Derivation of the equation for the envelope Suppose we have a family of curves on the plane. The equation of the curves is given by 
$$ f(x ,y ;t) = 0  . $$
Here $t$ is the parameter. On Wiki, the equations determining the envelope of this family are 
$$ f(x, y ; t ) = f_t(x, y; t) =0 . $$
The derivation on wiki sounds reasonable, but I still find it not so intuitive. 
Any one has some other method? 
 A: Let a family $\Gamma$ of curves $\gamma_c$ in the $(x,y)$-plane be given by an equation of the form
$$F(x,y,c)=0\ .\tag{1}$$
A triple $(x_0,y_0,c_0)$ satisfying $(1)$ is regular if
$$\eqalign{\bigl(F_x(x_0,y_0,c_0),F_y(x_0,y_0,c_0)\bigr)&\ne(0,0),\cr F_c(x_0,y_0,c_0)&\ne0\ .\cr}\tag{2}$$
The first condition guarantees that the the curve $\gamma_{c_0}$ is a nice level line of the function $(x,y)\mapsto F(x,y,c_0)$ passing through the point $(x_0,y_0)$. The second condition is more crucial. It guarantees that the curves $\gamma_c$ with $c_0-\delta<c<c_0+\delta$ form a "homogeneous curve field in the neighborhood of $(x_0,y_0)$": Up to a differentiable distortion it looks like a field of parallels. For an explanation of this fact see below.
Now to the envelopes: If the point $(x_0,y_0)$ is lying not only on the curve $\gamma_{c_0}$, but also on the envelope $\epsilon$ of the family $\Gamma$ then the curve field in the neighborhood of $(x_0,y_0)$ does definitely not look like a field of parallels, because all $\gamma_c$ with $c$ near $c_0$ intersect $\gamma_{c_0}$ in points near $(x_0,y_0)$. It follows that in all points of $\epsilon$ the second condition of $(2)$ is violated; in other words: The triples $(x,y,t)$ with $(x,y)\in\epsilon$ necessarily satisfy $F_c(x,y,c)=0$.
Coming back to the regular triples: The  second condition $(2)$ guarantees that we can solve $(1)$ for $c$ in the neighborhood of $(x_0,y_0,c_0)$: There is a differentiable function $$g:\quad (x,y)\mapsto c=g(x,y)$$ with $g(x_0,y_0)=c_0$ such that in a neighborhood of $(x_0,y_0,c_0)$ the equation $(1)$ is equivalent with
$$g(x,y)=c\ .$$
This can be interpreted as follows: The family curves $\gamma_c$ appear as level lines of the function $g$. In order to check that these level lines behave as desired in the neighborhood of $(x_0,y_0)$ we have to make sure that $\nabla g(x_0,y_0)\ne(0,0)$. Now from the formula for the derivative of an implicitly given function we get
$$\nabla g(x_0,y_0)=\left(-{F_x\over F_c}, {F_y\over F_c}\right)_{(x_0,y_0,c_0)}\ne(0,0)\ .$$
A: The envelope is tangent to each member of the family of curves. So regard the envelope as a map from parameter $t$ to the point of tangency $(\hat x(t),\hat y(t))$. Since the envelope meets every curve,
$$F(\hat x,\hat y, \hat t)=0\qquad\text{for all $\hat t$}\tag1.$$
In particular (1) has zero derivative as a function of $\hat t$, so by the chain rule
$$
0=F_x(\hat x,\hat y, \hat t)\,\hat x'(\hat t)+
F_y(\hat x,\hat y, \hat t)\,\hat y'(\hat t)+
F_t(\hat x,\hat y, \hat t).\tag2
$$
Recall the envelope is tangent at $(\hat x,\hat y)$ to the level curve defined by
$$
F(x,y,\hat t)=0.\tag3$$
At $\hat t$ the envelope has slope $$\frac{dy}{dx}(\hat t)=\frac{\hat y'(\hat t)}{\hat x'(\hat t)},\tag4$$ but by tangency and (3) the slope at $\hat t$ also satisfies
$$
F_x(\hat x,\hat y,\hat t)+
F_y(\hat x,\hat y,\hat t)\frac{dy}{dx}(\hat t)=0.\tag5$$
Combining (4) and (5) yields
$$
F_x(\hat x,\hat y,\hat t)\,\hat x'(\hat t)+
F_y(\hat x,\hat y,\hat t)\,\hat y'(\hat t)=0$$
and comparing to (2), this implies
$$F_t(\hat x,\hat y,\hat t)=0\qquad\text{for all $\hat t$}.$$
A: There are a number of explanations here:
https://en.wikipedia.org/wiki/Envelope_(mathematics)
The mathworld explanation
is much too short;
I can see how you 
would be puzzled 
if you looked there.
A: Make a simple construction and all will be clear. ( That what I did way back) 
Plot 3 lines (  3 graphs)  for three values of c for three straight lines in (2).
$$ y = 2 x c + c^2      \tag{1} $$
Differentiate with respect to  $ c $ 
$$ 0 = 2 x + 2 c  \tag{2} $$
So plug in (1) $ x=-c $
Draw a smooth parabola touching the three straight lines, as an envelope or singular solution of (1).
Called C- discriminant method using Clairaut's differential equation. 
A: I was troubled with the same question, but at the end developed an intution. You can consider envelope as the locus of all the points when you keep fixing $x$ but vary $t$ to obtain maximum $y$ at that $x$. Then again change $x$ and find maximum $y$, varying $t$. The obtained locus of $(x,y)$ will be the envelope.
mathematically,differentiate $y(x,t)$ w.r.t. $t$ (keeping $x$ constant), put it equal to zero to obtain $t$ in terms of $x$ or $x$ in terms of $t$, i.e. $x(t)$. Now $y$ is maximum at that $t$ for fixed $x$. Find $y$ at that $x$ by putting $x$ in $y=f\big(x(t),t\big)$ and obtain $y(t)$. Now $x(t)$ and $y(t)$ will give the required locus.
