Expert: Paul Klarreich - 4/1/2008

Question
Two friends live 4 miles apart and on the same east-west street, and you live halfway between them. You are having a 3-way conversation when you hear an explosion. 6 seconds later, your friend to the east hears it, and your friend to the west hears it 8 seconds after you do. Find equations of 2 hyperbolas that would locate the explosion. Assume coordinate system is measured in feet, and sound travels 1100 feet per second.

Answer
Questioner:   mindi
Category:  Advanced Math
Private:  No
 
Subject:  precalc
Question:  Two friends live 4 miles apart and on the same east-west street, and you live halfway between them. You are having a 3-way conversation when you hear an explosion. 6 seconds later, your friend to the east hears it, and your friend to the west hears it 8 seconds after you do. Find equations of 2 hyperbolas that would locate the explosion. Assume coordinate system is measured in feet, and sound travels 1100 feet per second.

I shall change the units to 'sound-seconds', analogous to light-years in astronomy.  

One sound-second = 1100 feet or so, and one mile is 5 S-S (or so).
 
Given two points f1,f2, the set of all points P such that

Pf2 - Pf1  = c

is a hyperbola. [or 'an' hyperbola]

So lets let you be at f1(0,0) and a friend be at f2(10,0). [in S-S].

Let P be (x,y) and take c = 6 sound-seconds.  [Your friend heard it 6 seconds later, so his distance, Pf2, is greater.]

Pf2 - Pf1  = 6

sqrt((x - 10)^2 + y^2) - sqrt(x^2 + y^2) = 6

[Believe it or not, that is one of your equations.  It might take some simplifying to make it look like a hyperbola equation, but it is.]

Here we go:

sqrt((x - 10)^2 + y^2) =  sqrt(x^2 + y^2) + 6 << bring term over.

(x - 10)^2 + y^2  =  x^2 + y^2 + 36 + 12 sqrt(x^2 + y^2) <<sq both sides.

x^2 - 20x + 100 + y^2  =  x^2 + y^2 + 36 + 12 sqrt(x^2 + y^2)

- 20x + 64   =  12 sqrt(x^2 + y^2) << some stuff cancels.

- 5x + 16   =   3 sqrt(x^2 + y^2)

25x^2 - 160x + 256   =   9x^2 + 9y^2  << sq both sides AGAIN.

16x^2 - 160x - 9y^2 = - 256  << const on right, other stuff on left.

16(x^2 - 10x     ) - 9y^2 = - 256 << completing the square.

16(x^2 - 10x + 25) - 9y^2 = - 256 + 400 << finished.

16(x - 5)^2  - 9y^2 = 144  

(x - 5)^2     y^2
----------- - --- = 1  << looks beautiful, doesn't it?
     9        16
.......................................
Now your second friend is at f3, which will be (-10,0) [You DID say you are between them, right?]

And you are still at f1(0,0)

Let P be (x,y) and take c = 8 sound-seconds.  [THIS friend heard it 8 seconds later, so his distance, Pf3, is greater.]

Pf3 - Pf1  = 6

Ok, now.  I'll let you have the fun of doing the second one.  Set up the same stuff, do the same simplifications. Enjoy.