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Analytic geometry hyperbola

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A Presentation for Analytic Geometry Topic

A Presentation for Analytic Geometry Topic

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  • 1. Hyperbola
  • 2. A hyperbola is the set of points in a plane, the absolute value of the difference of whose distances from two fixed points, called foci , is a constant. Hyperbola F 2 F 1
  • 3. Hyperbola F 2 F 1 d 1 d 2 P d 2 – d 1 is always the same.
  • 4. Hyperbola F F V V
  • 5. Hyperbola F F V V C
  • 6. Hyperbola F F V V C
  • 7. Equation of a Hyperbola The center is at the point (0, 0) c 2 = a 2 + b 2 c is the distance from the center to a focus point. The foci are at (c, 0) and (-c, 0)
  • 8. Equation of a Hyperbola The conjugate points are at (0, b) and (0, -b) The vertices are at (a, 0) and (-a, 0) Length of the latus rectum is 2b 2 a
  • 9. Equation of a Hyperbola Ends of the latus rectum :
  • 10. Equation of a Hyperbola Horizontal Hyperbola Equation of the directrix
  • 11. Equation of a Hyperbola The center is at the point (0, 0) c 2 = a 2 + b 2 c is the distance from the center to a focus point. The foci are at (0, c) and (0, -c)
  • 12. Equation of a Hyperbola The conjugate points are at (b, 0) and (-b, 0) The vertices are at (0, a) and (0, -a) Length of the latus rectum is 2b 2 a
  • 13. Equation of a Hyperbola Ends of the latus rectum :
  • 14. Equation of a Hyperbola Vertical Hyperbola Equation of the directrix
  • 15. Example 1. a 2 = 9; a = 3 b 2 = 16; b = 4 c 2 = 25; c = 5
  • 16. Example 1. a 2 = 9; a = 3 b 2 = 16; b = 4 c 2 = 25; c = 5 V 1 (3, 0), V 2 (-3, 0) F 1 (5, 0), F 2 (-5, 0) Center at (0, 0) Center at (0, 0) V 1 (a, 0), V 2 (-a, 0) F 1 (c, 0), F 2 (-c, 0)
  • 17. Example 1. a 2 = 9; a = 3 b 2 = 16; b = 4 c 2 = 25; c = 5 LR = 2b 2 = 2(4) 2 = 32 a Conjugate points (0, 4), (0, -4) 3 3 Conjugate Points (0, b), (0, -b) LR = 2b 2 a
  • 18. Example 1. LR = 2b 2 = 32 a 3 c 2 = 25; c = 5 Endpoints of LR
  • 19. Example 1. Asymptotes c 2 = 25; c = 5 b 2 = 16; b = 4 a 2 = 9; a = 3 Asymptotes
  • 20. Example 1. V 1 (3, 0), V 2 (-3, 0) F 1 (5, 0), F 2 (-5, 0) Center at (0, 0) Conjugate points (0, 4), (0, -4) Endpoints of LR Asymptotes Symmetric at x-axis
  • 21. Example 1. 4x + 3y = 0 4x – 3y = 0
  • 22. Example 2. a 2 = 9; a = 3 b 2 = 16; b = 4 c 2 = 25; c = 5
  • 23. Example 2. a 2 = 9; a = 3 b 2 = 16; b = 4 c 2 = 25; c = 5 V 1 (0, 3), V 2 (0, -3) F 1 (0, 5), F 2 (0, -5) Center at (0, 0) Center at (0, 0) V 1 (0, a), V 2 (0, -a) F 1 (0, c), F 2 (0, -c)
  • 24. Example 2. a 2 = 9; a = 3 b 2 = 16; b = 4 c 2 = 25; c = 5 LR = 2b 2 = 2(4) 2 = 32 a Conjugate points (4, 0), (-4, 0) 3 3 Conjugate Points (b, 0), (-b, 0) LR = 2b 2 a
  • 25. Example 2. LR = 2b 2 = 32 a 3 c 2 = 25; c = 5 Endpoints of LR
  • 26. Example 2. Asymptotes c 2 = 25; c = 5 b 2 = 16; b = 4 a 2 = 9; a = 3 Asymptotes
  • 27. Example 2. V 1 (0, 3), V 2 (0, -3) F 1 (0, 5), F 2 (0, -5) Center at (0, 0) Conjugate points (4, 0), (-4, 0) Endpoints of LR Asymptotes Symmetric at y-axis
  • 28. Example 2. 3x + 4y = 0 3x – 4y = 0
  • 29.  
  • 30. Standard Equation of a Hyperbola C (h, k)
  • 31. Center (h, k) F 1 (h+c, k) F 2 (h-c, k) V 1 (h+a, k) V 2 (h-a, k)
  • 32. Example 1 .
  • 33. Graphing a Hyperbola Graph: ( x + 2) 2 (y – 1) 2 9 25 c 2 = 9 + 25 = 34 c =  34 = 5.83 Foci: (-7.83, 1) and (3.83, 1) – = 1 Center: (-2, 1) Horizontal hyperbola Vertices: (-5, 1) and (1, 1) Asymptotes: y = (x + 2) + 1 5 3 y = (x + 2) + 1 5 3 -
  • 34. Example 2 .
  • 35. Properties of this Hyperbola
    • Center ((1,2)
  • 36. Graphing the Hyperbola Foci: (1,7), (1, -3) Vertices: (1,5), (1, -1) The hyperbola is vertical Transverse Axis: parallel to y-axis
  • 37. Properties of this Hyperbola
    • Asymptotes:
  • 38. Latera Recta
  • 39.  
  • 40. General Equation of a Hyperbola Ax 2 + By 2 + Cx + Dy + E = 0
    • Group the x terms together and y terms together.
    • Complete the square.
    • Express in binomial form.
    • Divide by the constant term, where the first term has a positive sign.
  • 41. Example 1 . 9x 2 – 4y 2 – 18x – 16y + 29 = 0
  • 42. Converting an Equation (y – 1) 2 (x – 3) 2 4 9 c 2 = 9 + 4 = 13 c =  13 = 3.61 Foci: (3, 4.61) and (3, -2.61) – = 1 Center: (3, 1) The hyperbola is vertical Graph: 9y 2 – 4x 2 – 18y + 24x – 63 = 0 9(y 2 – 2y + ___) – 4(x 2 – 6x + ___) = 63 + ___ – ___ 9 1 9 36 9(y – 1) 2 – 4(x – 3) 2 = 36 Asymptotes: y = (x – 3) + 1 2 3 y = (x – 3) + 1 2 3 -
  • 43. Finding an Equation Find the standard form of the equation of a hyperbola given: 49 = 25 + b 2 b 2 = 24 Horizontal hyperbola Foci: (-7, 0) and (7, 0) Vertices: (-5, 0) and (5, 0) 10 8 F F V V Center: (0, 0) c 2 = a 2 + b 2 (x – h) 2 (y – k) 2 a 2 b 2 – = 1 x 2 y 2 25 24 – = 1 a 2 = 25 and c 2 = 49 C
  • 44. Center: (-1, -2) Vertical hyperbola Finding an Equation Find the standard form equation of the hyperbola that is graphed at the right (y – k) 2 (x – h) 2 b 2 a 2 – = 1 a = 3 and b = 5 (y + 2) 2 (x + 1) 2 25 9 – = 1
  • 45. Applications M 2 M 1 An explosion is recorded by two microphones that are two miles apart. M1 received the sound 4 seconds before M2. assuming that sound travels at 1100 ft/sec, determine the possible locations of the explosion relative to the locations of the microphones. (5280, 0) (-5280, 0) E(x,y) Let us begin by establishing a coordinate system with the origin midway between the microphones Since the sound reached M 2 4 seconds after it reached M 1 , the difference in the distances from the explosion to the two microphones must be d 2 d 1 1100(4) = 4400 ft wherever E is This fits the definition of an hyperbola with foci at M 1 and M 2 Since d 2 – d 1 = transverse axis, a = 2200 x 2 y 2 4,840,000 23,038,400 – = 1 x 2 y 2 a 2 b 2 – = 1 c 2 = a 2 + b 2 5280 2 = 2200 2 + b 2 b 2 = 23,038,400 The explosion must be on the hyperbola