Business Calc Homework w answers_Part_93

X x2 c a 5 4 2pd y2 4 y2 y 1 x 4 x 22 1 2 2 1 2 20

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Unformatted text preview: 45 3 y 2 1 2 asymptotes are y x 16. y 2 c 0 x2 a e 16 . 5 3 x; F( 5, 0); directrices are 4 x2 x2 8 6y 60y 5(y 2 9 4 12y 9 2 (y 4 4y 4) 8 a2 y2 8 1 8 8 4e c a 4 8 4x 2 2; 4(x 2 24. c 2 1 c2 thus, 10 2x; 2 5 2 10 x a2 2 5y 2 2x 2 0 36) 4 4 180 b2 (y asymptotes are y y 17. 8x 2 c e F( 18. 8y 2 c e F(0, c a c a x; F(0, 8 2 x2 2 4); directrices are 6) 36 2 1 c2 e2a 2 x a2 2 0 a e 2 y2 8 a2 b2 b2 e2a 2 b2 y b2 2 a 2; e a2 y a 2(e2 2 c c a ea 1); 2y 2 a2 16 b2 10 2 a 2(e2 1) 2 8 1 1; the asymptotes of this hyperbola are y 5; asymptotes are y 0 1 8 10 x ; 2 2 5 c a a e e2 1x. As e increases, the slopes of the asymptotes increase and the hyperbola approaches a single straight line. 25. The ellipse must pass through (0, 0) c ( 1, 2) lies on the ellipse a 2b 0; the point 8. The ellipse is 0 10, 0); directrices are x 2x 2 a2 16 b2 10 2 y 2 2 x 8 2 2 5; asymptotes are y 0 a e tangent to the x-axis its center is on the y-axis, so a 2 10 10); directrices are y 1) and e 3 b2 1 3c a2 9 3a c2 a2 and b 4x 2 x2 y2 (y 4 the equation is 4x 2 4y 2) 4 2 y2 2)2 4y 4 0. Next, 19. Vertices (0, c y2 3a x2 8 1 and e 9 1 8 3 4 4 4x 2 1a (y 2 and b a2 3 1 (now using the 4 1 3 standard symbols) c2 c 3 and e 1 c a b2 3 2 3e 20. Foci ( 3, 0) and e a 1 b2 c2 3c y 8 2 c a 2 . 3a 1 8 x2 ae 2. (y 0)2 21. Focus (4, 0) and directrix x ae e2 2c 2e 4)2 2 4 and a e 2 2 4 e2 2 e2 (x 2) Then PF (x x2 x2 4) 2 2PD y 2 2x 2 2(x y2 8 x 8 2 8x y2 16 2(x 2 y 8 2 4x 1 4) Appendix A5.2 26. We first prove a result which we will use: let m1 and m2 be the slopes of two nonparallel, nonperpendicular lines. Let be the acute angle between the lines. Then tan y 463 Similarly, tan b2 . Since tan cy0 tan , and . and are both less than 90 , we have m1 1 m2 m1m2 . (To see this result for positive-slope lines, P(x0, y0) x let and 1 be the angle of inclination of the line with slope m1, 2 be the angle of inclination of the line with slope m2. m2. Then tan ( tan 1 1 1 2 F1(c, 0) F2(c, 0) Assume m1 Then tan since m1 and we have 1 1 2 . , 2 ) t...
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