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Stitz-Zeager_College_Algebra_e-book

# The reader is 2 3 3 3 3 invited to check that mixing

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Unformatted text preview: ince the foci units above and below (−3, 0) to arrive at −3, √ 10 3 and . To determine the asymptotes, recall that the asymptotes go through the center of b the hyperbola, (−3, 0), as well as the corners of guide rectangle, so they have slopes of ± a = ± 1 . 3 1 1 Using the point-slope equation of a line, Equation 2.2, we get y = 3 x + 1 and y = − 3 x − 1. Putting it all together, we get y 1 −6 x −1 −1 Hyperbolas can be used in positioning problems, as the next example illustrates. Example 7.5.4. Jeﬀ is stationed 10 miles due west of Carl in an otherwise empty forest in an attempt to locate an elusive Sasquatch. At the stroke of midnight, Jeﬀ records a Sasquatch call 9 seconds earlier than Carl. If the speed of sound that night is 760 miles per hour, determine a hyperbolic path along which Sasquatch must be located. Solution. Since Jeﬀ hears Sasquatch sooner, it is closer to Jeﬀ than it is to Carl. Since the speed of sound is 760 miles per hour, we can determine how much closer Sasquatch is to Jeﬀ by multiplying miles 1 hour 760 × × 9 seconds = 1.9 miles hour 3600 seconds This means that Sasquatch is 1.9 miles closer to Jeﬀ than it is to Carl. In other words, Sasquatch must lie on a path where (the distance to Carl) − (the distance to Jeﬀ) = 1.9 This is exactly the situation in the deﬁnition of a hyperbola, Deﬁnition 7.6. In this case, Jeﬀ and Carl are located at the foci, and our ﬁxed distance d is...
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