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Homework Help: Anyone Else See A Problem With This Example From My Book?

  1. May 25, 2010 #1
    1. The problem statement, all variables and given/known data

    My book gives the theorem for parabolas as:

    The graph of the equation:

    [tex] y = ax^{2} [/tex]

    (where [tex] a \neq 0 [/tex] ) is the parabola with focus [tex] F(0,\frac{1}{4}a) [/tex] and the directrix [tex] y = -(\frac{1}{4}a)[/tex]. Its vertex is [tex] (0,0) [/tex], and its axis is the y-axis.

    It then goes on to use these equations in an example like so:

    PROOF: Let us find the equation of the parabola with focus F(0,d) and directrix y= -d.

    Where [tex] d= \frac{1}{4}a [/tex]

    *Rest of proof omitted as it has nothing to do with what I am asking*

    so then it moves on to an example where it asks:

    Find the focus and directrix of the parabola:

    [tex] y = -\frac{1}{2}x^{2} [/tex]

    *straight from the book*=

    Using Theorem 1 (the theorem posted above):

    [tex] a = -\frac{1}{2} \;\;\;\; and \;\;\;\; d= \frac{1}{4}a [/tex]

    so in this problem:

    [tex] d = -\frac{1}{2} [/tex] ?????

    *End from book*

    Shouldn't d = -1/8 not -1/2? If d = (1/4)a and a = -1/2, then isn't (-1/2)(1/4) = -1/8?

    I ask this only because every other point in this chapter builds from this point and I want to make sure I am not just stupid and there is actually a problem here.
    Last edited: May 25, 2010
  2. jcsd
  3. May 25, 2010 #2


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    Homework Helper

    The problem is that

    [tex]d = \frac{1}{4a}[/tex]

    Perhaps you copied down out of the book incorrectly? Was it written like d = 1/4a? Without parentheses it's kind of ambiguous if the a is in the numerator or denominator.
  4. May 25, 2010 #3
    Well here is an image right out of the textbook:


    I am very confident when I say that it has been presenting it as:

    [tex] d = \frac{1}{4}a [/tex]

    In the image it looks, quite clearly to me, that the a is not in the denominator of the fraction.

    I am not sure I did the image right, if it doesnt show here is a link:

    Last edited by a moderator: Apr 25, 2017
  5. May 25, 2010 #4


    Staff: Mentor

    That's definitely a typo in the book. If a = (1/4)d, then d = 4a.

    I suspect the author really meant d to be the distance from the vertex of the parabola to the directrix (or from the vertex to the focus) and a is the distance across the parabola through the focus. The relation then would be a = 4d, or d = (1/4)a.

    Some books define parabolas geometrically as the locus of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

    Who is the author of this book?
  6. May 25, 2010 #5
    That is the definition of parabola they gave in this book, the book is:

    Elementary Calculus: An Infinitesimal Approach
    by H. Jerome Keisler
  7. May 25, 2010 #6
    I had his email on file so I sent him an email about it already, if that's why you asked...
  8. May 25, 2010 #7

    George Jones

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  9. May 25, 2010 #8
    Yeah, thats where the image came from, although I downloaded it as pdf, not that it matters. As I said, H Jerome Keisler has been notified. All I can do is move on and not get mixed up by his typo.
  10. May 25, 2010 #9

    George Jones

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    I am confused. When I look at EXAMPLE 2 form the link that I gave, I see

    [tex]y = -2x^2[/tex]

    In Theorem 1, [itex]a = -2[/itex] and [itex]d = \frac{1}{4} a = - \frac{1}{2}[/itex]
  11. May 25, 2010 #10
    Oh yeah, ok I didn't bother to look as I downloaded the full pdf from that same site. Evidently it has been updated and I have an older version of the book. I wouldn't just make up a false picture :) lol.
  12. May 25, 2010 #11
    Also if you look the problem with:

    [tex] d = \frac {1}{4} a \;\;\;\;and\;\;\;\; a = \frac{1}{4}d [/tex]

    is still there, right above EXAMPLE 2.
  13. May 25, 2010 #12

    George Jones

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    Yes, this what I meant in my first post in this thread.

    Good luck.
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