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Rusty - fill in missing steps, please

  1. Sep 8, 2006 #1
    Hello everyone,

    I'm taking a high level Geometry course, and as it turns out, I'm a tad out of touch on at least one basic aspect learned years passed.

    Parabola with equation [tex] y^2 = 2x[/tex], and parametric equation [tex] x = \frac{1}{2}t^2, y = t (t \in \Re) [/tex] etc.

    The problem is concered with finding the equation of the chord that joins the distinct points P and Q on the parabola w/ parameters [tex] t_1 [/tex] and [tex] t_2 [/tex] respectively, etc.

    No that much of that was relevant to my quandray, which is in finding the gradient of PQ.
    [tex] m = \frac{t_1 - t_2}{{\frac{1}{2}(t_1^2 - t_2^2)} = \frac{2}{t_1 + t_2} [\tex]
    edit: sorry, I guess I didn't get that tex tag right (what is wrong with it? Can I nest fractions like that?)

    m = (t1 - t2)/[.5(t1^2 - t2^2)] = 2/(t1 + t2).

    I can't remember (if I had learned it that is) how they arrived from the first to the latter fraction.

    An explanation would be quite welcome, thanks.
    Last edited: Sep 8, 2006
  2. jcsd
  3. Sep 8, 2006 #2


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    Are you asking why:

    [tex]\frac{t_1-t_2}{\frac{1}{2} ({t_1}^2-{t_2}^2)}=\frac{2}{t_1+t_2}[/tex]

    ? If so, remember that [itex]a^2-b^2=(a-b)(a+b)[/itex]. And you can click on the equation above to see how I got it working. I think you have an extra {.
  4. Sep 8, 2006 #3

    Thanks, I guess I wasn't looking at it the right way.

    Did you need the {'s around each t in the denominator?
  5. Sep 8, 2006 #4


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    No, but it makes it look nicer. {...}^2 puts the two above and to the right of the whole expression. If you just wrote t_1^2, you get some looks like a tensor element:

    [tex] {t_1}^2 [/tex] vs [tex] t_1^2[/tex]
  6. Sep 8, 2006 #5
    Yoss, your last equation... you used a backslash to close the tex tag is all you got wrong.
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