Envelope of a family of curves 
Show that the envelope of the family of curves $$\frac{x}{a}+\frac{y}{4-a}=1$$ is the parabola $$\sqrt{y}+\sqrt{x}=2$$

I know how we can get the envelope, but I could not get the required relation. I differentiated the family of curves w.r.t $a$. I got 
$$a=\frac{1}{2}(x-y+4)$$
I then substituted by $a$ in the family of curves and got
$$-x^2+2xy-y^2+8y+8x-16=0$$
How to complete to get $\sqrt{x}+\sqrt{y}=2$?

Maybe this is a simpler way to approach the problem.
I will assume that ${0\leq a \leq 4}$. 
Define $D$ as the union of the family of curves $\begin{align*}\frac{x}{a}+\frac{y}{4-a}=1\end{align*}$. Then there exists a function $~{f(x)}$ defined on the interval $~x\in [0,~4]$ such that the region $~D$ can be written as $\{(x,y)~|0\leq x \leq 4,~0\leq y\leq f(x)\}$.
Rearrange the given equation $\begin{align*}\frac{x}{a}+\frac{y}{4-a}=1\end{align*}$ as $\begin{align*}y=4+x-\frac{4x}{a}-a\end{align*}$. By AM-GM inequality, we have $\begin{align*}4+x-\frac{4x}{a}-a\end{align*}\leq4+x-4\sqrt{x},~$ where the equality holds if and only if $a=2\sqrt{x}\cdots$(1). 
Since $0\leq x\leq 4$, by the equation (1) we have $0 \leq a \leq 4$ and the equality condition is satisfied in the given interval of $x$. 
Therefore the desired envelope is $y=4+x-4\sqrt{x}$, or $\sqrt{x}+\sqrt{y}=2$
