In this situation, we say that this curve is part of the

In general, suppose we have a family of curves C_{t} is defined by an equation F(t,x,y)=0. A curve E
is said to be part of the envelope of the family if it shares a
tangent line with each curve C_{t}.
We would like to be able to compute an equation for curves in
the envelope for this family, but this may be difficult. It is
also possible that no such envelope curves exist.

However, when F(t,x,y) is a polynomial function of
all 3 variables t,x,y, there is a way to use elimination of
variables to compute the equation of a curve containing an
envelope to the family C_{t}. This is explained in
Projection, profiles, envelopes.

The idea goes like this. The equation F(t,x,y)=0 defines a
surface in 3-dimensional *(x,y,t)*-space.
In the case of our family of reflected line this surface
is built geometrically as a "ruled surface" by stacking up each
reflected light ray in sequence. Click here to see an
MPEG movie (478 Kbytes) of this process.
Of course, the complete surface is not merely the union of the
reflected rays that come in from the left (as in Figure 5),
but also includes the union of rays that reflect off the
left-hand side of the coffee cup.

As explained in Projection, profiles, envelopes, the points of the envelope are the projections onto the xy-plane of those points on this surface F(t,x,y)=0 for which the tangent plane is parallel to the t-axis. This leads us to look for the equation of the envelope curves by eliminating the variable t from the equations

F(t,x,y) = 0

d/dt F(t,x,y) = 0

In order to do this for our family of reflected lines C_{t}, we
must first express their equations in the form F(t,x,y) = 0,
i.e. we must get rid of the square root occurring in the equation
that we already found, and then bring all the terms to one side.

- Show that you can manipulate the equation we found for the line reflected from y=t so that there is only a single product involving a square root sitting alone on one side of the equation.
- Now square both sides of the equation to get rid of the square
root, and bring all the terms to one side to obtain something
of the form F(t,x,y)=0 for some polynomial F(t,x,y) in the
variables t,x,y. Compare your answer with the one we got:
F(t,x,y) = ((y-t)(2-t

^{2}) + t(4-t^{2}))^{2}- t^{2}x^{2}(4-t^{2})

Using Maple, we can eliminate the

> F:=((y-t)*(2-t^2)+t*(4-t^2))^2-t^2*x^2*(4-t^2); 2 2 2 2 2 2 F := ((y - t) (2 - t ) + t (4 - t )) - t x (4 - t ) > GB:=gbasis({F,diff(F,t)},[t,x,y],plex):Only one polynomial (the 9th and last) in the resulting Gröbner basis does not involve the variable t, so this gives the equation for the envelope:

> factor(GB[9]); 2 2 2 2 4 2 2 4 6 4 2 x y (- 4 + 12 x - 15 y - 12 x - 24 x y - 12 y + 4 x + 12 x y 4 2 6 + 12 y x + 4 y )

Vic Reiner <reiner@math.umn.edu> Frederick J. Wicklin <fjw@geom.umn.edu> Last modified: Tue Apr 16 07:43:57 1996