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-   -   How do you find the vertex, focus, and directrix of parabolas? (https://www.askmehelpdesk.com/showthread.php?t=247995)

  • Aug 11, 2008, 09:14 PM
    cmy89
    How do you find the vertex, focus, and directrix of parabolas?
    I don't really understand this concept, so thorough examples would be much appreciated. :D

    1. y^2=8x


    2. x^2 + 8y + 16 = 0

    ... and
    Find an equation of the parabola that satisfies the given conditions.
    3. Focus (0,1), directrix y = 7

    Thanks!
  • Aug 12, 2008, 02:26 PM
    galactus
    Quote:

    Find an equation of the parabola that satisfies the given conditions.
    3. Focus (0,1), directrix y = 7

    Thanks!
    The vertex of the parabola is halfway between the directrix and the focus.

    So, the vertex is at (0,4)

    The parabola opens down.

    p is the distance (absolute value because distance is always positive) between the focus and the vertex and between the vertex and the directrix.

    Therefore, p=3 because 7-4=3 and 4-1=3.

    So, we have





  • Aug 14, 2008, 07:55 AM
    galactus
    Quote:

    1.
    |p| is the distance from the focus to the vertex and also the distance from the vertex to the

    directrix. This one is of the form .

    We can use



    or

    In this case,



    Since , then p=2.

    The vertex is at the origin and it opens to the right, so it has focus 2 units to the right of

    the origin, F(2,0) and

    directrix 2 units to the left of the origin at x=-2


    Quote:

    2.
    Can you do this one? Remember, you have the form:

    From your given equation, can you see what V(h,k) and p are?

    I don't know if this matters or not. Have you even read, or even care about, anything I have posted to help you along? I posted the solutions to two of them so you would have an example to work from for other problems.
  • Jun 17, 2011, 06:30 AM
    joele12
    if your problem is x^2=8y what is your p vaule.
  • Jun 17, 2011, 07:38 AM
    Unknown008

    Put it in the form that galactus gave.





    So, you see clearly that h = 0. And if you expand;



    So, you see again that p = 2 and k = 0
  • Nov 28, 2011, 02:15 PM
    junemarie15
    equation of the parabola that satisfies the given conditions.Focus (6, 0), directrix x = 5
  • Nov 28, 2011, 02:52 PM
    ebaines
    Quote:

    Originally Posted by junemarie15 View Post
    equation of the parabola that satisfies the given conditions.Focus (6, 0), directrix x = 5

    Please do not double post questions. See:
    https://www.askmehelpdesk.com/mathem...ns-615388.html
  • May 20, 2012, 07:11 PM
    maribel95
    what is the directrix and focus of y=x^2 and of x=(1/36) y^2?
  • May 21, 2012, 10:14 AM
    ebaines
    1 Attachment(s)
    For each of these first put the equation into standard form - either



    or



    Once you have this the vertex is at , and the focus is a distance 'p' towards the "inside" of the parabola, and the directrix is a line that us distance 'p' from the cvertex towards the outside.

    I'll do an example for you: consider



    The vertex is at (2,0), The value of 'p' is 3 (since 1/(4*3) = 1/12), and hence the focus is at (2,3). The directrix is a horizontal line a distance 3 below the vertex, or at y= -3.

    Hope this helps!
  • Jun 19, 2012, 08:25 PM
    Teresaw3
    What is the directrix of (y-(-2))^2=3(x-(-1))? I do not understand how to find the answer for this please help!

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