SECTION 10.4 The Hyperbola 715 Lightning Strikes Suppose that two people standing 1 mile apart both see a flash of lightning. After a period of time, the person standing at point A hears the thunder. One second later, the person standing at point B hears the thunder. If the person at B is due west of the person at A and the lightning strike is known to occur due north of the person standing at A, where did the lightning strike occur? EXAMPLE 10 Figure 50 Lightning strike model North 1 mile 5 5280 feet East B 5 (22640, 0) (2a, o) (a, o) A 5 (2640, 0) (x, y) Solution See Figure 50, where the point x y , ( ) represents the location of the lightning strike. Sound travels at 1100 feet per second, so the person at point A is 1100 feet closer to the lightning strike than the person at point B. Since the difference of the distance from x y , ( ) to B and the distance from x y , ( ) to A is the constant 1100, the point x y , ( ) lies on a hyperbola whose foci are at A and B. An equation of the hyperbola is x a y b 1 2 2 2 2 − = where a2 1100, = or a 550. = Because the distance between the two people is 1 mile (5280 feet) and each person is at a focus of the hyperbola, then c c 2 5280 5280 2 2640 = = = Since b c a 2640 550 6,667,100, 2 2 2 2 2 = − = − = the equation of the hyperbola that describes the location of the lightning strike is x y 550 6,667,100 1 2 2 2 − = Refer to Figure 50. Since the lightning strike occurred due north of the individual at the point A 2640, 0 , ( ) = let x 2640 = and solve the resulting equation. y y y y 2640 550 6,667,100 1 6, 667,100 22.04 146,942,884 12,122 2 2 2 2 2 − = − = − = = The lightning strike occurred 12,122 feet north of the person standing at point A. (continued) =x 2640 Subtract 2640 550 2 2 from both sides. Multiply both sides by −6,667,100. >y 0 since the lightning strike occurred in quadrant I.
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