The document provides information about hyperbolas including their key properties and equations. It gives the standard form of the hyperbola equation and provides two examples of converting equations to standard form and graphing hyperbolas. It also gives an example of using the properties of a hyperbola to find the possible locations of an explosion based on the time difference sound reached two microphones.

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

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

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