Finding the curve $C$ whose every point in the first quadrant is the midpoint of $L(x,y) $? Let $C$ be a curve in $\Bbb{R}^2$  passing through $(3, 5)$ and $L(x, y)$ denote the segment of the tangent line to $C$ at $(x, y)$ lying in the first quadrant.
Assuming that each point $(x, y)$ of $C$ in the first quadrant is the
midpoint of $L(x,y)$.

So I know that $(3,5)$ satisfies the curve and that every point of $C$ in the first quadrant is the midpoint of $L(x,y)$. But how can I deduce from this information about the curve.

Is there some concept needs to be known which will make it trivial to find the curve?  

Can we find the curve with these conditions?
The information is sufficient.  Let $x_0$ and $y_0$ be any point on C.  Then we have $$y-y_0 = (x-x_0)y'(x_0)$$

Then the x intercept of the line segment in the first quadrant is 
$$(\frac{-y_0}{y'(x_0)}+x_0, 0)$$
The y intercept is $$(0, y_0-x_0y'(x_0))$$

As the mid point is $x_0$ and $y_0$ and replace $x_0$ by x and $y_0$ by y, we have xy' + y = 0 or $$xdy + ydx =0$$ or $$d(xy) = 0$$ or $xy=K$

Since (3, 5) is on C, we have K = 15.  Therefore the solution is xy = 15
The information is sufficient.  Let $x_0$ and $y_0$ be any point on C.  Then we have $$y-y_0 = (x-x_0)y'(x_0)$$

Then the x intercept of the line segment in the first quadrant is 
$$(\frac{-y_0}{y'(x_0)}+x_0, 0)$$
The y intercept is $$(0, y_0-x_0y'(x_0))$$

As the mid point is $x_0$ and $y_0$ and replace $x_0$ by x and $y_0$ by y, we have xy' + y = 0 or $$xdy + ydx =0$$ or $$d(xy) = 0$$ or $xy=K$

Since (3, 5) is on C, we have K = 15.  Therefore the solution is xy = 15