

The Coordinate Planes Divide Space into Eight Octants

Distance Formula (Cartesian Space)

1 1 1 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

Midpoint Formula (Cartesian Space)

1 2 1 2 1 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

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 .

( , ) 1 4 2 5 3 6

1 4 2 5 3 6 5 7 9The midpoint is , , , , .

2 2 2 2 2 2

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

Drawing Lesson

Drawing Lesson (cont’d)

Example Finding the Standard Equation of a Sphere

Find the standard equation of the sphere with center (1,2,3)

and radius 4.

1 2 3 16x y z

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

The Vector v = <v1,v2,v3>

Vector Relationships in Space

1 2 3 1 2 3

1 1 2 2 3 3

For vectors , , and , , ,

= if and only if , ,

+ = , ,

v v v w w w

v w v w v w

Equality :

Addition :

Subtraction :

Magnitude :

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 :

u v v v v

Equations for a Line in Space

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

O O O O

P x y z

a b c P x y z

t x y z x y z

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

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

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

Chapter Test

1. Find the vertex, focus, directrix, and focal width of the parabola

2 12. Given 1. Identify the type of conic, find

16 7the center, vertices, and foci.

3. Given 6 3 0. Identify the

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).

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

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

ame. If one of its microphones has a parabolic surface

generated by the parabola 18 , locate the focus (the electronic

receiver) of the parabola.

Chapter Test Solutions

vertex (0,0), focus (3,0), directrix 3, foca

1. Find the vertex, focus, directrix, and focal width of the parabola

2 12. Given 1. Identify the type of conic, find

6 7the ce

idth:12

ter, vertices, and foci.

3. Given 6 3 0. Identify the

Ellipse, center (2, 1), vert

conic and complete the

ices (6, 1)

( 2, 1), foci

re to writ

) ( 1, 1)

parabola ( in standard form 3.

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

hyperbola

ndard form for the el

30 4lipse with c n

r (0,2),

semimajor axis = 3, and one focus at (2,22

Chapter Test Solutions

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

n the dir

x t y ty

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

3 7x t y t z t

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)

Slide 8- 1. Chapter 8 Analytic Geometry in Two and Three Dimensions.

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