Grade 12
Mathematics Project
Conic Sections
Term 2
Student Name: Ahmed Mohamed Obaid AlKaabi
Id#: st1061110072
Section: 12 -04
Teacher: Mr.Abdul Salam
My wiki: http://mathcore-term2-project.weebly.com
2014 - 2015
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.
2
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/
3
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 + 0
𝑎 = 0.12
The equation will be:
x = 0.12(y − 0)2 + 0
So, x= (0.12)𝑦 2
1
1
Focus=(ℎ + 4𝑎 , 𝑘) → (0 + 4(0.12) , 0) → (2.08, 0)
25
𝑠𝑜 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑒𝑥 𝑤𝑖𝑙𝑙 𝑏𝑒 12 ft to the right of the vertex and a=0.12
4
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.
79.6
𝑎=
= 39.8 𝐴𝑈
2
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
( x  c )2 y 2
 2 1
a2
b
C.
Write an equation of the form
to model Pluto’s orbit. (Find b first.)
2
2
2
Equation:
𝑐 =𝑎 −𝑏
(9.87)2 = (39.8)2 − 𝑏 2
𝑏 2 ≈ 1486.62
D.
(𝑥 − 9.87)2
𝑦2
+
=1
1584.04
1486.62
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
5
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.
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
𝟐𝟕 ft
Write an expression for the height of the bridge.
𝒃 = √𝟏𝟑𝟐. 𝟐𝟓 = 𝟏𝟏. 𝟓 ft
So the height will = 𝟏𝟏. 𝟓 ft
6
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)
𝐶0 − 𝑉𝑒𝑟𝑡𝑖𝑐𝑒𝑠: (0,23), (0,23)
b.
Write an equation for the design of Salim’s bridge using his minimum
dimensions
𝑥2
𝑦2
+
=1
308.00 529.00
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
y
x
7
i) Find an equation of the parabola.
1) Let the vertex (h, k)  (0,0) “ at origin”
1
2) Letus Rectum = |𝑎| = 10𝑐𝑚
3) a is positive since it is opening right.
1
1
a=|𝑙𝑒𝑡𝑢𝑠 𝑟𝑒𝑐𝑡𝑢𝑚| = |10| = 0.1
1
1
4) Focus: (ℎ + 4𝑎 , 𝑘)  (0 + 4(0.1) , 0)  focus (4,0)
1
1
5) 𝑥 = 𝑎 (𝑦 − 𝑘)2 + ℎ  𝑥 = 10 (𝑦 − 0)2 + 0  𝑥 = 10 𝑦 2
1
The equation of parabola is 𝑥 = 10 𝑦 2
ii) Find the diameter of the opening |CD|, 11 cm from the vertex.
C= (11, 𝑦), D= (11, −𝑦)  to find y1 , y2 we should substitute:
1
1) x1 = 10 y1 2 ; y1 2 = 10 x → y1 = √10x  √(10)(11) → √110
C= (11,110), D=(11, −110)
CD = √(y2 − y1 )2 + (x 2 − x1 )2 → √(−√110 − √110 )2 + (11 − 11)2 → 2√110 cm
The diameter is 2√110 cm
8
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.
(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 =
𝟒𝟎𝟎
𝟐
= 𝟐𝟎𝟎 𝒎𝒊
6- 𝒅 = 𝒔. 𝒕 = (𝟏𝟐𝟎𝟎 × 𝟗𝟖𝟎) = 𝟏𝟏𝟕𝟔𝟎𝟎𝟎 𝒇𝒕 =
𝟐𝟒𝟓𝟎
𝟏𝟏
𝒎𝒊
7- According to definition 𝑷𝑭𝟏 − 𝑷𝑭𝟐 = 𝟐𝒂
8- 𝟐𝒂 =
𝟐𝟒𝟓𝟎
𝟏𝟏
𝟐
𝒎𝒊
𝒂=
𝟏𝟐𝟐𝟓
𝟏𝟏
𝒎𝒊
9- 𝒃𝟐 = 𝒄 − 𝒂𝟐
𝒃𝟐 = 𝟒𝟎𝟎𝟎𝟎 − (
𝟏𝟐𝟐𝟓 𝟐
𝟏𝟏
) = 𝟐𝟕𝟓𝟗𝟖. 𝟏𝟒𝟎𝟓 𝒎𝒊
𝒙𝟐
𝒚𝟐
The final equation is 𝟏𝟐𝟒𝟎𝟏.𝟖𝟓𝟗𝟓 − 𝟐𝟕𝟓𝟗𝟖.𝟏𝟒𝟎𝟓 = 𝟏
9
(b) If the ship is due north of B, how far of the coastline is the ship?
𝒙𝟐
𝒚𝟐
(𝟐𝟎𝟎)𝟐
𝒚𝟐
− 𝟐𝟕𝟓𝟗𝟖.𝟏𝟒𝟎𝟓 = 𝟏
𝟏𝟐𝟒𝟎𝟏.𝟖𝟓𝟗𝟓
− 𝟐𝟕𝟓𝟗𝟖.𝟏𝟒𝟎𝟓 = 𝟏
𝟏𝟐𝟒𝟎𝟏.𝟖𝟓𝟗𝟓
𝒚𝟐
− 𝟐𝟕𝟓𝟗𝟖.𝟏𝟒𝟎𝟓 = 𝟏 −
−
(𝟐𝟎𝟎)𝟐
𝟏𝟐𝟒𝟎𝟏.𝟖𝟓𝟗𝟓
𝒚𝟐
= −𝟐. 𝟐𝟐𝟓𝟑𝟐𝟐𝟕𝟖𝟑
𝟐𝟕𝟓𝟗𝟖. 𝟏𝟒𝟎𝟓
𝒚 = √(−𝟐𝟕𝟓𝟗𝟖. 𝟏𝟒𝟎𝟓)(−𝟐. 𝟐𝟐𝟓𝟑𝟐𝟐𝟕𝟖𝟑) = 𝟐𝟒𝟕. 𝟖𝟐𝟎 𝒎𝒊
𝒚 ≈ 𝟐𝟒𝟖 𝒎𝒊
10
Rubrics
Criteria
4
Completeness of
Tasks
Tasks are totally
completed and
correct. (100%)
20%
Presentation and
Integration of
Technology
70%
Creativity&
enrichment
Points
3
2
1
Tasks are partially
completed,
OR
Partially wrong.(75%)
Tasks are
partially
completed,
AND
Partially wrong
(50%).
Tasks are
Attempted (25% or
less)
____
____
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.
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.
____
10%
This rubric is out of 100, percentage orientation.
Total ---->
____
To make the mark out of 30 (Student’s Mark/10*3)
11