rev.s08 mac 1140 module 11 conic sections. 2 rev.s08 learning objectives upon completing this...

Rev.S08

MAC 1140

Module 11

Conic Sections

2Rev.S08

Learning Objectives

• Upon completing this module, you should be able to

• find equations of parabolas.• graph parabolas.• use the reflective property of parabolas.• translate parabolas.• find equations of ellipses.• graph ellipses.• use the reflective property of ellipses.

http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

http://faculty.valenciacc.edu/ashaw/

3Rev.S08

Learning Objectives (Cont.)

• 8. find the center and radius of a circle.• 9. solve systems of nonlinear equations and inequalities.• 10. find equations of hyperbolas.• 11. graph hyperbolas.• 12. use the reflective property of hyperbolas.• 13. translate hyperbolas.• 14. solve systems of nonlinear equations.



4Rev.S08

Conic Sections


10.110.1 ParabolasParabolas

10.210.2 EllipsesEllipses

10.3 Hyperbolas10.3 Hyperbolas

There are three sections in this module:


5Rev.S08

What are Conic Sections?


Conic Sections are named after the different ways a plane can intersect a cone. The three basic conic sections are parabolas, ellipses, and hyperbolas. A circle is an example of an ellipse.


6Rev.S08

What are Focus and Directrix?


In previous module, we learned that a parabola with vertex (0,0) can be represented symbolically by y = a x2 .

A parabola is the set of points in a plane that are equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.

In this graph, the parabolas has a

vertical axis of symmetry;

the fixed point is (0, -1) and

the fixed line is y = 1.

y = 1

Focus


7Rev.S08

Equation of a Parabola with Vertex (0,0)


If the value of p is known, then the equation of a parabola with vertex (0,0) can be written as one of the following equations:

Vertical axis of symmetry: The parabola with a focus at (0,p) and directrix y = -p has equation x2 = 4py. The parabola opens upward if p > 0 and downward if p < 0.

Horizontal axis of symmetry: The parabola with a focus at (p,0) and directrix x = -p has equation y2 = 4px. This parabola opens to the right if p > 0 and to the left if p < 0.


8Rev.S08

Example of Finding Focus and Directrix


Let’s try to find the focus and directrix for x2 = -4y.

Since this parabola has

a vertical axis of symmetry, it

has equation

x2 = 4py

so, −4 = 4p

−1 = p

Since p < 0, the parabola opens downward, and

Focus: (0, p) = (0, −1) Directrix: y = - p

y = -(-1)=1

y = 1

Focus


9Rev.S08

Example of Finding Focus and Directrix


Let’s try to find the focus and directrix for y2 = 8x.

Since this parabola has

a horizontal axis of symmetry, it

has equation

y2 = 4px

so, 8 = 4p

= p

Since p > 0, the parabola opens to the right, and

Focus: (p, 0) = (2, ) Directrix: x = - p

x = - 2

x = −2

F(2, 0)

V(0, 0)


10

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How to Find the Equation of the Parabola?


Find the equation of the parabola with focus (0.5, 0) and vertex at the origin.

Solution

The given focus is at (p, 0), so the parabola has a horizontal axis of symmetry. The parabola opens to the right since p = 0.5 (p > 0.) The equation of this parabola:

y2 = 4px

y2 = 4(0.5)x

y2 = 2x


11

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Reflective Property of Parabola


When a parabola is rotated about its axis, it sweeps out a shape called a paraboloid.

Paraboloids have a special reflective property. When incoming parallel rays of light from the sun or distant star strike the surface of a paraboloid, each ray is reflected toward the focus.


12

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How to Use Translations of Graphs to Find the Equation of a Parabola?


If the equation of a parabola is given by x2 = 4py or y2 = 4px, we know the vertex is (0,0). If the vertex is (h,k), we can use translations of graphs to find the equation of the parabola. This equation can be obtained by replacing x with (x - h) and replacing y with (y - k).


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Example


Graph the parabola

Solution

Rewrite the equation in standard form.

The equation of this parabola:

(y - k)2 = 4p(x - h)

Vertex: (h, k) = (1, 3) 4p = 12

p > 0 opens right p = 3

Focus: (h+p,k) = (1+3,3) = (4, 3)

Directrix: x = h - p = 1 - 3 = −2


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Example


Find the equation of the parabola with focus (−2, 3) and directrix y = −3.

Solution

Sketch the focus and directrix.Parabola opens up with vertical axis of symmetry.

Focus: (h, k + p) = (-2, 3), so h = -2, k + p= 3 (1)

Directrix: y = k - p = -3 (2)

Add Equation (1) to Equation (2) => k = 0, p = 3

Vertex: (h, k) = (−2, 0)

Distance from focus to vertex is 3.

Equation: (x - h)2 = 4py => (x + 2)2 = 12y


15

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How to Find the Equation of a Parabolaby Completing the Square?


Write x2 + 6x + 4y + 5 = 0 in the form (x − h)2 = a(y − k).

Solution

Complete the square


16

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What is an Ellipse?


An ellipse is the set of points in a plane, the sum of whose distances from two fixed points is constant. Each fixed point is called a focus (plural foci) of the ellipse.

The major axis is the longer axis, it can be either horizontal or vertical.


17

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What are the Standard Equations for Ellipses Centered at the Origin?



18

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How to Graph Ellipses?


Horizontal Major Axis Vertical Major Axis


19

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Example


Sketch the graph of the ellipse 9x2 + 4y2 = 36.

SolutionFoci:

Endpoints of minor axis (2, 0)

Vertical major axis a = 2 and b = 3

c2 = 32 − 22 = 5 or c =

Vertices: (0, 3)

foci


20

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Example of Finding the Standard Equation of an Ellipse


Find the standard equation of the ellipse with foci (0, 3) and major axis length 8.

Solution

2a = 8

a = 4

c = 3

b2 = a2 – c2

= 42 – 32

= 7

Focus on y-axis use larger intercept for denominator of y2.


21

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Reflective Property of Ellipses


Like parabolas, ellipses also have an important reflective property. If an ellipse is rotated about the x-axis, an ellipsoid is formed, which resembles the shell of an egg. If a light source is placed at focus F1, then every beam of light emanating from the light source, regardless of direction, is reflected at the surface of an ellipsoid toward focus F2.


22

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What are the Standard Equations for Ellipses?


As we can see here, we use translations of graphs to write the equation of ellipses centered at (h,k) by replacing x with (x - h) and replacing y with (y - k) from the standard equations for ellipses centered at the origin.


23

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Example


Graph the ellipse whose equation is

Solution

Vertical major axis

Center: (2, −1)

a2 = 16, b2 = 9, c2 = 7

Vertices are 4 units above and below the center. (2, 3)(2, −5).

Foci:

center (2, −1)


24

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Example of Using Completing the Square to find the Standard Form for an Ellipse


Write 4x2 + 9y2 − 16x + 18y − 11 = 0 in the standard form for an ellipse centered at (h, k). Identify the center and vertices.

Solution Complete the square:

Center (2, −1)

Vertices: (−1, −1) and (5, −1)

25

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What is the Standard Form of a Circle?


The standard form of a circle with center (h, k) and radius r is (x – h)2 + (y – k)2 = r2.

Example: Find the center and radius of a circle given by x2 + y2 − 14x + 4y = 11.

Solution

center: (7, −2)

radius: 8

26

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How to Find the Area Inside an Ellipse?


Given the standard equation of an ellipse, the area A of the region contained inside is given by A = πab.

Example: Shade the region in the xy-plane that satisfies the inequality 9x2 + 4y2 < 36. Find the area of this region if units are in inches.

Solution

Write in standard form:

27

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How to Find the Area Inside an Ellipse? (cont.)


The region inside the ellipse satisfies the inequality.

A = πab

A = π(2)(3) ≈ 18.85 square inches

28

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What is a Hyperbola?


A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points is constant. Each fixed point is called a focus of the hyperbola.

The transverse axis is the line segment connecting the vertices, and its length equals 2a.

There are two branches and asymptotes to a hyperbola.

29

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What is a Hyperbola? (cont.)


The top hyperbola is having a horizontal transverse axis, and its length equals 2a. There are one left branch and one right branch.

The bottom hyperbola is having a vertical transverse axis, and its length equals 2a. There are one upper branch and one lower branch.

A hyperbola consists of two solid curves or branches. The other parts are aids for sketching its graph. Note: The dashed diagonal axes are asymptotes. The dashed rectangle is sometimes called the fundamental rectangle.

30

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What are the Standard Equations for Hyperbolas centered at the Origin?


31

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How to Graph a Hyperbola?


Sketch the graph of

Solution

a = 5 and b = 7

c2 = a2 + b2

c2 = 25 + 49 = 74

Asymptotes:

Vertices: (5, 0)

Foci:

32

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How to Find the Equation of the Hyperbola centered at the Origin?


Find the equation of the hyperbola centered at the origin with vertices (0, 4) and foci (0, 6).

Solution

a = 4, c = 6

b2 = c2 – a2

= 62 – 42

= 20

33

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Reflective Property of Hyperbolas


Hyperbolas have an important reflective property. If a hyperbola is rotated about the x-axis, a hyperboloid is formed. Any beam of light that is directed toward focus F1, will be reflected by the hyperboloid toward focus F2.

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What are the Standard Equations for Hyperbolas Centered at (h, k)?


35

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Example


Graph the hyperbola

Solution

a2 = 9, b2 = 4

c2 = a2 + b2

c2 = 9 + 4 = 13

Center: (−3, −2)

Vertices: (−3, −23)

Foci:

Asymptotes:

36

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Example


Write 4x2 + 8x – y2 – 4y = 4 in standard form for a hyperbola centered at (h, k). Identify the center and the vertices.

Solution

Center: (−1, −2)

Vertices:

37

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What have we learned?

• We have learned to

• find equations of parabolas.• graph parabolas.• use the reflective property of parabolas.• translate parabolas.• find equations of ellipses.• graph ellipses.• use the reflective property of ellipses.


38

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What have we learned? (Cont.)

• 8. find the center and radius of a circle.• 9. solve systems of nonlinear equations and inequalities.• 10. find equations of hyperbolas.• 11. graph hyperbolas.• 12. use the reflective property of hyperbolas.• 13. translate hyperbolas.• 14. solve systems of nonlinear equations.


39

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Credit

Some of these slides have been adapted/modified in part/whole from the slides of the following textbook:

• Rockswold, Gary, Precalculus with Modeling and Visualization, 3th Edition


rev.s08 mac 1140 module 11 conic sections. 2 rev.s08 learning objectives upon completing this...

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