How to find a curve which satisfies some properties? 
Let $C$ be a curve in $\mathbb R^2$ 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 $L(x,y)$, find the curve.

First I found the equation of the line whose line segment in first quadrant is $L(3,5)$, which is $5x+3y=30 \ [\text{ It is easy to find as (3,5) will be mid-point of the line}].$ 
From here how can we find the equation of the curve?
Any help is appreciated. Thnak you.
 A: Let the curve be parametrized by $c(t) =\big(x(t), y(t)\big)$, where $c(0) = (3,5)$.
The tanget $T_t$ at $c(t)$ is given by the parametrization
\begin{align}
T_t(s)
&=
c(t) + s\cdot \big(x'(t), y'(t)\big)
\\&=
\Big(x(t) + s\cdot x'(t)\,, \,\,\,y(t) + s\cdot y'(t)\Big)
.\end{align}
Notice that near $t=0$, the curve lies on the first quadrant.
Near $t=0$, we may hence not have $x'=0$ or $y'=0$, for then the midpoint condition would be impossible to satisfy.
Then, for $t$ near $0$, $T_t$ crosses the $y$-axis when
$$s_{y, t} =  \frac{-y(t)}{y'(t)},$$
and $T_t$ crosses the $x$-axis when
$$s_{x, t} =  \frac{-x(t)}{x'(t)}.$$
The midpoint between them is $T_t\left(\frac12(s_{y,t} + s_{x,t})\right)$, and we have
\begin{align}
s_{y,t} + s_{x,t}
&=
\frac{-y(t)x'(t) -x(t)y'(t)}{x'(t)y'(t)}
\\
\\&=
-\,\frac{\frac{d}{dt}\big[x(t)y(t)\big]}{x'(t)y'(t)}.
\end{align}
For $t$ near $0$, we need to have
$$T_t\left(\frac12(s_{y,t} + s_{x,t})\right) = c(t),$$
which simplifies to
$$(s_{y,t} + s_{x,t}) \cdot \Big(x'(t), y'(t)\Big) = 0.$$
Since neither $x'$ nor $y'$ may be $0$, it follows that for $t$ near $0$ we must have
$$s_{y,t} + s_{x,t} = 0 \iff \frac{d}{dt}\big[x(t)y(t)\big] = 0
\iff x(t)y(t) \,\text{ is constant}$$
Substituting $t=0$ we get $x(0)y(0) = 15$, so the curve looks like
$$y = \frac{15}x,$$
at least in its domain of definition in the first quadrant.
A: The segment L, by definition, must cut the axes at points $(2x, 0)$ and $(0,2y)$ and the tangent at $(x, y)$ is given by the differential equation $\dfrac{dy}{dx}= \dfrac{ 2y}{-2x}$ whose solution is $y=\dfrac cx$. The constant $c$ is clearly $ 15$ so the curve $C$ has equation
$$xy=15$$  which is an equilateral hyperbola, symmetric with respect to the diagonal $y = x$ in which it is easy to verify that the tangential segment $L$ corresponding to the arbitrary point $(x, y)$ of the curve $C$ has extremes $(2x, 0)$ and $(0,2y)$.   This curve, in turn, it is well defined on the third quadrant.
