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Stitz-Zeager_College_Algebra_e-book

Example 1194 let v 1 8 and w 1 2 find p projw

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Unformatted text preview: ess outlined in Section 11.4 gives6 1 − e2 e2 d2 2 x− e2 d 1 − e2 2 + 1 − e2 e2 d2 y2 = 1 e2 d ,0 1−e2 2 d2 e and (1−e2 )2 We leave it to the reader to show if 0 < e < 1, this is the equation of an ellipse centered at with major axis along the x-axis. Using the notation from Section 7.4, we have a2 = 22 e b2 = 1−d 2 , so the major axis has length 12ed2 and the minor axis has length √2ede2 . Moreover, we find e −e 1− that one focus is (0, 0) and working through the formula given in Definition 7.5 gives the eccentricity 2d to be e, as required. If e > 1, then the equation generates a hyperbola with center 1e e2 , 0 whose − 2 2 h transverse axis lies along the x-axis. Since such hyperbolas have the form (x−2 ) − y2 = 1, we need a b 2 d2 2 d2 to take the opposite reciprocal of the coefficient of y 2 to find b2 . We get7 a2 = (1e e2 )2 = (ee −1)2 and 2 − 22 22 e ed b2 = − 1−d 2 = e2 −1 , so the transverse axis has length e2ed1 and the conjugate axis has length √2ed 1 . 2− e e2 − Additional...
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