Necessary equation for envelopes Consider a family of curves parameterised by $t$, where each member of the family is described by
$$F(t,x,y) = 0.$$
Define an envelope of the family to be a curve $E$ such that every point on $E$ is tangent to some member of the family. When do we have that
$$\frac{\partial F}{\partial t}(t,x,y) = 0$$
for all $(x,y) \in E$, and in those cases, why?

Yes, the envelope $E$ is precisely the solution set of the system
\begin{align*}
F(t,x,y) &= 0 \\
\frac{\partial F}{\partial t}(t,x,y) &= 0\,,
\end{align*}
if we have a sufficient condition given by the Implicit Function Theorem. That is, in order for these equations to be guaranteed to define $(x,y)$ locally as a $C^1$ function of $t$, we need the $2\times 2$ matrix
$$\begin{bmatrix} \frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\
\frac{\partial^2 F}{\partial x\partial t} & \frac{\partial^2 F}{\partial y\partial t}
\end{bmatrix}$$
to be nonsingular.
Here's a hint to explain where the condition $\frac{\partial F}{\partial t}=0$ comes from. At the point $(x(t),y(t))\in E$ the tangent vector to $E$ should be tangent to the curve 
$$F(x,y,t)=0\tag{$\star$}$$
(with $t$ fixed), hence orthogonal to the gradient vector $\big(\frac{\partial F}{\partial x}(x(t),y(t)),\frac{\partial F}{\partial y}(x(t),y(t))\big)$. Now differentiate equation $(\star)$ with $t$ varying and $x=x(t)$, $y=y(t)$.
