$\sqrt{y}+\sqrt{x}=\sqrt{A}$. prove that x-intercept + y-intercept of any tangent = A 
This  is equation of a  curve
$\sqrt{y}+\sqrt{x}=\sqrt{A}$
$A$  is  a positive constant
$T$  is a tangent of  the  curve from any point on it
$B$  is the y-intercept  of $T$
$C$  is the x-intercept  of  $T$
Prove that  $B+C=A.$
 A: Hint
Assume a point $(h,k)$ on the curve.
Differentiate given equation to get
$y^{'} = -\sqrt{\frac yx}$
Therefore equation of tangent using slope point form is 
$y-k = m(x-h)$ , where  m is $y^{'}$
Put $x=0 , y=0$ for C and B , and then add the two
A: The tangent at the point $(x_0,y_0)$ of the curve $f(x,y)=0$ has equation
$$
(x-x_0)\frac{\partial f}{\partial x}(x_0,y_0)+(y-y_0)\frac{\partial f}{\partial x}=0
$$
and in this case we get
$$
\frac{x-x_0}{2\sqrt{x_0\mathstrut}}+\frac{y-y_0}{2\sqrt{y_0\mathstrut}}=0
$$
Thus we can rewrite it as
$$
\frac{x}{\sqrt{x_0\mathstrut}}+\frac{y}{\sqrt{y_0\mathstrut}}=\sqrt{A}
$$
or
$$
\frac{x}{\sqrt{A}\,\sqrt{x_0\mathstrut}}+
\frac{y}{\sqrt{A}\,\sqrt{y_0\mathstrut}}=1
$$
So the sum of the intercepts is
$$
\sqrt{A\mathstrut}\,\sqrt{x_0\mathstrut}+
\sqrt{A\mathstrut}\,\sqrt{y_0\mathstrut}=A
$$
A: It is also interesting to go in the opposite direction, i.e. to find the envelope of the segments for which $B+C=1$. Let $S_t$ be the segment with endpoints in $(0,t)$ and $(1-t,0)$, for any $t$ in the interval $(0,1)$. For any $\varepsilon>0$, $S_t$ and $S_{t+\varepsilon}$ meet in:
$$ S_t\cap S_{t+\varepsilon} = \left((1-t)(1-t-\varepsilon),t\,(\varepsilon+t)\right) \tag{1} $$
hence by letting $\varepsilon\to 0$ we get that the parametric equation of the envelope is given by:
$$ \left((1-t)^2,t^2\right) \tag{2}$$
that is a parabola with focus in $\left(\frac{1}{2},\frac{1}{2}\right)$ and directrix given by the line $x+y=0$.
In general, if we have a projective map $\varphi$ between two lines $l_1,l_2$, the envelope of the lines joining $P\in l_1$ with $\varphi(P)\in l_2$ is always a conic (I cannot recall the name of this theorem at the moment, but I am sure it is well-known).
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