1379. Find the curve passing through the point $(2 ; 3)$ and having the property that the segment of any of its tangents, enclosed between the coordinate axes, is bisected at the point of tangency.
Solution. Let $M(x ; y)$ be an arbitrary point on the desired curve, then by condition $A M=M B$ (Fig. 98), hence $O C=C B, O B=2 O C=$ $=2 x$ and similarly $O A=2 O D=2 y$.

Further, since $\varphi=\pi-\alpha$, then $\operatorname{tg} \varphi=\operatorname{tg}(\pi-\alpha)=-\operatorname{tg} \alpha=\frac{O A}{O B}=\frac{y}{x}$, from which

$$
\operatorname{tg} \alpha=-\frac{y}{x} \text { or } y^{\prime}=-\frac{y}{x}
$$

Integrating the differential equation (a), we get

$$
\begin{gathered}
\frac{d y}{d x}=-\frac{y}{x}, \frac{d y}{y}+\frac{d x}{x}=0, \int \frac{d y}{y}+\int \frac{d x}{x}=\ln C \\
\ln |y|+\ln |x|=\ln C, \ln |x y|=\ln C \\
x y=C \text { or } y=\frac{C}{x}(C \neq 0) .
\end{gathered}
$$

Thus, the general solution of equation (a) is a family of equilateral hyperbolas with asymptotes coinciding with the coordinate axes.

Now let's find the hyperbola passing through the given point (2; 3). Substituting the coordinates of this point into equation (b), we get $2 \cdot 3=C$, from which $C=6$. Therefore, the desired curve is $x y=6$.