Chapter 8 - SECTION 8.1 (PAGE 443) R. A. ADAMS: CALCULUS...

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SECTION 8.1 (PAGE 443) R. A. ADAMS: CALCULUS CHAPTER 8. CONICS, PARAMETRIC CURVES, AND POLAR CURVES Section 8.1 Conics (page 443) 1. The ellipse with foci ( 0 , ± 2 ) has major axis along the y -axis and c = 2. If a = 3, then b 2 = 9 4 = 5. The ellipse has equation x 2 5 + y 2 9 = 1 . 2. The ellipse with foci ( 0 , 1 ) and ( 4 , 1 ) has c = 2, centre ( 2 , 1 ) , and major axis along y = 1. If ± = 1 / 2, then a = c = 4 and b 2 = 16 4 = 12. The ellipse has equation ( x 2 ) 2 16 + ( y 1 ) 2 12 = 1 . 3. A parabola with focus ( 2 , 3 ) and vertex ( 2 , 4 ) has a =− 1 and principal axis x = 2. Its equation is ( x 2 ) 2 4 ( y 4 ) = 16 4 y . 4. A parabola with focus at ( 0 , 1 ) and principal axis along y 1 will have vertex at a point of the form (v, 1 ) . Its equation will then be of the form ( y + 1 ) 2 4 v( x v) . The origin lies on this curve if 1 4 ( v 2 ) . Only the sign is possible, and in this case v 1 / 2. The possible equations for the parabola are ( y + 1 ) 2 = 1 ± 2 x . 5. The hyperbola with semi-transverse axis a = 1 and foci ( 0 , ± 2 ) has transverse axis along the y -axis, c = 2, and b 2 = c 2 a 2 = 3. The equation is y 2 x 2 3 = 1 . 6. The hyperbola with foci at ( ± 5 , 1 ) and asymptotes x ( y 1 ) is rectangular, has centre at ( 0 , 1 ) and has transverse axis along the line y = 1. Since c = 5 and a = b (because the asymptotes are perpendicular to each other) we have a 2 = b 2 = 25 / 2. The equation of the hyperbola is x 2 ( y 1 ) 2 = 25 2 . 7. If x 2 + y 2 + 2 x 1, then ( x + 1 ) 2 + y 2 = 0. This represents the single point ( 1 , 0 ) . 8. If x 2 + 4 y 2 4 y = 0, then x 2 + 4 ± y 2 y + 1 4 ² = 1 , or x 2 1 + ( y 1 2 ) 2 1 4 = 1 . This represents an ellipse with centre at ± 0 , 1 2 ² , semi-major axis 1, semi-minor axis 1 2 , and foci at ± ± 3 2 , 1 2 ² . y x 1 2 1 x 2 + 4 y 2 4 y = 0 1 Fig. 8.1.8 9. If 4 x 2 + y 2 4 y = 0, then 4 x 2 + y 2 4 y + 4 = 4 4 x 2 + ( y 2 ) 2 = 4 x 2 + ( y 2 ) 2 4 = 1 This is an ellipse with semi-axes 1 and 2, centred at ( 0 , 2 ) . y x 4 4 x 2 + y 2 4 y = 0 ( 1 , 2 ) 2 ( 1 , 2 ) Fig. 8.1.9 10. If 4 x 2 y 2 4 y = 0, then 4 x 2 ( y 2 + 4 y + 4 ) 4 , or x 2 1 ( y + 2 ) 2 4 1 . This represents a hyperbola with centre at ( 0 , 2 ) , semi- transverse axis 2, semi-conjugate axis 1, and foci at ( 0 , 2 ± 5 ) . The asymptotes are y 2 x 2. 312
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INSTRUCTOR’S SOLUTIONS MANUAL SECTION 8.1 (PAGE 443) y x 2 4 x 2 y 2 4 y = 0 Fig. 8.1.10 11. If x 2 + 2 x y = 3, then ( x + 1 ) 2 y = 4. Thus y = ( x + 1 ) 2 4. This is a parabola with vertex ( 1 , 4 ) , opening upward. y x ( 1 , 4 ) x 2 + 2 x y = 3 Fig. 8.1.11 12. If x + 2 y + 2 y 2 = 1, then 2 ± y 2 + y + 1 4 ² = 3 2 x x = 3 2 2 ± y + 1 2 ² 2 . This represents a parabola with vertex at ( 3 2 , 1 2 ) , focus at ( 11 8 , 1 2 ) and directrix x = 13 8 . y x ³ 3 2 , 1 2 ´ x + 2 y + 2 y 2 = 1 Fig. 8.1.12 13. If x 2 2 y 2 + 3 x + 4 y = 2, then ± x + 3 2 ² 2 2 ( y 1 ) 2 = 9 4 ( x + 3 2 ) 2 9 4 ( y 1 ) 2 9 8 = 1 This is a hyperbola with centre ( 3 2 , 1 ) , and asymptotes the straight lines 2 x + 3 2 2 ( y 1 ) .
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Chapter 8 - SECTION 8.1 (PAGE 443) R. A. ADAMS: CALCULUS...

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