TITLE: How to determine the curve?
QUESTION [7 upvotes]: 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:
$$y=\frac{t(x-t)}{1-t}$$
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 $\frac{1}{6}$?

REPLY [2 votes]: I have a simpler solution.

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

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.