Understanding the derivation of the equation for envelopes.

Question

Given a family of curves, an envelope is defined as a curve that it is tangent to every curve in the family of curves at some point on it.

To derive the equation for it, the first step is to parameterize the envelope with parameter $t$, which also indexes which curve in the family.

But what if there is a curve tangent to the envelope at two or more points? Which points should $x(t),y(t)$ be then?

And my second question is that whether the equations $F(x,y,p)=0$ and $F_p(x,y,p)=0$ is just a necessary condition for a point to be on the envelope or it is actually both necessary and sufficient.

Edit:

Now I know that there's actually three definitions:

1. synthetic: the union of characteristic points (aka "limit points" between two nearby curves)
2. impredicative: a curve that it is tangent to every curve in the family of curves at every point on it
3. analytic: the union of the solution set of $F(x,y;p)=0$ and $F_p(x,y;p)=0$ of every $p$

And they're somehow equivalent. (source: https://www.emis.de/journals/BAG/vol.48/no.2/b48h2koc.pdf)

Is there an easy way to show this?

Answer 1

There is in projective geometry a notion of dual, that extends to this situation for the special case of a family of lines. The dual projective plane of lines in the usual projective plane, is just that, a projective plane. The dual curve is the envelope seen as a curve in the dual projective plane. The dual curve of a curve with a bitangent has a corresponding node. Traversing a nodal curve a curve you come once to the node for one value of $t$ and back to it at a different value of $t$. If a line of your envelope touches a curve twice, it's the same story.

To continue with a family of lines, let a specific one be $xt^2+yt+1=0,$ then the second equation is $2tx+y=0$ which on eliminating $t$ gives $y^2-4x=0.$

For the more general family of lines $p(t)x+q(t)y+1=0, p'(t) x+ q'(t) y=0,$ the usual formula for the dual emerges from solving the linear system: $(x,y)=(\frac{-q'}{pq'-p'q},\frac{p'}{pq'-p'q}).$

To your second question, parametrized curves are irreducible and the elimination map is closed.