copyright © 2014, 2010 pearson education, inc. chapter 8 analytical geometry copyright © 2014,...

Copyright © 2014, 2010 Pearson Education, Inc.

Chapter 8

Analytical Geometry



Section 8.2 The Parabola

1. Define a parabola geometrically.2. Find an equation of a parabola.3. Translate a parabola.4. Use the reflecting property of parabolas.

SECTION 1.1

Copyright © 2014, 2010 Pearson Education, Inc. 3

PARABOLA

Let l be a line and F a point in the plane not on the line l. The the set of all points P in the plane that are the same distance from F as they are from the line l is called a parabola.

Thus, a parabola is the set of all points P for which d(F, P) = d(P, l), where d(P, l) denotes the distance between P and l.


PARABOLA

Line l is the directrix.

the directrix is the axis or axis of symmetry.

The line through the focus, perpendicular to

Point F is the focus.

The point at which the axis intersects the parabola is the vertex.


EQUATION OF A PARABOLA

The equation y2 = 4ax is called the standard equation of a parabola with vertex (0, 0) and focus (a, 0). Similarly, if the focus of a parabola is placed on the negative x-axis, we obtain the equation y2 = – 4ax as the standard equation of a parabola with vertex (0, 0) and focus (–a, 0).

y2 4ax


EQUATION OF A PARABOLA

By interchanging the roles of x and y, we find that the equation x2 = 4ay is the standard equation of a parabola with vertex (0, 0) and focus (0, a). Similarly, if the focus of a parabola is placed on the negative x-axis, we obtain the equation x2 = – 4ay as the standard equation of a parabola with vertex (0, 0) and focus (0, –a).

x2 4ay


MAIN FACTS ABOUT A PARABOLA WITH a > 0



2 4x ay


Graph each parabola and specify the vertex, focus, directrix, and axis.

a. x2 = −8y b. y2 = 5x

a. The equation x2 = −8y has the standard form x2 = −4ay; so

Example: Graphing a Parabola


The parabola opens down. The vertex is the origin, and the focus is (0, −2). The directrix is the horizontal line y = 2; the axis of the parabola is the y-axis.

Since the focus is (0, −2) substitute y = −2 in the equation x2 = −8y of the parabola to obtain



Thus, the points (4, −2) and (−4, −2) are two symmetric points on the parabola to the right and the left of the focus.



b. The equation y2 = 5x has the standard form y2 = 4ax.

The parabola opens to the right. The vertex is the origin, and the focus is The directrix is the vertical line the axis of the parabola is the x-axis.


5,0 .

4 5

,4

x a


To find two symmetric points on the parabola that are above and below the focus, substitute in the equation of the parabola.


5

4x


Plot the two additional

points and

on the

parabola.



LATUS RECTUM

The line segment passing through the focus of a parabola, perpendicular to the axis, and having endpoints on the parabola is called the latus rectum of the parabola.

The following figures show that the length of the latus rectum for the graphs of y2 = ±4ax and x2 = ±4ay for a > 0 is 4a.


LATUS RECTUM


Find the standard equation of a parabola with vertex (0, 0) and satisfying the given description.a. The focus is (–3, 0).b. The axis of the parabola is the y-axis, and the graph passes through the point (–4, 2).

a. Vertex (0, 0) and focus (–3, 0) are both on the x-axis, so parabola opens left and the equation has the form y2 = – 4ax with a = 3.

2

2 3

4

4

y x

y

a

x

2 12 .y xThe equation is

Example: Finding the Equation of a Parabola


b. Vertex is (0, 0), axis is the y-axis, and the point (–4, 2) is above the x-axis, so parabola opens up and the equation has the form x2 = – 4ay and x = –4 and y = 2 can be substituted in to obtain

2

2

4

4 2

16

4

8

2

ya

a

a

x

a

2

2

2

4

4

8

2

.

x y

x y

x y

a

The equation is

Example: Finding the Equation of a Parabola


Main facts about a parabola with vertex (h, k) and a > 0

Standard Equation (y – k)2 = 4a(x – h)

Equation of axis y = k

Description Opens right

Vertex (h, k)

Focus (h + a, k)

Directrix x = h – a



Standard Equation (y – k)2 = –4a(x – h)

Equation of axis y = k

Description Opens left

Vertex (h, k)

Focus (h – a, k)

Directrix x = h + a



Standard Equation (x – h)2 = 4a(y – k)

Equation of axis x = h

Description Opens up

Vertex (h, k)

Focus (h, k + a)

Directrix y = k – a



Standard Equation (x – h)2 = – 4a(y – k)

Equation of axis x = h

Description Opens down

Vertex (h, k)

Focus (h, k – a)

Directrix y = k + a


Find the vertex, focus, and the directrix of the parabola 2y2 – 8y – x + 7 = 0. Sketch the graph of the parabola.

Complete the square on y.

2

2

2

2

2 4 7

4 72 4 8

2 2 1

12 1

2

y y x

y y x

y x

y x



2 12 1

2y x

We have h = –1, k = 2, and 1 1

4 , or .2 8

a a

24 .y k a x h

Vertex: (h, k) = (–1, 2)

1 71 ,2 ,2

8 8

Focus: (h + a, k) =

1 91

8 8 Directrix: x = h – a =

Compare with the standard

form



REFLECTING PROPERTY OF PARABOLAS

A property of parabolas that is useful in applications is the reflecting property: If a reflecting surface has parabolic cross sections with a common focus, then all light rays entering the surface parallel to the axis will be reflected through the focus.

This property is used in reflecting telescopes and satellite antennas, because the light rays or radio waves bouncing off a parabolic surface are reflected to the focus, where they are collected and amplified. (See next slide.)



Conversely, if a light source is located at the focus of a parabolic reflector, the reflected rays will form a beam parallel to the axis. This principle is used in flashlights, searchlights, and other such devices. (See next slide.)


The parabolic mirror used in the Hubble Space Telescope has a diameter of 94.5 inches. Find the equation of the parabola if its focus is 2304 inches from the vertex. What is the thickness of the mirror at the edges?

(Not to scale)

Example: Calculating Some Properties of the Hubble Space Telescope


Position the parabola so that its vertex is at the origin and its focus is on the positive y-axis. The equation of the parabola is of the form



To find the thickness y of the mirror at the edge, substitute x = 47.25 (half the diameter) in the equation x2 = 9216y and solve for y.

Thus, the thickness of the mirror at the edges is approximately 0.242248 inch.



Section 8.3 The Ellipse

1. Define an ellipse.2. Find an equation of an ellipse.3. Translate ellipses.4. Use ellipses in applications.

SECTION 1.1


ELLIPSE

An ellipse is the set of all points in the plane, the sum of whose distances from two fixed points is a constant.

The fixed points are called the foci (the plural of focus) of the ellipse.


ELLIPSE


EQUATION OF AN ELLIPSE

is the standard form of the equation of an ellipse with center (0, 0) and foci (–c, 0) and (c, 0), where b2 = a2 – c2.

x2

a2 y2

b2 1


EQUATION OF AN ELLIPSE

Similarly, by reversing the roles of x and y, an equation of the ellipse with center (0, 0) and foci (0, −c) and (0, c) and on the y-axis is given by

.2 2

2 21

x y

b a


HORIZONTAL AND VERTICAL ELLIPSES

If the major axis of an ellipse is along or parallel to the x-axis, the ellipse is called a horizontal ellipse, while an ellipse with major axis along or parallel to the y-axis is called a vertical ellipse.


MAIN FACTS ABOUT AN ELLIPSE WITH CENTER (0, 0)


Find the standard form of the equation of the ellipse that has vertex (5, 0) and foci (±4, 0).

Since the foci are (−4, 0) and (4, 0) the major axis is on the x-axis. We know c = 4 and a = 5; find b2.

Substituting into the standard equation, we get

.

Example: Finding an Equations of an Ellipse


Sketch a graph of the ellipse whose equation is 9x2 + 4y2 = 36. Find the foci of the ellipse.First, write the equation in standard form:

Example: Graphing an Ellipse


Because the denominator in the y2-term is larger than the denominator in the x2-term, the ellipse is a vertical ellipse.

Here a2 = 9 and b2 = 4, so c2 = a2 – b2 = 5.

Vertices: (0, ±3)Foci:Length of major axis: 6Length of minor axis: 4

Example: Graphing an Ellipse

0, 5


TRANSLATIONS OF ELLIPSES

Horizontal and vertical shifts can be used to obtain the graph of an ellipse whose equation is

The center of such an ellipse is (h, k), and its major axis is parallel to a coordinate axis.

x h 2

a2 y k 2

b2 1


Main facts about horizontal ellipses with center (h, k)

Standard Equation

Center (h, k)

Major axis along the line y = k

Length of major axis 2a

Minor axis along the line x = h

Length of minor axis 2b

2 2

2 21

x h y k

a b


Main facts about horizontal ellipses with center (h, k)

Vertices (h + a, k), (h – a, k)

Endpoints of minor axis (h, k – b), (h, k + b)

Foci (h + c, k), (h – c, k)

Equation involving a, b, and c

c2 = a2 – b2

SymmetryThe graph is symmetric

about the lines x = h and y = k.


Graphs of horizontal ellipses


Main facts about vertical ellipses with center (h, k)

Standard Equation

Center (h, k)

Major axis along the line x = h

Length major axis 2a

Minor axis along the line y = k

Length minor axis 2b

2 2

2 21

x h y k

b a


Main facts about vertical ellipses with center (h, k)

Vertices (h, k + a), (h, k – a)

Endpoints of minor axis (h – b, k), (h + b, k)

Foci (h, k + c), (h, k – c)

Equation involving a, b, c

c2 = a2 – b2

SymmetryThe graph is symmetric

about the lines x = h and y = k


Graphs of vertical ellipses


Find an equation of the ellipse that has foci (–3, 2) and (5, 2), and has a major axis of length 10.

Foci lie on the line y = 2, so horizontal ellipse.

3 5 2 2, 1,2

2 2

Center is midpoint of foci

Length major axis =10, vertices at a distance of a = 5 units from the center.

Foci at a distance of c = 4 units from the center.

Example: Finding the Equation of an Ellipse


2 2

2 21

x h y k

a b

Major axis is horizontal so standard form is

Use b2 = a2 – c2 to obtain b2. b2 = (5)2 – (4)2 = 25 – 16 = 9 to obtain b2.

2 21 2

125 9

x y

Replace: h = 1, k = 2, a2 = 25, b2 = 9

Center: (1, 2) a = 5, b = 3, c = 4



Vertices: (h ± a, k) = (1 ± 5, 2) = (–4, 2) and (6, 2)

Endpoints minor axis: (h, k ± b) = (1, 2 ± 3) = (1, –1) and (1, 5)



Find the center, vertices, and foci of the ellipse with equation 3x2 + 4y2 +12x – 8y – 32 = 0.

Complete squares on x and y.

2 2

2 2

2 2

2 2

3 12 4 8 32

3 4 4 2 32

4 2 32

3 2

4 1

4 1 4

3 1

8

4 2 4

x x y y

x x y y

x x y y

x y

Example: Converting to Standard Form


2 2

2 2

3 2 4 1 48

2 11

16 12

x y

x y

Length of major axis is 2a = 8.

This is standard form. Center: (–2, 1), a2 = 16, b2 = 12, and c2 = a2 – b2 = 16 – 12 = 4. Thus, a = 4, and c = 2.b 12 2 3,

Length of minor axis is 2b 4 3.



Center: (h, k) = (–2, 1)

Foci: (h ± c, k) = (–2 ± 2, 1) = (–4, 1) and (0, 1)

, 2,1 2 3

2,1 2 3 and 2,1 2 3

2,4.46 and 2, 2.46

h k b

Endpoints of minor axis:

Vertices: (h ± a, k) = (–2 ± 4, 1)= (–6, 1) and (2, 1)



APPLICATIONS OF ELLIPSES

1. The orbits of the planets are ellipses with the sun at one focus.

2. Newton reasoned that comets move in elliptical orbits about the sun.

3. We can calculate the distance traveled by a planet in one orbit around the sun.

4. The reflecting property for an ellipse says that a ray of light originating at one focus will be reflected to the other focus.


REFLECTING PROPERTY OF ELLIPSES


An elliptical water tank has a major axis of length 6 feet and a minor axis of length 4 feet.

The source of high-energy shock waves from a lithotripter is placed at one focus of the tank.

To smash the kidney stone of a patient, how far should the stone be positioned from the source?

Example: Lithotripsy


Since the length of the major axis of the ellipse is 6 feet, we have 2a = 6; so a = 3.

Similarly, the minor axis of 4 feet gives 2b = 4 or b = 2. To find c, we use the equation c2 = a2 – b2. We have

c2 = 32 – 22 = 5. Therefore,


5.c


If we position the center of ellipse at (0, 0) and the major axis along the x-axis, then the foci of the ellipse are and .

The distance between these foci is ≈ 4.472 feet. The kidney stone should be positioned 4.472 feet from the source of the shock waves.


5,0 5,0

2 5


Section 8.4 The Hyperbola

1. Define a hyperbola.2. Find the asymptotes of a hyperbola.3. Graph a hyperbola.4. Translate hyperbolas.

SECTION 1.1


HYPERBOLA

A hyperbola is the set of all points in the plane, the difference of whose distances from two fixed points is constant.

The fixed points are called the foci of the hyperbola.


HYPERBOLA

Here is a hyperbola in standard position, with foci F1(–c, 0) and F2(c, 0) on the x-axis at equal distances from the origin. The two parts of the hyperbola are called branches.


EQUATION OF A HYPERBOLA

is called the standard form of the equation of a hyperbola with center (0, 0).

The x-intercepts are −a and a. The points corresponding to these x-intercepts are the vertices of the hyperbola.

2 2

2 21

x y

a b


PARTS OF A HYPERBOLAThe line segment joining the two vertices is the transverse axis. The center is the midpoint of the transverse axis. The line segment joining (0, –b) and (0, b) is the conjugate axis.


EQUATION OF A HYPERBOLA

Similarly, an equation of a hyperbola with center (0, 0) and foci (0, –c) and (0, c) on the y-axis is given by:

2 22 2 2

2 21, where

y xb c a

a b

Here the vertices are (0, − a) and (0, a). The transverse axis of length 2a of the graph of the equation lies on the y-axis, and its conjugate axis is the segment joining the points (−b, 0) and (0, b) of length 2b that lies on the x-axis.


MAIN FACTS ABOUT HYPERBOLAS CENTERED AT (0, 0)


Does the hyperbola have its transverse

axis on the x-axis or y-axis?

The orientation (left–right branches or up–down branches) of a hyperbola is determinedby noting where the minus sign occurs in the standard equation.

In the equation in this example, the minus sign precedes the y2-term, so the transverse axis is on the x-axis.

Example: Determining the Orientation of a Hyperbola

2 2

14 10

x y


Find the vertices and foci for the hyperbola28y2 – 36x2 = 63.

First, convert the equation to the standard form.

Example: Finding the Vertices and Foci from the Equation of a Hyperbola


Since the coefficient of x2 is negative, the transverse axis lies on the y-axis.

and

and therefore


2 9

4a 2 7

4b

2 2 2

9 7

4 416

44

c a b


Since we require a > 0 and c > 0, and c = 2.

The vertices of the hyperbola are and

The foci of the hyperbola are (0, −2) and (0, 2).


3

2a

30, .

2

30,

2


Find the standard form of the equation of a hyperbola with vertices (±4, 0) and foci (±5, 0).

Since the foci of the hyperbola are on the x-axis, the transverse axis lies on the x-axis.

The center of the hyperbola is midway between the foci, at (0, 0). The standard form of such a hyperbola is

Example: Finding the Equation of a Hyperbola


The distance a between the center (0, 0) to either vertex, (−4, 0) or (4, 0) is 4; so a = 4 and a2 = 16.

The distance c between the center to either focus, (5, 0) or (−5, 0) is 5; so c = 5 and c2 = 25.

Use b2 = c2 – a2: b2 = 25 − 16 = 9.

Substitute a2 = 16 and b2 = 9 into the standard

form to get

Example: Finding the Equation of a Hyperbola

2 2

1.16 9

x y


THE ASYMPTOTES OF A HYPERBOLA WITH CENTER (0, 0)

1. The graph of the hyperbola has

transverse axis along the x-axis and has the following two asymptotes:

2 2

2 21

x y

a b

and b b

y x y xa a


THE ASYMPTOTES OF A HYPERBOLA WITH CENTER (0, 0)

2. The graph of the hyperbola has

transverse axis along the y-axis and has the following two asymptotes:

2 2

2 21

y x

a b

and a a

y x y xb b


Determine the asymptotes of each hyperbola.

a. The hyperbola is of the form so a = 2 and b = 3.

Substituting these values into and

we get the asymptotes

Example: Finding the Asymptotes of a Hyperbola

2 2

2 21;

x y

a b

by x

a ,

by x

a

3 3 and .

2 2y x y x


b. The hyperbola has the form so a = 3 and b = 4.

Substituting these values into and we get the asymptotes

Example: Finding the Asymptotes of a Hyperbola

2 2

2 21;

y x

a b

ay x

b

ay x

b

3 3 and .

4 4y x y x


PROCEDURE FOR GRAPHING A HYPERBOLA CENTERED AT (0, 0)

Step 1 Write the equation in standard form.Determine transverse axis and the orientation of the hyperbola.

Step 2 Locate vertices and the endpoints ofthe conjugate axis.

Step 3 Lightly sketch the fundamental rectangle by drawing dashed lines parallel to the coordinate axes through the points in Step 2.


PROCEDURE FOR GRAPHING A HYPERBOLA CENTERED AT (0, 0)

Step 4 Sketch the asymptotes. Extend thediagonals of the fundamental rectangle. These are the asymptotes.

Step 5 Sketch the graph. Draw both branchesof the hyperbola through the vertices, approaching the asymptotes. The foci are located on the transverse axis, c units from the center, where c2 = a2 + b2.


MAIN PROPERTIES OF HYPERBOLAS CENTERED AT (h, k)

Standard Equationa > 0, b > 0

Equation of transverse axis y = k

Length of transverse axis 2a

Equation of conjugate axis x = h

Length of conjugate axis 2b

Center (h, k)

2 2

2 21;

x h y k

a b


Vertices (h − a, k), (h + a, k)

Endpoints of conjugate axis (h, k − b), (h, k + b)

Foci (h − c, k), (h + c, k)


c2 = a2 + b2

Asymptotes by k x h

a



Standard Equationa > 0, b > 0

Equation of transverse axis x = h

Length of transverse axis 2a

Equation of conjugate axis y = k

Length of conjugate axis 2b

Center (h, k)

2 2

2 21;

y k x h

a b



Vertices (h, k – a), (h, k + a)

Endpoints of conjugate axis (h – b, k), (h + b, k)

Foci (h, k – c), (h, k + c)


c2 = a2 + b2

Asymptotes ay k x h

b



PROCEDURE FOR SKETCHING THE GRAPH OF A HYPERBOLA CETNERED AT (h, k)

Step 1 Plot the center (h, k), and draw horizontal and vertical dashed lines through the center.

Step 2 Locate the vertices and the endpoints of the conjugate axis. Lightly sketch the fundamental rectangle, with sides parallel to the coordinate axes, through these points.


PROCEDURE FOR SKETCHING THE GRAPH OF A HYPERBOLA CETNERED AT (h, k)

Step 3 Sketch dashed lines through opposite vertices of the fundamental rectangle. These are the asymptotes.

Step 4 Draw both branches of the hyperbola, through the vertices and approaching the asymptotes.

Step 5 The foci are located on the transverse axis, c units from the center, where c2 = a2 + b2.


Show that 9x2 – 16y2 + 18x + 64y –199 = 0 is an equation of a hyperbola, and then graph the hyperbola.

Complete the squares on x and y.

2 2

2 2

2 2

2 2

9 18 16 64 199

9 2 16 4 199

2 4 199

9 1 1

9 1

6 2 1 4

91 4 64

4

6

x x y y

x x y y

x x y y

x y

Example: Graphing a Hyperbola


2 2

2 2

9 1 16 2 144

1 21

16 9

x y

x y

Steps 1-2 Locate the vertices.

Center (–1, 2); a2 = 16, a = 4; b2 = 9, b = 3

(h – a, k) = (–1– 4, 2) = (–5, 2)

(h + a, k) = (–1+ 4, 2) = (3, 2)

Vertices: (3, –1), (3, 5), (–5, 5), (–5, –1).

Draw the fundamental rectangle.



Step 3 Sketch the asymptotes. Extend the diagonals of the rectangle obtained in Step 2 to sketch the asymptotes:

3 32 1 and 2 1

4 4y x y x

Step 4 Sketch the graph. Draw two branches opening left and right, starting from the vertices (–5, 2) and (3, 2) and approaching the asymptotes.


copyright © 2014, 2010 pearson education, inc. chapter 8 analytical geometry copyright © 2014,...

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