# Hyperbola Application?

Suppose that two people are standing 1 miles apart both see the burst from a fireworks display. After a period of time, the first person standing at point A hears the burst. Two seconds later, the second person standing at point B hears the burst. If the person at point B is due west of the person at point A and if the display is known to occur due north of the person standing at point A, where did the fireworks display occur? Please, show solution steps.

Relevance
• 4 months ago

I don't get the "hyperbola" part; and you can't convert the 2 s difference in time to any kind of distance unless you know the speed of sound. That varies with temperature, humidity and altitude. I'll use v as the speed of sound in miles per second. Using a figure of about 767 mi/hr, divide by 3600 to get about 0.213 mi/s for v.

The geometry I see here is a right triangle ABX, right angled at A with the explosion at point X. You're told that one leg (AB) is 1 mile long and can deduce that the hypotenuse BX is 2 seconds of sound travel distance longer than the other leg AX. That's about 0.428 miles longer. Let h be the length of AX and apply the Pythagorean Theorem. The hypotenuse is 2 sec longer in travel time, which is 2v longer in distance:

1² + h² = (h + 2v)²

1 + h² = h² + 4vh + 4v²

0 = 4vh + 4v² - 1

4vh = 1 - 4v²

h = (1 - 4v²)/(4v)

With v=0.213 you get a distance of about h=0.961. That's 0.961 miles north of point A.

• Pope
Lv 7
4 months ago

Through air, at sea level, sound travels about 0.4 miles in two seconds. The locus of points 0.4 miles further from B than from A is one branch of a hyperbola having foci A and B. Give A and B coordinates in the Cartesian plane.

A(1/2, 0), B(-1/2, 0)

Let P(x, y) trace the locus of points 0.4 miles further from B than from A.

PB = PA + 2/5

√[(x + 1/2)² + y²] = √[(x - 1/2)² + y²] + 2/5

The burst also is due north of A, so P would have to satisfy this condition:

x = 1/2, y > 0

Solve that system.