Finding the equation of a function 
Find the function $y=f(x)$ whose graph is the curve $C$ passing through the point $(2,1)$ and satisfying the following property: each point $(x,y)$ of $C$ is the midpoint of $L(x,y)$ where $L(x,y)$ denotes the segment of the tangent line to $C$ at $(x,y)$ which lies in the first quadrant.

Okay, so with $(2,1)$ being the midpoint of the line segment $L(2,1)$, we have that the $L(2,1)$ must lie on the line $y=2-x/2$. This gives us that that slope of $C$ at $(2,1)$ is $-\frac{1}{2}$. Now I'm not really sure where to go from here.

From the bisecting similar triangles shown
slope of ATB is $ - y^{'}$
$ AT = TB, OS = SB $
$$ x = u = \dfrac {-y}{y^{'}} $$
$$ x dy + y dx =0 $$
Integrating 
$$ x \ y = const = 2, $$ using given boundary conditions. It is one of the properties of the rectangular hyperbola.

