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.
 A: At any point of your curve you have information about the slope of the curve at your point. This means that you have to solve a differential equation.
Suppose $(x_0,y_0) \in C$ is a point of your curve. Then (I skip boring computations) you have that the slope of the curve at $x_0$ is 
$$f'(x_0)=-\frac{y_0}{x_0}$$
so you have to solve the Cauchy problem
$$\left\{ \begin{matrix}
y'=-\frac{y}{x} \\
y(2)=1
\end{matrix} \right.$$
with constraints $y >0, x>0$.
The solution is $f(x)=\frac{2}{x}$
EDIT: How did I find $f'(x_0)=-\frac{y_0}{x_0}$? Here I show my boring (actually there is nothing smart in here) computations.
Let $m= f'(x_0)$. The line $L$ passing through $(x_0,y_0)$ of slope $m$ has equation
$$L: \ \ y-y_0=m(x-x_0)$$
Intersecting $L$ with the $x$ axis we get the point $(-\frac{y_0}{m}+x_0, 0)$.
Intersecting $L$ with the $y$ axis we get the point $(0, y_0-mx_0)$.
So we get the conditions
$$\left\{\begin{matrix}2x_0 = -\frac{y_0}{m}+x_0 \\
2y_0 = y_0-mx_0\end{matrix}\right.$$
which are equivalent to $m= -\frac{y_0}{x_0}$.
A: 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.

