<s>[INST] 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 $\sqrt{x}+\sqrt{y}=2$?
 [/INST] Calculate $y$ from last equation:
$$
 y = \frac{2x+8 \pm \sqrt{(2x+8)^2 - 4(16+x^2-8x)}}{2}
 =
 \frac{2x+8 \pm 8\sqrt{x}}{2} = x \pm 4\sqrt x + 4.
$$
Now if you want to get form $\sqrt x + \sqrt y = 2$, you need to assume that $0 \leq x,y \leq 4$, hence $y = x - 4\sqrt x + 4 = (2 - \sqrt x)^2$. Now taking square root of both sides yields desired result. 
</s>