Geometry of Envelope form definition I had read about the envelope of the family of the curve.
It is defined as a curve which is tangent to each member of the family at a single point and it is union of all such points.
To find envelope they had provided condition:
The envelope is the simultaneous solution of following 
$F(x,y,t)=0$ and $\frac{\partial}{\partial t}F(x,y,t)=0$
I do not understand how the definition and above conditions are equivalent.
Please Help me.
Any Help will be appreciated.
 A: Consider the curve $t\mapsto (x(t),y(t))$ and let's see what are the necessary conditions it must fullfil to be an envelope.

I had read about the envelope of the family of the curve. It is
  defined as a curve which is tangent to each member of the family at a
  single point and it is union of all such points.



*

*First of all, the curve $t\mapsto (x(t),y(t))$ must have a point of contact with the one-parameter $t$ family of curves implicitly defined by $F(x,y,t)=0$. So you must have $F(x(t),y(t),t)=0$ which is your first equation.

*Then the curve $t\mapsto (x(t),y(t))$ must be tangent to the curves implicitly defined by $F(x,y,t)=0$. 
On one side, the vector $\left(\begin{array}{c} x'(t^\star) \\ y'(t^\star) \end{array}\right)$ is the tangente direction at $t^\star$ of the curve $t\mapsto (x(t),y(t))$. 
On the other side, for a fixed $t=t^\star$ the vector $\nabla F=\left(\begin{array}{c} \partial_x F \\ \partial_y F  \end{array}\right)$ is a vector orthogonal to the tangent of the level-set of $F(x,y,t^\star)=0 $ (if this last point is not clear, you can read  this) 
Like we want to have an identical tangent direction at point $(x(t^\star),y(t^\star))$ we must impose a zero scalar product
$$
\left(\begin{array}{c} x'(t^\star) \\ y'(t^\star) \end{array}\right).\nabla F = 0 = \partial_x F(x(t^\star),y(t^\star),t^\star) x(t^\star)+\partial_y F(x(t^\star),y(t^\star),t^\star) y(t^\star)\ \ \ \ \ (1)
$$
However, from point 1/, we also have $F(x(t),y(t),t)=0$, thus $\frac{d}{dt}F(x(t),y(t),t)=0$ which is
$$
\partial_x F(x(t^\star),y(t^\star),t^\star) x(t^\star)+\partial_y F(x(t^\star),y(t^\star),t^\star)y(t^\star)+\partial_tF(x(t^\star),y(t^\star),t^\star) = 0\ \ \ \ \ (2)
$$
Equations (1) & (2) implies that $\partial_tF(x(t^\star),y(t^\star),t^\star) = 0$
To recap, a necessary condition for $(x(t),y(t))$ being your envelope is that:


*

*$F(x(t),y(t),t)=0$

*$\partial_t F(x(t),y(t),t)=0$ 
which is the expected result.
