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Distance from vertex to focus in a parabola

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data
    I am a bit rusty on parabolae.
    I am doing a question on projectiles and have found the coordinates of the vertex of a parabola as:
    [tex]
    (\frac{(v_0)^2\sin\alpha\cos\alpha}{g},\frac{(v_0)^2(\sin^2\alpha)}{2g})
    [/tex]


    The question now requires you to show that the distance from the vertex to the focus is given by:
    [tex]
    \frac{(v_0)^2(\cos^2\alpha)}{2g}
    [/tex]

    and that the equation of the directrix is
    [tex]
    y=\frac{(v_0)^2}{2g}
    [/tex]

    John


    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 19, 2009 #2

    Mentallic

    User Avatar
    Homework Helper

    The distance from the vertex to the focus is equal to the distance from the vertex to the directrix. So all you have to do is find the perpendicular distance between the directrix and vertex (thus, the directrix is going to be vertically above the vertex and focus in this case).
     
  4. Aug 19, 2009 #3
    Many thanks Mentallic
    My apologies but I did not include the original equation in my original post.
    However, I have now solved the problem by converting the original equation by completing the square etc into the form:
    [tex]
    (x-h)^2=4p(y-k)
    [/tex]
    where (h,k) are the vertex coordinates and p is the distance from the vertex to the focus.
    I was then able to find p and then the equation of the directrix.
    John
     
  5. Aug 19, 2009 #4

    Mentallic

    User Avatar
    Homework Helper

    Ahh hehe I was gonna say! For a second there I thought you were going to take a very complicated and winding road when all that needs to be done is simple distance calculations :smile:
     
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