Grade 12
Mathematics Project
Term 2
Student’s Name: HAMDAN MUBARAK SAEED ALI AL ZAABI
ID #: ST1041090081
Section: 12-07
Your project should be uploaded on your wiki. (Good Luck)
2011 - 2012
Part I:
The figure below shows the graph of a parabola.
1. Plot the point A (3, 4.5), and draw the straight line () with equation 𝑦 =
−
1
2
on the same graph above.
2. Calculate the distance between the points A and S (0, 0.5).
A (3, 4.5) S (0, 0.5)
= √9 + 16 = 5
3. Find the distance between the point A and the straight line ().
General equation of a line is
ax + by + c = 0
Our line is: y = 0.5 there for a= 0, b = 1, c = -0.5
Distance =
0(3) + 1(4.5) +(−0.5)
√0+1
d=
ax1 + by1 + c
√𝑎2 + 𝑏2
=4
2
4. Compare the distances found in 2 and 3.
Distance found between the two points A and S are greater than
the distance between the point A and the line 
5. What is the name of the:
i.
Point S: ______________
ii.
Straight line (): __________________
3
Generalization:
Let 𝑃(𝑥, 𝑦) is a point of the drawn parabola, and 𝐹(0, 𝑝) the focus point of
the parabola ( 𝑝 > 0 ) as illustrated in the figure below.
6. Find the distance d1 between P and F, and the distance d2 between P and the
straight line with equation 𝑦 = −𝑝
d1= √𝑥 2 + (𝑦 − 𝑝)2
√𝑥 2 + (𝑦 − 𝑝)2 = y + p
d2 = y + p
7. Find the relation between x and y when d1 = d2.
X=y
4
Application
8. 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.
Choose V to be the origin, with x-axis V and F, then F is (p, 0),
A is (p, 5) so: substituting A into the equation y2 = 4px
Gives
25 = 4p2 , p =
ii.
5
2
and y2 = 10x
Find the diameter of the opening |CD|, 11 cm from the vertex.
X = 11 then y = √110
then |CD| = 2√110
5
Part II
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.
PA - PB = (1200) (980) = 1,176,000 ft =
1225
Then a =
and c = 200,
11
b2 = c2 – a2
=
3339375
then
121
2450
11
mi = 2a
so
121𝑥 2
1500625
-
121𝑦2
3339375
=1
(b) If the ship is due north of B, how far of the coastline is the ship?
Due north of B
(121)(200)2
1500625
y=
133575
539
-
x = 200
121𝑦 2
3339375
≈ 248 mi
then
= 1
6