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Math1.4

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  • 1. A hyperbola is the collection of points in the plane the difference of whose distances from two fixed points, called the foci , is a constant. HYPERBOLA
  • 2.  
  • 3. Theorem Equation of a Hyberbola; Center at (0, 0); Foci at ( + c , 0); Vertices at ( + a , 0); Transverse Axis along the x -Axis An equation of the hyperbola with center at (0, 0), foci at ( - c , 0) and ( c , 0), and vertices at ( - a , 0) and ( a, 0) is The transverse axis is the x -axis.
  • 4.  
  • 5. Theorem Equation of a Hyberbola; Center at (0, 0); Foci at ( 0, + c ); Vertices at (0, + a ); Transverse Axis along the y -Axis An equation of the hyperbola with center at (0, 0), foci at (0, - c ) and (0, c ), and vertices at (0, - a ) and (0, a ) is The transverse axis is the y -axis.
  • 6.  
  • 7. Theorem Asymptotes of a Hyperbola The hyperbola has the two oblique asymptotes
  • 8.  
  • 9. Theorem Asymptotes of a Hyperbola The hyperbola has the two oblique asymptotes
  • 10. Find an equation of a hyperbola with center at the origin, one focus at (0, 5) and one vertex at (0, -3). Determine the oblique asymptotes. Graph the equation by hand and using a graphing utility. Center: (0, 0) Focus: (0, 5) = (0, c ) Vertex: (0, -3) = (0, - a ) Transverse axis is the y -axis, thus equation is of the form
  • 11. = 25 - 9 = 16 Asymptotes:
  • 12. V (0, 3) V (0, -3) (4, 0) (-4, 0) F (0, 5) F (0, -5)
  • 13. Hyperbola with Transverse Axis Parallel to the x -Axis; Center at ( h , k ) where b 2 = c 2 - a 2 .
  • 14.  
  • 15. Hyperbola with Transverse Axis Parallel to the y -Axis; Center at ( h , k ) where b 2 = c 2 - a 2 .
  • 16.  
  • 17. Find the center, transverse axis, vertices, foci, and asymptotes of
  • 18. Center: ( h , k ) = (-2, 4) Transverse axis parallel to x -axis. Vertices: ( h + a , k ) = (-2 + 2, 4) or (-4, 4) and (0, 4)
  • 19. Asymptotes: ( h , k ) = (-2, 4)
  • 20. C (-2,4) V (-4, 4) V (0, 4) F (2.47, 4) F (-6.47, 4) (-2, 8) (-2, 0) y - 4 = -2( x + 2) y - 4 = 2( x + 2)
  • 21. Sketch the curve represented by the equation: Exercise :
  • 22. Solution:
  • 23. Note:- To understand what this curve might look like, we have to work towards a standard form. This is best accomplished by completing the square in the x terms and in the y terms. From this, we see that the curve is a hyperbola centered at (1, 4). When y = 4 we have:
  • 24.
    • So,
    • Thus, or
    Therefore, (3, 4) and are both on the curve. The asymptotes are the lines and and they pass through the centre (1, 4).

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