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Problem: Let $u$ denote the distance of a moving a point $P$ on the parabola $y^{2}=4px$ from the directrix $x=-p$ and from the focus $\left(p,0\right)$. If the point moves in such a way that $u^{\prime}=x^{\prime}=1$ (unit horizontal speed), show that the tangent vector $\overline{t}$ show in the figure is $\overline{t}=\left(2x/\left(x+p\right),y/\left(x+p\right)\right)$ , while the actual velocity vector of $P$ is $\overline{v}=\left(1,\sqrt{p/ x}\right)$. Then show that $\overline{t}$ and $\overline{v}$ point in the same direction, but $$ \left\Vert \overline{t}\right\Vert =2\sqrt{\frac{x}{x+p}}\quad\text{while}\quad\left\Vert \overline{v}\right\Vert =\sqrt{\frac{x+p}{x}}. $$

enter image description here

Here's what I have so far: we parametrize $y^{2}-4px=0$, by letting $x=pu^{2}$ and $y=2pu$ we have parametrized the parabola with respect to parameter $u$. Now, let the curve $\boldsymbol{c}\left(u\right)=\left(pu^{2},2pu\right)$. Taking, the derivative, we have $$ \mathbf{c}^{\prime}\left(u\right)=\left(2pu,2p\right). $$ I want to obtain $\overline{v}$ from this, but I'm not sure how, so here's what I did. $$ \mathbf{c}^{\prime}\left(u\right)=2p\underbrace{\left(u,1\right)}_{\overline{v}???}. $$ So, I just guess (but I'm note sure why) \begin{align*} \overline{v} & =\left(u,1\right)\\ & =\left(\sqrt{\frac{x}{p}},1\right)\quad\left(\text{since }x=pu^{2}\right) \end{align*} then, \begin{align*} \left\Vert \overline{v}\right\Vert & =\sqrt{\frac{x}{p}+1}\\ & =\sqrt{\frac{x+p}{p}} \end{align*} which isn't quite what's expected in the problem above. Where am I going wrong? Also, I'm not sure how to derive $\overline{t}$. I'm assuming that it's obtained by first normalizing the vector from the focal point $F$ to $P$ as well as the vector from the directrix to $P$, then summing these two vectors gives us $\overline{t}$, but I'm not sure how to formulate these vectors. Any ideas?

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2 Answers 2

up vote 1 down vote accepted

The aim of the task is to show that Roberval's method gives a tangent vector in the sense of modern calculus. Roberval determines two forces that keep a point on the parabola and claims that the sum of those two forces $\bar t$ is tangent to the parabola. Quite clever.

The first force is the vector emanating from $P$ in $x$-direction of length $1$, thus $(1,0)$. The second one also emanates from $P$ in direction $FP=(x-p,y)$. Now to keep the point on the parabola, the second vector must have the same length as the first, i.e., $1$. You'll easily calculate (using $y^2=4px$) it's length to be $x+p$. Finale: adding both forces gives the Roberval-vector $\bar t$. From here you may calculate $\|\bar t\|$.

The last step is to show that Roberval's $\bar t$ is indeed tangent in the sense of modern calculus. Now your velocity vector $c'(u)$ luckily satisfies $$c'(u)=(2pu,2p)\parallel(2pu^2,2pu)=(2x,y)\parallel\frac{1}{x+p}(2x,y)=\bar t.$$

We are essentially done! Good ole Rob was right. (Of course we cannot hope that $\|\bar t\|=\|c'(u)\|$, because $\|c'(u)\|$ heavily depends on the parametrization choosed. So comparing the lengths is somewhat ... unnecessary.)

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something doesn't add up here. I believe the the first force vector, call it $\overline{w}$ should be $(1,0)$. Now, for the second force vector, call it $\overline{v}= FP/|FP|$, we have $\overline{v}=((p-x)/(x+p),y/(x+p))$, then summing $\overline{v}$ and $\overline{w}$, we have $(2p/(x+p),y/(x+p))$, which isn't quite what we want. –  Black Milk Nov 8 '13 at 13:21
    
changing the first force vector to $(x-p,y)$ seems to do the trick. –  Black Milk Nov 8 '13 at 13:29
    
Thank you, I mixed up the coordinates. Very nice question anyway. I've corrected my answer. –  Michael Hoppe Nov 8 '13 at 14:22

The point moves with unit speed in the horizontal direction, so $\mathbf{c}'(u) = 2pu(1, 1/u)$, and $\bar{v} = (1, 1/u)$.

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