Grade 12Mathematics Project

Student Name: Ahmed Mohamed Obaid AlKaabi

Id#: st1061110072

Section: 12 -04

Teacher: Mr.Abdul Salam

My wiki: http://mathcore-term2-project.weebly.com

Term 2

2014 - 2015

Conic Sections

2

Goals

- Students will interpret their understanding of applications of the conic sections.

- Students will use technology for research and communication: wiki/blog.

- Students will present their findings to the class through digital Media (Video/Booklet).

Theme

This project covers the application of topics in conic sections; students will apply the concepts and theorems they have learned on circles, parabolas, ellipses and hyperbolas.

Moreover students will be able to apply their knowledge in relation to various Engineering problems.

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Part 1: Parabola:-

i) Find at least three real life application pictures representing the parabola.

1- My Own Pictures:

2- Web Pictures:

These are five real-world structures: a roller coaster, the reflector from a flashlight, the base of the Eiffel Tower, the McDonald's arches, and the flight trajectory of NASA's zero-G simulator, known as the "vomit comet".

Prezi: http://prezi.com/h4yjttdy0cpn/real-world-examples-of-parabolas/

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ii) To hear what athletes are saying on the field or on the course, sports reporters use microphones with parabolic shields to concentrate the sound at a microphone.

For optimum audio reception, the microphone should be placed at the focus of this parabola. Determine the coordinates of the focus, then write the equation of the parabola.

(Hint: Write an equation for a parabola based on ordered pairs.)

Answer:

- Vertex at origin (0,0)- The point (0.75,2 .5 )belong to the graph.- The direction is horizontal right - General equation:x=a ( y−k)2+h

To find a:0.75=a (2.5−0 )2+0a=0.12

The equation will be:x=0.12( y−0)2+0

So, x=(0.12) y2 Focus=(h+ 1

4a, k)→(0+ 1

4 (0.12 ),0)→(2.08 ,0)

so the vertex will be 2512 ft to the right of the vertex and a=0.12

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Part 2:Solar System:

The elliptical orbit of Pluto has the greatest eccentricity among all planets of our solar system (𝑒 = 0.248). Pluto’s orbit has the sun at one focus and a major axis of 79.6 AU (astronomical units).

Consider Pluto’s orbit on a giant coordinate grid where x and y are measured in AU and the sun is at the origin. (As shown in the figure above)

A. Use the length of the major axis to determine a.

a=79.62

=39.8 A U

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B. Use the eccentricity and a to determine the focal radius c. (Round to 2 decimal places.)

c=ea=0.248 (39.8 )=9.87 AU

C. Write an equation of the form to model Pluto’s orbit. (Find b first.)c2=a2−b2

(9.87)2=(39.8)2−b2

b2≈1486.62

Equation:

(x−9.87)2

1584.04+ y2

1486.62=1

D. The perigee is when the Pluto is closest to the sun. The apogee is when it is furthest from the sun. What are the distances from the sun at these points?

Perigee=( a+c )=29.93 AU Apogee=(a−c )=49.67

Part 3: Bridge:-

Mohamed and Salim study a drawing of an ornamental bridge such as the one shown at the right. It shows an elliptical arch that spans a narrow strait of water. The arch they are studying can be modeled by the following formula

1. Mohamed wants to know the dimensions of the arch

a. Which term in the equation defines the horizontal axis?

x2

182.25

b. Write a comparative statement to show whether the major axis of the arch is horizontal or vertical

Since 182.5>132.25, the major axis is horizontal.

1)(

2

2

2

2

by

acx

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c. Mohamed says that the width of the bridge is 13.5 feet. Is he correct? Explain.

No; 13.5 ft is the distance from the center of the bridge to one side, so the width is 27ft 

Write an expression for the height of the bridge.

b=√132.25=11.5 ft

So the height will = 11.5 ft

2. Salim notes that this bridge is hardly high enough to pass under while standing up in a moderate-size boat. If he were to build a bridge, it would be at least 1.3 times as wide and twice as high.

a. Find the vertices and co-vertices of Salim’s bridge design.

vertices : (17.55,0 ) ,(−17.55,0)

C 0−Vertices : (0,23 ) ,(0,23)

b. Write an equation for the design of Salim’s bridge using his minimum dimensions

x2

308.00+ y2

529.00=1

Part 4: Application:-

A cross-section of a parabolic reflector is shown in the figure. The bulb is located at the focus and the opening at the focus is 10 cm

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y

x

i) Find an equation of the parabola.

1) Let the vertex (h, k) (0,0) “ at origin”

2) Letus Rectum = |1a|=10 cm

3) a is positive since it is opening right.

a=| 1letus rectum|=¿ | 110|=0.1

4) Focus: (h+ 14a

, k) (0+ 14(0.1)

,0) focus (4,0)

5) x=a¿ x=110

¿ x=110

y2

The equation of parabola is x=110

y2

ii) Find the diameter of the opening |CD|, 11 cm from the vertex.

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C= (11 , y ), D= (11 ,− y) to find y1 , y2we should substitute:

1) x1=110

y12; y1

2=10 x→ y1=√10 x √ (10 ) (11)→√110C=(11,110), D=(11 ,−110)

CD=√¿¿

The diameter is 2√110 cm

Part 5:

In the LORAN (LOng RAnge Navigation) radio navigation system, two radio stations located at A and B transmit simultaneous signals to a ship located at P. The onboard computer converts the time difference in receiving these signals into a distance difference |PA| |PB|, and this, according to the definition of a hyperbola, locates the ship on one branch of a hyperbola (see the figure).

Suppose that station B is located 400 mi due east of station A on a coastline. A ship received the signal from B 1200 microseconds (s) before it received the signal from A.

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(a) Assuming that radio signals travel at a speed of 980 ft/s, find an equation of the hyperbola on which the ship lies.

1- Horizontal 2- Center (0 ,0)3- F1 (−200 ,0 )& F2 (200,0 )4- d=√(−200−200)2+(0−0)2=400

5- c = 4002

=200mi

6- d=s . t=(1200×980 )=1176000 ft=245011

mi

7- According to definition PF 1−PF 2=2a

8- 2a=245011

mia=122511

mi

9- b2=c2−a2

b2=40000−( 122511 )2

=27598.1405mi

The final equation is x2

12401.8595− y2

27598.1405=1

(b) If the ship is due north of B, how far of the coastline is the ship?

x2

12401.8595− y2

27598.1405=1

(200)2

12401.8595− y2

27598.1405=1

− y2

27598.1405=1− (200)2

12401.8595

− y2

27598.1405=−2.225322783

y=√(−27598.1405 ) (−2.225322783 )=247.820mi y ≈248mi

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Rubrics

Criteria Points

12

4 3 2 1

Completeness of Tasks

20%

Tasks are totally completed and correct. (100%)

Tasks are partially completed,

ORPartially wrong.

(75%)

Tasks are partially

completed,AND

Partially wrong (50%).

Tasks are Attempted (25%

or less)____

Presentation and

Integration of Technology

70%

Students used one mean of technology. The tool used helped the student and was useful to support his project. Moreover, the student was able to explain the work he/she submitted confidently and fluently; he/she was able to answer all of colleagues and instructor’s questions

Student used a mean of technology but it was not that supportive to the topic. In addition, student was able to explain the work he/she submitted confidently and fluently and he/she reflected an understanding of his/her works. The student was able to answer most of colleagues and instructor’s questions.

Student was able to explain the work he/she submitted. Student reflected a shallow understanding of his/her work; she was able to answer some of colleagues and instructor’s questions,

Student use of technology was primitive and way below the level of other IAT students.Student was unable to explain the work he/she submitted. Student reflected no understanding of his/her work; he/she was unable to answer any of colleagues and instructor’s questions.

____

Creativity& enrichment

10%

Student had an outstanding addition in all aspects of his/her project.

Student had an outstanding addition in some aspects of his/her project.

Student had an outstanding addition in very few aspects of his/her project.

Student had an outstanding addition in no aspects of his/her project.

____

 This rubric is out of 100, percentage orientation.

To make the mark out of 30 (Student’s Mark/10*3)Total ----> ____