Chapter 4 164 - 426 CHAPTER 4 PROBLEMS PLUS The left—hand...

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Unformatted text preview: 426 CHAPTER 4 PROBLEMS PLUS The left—hand inequality holds if and only if 2333; — ac + y S 0 4:) :cy — 53c + %y S O 4:) (a: + %) (y — g) S ’1 4:) (cc y) lies on or below the hyperbola (a: + i) (y — é) : —fi. which passes through the origin and approaches the lines y = % and :1: = —% asymptotically. Case 2: y 2 .7: This is the case in which (a: y) lies on or above the line y : :c. The double inequality becomes 2ccy S y — a: S x2 + 31°. The right—hand inequality holds if and only if 3:2 + a: + y2 — y 3 0 4:} (m + %)2 + (y — %)2 2 § 4:} (cc,y) lies on or outside the circle of radius % centered at (— lob—- 7%).The left—hand inequality holds ifand only if2acy+miy S 0 <2> :L'y+ fix 7 $3; 3 0 {it (x 1 %) (y —l— g) g 7% 4:} (cc. 3/) lies on or above the left—hand branch of the hyperbola (x — %) (y + %) : ’1‘ which passes through the 1 origin and approaches the lines y : 7% and ac : 5 asymptotically. Therefore the region of interest consists of the points on or above the left branch of the hyperbola (ac — %) (y + g) : 7% that are on or outside the circle (1: + 92 + (y i %)2 = 5 together with the points on or below the right branch of the hyperbola (a: + %) (y * é) : —711 that are on or outside the circle (cc 7 a2 + (y —l— i)2 : % Note that the inequalities are unchanged when a: and y are interchanged. so the region is symmetric about the line y : cc. So we need only have analyzed case 1 and then reflected that region about the line y : x. instead of considering case 2. 13. (a) Lety : |AD|. cc : |AB|. and 1/1: : |AC|. so that lAB| - IACI : 1. B We compute the area A of AABC in two ways. First. D A:%|AB||AC|sin%":%~1~§:§.Second. X I A : (area of AABD) —l— (area of AAC'D) A 1 C = g |AB| |AD| sin g + é |AD| |AC| sin g " : mi + aye/w) é : @ytw + was) E at'n the two ex ressions for the area we et fl 3: + l : fl 4:> y = 1 : 30 ac > 0. qu1g P - g 4y 1: 4 $+1/$ m2+1i Another method: Use the Law of Sines on the triangles ABD and ABC. In AABD. we have AA+ AB + AD : 180° <=> 60° + a + 1D : 180° 4:) 1D : 120° 7 (1. Thus. cc sin(120° # a) sin 120° cosa — cos 120° sina g 6086! + éSiHOt ac _ \/§ 1 ———.——— . — _ — Tcota+§, y sma sma sma y and by a similar argument with AABC, é cot oz : m2 + % Ellmmating cota gives a = (9:2 —l— %) + % :> y : cc 3: > 0 ...
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