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

[Asphyxiated]

Homework Statement

My book gives the theorem for parabolas as:

The graph of the equation:

y = ax^{2}

(where a \neq 0 ) is the parabola with focus F(0,\frac{1}{4}a) and the directrix y = -(\frac{1}{4}a). Its vertex is (0,0), 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 d= \frac{1}{4}a

*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:

y = -\frac{1}{2}x^{2}

*straight from the book*=

Using Theorem 1 (the theorem posted above):

a = -\frac{1}{2} \;\;\;\; and \;\;\;\; d= \frac{1}{4}a

so in this problem:

d = -\frac{1}{2} ?

_________________________________
*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.

 

[Mute]

The problem is that

d = \frac{1}{4a}

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.

 

[Asphyxiated]

Well here is an image right out of the textbook:

[URL]http://images2e.snapfish.com/232323232%7Ffp53834%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D32%3B5328%3B%3C%3A347nu0mrj[/URL]

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

d = \frac{1}{4}a

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 doesn't show here is a link:

http://www2.snapfish.com/snapfish/slideshow/AlbumID=2046768024/PictureID=67575340024/a=2078147024_2078147024/otsc=SHR/otsi=SPIClink/COBRAND_NAME=snapfish/"

 

[Mark44]

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?

 

[Asphyxiated]

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).

That is the definition of parabola they gave in this book, the book is:

Elementary Calculus: An Infinitesimal Approach
by H. Jerome Keisler

 

[Asphyxiated]

I had his email on file so I sent him an email about it already, if that's why you asked...

 

[George Jones]

It seems that something like a typo occurred. Look at EXAMPLE 2 on page 259 from chapter 5 at

http://www.math.wisc.edu/~keisler/calc.html.

 

[Asphyxiated]

Yeah, that's 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.

 

[George Jones]

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

y = -2x^2

In Theorem 1, a = -2 and d = \frac{1}{4} a = - \frac{1}{2}

 

[Asphyxiated]

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.

 

[Asphyxiated]

Also if you look the problem with:

d = \frac {1}{4} a \;\;\;\;and\;\;\;\; a = \frac{1}{4}d

is still there, right above EXAMPLE 2.

 

[George Jones]

Evidently it has been updated and I have an older version of the book.

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

Good luck.