Let's remember something simple: when you have a 2-variable ($x$ and $y$) family of curves in $\mathbb{R} ^{3}$ with 1 parameter ($\mu$) expressed in the following implicit equation: $f(x,y,\mu)=0$, you can find its envelope with the 2 following system of equations: \begin{cases} f(x,y,\mu)=0 \\ f_\mu(x,y,\mu)=0 \end{cases}
Now here comes my problem:
Consider we have a 2-variable ($x$ and $y$) family of curves in $\mathbb{R} ^{3}$ with 2 parameters ($\mu$ and $\nu$) related with the implicit function $g(\mu,\nu)=0$ expressed in the following implicit equation: $f(x,y,\mu,\nu)=0$.
I don't know how to find the equations what determinates the envelope.
I think we could use the Lagrange multiplier, but I don't know how to use it correctly in this context.
This was my try:
\begin{cases} Optimize:\space f(x,y,\mu,\nu) \\ Subject\space to:\space g(\mu,\nu)=0 \end{cases}
Let $h(x,y,\mu,\nu)=f(x,y,\mu,\nu)-\lambda g(\mu,\nu)$ the lagrangian function, then:
$$\begin{cases} \frac{\partial h}{\partial x}=f_x(x,y,\mu,\nu)=0\\ \frac{\partial h}{\partial y}=f_y(x,y,\mu,\nu)=0\\ \frac{\partial h}{\partial \mu}=f_\mu(x,y,\mu,\nu)-\lambda g_\mu(\mu,\nu)=0\\ \frac{\partial h}{\partial \nu}=f_\nu(x,y,\mu,\nu)-\lambda g_\nu(\mu,\nu)=0\\ \frac{\partial h}{\partial \lambda}=g(\mu,\nu)=0\\ \end{cases}$$
In parallel, let's remember that the derivative of an implicit function for 2 variables has the following expression:
$$\frac{du}{dv}=-\frac{\frac{\partial g}{\partial v}}{\frac{\partial g}{\partial u}}$$
I think we can use both of them in some way, but I don't get any idea.