Results 1 to 3 of 3

Math Help - Parabola directrix

  1. #1
    Junior Member
    Joined
    Jan 2007
    Posts
    40

    Parabola directrix

    Could someone help me with these two types of problems?

    1) Find the directrix of the parabola (y-7)[squared]=4(x+3).

    2) Find the equation of the parabola whose focus is (2,3) and the directrix is x=12.

    Thanks,
    Gretch
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by gretchen View Post
    Could someone help me with these two types of problems?

    1) Find the directrix of the parabola (y-7)[squared]=4(x+3).

    2) Find the equation of the parabola whose focus is (2,3) and the directrix is x=12.

    Thanks,
    Gretch
    Here is one way.

    The directrix of a parabola is "p" away from the vertex along the axis of symmetry. [Also, the focus of a parabola is "p" away from the vertex.]

    1) The directrix of parabola y^2 = 4(x+3).

    The y is squared, so the parabola is horizontal, or the axis of symmetry is horizontal.
    A standard form of the equation of a horizontal parabola, in terms of "p", is
    (x- h) = [1/(4p)](y -k)^2 -------------(1)
    where
    (h,k) is the vertex
    p is vertex to focus, or vertex to direcrix along the axis of symmetry.
    If the 1/(4p) is negative, then the parabola opens to the left.
    If the 1/(4p) is positive, then the parabola opens to the right.
    So,
    y^2 = 4(x+3) -----------------------(i)
    Convert that into the form of (1),
    (x+3) = [1/(4*1)](y -0)^2 ---------------(ii)
    Therfore,
    vertex is (-3,0)
    p = 1 unit.

    The 1/(4p) is 1/(4*1) is positive, so the directrix is to the left of the vertex.
    The vertex (-3,0) is on the verical line x = -3. One unit to the left of that is x = -4.

    Therefore, the directrix is the vertical line x = -4. -----------answer.

    ===========================================
    2) Find the equation of the parabola whose focus is (2,3) and the directrix is x=12.

    The vertex is halfway of the distance between the focus and the directrix along the axis of symmetry.

    The directrix x = 12 is a vertical line, therefore the parabola is horizontal. The axis of symmetry is the horizontal line y = 3 because the focus is always on the axis of symmetry.

    The focus (2,3) is on the vertical line x = 2.
    So distance from focus to directrix along the axis of symmetry is 12 -2 = 10 units.
    Hence the "p" is 10/2 = 5 units.
    And so the vertex is (2+5,3) = (7,3)

    The focus is at (2,3) and the directrix is the vertical line x=12. Therefore, the parabola opens to the left. Hence the 1/(4p) is negative.

    Then, the parabola is:
    (x -h) = -[1/(4p)](y -k)^2
    (x -7) = -[1/(4*5)](y -3)^2
    x -7 = -(1/20)(y -3)^2 --------------------answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2007
    Posts
    40
    Thanks for your help. I see how you did the first one and I will try the others on my own.

    On the 2nd problem, I ended up with -20(x-7)=(y-3)[squared] which is what you came up with. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Parabola and Directrix and Focus.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 15th 2010, 11:19 PM
  2. Parabola - vertex, focus, directrix
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 7th 2010, 07:04 AM
  3. Focus, directrix, and axis of the parabola
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 2nd 2010, 02:06 PM
  4. Equation of a parabola Given Focus and Directrix
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 20th 2008, 04:27 PM
  5. Parabola directrix
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 31st 2007, 10:15 PM

Search Tags


/mathhelpforum @mathhelpforum