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 $L(x,y),$ find the curve.
My attempt $:$
Let the curve be $y = f(x).$ Since the segment of the tangent line to any point $(x,y)$ on $C$ lying in the first quadrant bisected at $(x,y),$ it follows that the tangent line can't pass through the origin (i.e. the curve $C$ cannot pass through the origin), all the tangent lines to $C$ at a point $(x,y)$ lying in the first quadrant always have positive $x$ and $y$ intercepts and also they can't be parallel to the coordinate axes for otherwise their segment lying in the first quadrant cannot have any valid midpoint as they are infinite line segments. Now the equation of the tangent line to $C$ at $(x,y)$ can be given by $$Y - y = f'(x) (X - x).$$ On simplifying we get $$X f'(x) - Y = x f'(x) - y.$$ Clearly $x f'(x) - y \neq 0,$ for otherwise the tangent line will pass through the origin. Also we note that $f'(x) \neq 0,$ for all $x$ for otherwise the corresponding tangent line would be parallel to $x$-axis. So it's $x$-intercept and $y$-intercept are respectively $\dfrac {x f'(x) - y} {f'(x)}$ and $y - x f'(x).$ Since $(x,y)$ is the midpoint of $L(x,y)$ for every $(x,y) \in \Bbb R^2$ lying in the first quadrant, it follows that \begin{align*} x & = \dfrac {x f'(x) - y} {2f'(x)}; \\ y & = \dfrac {y - x f'(x)} {2}. \end{align*} Both the equations yield $$x f'(x) + y = 0.$$ Now if $x = 0$ then the corresponding curve becomes $y = 0$ which doesn't pass through $(3,5).$ Hence $x \neq 0.$ So dividing both sides by $x$ and writing $f'(x) = \dfrac {dy} {dx}$ we have $$\dfrac {dy} {dx} + \dfrac {y} {x} = 0$$ Which is a linear ordinary differential equation in $y.$ Solving we have $$xy = c$$ where $c$ is some arbitrary constant. Since it passes through $(3,5)$ we have $c = 15$ and hence the required equation of the curve $C$ lying in the first quadrant turns out to be $$xy = 15$$
Can anybody please check my work whether it holds good or not? Thanks for reading.
NOTE $:$ I can only determine the portion of the curve $C$ passing through $(3,5)$ lying in the first quadrant. Also here I am assuming that $C$ can be written down as a function of $x.$ Not all curves can have such explicit representation at each of it's points e.g. $x^2 + y^2 = 1$ at the point $(1,0).$