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Geometry Question

  1. Sep 19, 2010 #1
    A point P(a, b) is equidistant from the y-axis and from the point (4, 0). Find a relationship between a and b.

    Any hints on how to go about this appreciated.

  2. jcsd
  3. Sep 19, 2010 #2
    Write down the appropriate equation. Expand, simplify.
  4. Sep 19, 2010 #3

    I've tried numerous ways of tackling this but I can't seem to get the answer that I have in my solutions booklet. Looking to see if anyone can give an alternative starting. Anyway here's one approach I used...

    Let Q be the point the line A(4, 0) -> P(a, b) cuts the y-axis.

    [tex]AP^2 = (4 - a)^2 + b^2[/tex]

    [tex]PQ^2 = a^2 + (b - (4b/(4-a)))^2[/tex]

    but [tex]AP^2=PQ^2[/tex]. In trying to simplify that I get something that isn't even close... which is:

    [tex](b^2(6-a))/((4-a)^2) = 2 - a[/tex]

  5. Sep 19, 2010 #4
    Take several points P(a,b), draw them on a piece of paper, and for each one think what is its "distance from the y-axis". Draw this distance. Perhaps you will notice something?
  6. Sep 19, 2010 #5


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    That's the basic idea but I see no reason to introduce "Q". The distance from (a, b) to the y-axis is the distance from (a, b) to (0, b) and is equal to |a|. The distance from (a, b) to (4, 0) is [itex]\sqrt{(a- 4)^2+ b^2}[/itex]. (a, b) is "equidistant from the y-axis and (4, 0)" if and only if those are equal:
    [tex]|a|= \sqrt{(a- 4)^2+ b^2}[/itex]
    to get rid of the square root on the right, square both sides. Fortunately [itex]|a|^2= a^2[/itex] so that also gets rid of the absolute value:
    [tex]a^2= (a- 4)^2+ b^2[/tex]
    Multiplying out [itex](a- 4)^2[/itex] will also give an [itex]a^2[/itex] on the right which will cancel the one on the left. You will have an equation with a and [itex]b^2[/itex] but no [itex]a^2[/itex]- a parabola.

    Personally, I would have used (x, y) rather than (a, b) but its the same thing.
  7. Sep 19, 2010 #6
    Ah, I see...

    That makes perfect sense. The problem I had was always ending up with an a^2 after doing all the algebraic manipulation.

    Arkajad and HallsofIvy, many thanks. :)
  8. Apr 1, 2011 #7


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    Gold Member

    Isn't this a simple parabola with the focus at (4,0) and the directrix the Y-axis?
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