Finding the gradient of a chord on the parabola y² = 2x

[Yoss]

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 $$y^2 = 2x$$, and parametric equation $$x = \frac{1}{2}t^2, y = t (t \in \Re)$$ 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 $$t_1$$ and $$t_2$$ respectively, etc.

No that much of that was relevant to my quandray, which is in finding the gradient of PQ.
$$m = \frac{t_1 - t_2}{{\frac{1}{2}(t_1^2 - t_2^2)} = \frac{2}{t_1 + t_2} [\tex]<br /> edit: sorry, I guess I didn't get that tex tag right (what is wrong with it? Can I nest fractions like that?)<br /> <br /> m = (t1 - t2)/[.5(t1^2 - t2^2)] = 2/(t1 + t2).I can't remember (if I had learned it that is) how they arrived from the first to the latter fraction. <br /> <br /> An explanation would be quite welcome, thanks.$$

 

[StatusX]

Are you asking why:

$$\frac{t_1-t_2}{\frac{1}{2} ({t_1}^2-{t_2}^2)}=\frac{2}{t_1+t_2}$$

? If so, remember that $a^2-b^2=(a-b)(a+b)$. And you can click on the equation above to see how I got it working. I think you have an extra {.

 

[Yoss]

Are you asking why:

$$\frac{t_1-t_2}{\frac{1}{2} ({t_1}^2-{t_2}^2)}=\frac{2}{t_1+t_2}$$

? If so, remember that $a^2-b^2=(a-b)(a+b)$. And you can click on the equation above to see how I got it working. I think you have an extra {.

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

Did you need the {'s around each t in the denominator?

 

[StatusX]

Did you need the {'s around each t in the denominator?

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:

$${t_1}^2$$ vs $$t_1^2$$

 

[WhyIsItSo]

Yoss, your last equation... you used a backslash to close the tex tag is all you got wrong.