Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley CHAPTER 7: Conic Sections 7.1 The Parabola 7.2 The Circle and the Ellipse 7.3 The Hyperbola 7.4 Nonlinear Systems of Equations and Inequalities

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley 7.3 The Hyperbola  Given an equation of a hyperbola, complete the square, if necessary, and then find the center, the vertices, and the foci and graph the hyperbola.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Hyperbola A hyperbola is the set of all points in a plane for which the absolute value of the difference of the distances from two fixed points (the foci) is constant. The midpoint of the segment between the foci is the center of the hyperbola.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Standard Equation of a Hyperbola with Center at the Origin -- Horizontal Transverse Axis Horizontal Vertices: (–a, 0), (a, 0) Foci: (–c, 0), (c, 0) where c 2 = a 2 + b 2 Segment is the conjugate axis.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Standard Equation of a Hyperbola with Center at the Origin -- Vertical Transverse Axis Vertical Vertices: (0, –a), (0, a) Foci: (0, –c), (0, c) where c 2 = a 2 + b 2 Segment is the conjugate axis.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Find an equation of the hyperbola with vertices (0,  4) and (0, 4) and foci (0,  6) and (0, 6). Solution: We know that a = 3 and c = 5. We find b 2. Vertices and foci are on the y-axis, so the transverse axis is vertical.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Graphing a Hyperbola - Horizontal Transverse Axis Sketch the asymptotes: y = –(b/a)x and y = (b/a)x Draw a rectangle with vertical sides through the vertices and horizontal sides through the endpoints of the conjugate axis.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example For the hyperbola given by 9x 2  16y 2 = 144, find the vertices, the foci, and the asymptotes. Then graph the hyperbola. Solution: First, we find standard form:

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) The hyperbola has a horizontal transverse axis, so the vertices are (  4, 0) and (4, 0). Thus, the foci are (–5, 0) and (5, 0). For the standard form of the equation, we know that a 2 = 5 2, or 25, and b 2 = 4 2, or 16. We find the foci:

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) Next, we find the asymptotes: To draw the graph, sketch the asymptotes first. Draw the rectangle with horizontal sides through (0, 3) and (0, – 3) and vertical sides through (4, 0) and (–4, 0). Draw and extend the diagonals of the rectangle and these are the asymptotes. Next plot the vertices and draw the branches of the hyperbola outward from the vertices towards the asymptotes.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued)

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Standard Equation of a Hyperbola with Center at (h, k) -- Horizontal Transverse Axis Horizontal Vertices: (h – a, k), (h + a, k) Asymptotes: Foci: (h – c, k), (h + c, k) where c 2 = a 2 + b 2

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Standard Equation of a Hyperbola with Center at (h, k) -- Vertical Transverse Axis Vertical Vertices: (h, k – a), (h, k + a) Asymptotes: Foci: (h, k – c), (h, k + c) where c 2 = a 2 + b 2

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example For the hyperbola given by 4y 2  x y + 4x + 28 = 0, find the center, the vertices, and the foci. Then draw the graph. Solution: Complete the square to get standard form:

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) The center is (2, –3). Note: a = 1 and b = 2. The transverse axis is vertical, so vertices are 1 unit below and above the center: (2, –3 – 1) and (2, –3 + 1), or (2, –4) and (2, –2).

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) We know c 2 = a 2 + b 2, so c 2 = = = 5 and, the foci are unit below and above the center: The asymptotes are: Sketch the asymptotes, plot the vertices, draw the graph.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued)

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Applications Some comets travel in hyperbolic paths with the sun at one focus. Such comets pass by the sun only one time, unlike those with elliptical orbits, which reappear at intervals. A cross section of an amphitheater might be one branch of a hyperbola. A cross section of a nuclear cooling tower might also be a hyperbola.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Applications Another application of hyperbolas is in the long-range navigation system LORAN. This system uses transmitting stations in three locations to send out simultaneous signals to a ship or aircraft. The difference in the arrival times of the signals from one pair of transmitters is recorded on the ship or aircraft. This difference is also recorded for signals from another pair of transmitters. For each pair, a computation is performed to determine the difference in the distances from each member of the pair to the ship or aircraft. If each pair of differences is kept constant, two hyperbolas can be drawn. Each has one of the pairs of transmitters as foci,and the ship or aircraft lies on the intersection of two of their branches. See diagram on next slide.

Slide Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Applications