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In the figure above, segment $PQ$ is determined by two points: $P: (t,0)$ and $Q: (1,t)$, where $t\in [0,1]$ continuously increases and decreases between $0$ and $1$.

Then this gives a close region swept by $PQ$, the upper edge of which is a curve.

How to determine the equation of the curve (maybe implicit form)?

My own method

Suppose the curve is $y=y(x)$, then for any fixed $x_0\in (0,1)$, there is a vertical line $x=x_0$, which intersects a bundle of such $PQ(t)$ segments:


Easy to conclude that, the desired $y_0$ on the curve corresponding to $x_0\left(\in(0,1)\right)$ is the maximum of:

$$y_0= \max\limits_{t\in (0,1)}\frac{t(x_0-t)}{1-t}=2-2\sqrt{1-x_0}-x_0$$

How to use the envelope concept? Is there any elementary method since the area is $\frac{1}{6}$?

  • $\begingroup$ Do you know derivatives, tangent lines and integrals? $\endgroup$ – Tomas Feb 6 '15 at 9:47
  • $\begingroup$ Yes, I know; I am not sure whether calculus is necessary in order to determine the area, but I am pretty sure in order to determine the curve, calculus would be a help $\endgroup$ – LCFactorization Feb 6 '15 at 9:53
  • 3
    $\begingroup$ Look at the wiki entry for Envelope, it teach you how to determine the curve. $\endgroup$ – achille hui Feb 6 '15 at 10:29
  • $\begingroup$ I use optimization method and obtains $y=2-2\sqrt{1-x}-x$, then the area is $1/6$. Is there any elementary method in order to determine the area? $\endgroup$ – LCFactorization Feb 6 '15 at 11:12

I know that the result will be a conic section. I'll prove that later on, but start with it. I come from a background of projective geometry, so I'd homogenize your points by appending a $1$, then find coordinates for the line joining them by computing the cross product.

$$\begin{pmatrix}t\\0\\1\end{pmatrix}\times \begin{pmatrix}1\\t\\1\end{pmatrix}= \begin{pmatrix}-t\\1-t\\t^2\end{pmatrix}$$

This could also be written as $-tx + (1-t)y + t^2=0$ in usual Cartesian coordinates. It is a special case of a line equation $ax+by+c=0$. Now I want to describe a conic section in the dual sense, i.e. not as a set of points but instead a set of tangent lines. That means I need to find a homogeneous quadratic form in $a,b,c$ which is zero for the line given above. For that, consider all quadratic coefficients:

\begin{align*} a^2 &= t^2 & ab &= t^2-t & b^2 &= 1-2t+t^2 \\ ac &= -t^3 & bc &= t^2-t^3 & c^2 &= t^4 \end{align*}

The $c^2$ expression is the only one with degree $4$, and the $b^2$ expression is the only one with constant term. So these two can't be part of the quadratic equation, since they have nothing to cancel against. After removing them, the $ab$ expression is the only one with linear term, so we drop that as well. Now the relation is easy to see:

$$ a^2 + ac - bc = 0 $$

Written as a matrix:

$$ (a,b,c)\cdot\begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}a\\b\\c\end{pmatrix} = 0 $$

Now, as I said, that's the matrix for the dual conic. The primary conic is represented by the inverse matrix, or any multiple thereof:

$$ (x,y,1)\cdot\begin{pmatrix}1&1&0\\1&1&-2\\0&-2&0\end{pmatrix} \cdot\begin{pmatrix}x\\y\\1\end{pmatrix} = 0 $$

Since the determinant of the upper left $2\times2$ matrix is zero, this is a parabola. It can also be described by the equation

\begin{align*} x^2 + 2xy + y^2 &= 4y \\ (x+y)^2 &= 4y \end{align*}

Since this is the primal conic to the dual one we computed before, and that dual one was deduced from a relation between the terms of your lines, this ensures that your whole family of lines will be tangent to this curve. If you like, you can compute the point of tangency as

$$ \begin{pmatrix}2&0&1\\0&0&-1\\1&-1&0\end{pmatrix} \cdot\begin{pmatrix}-t\\1-t\\t^2\end{pmatrix} =-\begin{pmatrix}2t-t^2\\t^2\\1\end{pmatrix} $$

So the point $(2t-t^2, t^2)$ lies on both the parabola and the line.



I would say

$F(x,y,t)=0\equiv t(x-t)-y(1-t)=0,\quad t\in\langle 0,1 \rangle$

$F'_t(x,y,t)=0\equiv x-2t+y=0$

Solving this system of equations for the x, y:

$x=t(2-t), y=t^2, \quad t\in\langle 0,1 \rangle$ is a parametric curve of equation to search.


  • $\begingroup$ Very simple and neat answer. All three are acceptable. I choose the first one that also uses projective geometry. Thank you very much! @MvG $\endgroup$ – LCFactorization Feb 7 '15 at 2:40

I have a simpler solution.

  1. Find the equation of the line with $(t,0,1)\times(1,t,1)=(-t,1-t,t^2)$ or $$(-t)x+(1-t)*y+(t^2)=0$$
  2. Find the extema $t$ with $$\frac{{\rm d}}{{\rm d}t}\left(-t\,x+(1-t)\,y+t^2\right)=0$$ $$\left. -x-y+2 t=0 \right\}\; t=\frac{x+y}{2} $$
  3. Plug $t$ into line equation for $$-\frac{(x+y)^2}{4}+y=0$$ which is the resulting curve pic

Hint: The envelope of an implicit curve $f(x,y,t)=0$ is found by solving for $t$ in $\frac{\partial f}{\partial t}=0$ and using it into the original equation.


My other answer describes how I myself think about this. But I realize that there is a lot of background knowledge in there, which might make this less accessible to members of other communities than projective geometry. So here is my attempt at a more generic solution.

Suppose you have found the equation of the line to be $tx + (t-1)y = t^2$. Take a second line from your family, using $u$ instead of $t$ as the parameter, i.e. $ux + (u-1)y = u^2$. These two intersect in a point $(t+u-tu, tu)$. Now if you consider $\lim_{u\to t}$ then strictly speaking the point of intersection becomes undefined, but the above will simply become $(2t-t^2,t^2)$. So that's the one point that your line does not have in common with any other line of the family (extended from $t\in[0,1]$ to $t\in\mathbb R$). So at that point this line is the one defining the curve. Therefore that's your parametric equation of the envelope curve.


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