T2

Mathematics Project 2013-2014

Student’s Name:

Ali Mohammed Alnaqbi & Khalid Jamal.

ID #: AT1001315 & AT1001445

Section: 12-03

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. 3

Part 1:Parabola:

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

2) 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, and then write the equation of the parabola. (Hint: Write an equation for a parabola based on ordered pairs.)

SOLUTION

Let the vertex be at origins (0,0) , use (0.75,2 .5) from the following figure. Since the graph opens left the equation will be used is x=a ( y−k )2+h and a is positive.

Write the equation:x=a ( y−k)2+h

0.75=a (2.5−0 )2+0

0.75=6.25a

a= 325

x= 325

( y−0 )2+0

x= 325

y2

Handheld Bookshelf door design Beanie

Step 1:   Replace (h, k) with vertex and replace (x, y) with the given point (0.75, 2.5).Step 2:Substitute the values and calculate (2.5−0¿2+0.Step 3:Divide both sides by 6.25 .Step 4: Substitute again to get the correct equation.

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. The major axis is79.6 AU , and the distance of major axis is 2a

Meaning 2a=79.6 AU .

Divide both sides by 2 to find a

2a2

=79.62

a is 39.8

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

c=0.248(39.8)

c=9.87 AU

C) Write an equation of the form ( x−c )2

a2 + y2

b2 =1 to model Pluto’s orbit.

(Find 𝑏 first.)

c2=a2−b2

(9.87 )2=(39.8 )2−b2

b2=1486.62 b=√1486.62 b=38.56

The equation is

( 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−c39.8−9.87=29.93 AU

Apogee=a+c 39.8+9.87=49.67 AU

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

x2

182.25+ y2

132.25=1

1. Mohamed wants to know the dimensions of the arch

Which term in the equation defines the horizontal axis? X and y, if the number under x is greater than the number under y then it is vertical.

Write a comparative statement to show whether the major axis of the arch is horizontal or vertical.If number under x is greater than number under y then it is horizontal and major axis is the square root of number under x

If number under y is greater than number under x then it is vertical and major axis is the square root of number under y.

Mohamed says that the width of the bridge is 13.5 feet. Is he correct? Explain. He is correct because width is sometime known as x in math, therefore the square root of 182.25 is 13.5.

Write an expression for the height of the bridge. The height is square root number under y which is 132.25, the answer is 11.5

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.

Find the vertices and co-vertices of Salim’s bridge design.a=2(11.5)=23 b=1.3(13.5)=17.5

vertex 1 ¿(h , k+a) co-vertex 1 ¿(h+a , k )V1 ¿(0,0+23) co-v1 ¿(0+17.5,0)

Vertex 2 ¿(h , k−a) co-vertex 2¿(h−a , k)

V2 = (0,0−23) co-v2 ¿(0−17.5,0)

Write an equation for the design of Salim’s bridge using his minimum dimensions The ellipse is vertical

( x−h )2

b2 +( y−k )2

a2 =1

( x−0 )2

17.52 +( y−0 )2

232 =1

x2

306.25+ y2

529=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

i. Find an equation of the parabola. Vertex at origin (0,0), latus rectum is 10, a is positive since it opens to right and the equation is x = (y-k)^2+h.

Latus rectum=|1/a|

10=|1/a|

10a=|1|

a=1/10

Step 1:write the equation of latus rectum to find a.

Step 2:multiply "a" both sides.

Step 3:divide 10 from both sides.

Write an equation:

x=a ( y−k )2+h

x= 110 ( y−0 )2

+0

x= 110

y2

Step 1:write the equationStep 2:substitute the valuesStep 3:we found the final answer 

The final answer is x= 110 y2

2 ) Find the diameter of the opening |CD|, 11 cm from the vertex.Assign M as the midpoint of CD, M is (11,0) and C is (11 , y ) and the equation of

parabola is x= 110

y2

MC=y x= 1

10y2

11= 110

y2

11×10= 110×10

y2

√ y2=√110 y=MC=√110

CD=MC×2CD=2√110

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 (us) before it received the signal from A. (a) Assuming that radio signals travel at a speed of 980 ft . /u s, find an equation of the hyperbola on which the ship lies.

a=11760005280

a=222.72miles

222.72=2a2a=222.72a=111.36

The distance between a & b is 400 miles, therefore both of them are 200 miles away from center that is at originCenter is at origin, A is (200 , 0 ), B (-200 , 0 ) and c = 200

c2=b2+a2b2=c2−a2b2=2002−111.362

b2=27598.95b=166.1293

Step 4:Find b

( x−h )2

a2 −( y−k )2

b2 =1

x2

12401.85− y2

27598.14=1

Step 5:Write the equation of Hyperbola

(b) If the ship is due north of B, how far of the coastline is the ship? We need to find y. and since x is given which is the focus (200) we use the equation of hyperbola.

2002

12401.85− y2

27598.14=1

3.225− y2

27598.14=1

− y2

27598.14=1−3.225

− y2

27598.14=−2.225

Step1 :write the equationof hyperbola∧replace xwith 200

Step2 :Substitute .Step3 :subtract both sides .

Step 4 :multiply both sides thendivideboth sid es .

y2=61405.8615y=247.802Sois247.8miles, almost 248miles