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I've got two LORAN stations A and B that are 500 miles apart. A and B are also the Foci of a hyperbola. A ship at point P (which lies on the hyperbola branch with A as the focus) receives a nav signal from station A 2640 micro-sec before it receives from B. If the signal travels 980 ft/microsecond, how far away is P from A and B? Also, what are the values for a, b, and c?

I answered two of your questions. But it takes a while to get posted. I don't know why.

12/13/2012

Robert, I contacted wyzant about that, and it's because sometimes the answers have to be reviewed before they show up.

12/13/2012

Thanks for sharing.

I found that if you input "^", most likely your answer will be reviewed.

12/13/2012

I just posted an answer to this problem as well.  Last night I worked for an hour answering a questions posted with 4 problems, worked all of them and pluff!! my work just disappeared.  It was frustrating.  I hope it shows up later.  Patience my friends...

12/13/2012

Assuming the Transverse axis is horizontal and the center of the hyperbole is the origin, the foci are:

A(250, 0) and B(-250, 0) ; thus c = 250

Now, let's figure out how far appart is P from A and B.  This is the fun part.  Ready?  OK. Since the speed of the signal is given in feet/microsecond (ft/μs), we need to use the unit conversion 1 mile = 5,280 feet. (μs is the abreviation for microsecond... Fancy, huh?)

d = (2,640 μs)(980 ft/μs)

The signal travels 2,587,200 feet; or 490 miles in 2,640 μs

Now, let's think about this.  If the stations are 500 miles appart, and the ship receives the signal 2,640 μs sooner from A than from B, it means that the ship is very close to A because the signal traveled 490 additional miles from B before it reached the ship.  Let's put the ship P at the vertex of branch A and the vertices are 490 miles appart; or 245 miles from the origin...

Then... a = 245 and the vertices are (245, 0) and (-245, 0)

We find b from the fact: c2 = a2 + b2 → b2 = c2 - a2; or b2 = 2,475; thus b ≈ 49.75

Now the answers to the questions:

The distance from P to A is 5 miles → PA = 5; from P to B is 495 miles → PB = 495.

a = 245;  b ≈ 49.75; and c = 250

Let's take a minute to check our facts:

AP = 5 miles or 26,400 ft ÷ 980 μs/ft = 26.94 μs

BP = 495 miles or 2,613,600 ft ÷ 980 μs/ft = 2,666.94 μs

The difference 2,666.94 - 26.94 = 2,640 μs, is exactly the time P received the signal sooner from A than from B.

This was too much fun for a Thursday night.  Cheer up, tomorrow is Friday, finally!

2c = 500

c = 250 miles

2a = 980*2640/5280 = 490

a = 245 miles

b2 = c2 - a2 = 2475

Answer: x2/60025 - y2/2475 = 1