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Sullivan Algebra and Trigonometry: Section 10.4 The Hyperbola

Sullivan Algebra and Trigonometry: Section 10.4 The Hyperbola. Objectives of this Section Find the Equation of a Hyperbola Graph Hyperbolas Discuss the Equation of a Hyperbola Find the Asymptotes of a Hyperbola Work with Hyperbolas with Center at (h,k). F 1 : (-c,0). F 2 : (c,0).

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Sullivan Algebra and Trigonometry: Section 10.4 The Hyperbola

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  1. Sullivan Algebra and Trigonometry: Section 10.4The Hyperbola • Objectives of this Section • Find the Equation of a Hyperbola • Graph Hyperbolas • Discuss the Equation of a Hyperbola • Find the Asymptotes of a Hyperbola • Work with Hyperbolas with Center at (h,k)

  2. F1: (-c,0) F2: (c,0) Transverse Axis A hyperbola is the collection of all points in the plane the difference of whose distances from two fixed points, called the foci, is a constant.

  3. Equation of a Hyperbola: Center at (0,0); Foci at (c, 0) and (-c,0) where b2 = c2 - a2 The transverse axis is the x - axis. The vertices are at (-a, 0) and (a, 0)

  4. Find the equation of a hyperbola with center at the origin, one focus at (-5, 0), and a vertex at (4,0). Graph the equation. Since the given focus and vertex are on the x-axis, the transverse axis is the x-axis. The distance from the center to one of the foci is c = 5. The distance from the center to one of the vertices is a = 4. Use c and a to solve for b. b2 = c2 - a2 b2 = 52 - 42 = 25 - 16 = 9

  5. So, the equation of the hyperbola is: (-4,0) (4,0) F1: (-5,0) F2: (5,0)

  6. Discuss the equation: Since the equation is written in the desirable form, a2 = 9 and b2 = 7 Since b2 = c2 - a2, it follows that c2 = a2 + b2 or c2 = 9 + 7 = 16. So, the foci are at (4,0) and (-4,0) The vertices are at (-3, 0) and (3, 0)

  7. Since b2 = c2 - a2, it follows that c2 = a2 + b2 or c2 = 4 + 1 = 5. So, the foci are at and Discuss the equation: Since the x term is subtracted from the y term, the equation is that of a hyperbola with center at the origin and transverse axis along the y-axis The vertices are at (0,2) and (0,-2).

  8. The hyperbola has the two oblique asymptotes: Discuss the equation of the hyperbola Begin by dividing both sides of the equation by 16 to put the equation in the proper form.

  9. Since b2 = c2 - a2, it follows that c2 = a2 + b2 or c2 = 1 + 4 = 5. So, the foci are at and The center of the hyperbola is the origin. Since the x term comes first, the transverse axis is the x-axis. The vertices are at (1,0) and (-1,0).

  10. The asymptotes have the equation: To graph the hyperbola, form the rectangle containing the points (a,0), (-a,0), (0,b), and (0,-b). The extensions of the diagonals of this rectangle are the asymptotes.

  11. (0,2) (-1,0) (1,0) (0,-2) Now, graph the hyperbola using these guides

  12. If an ellipse with center at the origin and major axis coinciding with a coordinate axis is shifted horizontally h units and vertically k units, the resulting ellipse is centered at (h,k) and has the equation: Horizontal Transverse Axis Vertical Transverse Axis

  13. Find the equation of a hyperbola with center at (2, -3), one focus at (5, -3), and one vertex at (3, -3). The center is at (h,k) = (2, -3). So h = 2 and k = -3 The center, focus, and vertex all lie on the line y = -3, so the major axis is parallel to the x-axis and the hyperbola has a horizontal transverse axis and will have an equation in the form:

  14. The distance from the center to the vertex is a = 1. The distance from the center to the focus is c = 3. To solve for b, b2 = c2 - a2 b2 = 32 - 12 = 9 - 1 = 8 So, the equation of the ellipse is:

  15. The equations of the asymptotes can be found by shifting the equations for the asymptotes h units in the horizontal direction and k units in the vertical direction yielding: For our example:

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