Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.)

share|cite|improve this answer
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 $\bar{v} = (1, 1/u)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.