10'3 Hyperbolas - PowerPoint PPT Presentation

1 / 2
About This Presentation
Title:

10'3 Hyperbolas

Description:

sketch asymptotes as diagonals of auxiliary rectangle ... Sketch the graph of (x-3)2/25 - (y-6)2/4 = 1. Label the vertices, find the asymptote equations, ... – PowerPoint PPT presentation

Number of Views:54
Avg rating:3.0/5.0
Slides: 3
Provided by: james88
Category:

less

Transcript and Presenter's Notes

Title: 10'3 Hyperbolas


1
10.3 Hyperbolas
  • A hyperbola is the set of all points, whose
    distances from two fixed points (foci)
  • differ by a positive constant.
  • x2/a2 - y2/b2 1
    or y2/a2 - x2/b2 1
  • Where the center is the origin
  • the transverse (intercepted) axis has length of
    2a
  • the conjugate (untouched) axis has length of 2b
  • The focal points are located on the transverse
    axis a distance c units from the center.
  • The value of c is calculated using the equation
    c2 a2 b2, which ensures that the focal points
  • will exist beyond the vertices of the hyperbola.
  • Sketching the graph of hyperbolas identify
    transverse axis by finding intercepts
  • mark length of the conjugate axis
  • sketch asymptotes as diagonals of auxiliary
    rectangle

2
  • If y is the transverse axis
  • y2/a2 - x2/b2 1
  • y-intercepts y2/a2 1, y2 a2, y a
  • x-intercepts -x2/b2 1, x2 -b2, x imaginary
    number
  • The x-intercepts do not exist. However, the
    conjugate axis still marks a length of b units
    right and left.
  • The asymptote equations would be derived from y -
    y1 m (x-x1). In this case, y a/b x.
  • y2 - 4x2 36, y2/36 - x2/9 1
  • y-intercepts y2/36 1, y2 36, y 6
  • x-intercepts -x2/9 1, x2 -9, no intercepts
Write a Comment
User Comments (0)
About PowerShow.com