slide 5- 1 copyright © 2006 pearson education, inc. publishing as pearson addison-wesley


Analytic Geometry Topics Chapter 9

Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

9.1The Parabola

Given an equation of a parabola, complete the square, if necessary, and then find the vertex, the focus, and the directrix and graph the parabola.


Parabolas

A parabola is the set of all points in a plane equidistant from a fixed line (the directrix) and a fixed point not on the line (the focus).

The line that is perpendicular to the directrix and contains the focus is the axis of symmetry. The vertex is the midpoint of the segment between the focus and the directrix.

Directrix

Axis of symmetry

Vertex

Focus


Standard Equation of a Parabola with Vertex at the Origin

The standard equation of a parabola with vertex (0, 0) and directrix y = p is x2 = 4py.

The focus is (0, p) and the y-axis is the axis of symmetry.

The standard equation of a parabola with vertex (0, 0) and directrix x = p is y2 = 4px.

The focus is (p, 0) and the x-axis is the axis of symmetry.


Example

Find the focus, and the directrix of the parabola

Then graph the parabola.

Solution: We write

in the form x2 = 4py:

Thus, p = 4, so the focus is (0, p), or (0, 4). The directrix is y = p = (4).21

16 .y x

2116 .y x

2116

2

2

16

4(4)

x y

x y

x y


Example

Find an equation of the parabola with vertex (0, 0) and focus (3, 0). Then graph the parabola.

The focus is on the x-axis so the line of symmetry is the x-axis. Thus the equation is of the type y2 = 4px.

Since the focus (3, 0) is 3 units to the right of the vertex, p = 3 and the equation is y2 = 4(3)x, or y2 = 12x.


Standard Equation of a Parabola with Vertex (h, k) and Vertical Axis of Symmetry


Standard Equation of a Parabola with Vertex (h, k) and Horizontal Axis of Symmetry


Example

For the parabola x2 6x y + 11 = 0, find the vertex, the focus, and the directrix. Then draw the graph.

Solution: We first complete the square:

2

2

2

2

6 11 0

6 11

6 11

( 3)

9 9

2

x x y

x x y

x x y

x y


Example continued

We see that h = 3, k = 2 and p = ¼, so we have the following:

Vertex (h, k): (3, 2) Focus (h, k + p):

(3, 2 + ¼) or (3, 9/4) Directrix y = k p:

y = 2 ¼, or y = 7/4


Example

For the parabola y2 6y x + 4 = 0, find the vertex, the focus, and the directrix. Then draw the graph.

Solution: We first complete the square:

2

2

2

2

6 4 0

6 4

( 6 ) 4

(

9

3) 5

9

y y x

y y x

y y x

y x


Example continued

We see that h = 5, k = 3 and p = ¼, so we have the following:

Vertex (h, k): (5, 3) Focus (h + p, k):

(5 + ¼, 3) or (19/4, 3) Directrix x = h p:

x = 5 ¼, or y = 21/4


9.2The Circle and the Ellipse

Given an equation of a circle, complete the square, if necessary, and then find the center and the radius

and graph the circle. Given the equation of an ellipse, complete the square,

if necessary, and then find the center, the vertices, and the foci and graph the ellipse.


Circle

A circle is the set of all points in a plane that are at a fixed distance from a fixed point (the center) in the plane.

Standard Equation of a Circle

The standard equation of a circle with center (h, k) and radius r is

(x h)2 + (y k)2 = r2.


Example

For the circle

x2 + y2 + 2x + 6y 15 = 0 find

the center and the radius. Then graph the circle.

Solution: First, we complete the square twice:

Center = (1, 3) Radius = 5

2 2

2 2

2 2

2 2

2 2 2

2 6 15 0

2 + 6 + = 15

2 6 15

( 1) ( 3) 25

(

1 9

1) ( 3) 5

1 9

x y x y

x x y y

x x y y

x y

x y


Ellipse

An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points (the foci) is constant. The center of an ellipse is the midpoint of the segment between the foci.

The major axis of an ellipse is longer than the minor axis.


Standard Equation of an Ellipse with Center at the Origin (Major Axis Horizontal)


Standard Equation of an Ellipse with Center at the Origin (Major Axis Vertical)


Example

Find the standard equation of the ellipse with vertices (6, 0) and (6, 0) and foci (4, 0) and (4, 0). Then graph the ellipse.

Since the foci are on the x-axis and the origin is the midpoint of the segment between them, the major axis is horizontal and (0, 0) is the center of the ellipse. Thus the equation is of the form

2 2

2 21.

x y

a b


Example continued

Since the vertices are (6, 0) and (6, 0) and the foci are (4, 0) and (4, 0), we know that a = 6 and c = 4. These values can be used to find b2.

Thus the equation of the ellipse is

To graph, we plot the vertices, and the y-intercepts are (0, 4.47) and (0, 4.47). We plot these points and connect the four points with a smooth curve.

2 2

136 20

x y

2 2 2

2 2 2

2

2

16 36

20

2

4 6

0 4.47

c a b

b

b

b

b


Example

For the ellipse find the vertices and the foci. Then draw the graph.

a = 7 and b = 6. The major axis is vertical.

Vertices: (0, 7) and (0, 7)

2 2

2 2

2 2

2 2

2 2

1764 1764 1764

49 36 1764

49 36 1764

136 49

16 7

x y

x y

x y

x y

2 2 2

2 2 27 6 13

13 3.61

c a b

c

c


Example continued

The foci are (0, 3.61) and (0, 3.61).

The x-intercepts are (6, 0) and (6, 0).

We plot the vertices and the x-intercepts and connect the four points with a smooth curve.


Example

For the ellipse find the center, the vertices, and the foci. Then draw the graph.

First, we complete the square twice to get standard form.

2 29 4 54 8 49 0x y x y

2 2

2 2

2 2

2 2

2 2

2 2

9 4 54 8 49 0

9( 6 ) + 4( 2 ) = 49

( 6 ) ( 2 ) 49

9( 3) 4( 1) 36

9( 3) 4( 1) 36

36 36 36

( 3) (

9 9 4 1

1

9

81 4

)1

4

x y x y

x x y y

x x y y

x y

x y

x y


Example continued

Center is (3, 1) a = 3 and b = 2 The major axis is vertical so the vertices are 3 units

above and below the center. (3, 1 + 3) (3, 1 3) or (3, 4) and (3, 2).

We know that c2 = a2 b2, so c2 = 32 22 = 5 and

c = . Then the foci are 2.24 units above and below the center:

5 2.24

( 3,1 5)( 3,1 5)


Example continued

To graph, we plot the vertices. Note also that since b = 2, two other points on the graph are the endpoints of the minor axis, 2 units right and left of center.


9.3The Hyperbola

Given an equation of a hyperbola, complete the square, if necessary, and then find the center, the vertices, and the foci and graph the hyperbola.


Hyperbola

A hyperbola is the set of all points in a plane for which the absolute value of the difference of the distances from two fixed points (the foci) is constant. The midpoint of the segment between the foci is the center of the hyperbola.


Standard Equation of a Hyperbola with Center at the Origin -- Horizontal


Standard Equation of a Hyperbola with Center at the Origin -- Vertical


Example

Find an equation of the hyperbola with vertices (0, 3) and (0, 3) and foci (0, 5) and (0, 5).

Solution: We know that a = 3 and c = 5. We find b2.

Since the vertices and foci are on the y-axis, we know that the transverse axis is vertical. We can now write the equation of the hyperbola:

2 2 2

2 2 2

2

2

5 3

25 9

16

c a b

b

b

b

2 2

2 2

2 2

1

19 16

y x

a b

y x


Example

For the hyperbola given by 16x2 25y2 = 400, find the vertices, the foci, and the asymptotes. Then graph the hyperbola.

First, we find standard form:

2 2

2 2

2 2

2 2

2 2

1 1

400 4

1

0

6 25 400

16 25 400

125 16

14

0

5

x y

x y

x y

x y


Example continued

The hyperbola has a horizontal transverse axis, so the vertices are (5, 0) and (5, 0).

For the standard form of the equation, we know that a2 = 52, or 25, and b2 = 42, or 16. We find the foci:

2 2 2

2

2

25 16

41

41

c a b

c

c

c

Thus, the foci are 41,0 and 41,0 .


Example continued

Next, we find the asymptotes:

To draw the graph, sketch the asymptotes first.

and

4

54

5

by x x

ab

y x xa


Standard Equation of a Hyperbola with Center at (h, k) -- Horizontal


Standard Equation of a Hyperbola with Center at (h, k) -- Vertical


Example

For the hyperbola given by

9y2 4x2 54y 8x + 41 = 0, find the center, the vertices, and the foci. Then draw the graph.

First, we complete the square to get standard form:

2 2

2 2

2 2

2 2

2 2

9 54 4 8 41

9( 6 ) 4( 2 ) 41

9( 6 9) 4( 2 1) 41 81 4

9( 3) 4( 1) 36

( 3) ( 1)1

4 9

y y x x

y y x x

y y x x

y x

y x


Example continued

Center = (1, 3)a = 2 b = 3Vertices: (1, 3 2) or (1, 1)

(1, 3 + 2) or (1, 5)We know that c2 = a2 + b2, so c2 = 4 + 9 = 13 and

The foci are units below and above the center:

13.c

13

1,3 13 and 1,3 13 .


Example continued

The asymptotes are:

We sketch the asymptotes, plot the vertices, and draw the graph.

23 ( 1) and

32

3 ( 1)3

y x

y x


9.4Nonlinear Systems of Equations

Solve a nonlinear system of equations. Use nonlinear systems of equations to solve applied

problems. Graph nonlinear systems of inequalities.


Nonlinear Systems of Equations

The graphs of the equations in a nonlinear system of equations can have no point of intersection or one or more points of intersection.

The coordinates of each point of intersection represent a solution of the system of equations.

When no point of intersection exists, the system of equations has no real-number solution.

We can solve nonlinear systems of equations by using the substitution or elimination method.


Example

Solve the following system of equations:

We use the substitution method. First, we solve equation (2) for y.

2 2 The graph is a circle9 (1)

2 3 (2)

.

The graph is a line.

x y

x y

2 3

2 3

2 3

x y

y x

y x


Example continued

Next, we substitute y = 2x 3 in equation (1) and solve for x:

Now, we substitute these numbers for x in equation (2) and solve for y.

y = 2x 3y = 2(0) 3y = 3

(0, 3) and

Check each solution.

2 2

2 2

2

(2 3) 9

4 12 9 9

5 12 0

(5 12) 0

120 or

5

x x

x x x

x x

x x

x x

125

2 3

2( ) 3

9

5

y x

y

y

12 9,

5 5


Example continued

Check: (0, 3)

Check:

Visualizing the Solution

12 9,

5 5

2 2

2 3

9

0 3 ? 9

9 9

x y

2 2

2 29125 5

9

? 9

9 9

x y

2 3

2(0) ( 3)? 3

3 3

x y

9125 5

2 3

2( ) ( )? 3

3 3

x y


Example

Solve the following system of equations:

xy = 4

3x + 2y = 10

Solve xy = 4 for y.

Substitute into 3x + 2y = 10.

4

4

xy

yx

4

8

8

2

2

3 2 10

3 2( ) 10

3 10

3 10( )

3 8 10

3 10 8 0

x

x

x

x y

x

x

x x x

x x

x x


Example continued

Use the quadratic formula: 2

2

4

2

10 10 4(3 8)

2(3)

10 100 96

6

10 4 10 2

6 610 2 10 2

and 6 6

4 and 2

3

b b acx

a

x

x

x

x

x

23 10 8 0x x


Example continued

Substitute values of x to find y. 3x + 2y = 10

x = 4/3 x = 2

The solutions are (4/3, 3) and (2, 2).

433 2 10

4 2 10

2 6

3

y

y

y

y

3 2 2 10

6 2 10

2 4

2

y

y

y

y



Example

Solve the system of equations:

Solve by elimination. Multiply equation (1) by 2 and add to eliminate the y2 term.

2 2

2 2

5 2 13

3 4 39

x y

x y

2 2

2 2

2

2

10 4 26

3 4 39

13 13

1

1

x y

x y

x

x

x


Example continued

Substituting x = 1 in equation (2) gives us:

The possible solutions are (1, 3), (1, 3), (1, 3) and (1, 3).

2 2

2 2

2

2

3 4 39

3(1) 4 39

4 36

9

3

x y

y

y

y

y

2 2

2 2

2

2

3 4 39

3( 1) 4 39

4 36

9

3

x y

y

y

y

y


Example continued

All four pairs check, so they are the solutions.



Modeling and Problem Solving

Example: For a student recreation building at Southport Community College, an architect wants to lay out a rectangular piece of land that has a perimeter of 204 m and an area of 2565 m2. Find the dimensions of the piece of land.

1. Make a drawing and label it. 2. Translate. We now have the following:

Perimeter: 2l + 2w = 204 Area: lw = 2565

Area = lw

= 2565 m2ww

l

l


Modeling and Problem Solving continued

3. Carry out. Solve the system.

Solving the second equation for l gives us l = 2565 w. We then substitute into equation (1) and solve for w.

2

2

2

2 2 204

2 2(2565) 204

2 204 2(2565) 0

102 2565 0

( 57)( 45) 0

57

25

or 5

5

4

6w

w w

w w

w w

w

w

w

w w


Modeling and Problem Solving continued

If w = 57, then l = 2565 w = 2565 57 = 45.

If w = 45, then l = 2565 w = 2565 45 = 57.

Since the length is generally considered to be longer than the width, we have the solution l = 57 and w = 45.

4. Check. If l = 57 and w = 45, the perimeter is

2(45) + 2(57) = 204.

The area is 57(45) = 2565. The numbers check.

5. State. The length of the piece of land is 57 m and the width is 45 m.


Nonlinear Systems of Inequalities

Example: Graph the solution set of the system.

21

2

y x

y x


9.5Rotation of Axes

Use rotation of axes to graph conic sections. Use the discriminant to determine the type of conic

represented by a given equation.


Rotation of Axes

When B is nonzero, the graph of

is a conic section with an axis that is not parallel to

the x– or y–axis. Rotating the axes through a positive angle yields an

coordinate system.

Ax2 Bxy Cy2 Dx Ey F 0

x y


Rotation of Axes Formulas

If the x– and y–axes are rotated about the origin through a positive acute angle then the coordinates of and of a point P in the and coordinates system are related by the following formulas.

x x cos ysin : y x sin ycos,

x x cos y sin : y x sin y cos.

(x, y) ( x , y ) xy x y


Example

Suppose that the axes are rotated through an angle of 45º.

Write the equation in the coordinate system.

Solution:

Substitute 45º for in the formulas

x x cos y sin : y x sin y cos.

x x cos 45º y sin 45º : y x sin 45º y cos 45.

x x2

2 y

2

2

2

2x y

y x2

2 y

2

2

2

2x y

xy x y xy1


Example continued

Now, substitute these into the equation

2

2x y g 2

2x y 1

1

2x 2 y 2

1

x 22

y 221

x 2

2 2

y 2

2 21

This is the equation of a hyperbola in the coordinate system. x y

xy1.


Eliminating the xy-Term

To eliminate the xy-term from the equation

select an angle such that

Ax2 Bxy Cy2 Dx Ey F 0, B0,

cot 2 A C

B, 0º 2 180º ,

and use the rotation of axes formulas.


Example

Graph the equation

Solution: We have

3x2 2 3xy y2 2x 2 3y0.

cot 2 3 1

2 3

2

2 3

1

3, 2 120º , 60º

A3, B 2 3, C 1, D2, E 2 3, F 0.

x x cos60º y sin 60º , y x sin60º y cos60º

x x1

2 y

3

2x

2y 3

2

y x3

2 y

1

2x 3

2y

2


Example continued

Substitute these into

and simplify.

3x2 2 3xy y2 2x 2 3y0

Parabola with vertex at (0, 0) in the coordinate system and axis of symmetry

4 y 2 4 x 0

y 2 x

y 0.

x y


The Discriminant

The expression

Ax2 Bxy Cy2 Dx Ey F 0.

B2 4AC

is, except in degenerate cases,

is the discriminant of

the equation

The graph of the the equationAx2 Bxy Cy2 Dx Ey F 0

1. an ellipse or circle if

2. a hyperbola if and

3. a parabola if

B2 4AC 0,

B2 4AC 0,

B2 4AC 0.

jlb


Example

Graph the equation 3x2 2xy 3y2 16.

so

Solution:

We have A3, B2, C 3,

The discriminant is negative so it’s a circle or ellipse.

Determine

Substitute into the rotation formulas

B2 4AC 22 4g3g34 36 32.

cot 2 A C

B

3 3

20 2 90º 45º

x x cos 45º y sin 45º : y x sin 45º y cos 45º .


Example continued

After substituting into the equation and simplifying we have:

4 x 2 2 y 2 16

x 24

y 281.

x x cos 45º y sin 45º , y x sin 45º y cos 45º

x x2

2 y

2

2

2

2x y

y x2

2 y

2

2

2

2x y


Example continued

with vertices

0, 2 2 and 0,2 2 on the y -axis

and x intercepts

2,0 and 2,0 .

x 24

y 281

is an ellipse,


9.6Polar Equations of Conics

Graph polar equations of conics. Convert from polar to rectangular equations of

conics. Find polar equations of conics.


An Alternative Definition of Conics

Let L be a fixed line (the directrix); let F be a fixed point (the focus), not on L; and let e be a positive constant (the eccentricity). A conic is the set of all points P in the plane such that

where PF is the distance from P to F and PL is the distance from P to L. The conic is a parabola if e = 1, an ellipse if e < 1, and a hyperbola if e > 1.

PF

PLe,


Polar Equations of Conics

To derive equations position focus F at the pole, and the directrix L either perpendicular or parallel to the polar axis. In this figure L is perpendicular to the polar axis and p units to the right of the pole.

Note: PL p r cos PF r



For an ellipse and a hyperbola, the eccentricity e is given by e = c/a, where c is the distance from the center to a focus and a is the distance from the center to a vertex.

PL p r cos PF r

PF

PL

r

p r cose

r ep

1 ecosSimplified:


Example

Describe and graph the conic r 18

6 3cos.

Solution:

r 3

1 0.5 cos, e0.5, p6

Since e < 1, it’s an ellipse with vertical directrix 6 units to the right of the pole. The major axis lies along the polar axis.

Let = 0, and π to find vertices: (2, 0) and (6, π).

The center is (2, π).

The major axis = 8, so a = 4.

Since e = c/a, then 0.5 = c/4, so c = 2

The length of the minor axis is given by b:

a2 c2 42 22 16 4 12 2 3b =


Example continued

Sketch the graph r 18

6 3cos.



A polar equation of any of the four forms

is a conic section. The conic is a parabola if e = 1, an ellipse if 0 < e < 1, and a hyperbola if e >1.

r ep

1ecos, r

ep

1esin


Converting from Polar to Rectangular Equations

Use the relationships between polar and rectangular coordinates.

Remember:

x r cos, yr sin, r2 x2 y2


Example

Convert to a rectangular equation: r 2

1 sin.

Solution:

We have r 2

1 sinr r sin 2

r r sin 2 Substitute.

x2 y2 y 2

x2 y2 y2 4y 2

x2 4y 4 0

This is the equation of a parabola.


Finding Polar Equations of Conics

We can find the polar equation of a conic with a focus at the pole if we know its eccentricity and the equation of the directrix.


Example

Find a polar equation of a conic with focus at the pole, eccentricity 1/3 and directrix

r ep

1 esin

13 2

1 13sin

2

3 sin

r 2csc.

Solution:

r 2 csc 2

sin or r sin 2, which is y2.

Choose an equation for a directrix that is a horizontal line above the polar axis and substitutee = 1/3 and p = 2


9.7Parametric Equations

Graph parametric equations. Determine an equivalent rectangular equation for

parametric equations. Determine parametric equations for a rectangular

equation. Determine the location of a moving object at a

specific time.


Graphing Parametric Equations

We have graphed plane curves that are composed of sets of ordered pairs (x, y) in the rectangular coordinate plane. Now we discuss a way to represent plane curves in which x and y are functions of a third variable t.

One method will be to construct a table in which we choose values of t and then determine the values of x and y.


Example

Graph the curve represented by the equations

x 1

2t, yt 2 3; 3t 3.

The rectangular equation is y4x2 3, 3

2x

3

2.


Parametric Equations

If f and g are continuous functions of t on an interval I, then the set of ordered pairs (x, y) such that x = f(t) and y = g(t) is a plane curve.

The equations x = f(t) and y = g(t) are parametric equations for the curve.

The variable t is the parameter.


Determining a Rectangular Equation for Given Parametric Equations

Solve either equation for t.

Then substitute that value of t into the other equation.

Calculate the restrictions on the variables x and y based on the restrictions on t.


Example

Find a rectangular equation equivalent to

Solution

x t 2 , yt 1; 1t 4

The rectangular equation is:

yt 1

t y 1

Substitute t y 1 into x t 2 .

x y 1 2

Calculate the restrictions:

1t 4

x t 2; 0x 16

yt 1; 2y3

x y 1 2 ; 0x 16.


Determining Parametric Equations for a Given Rectangular Equation

Many sets of parametric equations can represent the same plane curve. In fact, there are infinitely many such equations.

The most simple case is to let either x (or y) equal t and then determine y (or x).


Example

Find three sets of parametric equations for the parabola y4 x 3 2 .

Solution

If x t, then y4 t 3 2 t 2 6t 5.

If x t 3, then y4 t 3 3 2 t 2 4.

If x t

3, then y4

t

3 3

2

t

9

2

2t 5.


Applications

The motion of an object that is propelled upward can be described with parametric equations. Such motion is called projectile motion. It can be shown that, neglecting air resistance, the following equations describe the path of a projectile propelled upward at an

angle with the horizontal from a height h, in feet, at

an initial speed v0, in feet per second:

x v0 cos t, yh v0 sin t 16t 2 .

We can use these equations to determine the location

of the object at time t, in seconds.


ExampleA baseball is thrown from a height of 6 ft with an initial speed of 100 ft/sec at an angle of 45º with the horizontal.

a) Find parametric equations that give the position of the ball at time t, in seconds.

b) Find the height of the ball after 1 sec, 2 sec and 3 sec.

c) Determine how long the ball is in the air.d) Determine the horizontal distance that the ball

travels.e) Find the maximum height of the ball.


Example continued

Solution

a)

b) The height of the ball at time t is represented by y.

x v0 cos t, yh v0 sin t 16t 2

x 100 cos 45º t y6 100sin 45º t 16t 2

x 50 2t y 16t 2 50 2t 6

If t 1, y 16 1 2 50 2 1 660.7 ft.

If t 2, y 16 2 2 50 2 2 683.4 ft.

If t 3, y 16 3 2 50 2 3 674.1 ft.


Example continued

Solution

c) The ball hits the ground when y = 0.

The ball is in the air for about 4.5 sec.

y 16t 2 50 2t 6

0 16t 2 50 2t 6

t 50 2 50 2 2 4 16 6

2 16 t 0.1 or t 4.5


Example continued

Solution

d) Substitute t = 4.5

So the horizontal distance the ball travels is 318.2 ft.

e) Find the maximum value of y (vertex).

So the maximum height is about 84.1 ft.

x 50 2t 50 2 4.5 318.2 ft.

t b

2a

50 2

2 16 2.2

y 16 2.2 2 50 2 2.2 684.1 ft.


Applications

The path of a fixed point on the circumference of a circle as it rolls along a line is called a cycloid. For example, a point on the rim of a bicycle wheel traces a cycloid curve.


Applications

The parametric equations of a cycloid are

x a t sin t , ya 1 cos t ,where a is the radius of the circle that traces the curve and t is in radian measure.


ExampleThe graph of the cycloid described by the parametric

x 3 t sin t , y3 1 cos t ; 0t 6is shown below. equations

slide 5- 1 copyright © 2006 pearson education, inc. publishing as pearson addison-wesley

Documents