Stitz-Zeager_College_Algebra_e-book

# 3 for the variable y to that end we need to get the

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Unformatted text preview: hyperbolas when the time comes. Suppose we wish to derive the equation of a hyperbola. For simplicity, we shall assume that the center is (0, 0), the vertices are (a, 0) and (−a, 0) and the foci are (c, 0) and (−c, 0). We label the 7.5 Hyperbolas 435 endpoints of the conjugate axis (0, b) and (0, −b). (Although b does not enter into our derivation, we will have to justify this choice as you shall see later.) As before, we assume a, b, and c are all positive numbers. Schematically we have y (0, b) (x, y ) (−c, 0) (−a, 0) (a, 0) (c, 0) x (0, −b) Since (a, 0) is on the hyperbola, it must satisfy the conditions of Deﬁnition 7.6. That is, the distance from (−c, 0) to (a, 0) minus the distance from (c, 0) to (a, 0) must equal the ﬁxed distance d. Since all these points lie on the x-axis, we get distance from (−c, 0) to (a, 0) − distance from (c, 0) to (a, 0) = d (a + c) − (c − a) = d 2a = d In other words, the ﬁxed distance d from the deﬁnition of the hyperbola is actually the length of the transverse axis! (Where have we seen that type of coincidence before?) Now consider a poin...
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