SECTION 11.5 Conic Sections - Cengage A To graph a parabola The conic sections are curves that can...

SECTION 11.5 • Conic Sections 1

S E C T I O N

OBJECTIVE A To graph a parabola

The conic sections are curves that can be constructed from the intersection of a plane anda right circular cone. The parabola, which was introduced earlier, is one of these curves.Here we will review some of that previous discussion and look at equations of parabolasthat were not discussed before.

Every parabola has an axis of symmetry and a vertex that is on the axis of symmetry. To understand the axis of symmetry, think of folding the paper along that axis. The twohalves of the curve will match up.

The graph of the equation , , is a parabola with the axis ofsymmetry parallel to the y-axis. The parabola opens up when and opens downwhen . When the parabola opens up, the vertex is the lowest point on the parabola.When the parabola opens down, the vertex is the highest point on the parabola.

The coordinates of the vertex can be found by completing the square.

Find the coordinates of the vertex of the parabola whose equation is .

The coefficient of is positive, so the parabola opensup. The vertex is the lowest point on the parabola, orthe point that has the least y-coordinate.

Because for all x, the least y-coordinateoccurs when , which occurs when .This means the x-coordinate of the vertex is 2.

To find the y-coordinate of the vertex, replace x inby 2 and solve for y.

The vertex is (2, 1).

� �2 � 2�2 � 1 � 1 y � �x � 2�2 � 1

y � �x � 2�2 � 1

x � 2�x � 2�2 � 0�x � 2�2 � 0

x2

y � �x � 2�2 � 1

y � �x2 � 4x � 4� � 4 � 5y � �x2 � 4x� � 5y � x2 � 4x � 5

y � x2 � 4x � 5

a � 0a � 0

a � 0y � ax2 � bx � c

x

y

Vertex

Axis ofsymmetry

Conic Sections11.5

The four conic sections(parabola, circle, ellipse, andhyperbola) are obtained byslicing a cone with planes ofvarious orientations.

y

x– 4 40

– 2

4

– 4

2

– 2 2

Vertex

Jenn

ifer W

adde

ll/Ho

ught

on M

ifflin

Com

pany

HOW TO • 1

• Group the terms involving x.• Complete the square on . Note that 4 is added

and subtracted. Because , the equation isnot changed.

• Factor the trinomial and combine like terms.

4 � 4 � 0x 2 � 4x

46951_21_11.5online.qxd 1/20/10 9:02 AM Page 1

2 CHAPTER 11 • Functions and Relations

By following the procedure of the last example and completing the square on the

equation , we find that the x-coordinate of the vertex is � . The

y-coordinate of the vertex can then be determined by substituting this value of x intoand solving for y.

Because the axis of symmetry is parallel to the y-axis and passes through the vertex, the

equation of the axis of symmetry is � .

Find the equation of the axis of symmetry and the coordinates of the vertex of the parabola whose equation is . Then sketch its graph.

x-coordinate: • Find the x-coordinate of the vertex and the axis of symmetry.

The x-coordinate of the vertex is 1.The equation of the axis of symmetry is .

To find the y-coordinate of the vertex, replace x by 1 and solve for y.

The vertex is (1, 4).

Because a is negative, the parabola opens down.

Find a few ordered pairs and use symmetry to sketchthe graph.


x-coordinate: • Find the x-coordinate of the vertex and the axis of symmetry.

The x-coordinate of the vertex is 0.

The equation of the axis of symmetry is .

To find the y-coordinate of the vertex, replace x by 0 and solve for y.

The vertex is .

Because a is positive, the parabola opens up.


�0, �2�

� 02 � 2 � �2

y � x2 � 2

x � 0

a � 1, b � 0�

b

2a� �

0

2�1�� 0

y � x2 � 2

� �3�1�2 � 6�1� � 1 � 4

y � �3x2 � 6x � 1

x � 1

a � �3, b � 6�

b

2a� �

6

2��3�� 1

y � �3x2 � 6x � 1

b2a

x �

y � ax2 � bx � c

b2a

y � ax2 � bx � cPoint of Interest

Golden Gate Bridge

The suspension cables forsome bridges, such as theGolden Gate bridge, hang inthe shape of a parabola.Parabolic shapes are alsoused for mirrors in telescopesand in certain antennadesigns.

Integrating

TechnologyThe instructions in theKeystroke Guide: Min andMax can be used to find thevertex of a parabola.

y

x– 4 40

4

– 4

– 2

2

2

– 2

y

x– 4

2

4

4

– 2 0

– 2

– 4

2

© G

alen

Row

ell/T

erra

/Cor

bis

HOW TO • 2

HOW TO • 3



The graph of an equation of the form ,, is also a parabola. In this case, the parabola opens

to the right when a is positive and opens to the left whena is negative.

For a parabola of this form, the y-coordinate of the vertexis � . The axis of symmetry is the line � .

Using the vertical-line test, the graph of a parabola of this form is not the graph of afunction. The graph of is a relation.


y-coordinate: • Find the y-coordinate of the vertex and the axis of symmetry.

The y-coordinate of the vertex is 2.


To find the x-coordinate of the vertex, replace y by 2 and solve for x.

The vertex is .

Since a is positive, the parabola opens to the right.



y-coordinate: • Find the y-coordinate of the vertex and the axis of symmetry.

The y-coordinate of the vertex is .


To find the x-coordinate of the vertex, replace y by and solve for x.

The vertex is .

Because a is negative, the parabola opens to the left.


��1, �1�

� �2��1�2 � 4��1� � 3 � �1

x � �2y2 � 4y � 3

�1

y � �1

�1

a � �2, b � �4�

b

2a� �

�4

2��2�� 1

x � �2y2 � 4y � 3

��3, 2�

� 2�2�2 � 8�2� � 5 � �3

x � 2y2 � 8y � 5

y � 2

a � 2, b � �8�

b

2a� �

�8

2�2�� 2

x � 2y2 � 8y � 5

x � ay2 � by � c

b2a

y �b

2a

a � 0x � ay2 � by � cTips for Success

Express in your own words the difference between theequation of a parabola thatopens up or down and theequation of a parabola thatopens right or left. Doing sowill help you to remember thedifference.

Point of Interest

Hypatia

Hypatia (c. 340–415) isconsidered the first prominentwoman mathematician. Shelectured in mathematics andphilosophy at the Museum in Alexandria, at that time themost distinguished place oflearning in the world. One ofthe topics on which Hypatialectured was conic sections.One historian has claimedthat with the death (actuallythe murder) of Hypatia, “thelong and glorious history ofGreek mathematics was at an end.”

x

y

Vertex

Axis ofsymmetry

y

x– 4 2 40– 2

– 2

4

– 4

2

y

x– 4

2

40

– 2

4

– 4

2– 2

© B

ettm

ann/

Corb

is

HOW TO • 4

HOW TO • 5



EXAMPLE • 1 YOU TRY IT • 1Find the equation of the axis of symmetry and thecoordinates of the vertex of the parabola whoseequation is . Then sketch its graph.

Solution

Axis of symmetry:

Vertex:

Find the equation of the axis of symmetry and thecoordinates of the vertex of the parabola whoseequation is . Then sketch its graph.

Your solution

x

y

– 4 2

4

– 4

– 20 4

2

– 2

y � x2 � 2x � 1

�2, �1�

� �1 y � 22 � 4�2� � 3 y � x2 � 4x � 3

x � 2

x

y

– 4

4

– 4

– 20 4

2

– 2

�b

2a� �

�4

2�1�� 2

y � x2 � 4x � 3

Solutions on p. S1


Solution

Axis of symmetry:

Vertex:

Find the equation of the axis of symmetry and thecoordinates of the vertex of the parabola whoseequation is . Then sketch its graph.

Your solution

x

y

2

– 4 4

– 4

0– 2– 2

4

2

y � x2 � 2x � 1

�0, 1�

� 1 y � 02 � 1 y � x2 � 1

x � 0

x

y

– 4 42

4

– 4

– 20– 2

2

�b

2a� �

0

2�1�� 0

y � x2 � 1


Solution

Axis of symmetry:

Vertex:

Find the equation of the axis of symmetry and thecoordinates of the vertex of the parabola whoseequation is . Then sketch itsgraph.

Your solution

x

y

– 4 2

4

– 4

– 20 4

2

– 2

x � �y2 � 2y � 2

��1, 1�

� �1 x � 2�1�2 � 4�1� � 1 x � 2y2 � 4y � 1

y � 1

x

y

– 4

4

– 4

– 20

2

– 2 2 4

�b

2a� �

�4

2�2�� 1

x � 2y2 � 4y � 1



A circle is a conic section formed by the intersection of a cone and a plane parallel to thebase of the cone.

A circle can be defined as all points in the plane that are a fixed distance from agiven point , called the center. The fixed distance is the radius of the circle.�h, k�

�x, y�

x

y

radiuscenter

r

( )h k,

Find the radius and the coordinates of the center of the circlewhose equation is . Then sketch a graph of the circle.

To find the radius and the coordinates of the center,write the equation in standard form. The standard formis . The center is ,and the radius is 3. The graph is shown at the right.

(1, �2)(x � 1)2 � [y � (�2)]2 � 32

(x � 1)2 � (y � 2)2 � 9

OBJECTIVE B To find the equation of a circle and then graph the circle

Take NoteAs the angle of the plane thatintersects the cone changes,different conic sections areformed. For a parabola, theplane was parallel to theside of the cone. For a circle,the plane is parallel to thebase of the cone.

The Standard Form of the Equation of a Circle

Let r be the radius of a circle and let be the coordinates of the center of the circle. Then theequation of the circle is given by

�x � h�2 � � y � k �2 � r 2

�h, k �

HOW TO • 6

x

y

– 4

4

– 20

2

2

– 4

– 2 4

Find the equation of the circle with radius 4 and center .Then sketch its graph.

• Use the standard form of the equation of a circle.

• Replace h by , k by 2, and r by 4.

• Sketch the graph by drawing a circle with centerand radius 4.��1, 2�

x

y

4

0

2

– 4

– 2 4– 4– 2

2

�x � 1�2 � � y � 2�2 � 16

�1 �x � ��1�]2 � � y � 2�2 � 42

�x � h�2 � � y � k�2 � r 2

��1, 2�HOW TO • 7

Take NoteApplying the vertical-line testreveals that the graph of acircle is not the graph of afunction. The graph of a circleis the graph of a relation.



Find the equation of the circle whose center is (2, 1) and thatpasses through the point whose coordinates are .

The distance between the given point on the circle and the center is the radius. Seethe figure at the left. Use the distance formula, where is the given point onthe circle and is the center (2, 1), to find the radius.

• Use the distance formula.

• ;

Use the equation of a circle in standard form with and.

�x � 2�2 � � y � 1�2 � 26

�x � h�2 � � y � k�2 � r 2

r 2 � ��26 �2 � 26�h, k� � �2, 1�

� �26

� �1 � 25

� �12 � ��5�2

�x2, y2� � (2, 1)�x1, y1� � (3, �4) � ��3 � 2�2 � ��4 � 1�2

r � ��x1 � x2�2 � � y1 � y2�2

�x2, y2��3, �4��x1, y1�

�3, �4�

y

x

(2, 1)

(x − 2)2 + (y − 1)2 = 26

(3, −4)

– 8 8

8

– 8

4

40– 4

– 4

EXAMPLE • 4 YOU TRY IT • 4

Sketch a graph of .

Solution

Center:Radius:

Sketch a graph of .

Your solution

x

y

– 4

4

2

2– 2– 2

0

– 4

4

�x � 2�2 � � y � 3�2 � 9

�4 � 2

x

y

2

– 4

4– 4

4

– 20

2

– 2

��2, 1�

�x � 2�2 � � y � 1�2 � 4

HOW TO • 8

EXAMPLE • 5 YOU TRY IT • 5Find the equation of the circle with radius 5 andcenter . Then sketch its graph.

Solution

•

Find the equation of the circle with radius 4 andcenter . Then sketch its graph.

Your solution

y

x– 8 8– 4

8

– 8

4

40

– 4

�2, �3�

x

y

4

4

6

0

2

– 2– 2

– 4 2

8

�x � 1�2 � � y � 3�2 � 25r � 5k � 3,h � �1, �x � ��1�� 2 � � y � 3�2 � 52

�x � h�2 � � y � k�2 � r 2

��1, 3�

Solution on p. S1


The orbits of the planets around the sun are “oval” shaped. This oval shape can bedescribed as an ellipse, which is another of the conic sections.

There are two axes of symmetry for an ellipse. Theintersection of these two axes is the center of theellipse.

An ellipse with center at the origin is shown at the right. Note that there are two x-intercepts and two y-intercepts.

Using the vertical-line test, we find that the graph of anellipse is not the graph of a function. The graph of anellipse is the graph of a relation.

x

y

By finding the x- and y-intercepts of an ellipse and using the fact that the ellipse is oval-shaped, we can sketch a graph of the ellipse.

Find the coordinates of the intercepts of the ellipse

whose equation is . Then sketch its graph.

The x-intercepts are (3, 0) and . •The y-intercepts are (0, 2) and .

• Use the intercepts and symmetry tosketch the graph of the ellipse.

x

y

4

0

– 4

4– 4

2

– 2– 2

2

�0, �2�a 2 � 9, b 2 � 4��3, 0�

� 1y 2

4�

x2

9

OBJECTIVE C To graph an ellipse with center at the origin

Point of InterestThe word ellipse comes from the Greek word ellipsis,which means “deficient.” Themethod by which the earlyGreeks analyzed the conicscaused a certain area in theconstruction of the ellipse to be less than another area (deficient). The wordellipsis in English, meaning“omission,” has the sameGreek root as the wordellipse.

x(a, 0)(−a, 0)

(0, −b)

(0, b)

y

The Standard Form of the Equation of an Ellipse with Center at the Origin

The equation of an ellipse with center at the origin is .

The x-intercepts are and . The y-intercepts are and .�0, �b ��0, b ��a, 0��a, 0�

� 1y 2

b 2�x 2

a 2

HOW TO • 9






The x-intercepts are (4, 0) and . •The y-intercepts are (0, 4) and .

• Use the intercepts and symmetry tosketch the graph of the ellipse.

The graph in this last example is the graph of a circle. A circle is a special case of an

ellipse. It occurs when in the equation .� 1y2

b2�x2

a2a2 � b2

x

y

0

2

– 2– 2

2

4

– 4

4– 4

�0, �4�a 2 � 16, b 2 � 16��4, 0�

� 1y2

16�

x2

16Point of Interest

For a circle, and thus

. Early Greek

astronomers thought thateach planet had a circularorbit. Today we know that theplanets have elliptical orbits.However, in most cases, theellipse is very nearly a circle.

For Earth, .ab

1.00014

ab

� 1

a � b

EXAMPLE • 6 YOU TRY IT • 6Find the coordinates of the intercepts of the ellipse


Solution

x-intercepts:(3, 0) and

y-intercepts:(0, 4) and



Your solution

x

y

– 4

4

2

– 20 42– 2

– 4

� 1y2

25�

x2

4

�0, �4�

��3, 0�

x

y

2 4– 4– 2

0

2

– 4

4

– 2

a2 � 9, b2 � 16

� 1y2

16�

x2

9

Solutions on p. S1

EXAMPLE • 7 YOU TRY IT • 7Find the coordinates of the intercepts of the ellipse


Solution

x-intercepts:(4, 0) and

y-intercepts:

and



Your solution

x

y

– 4

4

2

– 20 42– 2

– 4

� 1y2

9�

x2

18

�2�3 3.5��0, �2�3 �

�0, 2�3 �

��4, 0�

x

y

2

– 4

4

– 20

2

– 2 4– 4

a2 � 16, b2 � 12

� 1y2

12�

x2

16

HOW TO •



A hyperbola is a conic section that can be formed by the intersection of a cone and aplane perpendicular to the base of the cone.

The hyperbola has two vertices and an axis of symmetry that passes through thevertices. The center of a hyperbola is the point halfway between the vertices.

The graphs at the right show twopossible graphs of a hyperbola withcenter at the origin.

In the first graph, an axis of symmetry isthe x-axis, and the vertices are x-intercepts.

In the second graph, an axis ofsymmetry is the y-axis, and the verticesare y-intercepts.

Note that in either case, the graph of a hyperbola is not the graph of a function. The graphof a hyperbola is the graph of a relation.

To sketch a hyperbola, it is helpful to draw two lines thatare “approached” by the hyperbola. These two lines arecalled asymptotes. As a point on the hyperbola getsfarther from the origin, the hyperbola “gets closer to” theasymptotes.

Because the asymptotes are straight lines, their equationsare linear equations. The equations of the asymptotes for

a hyperbola with center at the origin are x and

x.b

ay � �

b

ay �

y

x

OBJECTIVE D To graph a hyperbola with center at the origin

Point of InterestHyperbolas are used inLORAN (LOng RAngeNavigation) as a method bywhich a ship’s navigator candetermine the position of theship. A minimal systemincludes three stations thattransmit pulses at preciselytimed intervals. A user getslocation information bymeasuring the very smalldifference in arrival times ofthe pulses. You can find moreinformation on LORAN on theInternet.

Station2Station

1

Station3

Point of InterestThe word hyperbola comesfrom the Greek wordyperboli, which means“exceeding.” The method bywhich the early Greeksanalyzed the conics caused acertain area in theconstruction of the hyperbolato be greater than (to exceed)another area. The wordhyperbole in English,meaning “exaggeration,” hasthe same Greek root as theword hyperbola.

The word asymptotecomes from the Greek wordasymptotos, which means “not capable of meeting.”

y

x

y

x

x

y

y = xba

y = − xba

− = 1x2

a2

y2

b2

The Standard Form of the Equation of a Hyperbola with Center at the Origin

The equation of a hyperbola for which an axis of symmetry is the x-axis is .

The coordinates of the vertices are and .

The equation of a hyperbola for which an axis of symmetry is the y-axis is .

The coordinates of the vertices are and .

For each equation, the equations of the asymptotes are x and x.ba

y �ba

y � �

�0, �b ��0, b �� 1

x 2

a 2�y 2

b 2

��a, 0��a, 0�� 1

y 2

b 2�x 2

a 2



Find the coordinates of the vertices and the equations of the

asymptotes of the hyperbola whose equation is . Then sketch its graph.

An axis of symmetry is the y-axis.The coordinates of the vertices are (0, 3) and . •

The equations of the asymptotes are x and x.

• Locate the vertices. Sketch the asymptotes. Use symmetry and the fact that the hyperbola will approach the asymptotes to sketch the graph.

y

x8

8

– 8 4

– 8

0– 4

4

– 4

3

2y � �

3

2y �

b 2 � 9, a 2 � 4�0, �3�

� 1x2

4�

y2

9

Tips for SuccessYou now have completed thelessons on the four conicsections. You need to be ableto recognize the equation ofeach. To test yourself, try theChapter 11 Review Exercises.

EXAMPLE • 8 YOU TRY IT • 8Find the coordinates of the vertices and the equationsof the asymptotes of the hyperbola whose equation is

. Then sketch its graph.

Solution

Axis of symmetry:x-axis

Vertices:(4, 0) and

Asymptotes:

x and x

Find the coordinates of the vertices and the equationsof the asymptotes of the hyperbola whose equation is


Your solution

y

x8

8

– 8

– 8

4

4

– 4

– 4 0

� 1y2

25�

x2

9

�1

2y �

1

2y �

��4, 0�

y

x8

8

– 8

– 8

0

4

– 4

4– 4

a2 � 16, b2 � 4

� 1y2

4�

x2

16

Solutions on p. S1

EXAMPLE • 9 YOU TRY IT • 9Find the coordinates of the vertices and the equationsof the asymptotes of the hyperbola whose equation is


Solution

Axis of symmetry:y-axis

Vertices:(0, 4) and

Asymptotes:

x and x

Find the coordinates of the vertices and the equationsof the asymptotes of the hyperbola whose equation is


Your solution

y

x8

8

– 8

– 8

4– 4 0

4

– 4

� 1x2

9�

y2

9

�4

5y �

4

5y �

�0, �4�

y

x8

8

– 8

– 8

0 4– 4

4

– 4

a2 � 25, b2 � 16

� 1x2

25�

y2

16

HOW TO • 11



For Exercises 1 to 6, a. state whether the axis of symmetry is a vertical or a horizontalline, and b. state the direction in which the parabola opens.

1. 2. 3.

4. 5. 6.

For Exercises 7 to 24, find the equation of the axis of symmetry and the coordinates ofthe vertex of the parabola given by the equation. Then sketch its graph.

7. 8. 9.

10. 11. 12.

13. 14. 15.

x

y

2

– 4

4

– 4

420– 2

– 2

4

y

x– 8 80

– 4

8

4– 4

– 8

x

y

2

4

– 4

4– 2 0– 2

2– 4

y �1

2 x2 � x � 3y � �

1

2 x2 � 2x � 6x � �

1

2 y2 � 2y � 3

x

y

2

– 4

4

– 4

4– 2 0– 2

2x

y

2

– 4

4

– 4

4– 2 0– 2

2x

y

2

– 4

4

– 4

4– 2 0– 2

2

x �1

2 y2 � y � 1x � �

1

4 y2 � 1x � �

1

2 y2 � 4

x

y

– 4

4

– 4

42– 2 0– 2

2

x

y

2

– 4

4

– 4

42– 2 0– 2

y

x– 8 80

– 4

8

4

4

– 8

– 4

y � x2 � 2y � x2 � 2x � y2 � 3y � 4

y �1

4x2 � 6x � 1x � �

1

2y2 � 4y � 7x � �3y2 � y � 9

x � y2 � 2y � 8y � �x2 � 5x � 2y � 3x2 � 4x � 7

11.5 EXERCISES

OBJECTIVE A To graph a parabola



16. 17. 18.

19. 20. 21.

22. 23. 24.

Science Parabolas have a unique property that is important in the design of telescopesand antennas. If a parabola has a mirrored surface, then all light rays parallel to the axisof symmetry of the parabola are reflected to a single point called the focus of theparabola. The location of this point is p units from the vertex on the axis of symmetry.

The value of p is given by , where is the equation of a parabola with vertex

at the origin. For the graph of shown at the right, the coordinates of the focus are

. For Exercises 25 and 26, find the coordinates of the focus of the parabola given bythe equation.

25. 26. y �1

10x2y � 2x2

�0, 1�x21

4y �

y � ax21

4ap �

y

x– 8 80– 4

8

– 8

4

4

– 4

x

y

2

– 4

4

– 4

420– 2

– 2x

y

2

– 4

4

– 4

420– 2

– 2

x � y2 � 2y � 5y � x2 � 5x � 6y � x2 � 5x � 4

x

y

2

– 4

4

– 4

420– 2

– 2x

y

2

– 4

4

– 4

420– 2

– 2

4

y

x– 8 80

– 4

8

4– 4

– 8

y � 2x2 � x � 3y � 2x2 � x � 5y � 2x2 � 4x � 5

x

y

2

– 4

4

– 4

420– 2

– 2

y

x– 8 8

8

– 8

4

4

– 4

0– 4

y

x– 8 80– 4

8

– 8

4

4

– 4

x � �y2 � 2y � 3x � y2 � y � 6x � y2 � 6y � 5

xpy = x2

y Focus

Parallel rays of light arereflected to the focus

–3 3

14

a = 14

p = = = 114a

14(1/4)

3



For Exercises 27 to 32, sketch a graph of the circle given by the equation.

27. 28. 29.

30. 31. 32.y

x8

8

– 8

4

0– 8

– 4

4– 4x

y

– 4

4

2

– 4

42– 2

– 2 0x

y

– 4

4

– 20

2

– 4

– 2 42

�x � 1�2 � � y � 2�2 � 25�x � 2�2 � � y � 2�2 � 4�x � 2�2 � � y � 3�2 � 4

y

x– 8 8– 4

8

– 8

4

40

– 4

y

x– 8 8– 4

8

– 8

4

40

– 4

x

y

– 4

4

2

2– 2

0

– 4

4– 2

�x � 3�2 � � y � 1�2 � 25�x � 2�2 � � y � 3�2 � 16�x � 2�2 � � y � 2�2 � 9

33. Find the equation of the circle with radius 2 andcenter .

35. Find the equation of the circle with radius andcenter .


39. Find the equation of the circle whose center isand that passes through the point whose

coordinates are .


coordinates are .


coordinates are .

45. What is the equation of a circle whose center is the origin and whose radius is r?


36. Find the equation of the circle with radius and center .



coordinates are .


coordinates are .


coordinates are .

46. Describe the graph of .�x � 3�2 � �y � 2�2 � 0

�4, �2��0, 0�

�2, 0��5, �4�

��1, 3��2, 1�

�0, �4�

�5, �2��13

��1, �2�

�1, �5��0, 3�

��3, �4��1, 0�

�1, 2��1, 1�

��3, 0�

�0, �4��7

�2, �1�

OBJECTIVE B To find the equation of a circle

and then graph the circle



For Exercises 47 to 55, find the coordinates of the intercepts of the ellipse given by theequation. Then sketch its graph.

47. 48. 49.

50. 51. 52.

53. 54. 55.

56. Let � � 1, where a � b. If b remains 57. Let � � 1, where a � b. If a remains

fixed and a increases, does the graph of the fixed and the values of b get closer to a, does the ellipse get flatter or rounder? graph of the ellipse get flatter or rounder?

y2

b2

x2

a2

y2

b2

x2

a2

y

x– 8 4 8

8

– 8

0– 4

4

– 4

y

x– 8 8

8

– 8

4

– 4

0 4– 4

y

x8

8

– 8

– 8

0

4

– 4

4– 4

x2

4�

y2

25� 1

x2

25�

y2

36� 1

x2

16�

y2

49� 1

y

x8– 8 0

4

– 4

4– 4

8

– 8

y

x8

8

– 8

– 8

0 4– 4

4

– 4

y

x8

8

– 8

– 8

0

4

– 4

4– 4

x2

49�

y2

64� 1

x2

36�

y2

16� 1

x2

16�

y2

9� 1

y

x8

8

– 8

– 8

0

4

– 4

4– 4

y

x8

8

– 8

– 8

0 4– 4

4

– 4

y

x8

8

– 8

– 8

4– 4 0

4

– 4

x2

25�

y2

9� 1

x2

25�

y2

16� 1

x2

4�

y2

9� 1

For Exercises 58 to 69, find the coordinates of the vertices and the equations of theasymptotes of the hyperbola given by the equation. Then sketch its graph.

58. 59. 60.

y

x8

8

– 8 4

– 8

0– 4

– 4

4

y

x8

8

4

– 8

– 4

– 8

0 4– 4

y

x8

8

– 8

4

– 8

– 4

– 4 40

y2

16�

x2

9� 1

x2

25�

y2

4� 1

x2

9�

y2

16� 1

OBJECTIVE C To graph an ellipse with center at the origin

OBJECTIVE D To graph a hyperbola with center at the origin



61. 62. 63.

64. 65. 66.

67. 68. 69.

70. Are the vertices of � � �1 on the x-axis or the y-axis?

Applying the Concepts

For Exercises 71 to 76, write the equation in standard form and identify the graph. Thengraph the equation.

71. 72. 73.

74. 75. 76.y

x8

8

– 8 4

– 8

– 4

4

0

– 4

y

x8

8

– 8

– 8 40

4

– 4

– 4

y

x8

8

4

– 8

– 4

– 8

4– 4 0

9y2 � 16x2 � 1444y2 � x2 � 369x2 � 25y2 � 225

y

x– 8 4 8

8

– 4

4

– 8

– 4

0

y

x4

4

– 4

– 4 20

2

– 2

– 2

y

x8

8

4

– 8

– 4

– 8

0 4– 4

25y2 � 4x2 � �1004x2 � 9y2 � 3616x2 � 25y2 � 400

x2

9

y2

4

y

x– 8 4 8

8

– 4

4

– 8

– 4

0

y

x– 8 8

8

– 8

4

– 4

4– 4 0

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

y2

25�

x2

4� 1

y2

9�

x2

36� 1

x2

36�

y2

9� 1

y

x8

8

– 8

– 8

– 4 40

4

– 4

y

x8

8

4

– 8

– 8

– 4

0– 4 4

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

x2

4�

y2

25� 1

y2

4�

x2

16� 1

y2

25�

x2

9� 1

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

y

x8

8

4

– 8

– 8

– 4

0– 4 4

y

x8

8

– 8

0– 8 4– 4

– 4

4

x2

9�

y2

49� 1

x2

16�

y2

4� 1

y2

16�

x2

25� 1


SOLUTIONS TO CHAPTER 11 “YOU TRY IT”

SECTION 11.5

You Try It 1

Axis of symmetry:

Vertex:

You Try It 2

Axis of symmetry:

Vertex:

You Try It 3

Axis of symmetry:

Vertex:

You Try It 4 • Center:Radius:

You Try It 5

You Try It 6

x-intercepts:and

y-intercepts:and

You Try It 7

x-intercepts:and

y-intercepts:and

You Try It 8

Axis of symmetry:x-axis

Vertices:and

Asymptotes:

x and x

You Try It 9

Axis of symmetry:y-axis

Vertices:and

Asymptotes:and y � �xy � x

�0, �3��0, 3�

y

x8

8

– 8

– 8

4– 4 0

4

– 4

a2 � 9, b2 � 9

�5

3 y �

5

3 y �

��3, 0��3, 0�

y

x8

8

– 8

– 8

4

4

– 4

– 4 0

a2 � 9, b2 � 25

3�2 41

4��0, �3��0, 3�

��3�2, 0��3�2, 0�

x

y

2

2– 2

0

4

– 2

– 4

– 4 4

a2 � 18, b2 � 9

�0, �5��0, 5�

��2, 0��2, 0�

x

y

– 4

4

2

– 20 42– 2

– 4

a2 � 4, b2 � 25

y

x– 8 8– 4

8

– 8

4

40

– 4

�x � 2�2 � � y � 3�2 � 16�x � 2�2 � � y � ��3�]2 � 42

�x � h�2 � � y � k�2 � r2

�9 � 3(2, �3)

x

y

– 4

4

2

2– 2– 2

0

– 4

4

�1, �2� � �2

y � 12 � 2�1� � 1x � 1

�b

2a� �

�2

2�1�� 1

x

y

2

– 4 4

– 4

0– 2– 2

4

2

y � x2 � 2x � 1

�3, �1� � 3

x � ��1�2 � 2��1� � 2y � �1

�b

2a� �

�2

2��1�� 1

x

y

– 4 2

4

– 4

– 20 4

2

– 2

x � �y2 � 2y � 2

��1, 0� � 0

y � ��1�2 � 2��1� � 1x � �1

�b

2a� �

2

2�1�� 1

x

y

– 4 2

4

– 4

– 20 4

2

– 2

y � x2 � 2x � 1

Solutions to You Try It S1

•

r � 4k � �3,h � 2,

46951_21_11.5online.qxd 1/20/10 9:02 AM Page S1

ANSWERS TO CHAPTER 11 SELECTED EXERCISES

SECTION 11.51. a. Vertical line b. The parabola opens up. 3. a. Horizontal line b. The parabola opens right.

5. a. Horizontal line b. The parabola opens left.

7. 9. 11.

Axis of symmetry: Axis of symmetry: Axis of symmetry:

Vertex: , Vertex: Vertex:

13. 15. 17.


Vertex: Vertex: , Vertex: ,

19. 21. 23. 25. 0,


Vertex: Vertex: , Vertex: ,

27. 29. 31. 33.

35. 37. 39. 41.

43. 45.

47. x-intercepts: and 49. x-intercepts: and 51. x-intercepts: and

y-intercepts: and y-intercepts: and y-intercepts: and

y

x8

8

– 8

– 8

0 4– 4

4

– 4

y

x8

8

– 8

– 8

0

4

– 4

4– 4

y

x8

8

– 8

– 8

4– 4 0

4

– 4

�0, 4��0, �4��0, 3��0, �3��0, 3��0, �3��6, 0��6, 0��5, 0��5, 0��2, 0��2, 0�

x2 � y2 � r2x2 � � y � 3�2 � 65

�x � 1�2 � y2 � 20�x � 1�2 � � y � 1�2 � 5�x � 3�2 � y2 � 81x2 � � y � 4�2 � 7

�x � 2�2 � � y � 1�2 � 4

x

y

– 4

4

2

– 4

42– 2

– 2 0

y

x4 8

8

– 8

4

0– 4– 8

– 4

x

y

– 4

4

2

2– 2 4– 2

0

– 4

�1

4�

5

2��25

8�

1

4��1, �7�

5

2x � �

1

4x �x � �1

�1

8

x

y

2

– 4

– 4

4– 2 0– 2

2

4

x

y

– 4

4

– 4

42– 2

2

0– 2

y

x– 8 80– 4

8

4

4

– 4

– 8

�1

2

25

4��7

2��1��1, 2�

1

2y �x � �1y � 2

y

x– 8 8

8

– 8

4

4

– 4

0– 4x

y

2

– 4

4

– 4

420– 2

– 2x

y

2

4

– 4

4– 2 0– 2

2– 4

��1, 0��0, 2��3

2

25

4�y � 0x � 03

2y �

x

y

2

– 4

4

– 4

4– 2 0– 2

2x

y

– 4

4

– 4

42– 2 0– 2

2

y

x– 8 80– 4

– 4

8

– 8

4

4

Answers to Selected Exercises A1

46951_21_11.5online.qxd 1/20/10 9:02 AM Page A1

A2 CHAPTER 11 • Functions and Relations

53. x-intercepts: and 55. x-intercepts: and 57. Rounder

y-intercepts: and y-intercepts: and

59. Vertices: and 61. Vertices: and 63. Vertices: and

Asymptotes: x and x Asymptotes: x and x Asymptotes: x and x

65. Vertices: and 67. Vertices: and 69. Vertices: and

Asymptotes: x and x Asymptotes: x and x Asymptotes: x and x

71. , ellipse 73. , hyperbola 75. , hyperbola

y

x8

8

– 8

– 8 40

4

– 4

– 4

y

x8

8

– 8

0– 8 4– 4

– 4

4

y

x8

8

4

– 8

– 4

– 8

0 4– 4

� 1x2

36�

y2

9� 1y2

4�

x2

25� 1y2

16�

x2

25

y

x– 8 4 8

8

– 4

4

– 8

– 4

0

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

y

x8

8

4

– 8

– 8

– 4

0– 4 4

5

2y � 5

2y � �1

2y � 1

2y � �1

2y � 1

2y � �

�0, 5��0, �5��6, 0��6, 0��0, 2��0, �2�

y

x8

8

4

– 8

– 8

– 4 4

– 4

0

y

x8

8

– 8

0– 8 4– 4

– 4

4

y

x8

8

4

– 8

– 4

– 8

0 4– 4

7

3y � 7

3y � �4

5y � 4

5y � �2

5y � 2

5y � �

�3, 0��3, 0��0, 4��0, �4��5, 0��5, 0�

y

x– 8 4 8

8

– 8

0– 4

4

– 4

y

x8

8

– 8

– 8

0

4

– 4

4– 4

�0, 5��0, �5��0, 7��0, �7�

�2, 0��2, 0��4, 0��4, 0�

46951_21_11.5online.qxd 1/20/10 9:02 AM Page A2

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