How do I find the equation of an envelope?

Question

I read that you must solve the two equations $$g(x,y,c)=0\\\frac{\partial g}{\partial c}=0$$ for $x$ and $y$ as a function of $c$, but how exactly do you go about doing this? The specific example I am trying to solve is where $$F(x,y,\alpha)= -t y \sin\alpha \tan{\alpha\over2} - t \sin\alpha \left(x - r \cos\alpha - t \sin\alpha \tan{\alpha\over2}\right)$$ and find the parametric equation for the envelope ($r$ and $t$ are constants). BTW, this is from this post if you are curious.

Answer 1

The curves given by $F(x,y,\alpha)=0$, are also given by $G(x,y,u)=0$ where $u=\tan\left(\frac\alpha2\right)$ and $$ \begin{align} G(x,y,u) &=\frac{F(x,y,\alpha)}{-t\sin(\alpha)}\tag{1a}\\ &=y\tan\left(\frac\alpha2\right)+\left(x-r\cos(\alpha)-t\sin(\alpha)\tan\left(\frac\alpha2\right)\right)\tag{1b}\\ &=yu+x-r\frac{1-u^2}{1+u^2}-t\frac{2u^2}{1+u^2}\tag{1c} \end{align} $$ Then, since $\frac{\partial}{\partial u}\frac{2u^2}{1+u^2}=\frac{4u}{\left(1+u^2\right)^2}$, $$ \frac{\partial}{\partial u}G(x,y,u)=y-(t-r)\frac{4u}{\left(1+u^2\right)^2}\tag{2} $$ Therefore, $\frac{\partial}{\partial u}G(x,y,u)=0$ when $$ y=(t-r)\frac{4u}{\left(1+u^2\right)^2}\tag3 $$ and then $G(x,y,u)=0$ when $$ x=t+(r-t)\frac{1+4u^2-u^4}{\left(1+u^2\right)^2}\tag4 $$

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Example: $r=2,t=1$