Results 1 to 5 of 5

Math Help - Find the graph of equation...

  1. #1
    Junior Member
    Joined
    Mar 2010
    Posts
    41

    Find the graph of equation...

    Hi, I'm having trouble figuring out the answer to this question:

    Find an equation whose graph consists of all points P(x,y) whose distance from the point F(0,p) is equal to its distance PQ (there should be a line above the PQ) from the horizontal line y = -p (p is a fixed positive number).

    This is my effort of reproducing the graph that came with it:



    I thought I could do it by plugging (x,y) and (0,p) into the standard equation of a circle, but I'm not getting the same answer as is listed in the book (they say that it's x^2=4py (parabola))

    Any help greatly appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jan 2010
    Posts
    43
    Do you know the Distance Formula?

    Play with that and show us what you get from it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2010
    Posts
    41
    I tried it, but I still can't seem to get it.

    I did:

    d = [(0-x)^2 + (p-y)^2]^1/2
    d = [x^2 + p^2 - 2py +y^2]

    I'm not sure what to do from there
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2010
    Posts
    43
    Yeah, you have the right idea, this is all you have to remember for the parabola;

    \overline{PF} = \overline{PQ}

    the following, which I'll show you how to do, should just be the manual work that follows from remembering the above property.

    \overline{PF} \ = \ \sqrt{(x \ - \ 0)^2 + (y \ - \ p)^2}  \ = \sqrt{(x)^2 + (y \ - \ p)^2}

    \overline{PQ} \ = \ \sqrt{(x \ - \ x)^2 + (y \ - \ (-p))^2}  \ = \sqrt{(0)^2 + (y \ + \ p)^2}

    \overline{PF} = \overline{PQ}

    \sqrt{(x)^2 + (y \ - \ p)^2} \ = \ \sqrt{(0)^2 + (y \ + \ p)^2}

    Square both sides and solve;

    (x)^2 \ + \ (y \ - \ p)^2 \ = \  (y \ + \ p)^2

    x^2 \ + \ y^2 \ - \ 2py \ + \ p^2 \ = \  y^2 \ + 2py \ + \ p^2

    Cancel terms;

    x^2 \  - \ 2py \ = \   2py

    x^2 \  \ = \   2py \ + \ 2py

    x^2 \ = \ 4py

    It's basically the same idea for the ellipse and hyperbola with a little change
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2010
    Posts
    41
    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Find an equation from this graph.
    Posted in the Calculus Forum
    Replies: 8
    Last Post: August 14th 2011, 02:41 PM
  2. How do i find the equation of this graph?
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 4th 2010, 08:34 PM
  3. How to find vertex of a graph or equation
    Posted in the Calculus Forum
    Replies: 14
    Last Post: June 6th 2008, 07:57 PM
  4. how to find equation of polynomial from a graph
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: September 16th 2007, 07:04 AM
  5. Find pattern, equation, then graph...
    Posted in the Pre-Calculus Forum
    Replies: 11
    Last Post: September 13th 2007, 02:11 PM

Search Tags


/mathhelpforum @mathhelpforum