Q:

How to find curve C given by the equation?

Consider the curve C given by the following equation 
$$
\sqrt{x}+\sqrt{y}=\sqrt{a} 
$$
 where $a$ is a constant with the condition $a > 0$.
Let $(x_0, y_0)$ with  be a point on C such that $x>0$ and $y>0$. 
Now,  assume that $(x_1, 0)$ and $(0, y_1)$ be considered $x$ and $y$ such that intercepts of the tangent line to C at $(x_0, y_0)$.
My question: How to proof that $x_1 + y_1 = a$?
My try:
Currently I just found the derivative using implicit differentiation, and used the slope for its tangent line, but now I'm lost and not sure what to don.
Thanks for any suggestion. 

A:

By intrinsic differention show $dy/dx = -\sqrt{y/x}$.
The tangent line at $(x_0,y_0)$ is $y - y_0 = -\sqrt{y_0/x_0}(x - x_0)$.
Can you now find the intercepts and finish the problem?

