The curve that passes through the point (2, 3) and that the property that the segment of any tangent to it lying between the coordinate axes is bisected by the point of contact, is given by:

Let $P(x,y)$ be any point on the curve. The equation of tangent at P is given by

$Y-y=\frac{dy}{dx}(X-x)$

This cuts coordinate axes at $A\left(x-y\frac{dx}{dy},0\right)$ 

and $B\left(0,y-x\frac{dy}{dx}\right)$

 It is given that $P(x,y)$ is the mid-point of AB.

$\therefore 2x=x-y\frac{dx}{dy}$ and $2y=y-x\frac{dy}{dx}$

$\begin{array}{l}\Rightarrow \frac{dy}{dx}=-\frac{y}{x}\\ \Rightarrow xdy+ydx=0\Rightarrow d\left(xy\right)=0\Rightarrow xy=c\end{array}$

The curve passes through (2, 3).

$\therefore 2\times 3=C\Rightarrow C=6.$

The equation or the curve is $xy = 6$ or, $y=\frac{6}{x}$.