Results 1 to 2 of 2

Math Help - Parabola Help

  1. #1
    Newbie
    Joined
    May 2007
    Posts
    3

    Parabola Help

    I need help finding vertex,foci, and directrix of this parabola. Thanks

    x^2+5x-4y-1=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by creatureman View Post
    I need help finding vertex,foci, and directrix of this parabola. Thanks

    x^2+5x-4y-1=0
    y = (1/4)(x^2 + 5x - 1)

    Complete the square inside the parenthesis:

    y = (1/4)([x^2 + 5x + (5/2)^2] - (5/2) - 1)

    y = (1/4)(x + 5/2)^2 - (1/4)(7/2)

    y = (1/4)(x + 5/2)^2 - 7/8

    4(y + 7/32) = (x + 5/2)^2

    The general form for an upward opening parabola with a vertex at (h, k) is:
    4p(y - k) = (x - h)^2
    where p is the distance of the vertex to the focus, or the distance of the vertex to the directrix.

    So the vertex of your parabola is
    (h, k) = (-5/2, -7/32)

    The focus is p units directly above the vertex, and p = 1 (according to the above form) so the focus is at
    (-5/2, -7/32 + 1) = (-5/2, 25/32)

    The directrix is, in this case, a horizontal line p = 1 units below the vertex, so this line is
    y = -7/32 - 1 = -39/32

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parabola
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 9th 2010, 01:20 AM
  2. parabola
    Posted in the Geometry Forum
    Replies: 1
    Last Post: January 19th 2010, 01:10 AM
  3. Parabola help please
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 16th 2010, 07:41 AM
  4. Parabola
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 25th 2008, 09:07 AM
  5. parabola
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 25th 2008, 02:52 PM

Search Tags


/mathhelpforum @mathhelpforum