Envelope of a family of curves

Question

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 $\sqrt{x}+\sqrt{y}=2$?

Answer 1

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.

Answer 2

You have done it is basically correct, but proceeding this way in simplification it is more easily done:

Letting $4=b$ we have

$$ \dfrac{x}{a} + \dfrac{y}{b-a} =1 \tag1$$ Differentiate w.r.t. $a$ and solve for $a,b-a$ $$ a= \dfrac{b\sqrt x}{\sqrt x+\sqrt y}; \quad b-a= \dfrac{b\sqrt y}{\sqrt x+\sqrt y};\tag2$$

Plug into (1) and simplify

$$ \dfrac {\sqrt x}{\sqrt b}+ \dfrac {\sqrt y}{\sqrt b}= \pm 1 \tag 3$$

Answer 3

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 $\sqrt{x}+\sqrt{y}=2$