How can I make sure that when we eliminate the parameter from the curve \begin{align*} F(t,x,y) &= 0 \\ \frac{\partial F}{\partial t}(t,x,y) &= 0\,, \end{align*} the equation obtained is the needed envelope?
I didn't find anything anywhere I searched for about why we proceeded like this to find an envelope and from where those conditions derived.
Also; how is $(x,y, dy/dx)$ same on both envelope and corresponding curve from that family of curves?
Fix a value of parameter $t$ and consider the curve with equation $F(t,x,y)=0$. Take then a slightly displaced curve $F(t+\Delta t,x,y)=0$ and suppose this curve meets the other one at some point $P$. If $\lim_\limits{\Delta t\to0}P$ exists, then we say that this limiting point (which we may call $P(t)$) belongs to the envelope of the pencil of curves.
The coordinates of $P$ satisfy the system $$ \cases{ F(t,x,y)=0 \\ F(t+\Delta t,x,y)=0 \\ } $$ But this system is equivalent to the system obtained substituting the second equation with the difference of the two equations, hence equivalent to: $$ \cases{ F(t,x,y)=0 \\ \\ \displaystyle{F(t+\Delta t,x,y)-F(t,x,y)\over \Delta t}=0 \\ } $$ We can now take the limit $\Delta t\to0$ in the second equation (provided $F$ is regular enough): $$ \cases{ F(t,x,y)=0 \\ \\ \displaystyle{\partial F(t,x,y)\over \partial t}=0 \\ } $$
The solutions $(x(t),y(t))$ of this system will then be the coordinates of limiting point $P(t)$: as $t$ varies, they describe a curve which is the envelope of the pencil of curves.
If you want a cartesian equation for the envelope, you can instead eliminate $t$ to obtain a single equation in $(x,y)$.
