Stitz-Zeager_College_Algebra_e-book

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Unformatted text preview: squatch call 18 seconds after Jeff did. Use this added information to locate Sasquatch. Solution. Kai and Jeff are now the foci of a second hyperbola where the fixed distance d can be determined as before 760 miles 1 hour × × 18 seconds = 3.8 miles hour 3600 seconds Since Jeff was positioned at (−5, 0), we place Kai at (−5, 6). This puts the center of the new hyperbola at (−5, 3). Plotting Kai’s position and the new center gives us 442 Hooked on Conics y Kai 6 5 4 3 2 Jeff 1 Carl −9 −8 −7 −6 −5 −4 −3 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6 2 2 The second hyperbola is vertical, so it must be of the form (y−23) − (x+5) = 1. As before, the b a2 distance d is the length of the major axis, which in this case is 2b. We get 2b = 3.8 so that b = 1.9 and b2 = 3.61. With Kai 6 miles due North of Jeff, we have that the distance from the center to the focus is c = 3. Since a2 + b2 = c2 , we get a2 = c2 − b2 = 9 − 3.61 = 5.39. Kai heard the Sasquatch call...
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