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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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, René Descartes, Euclidean geometry

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