Example 1. Find the curve passing through the point $M_{0}(1,4)$ 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, and $AB$ be the segment of the tangent to the curve at this point, enclosed between the coordinate axes. By the condition of the problem, $BM = MA$. If $OP$ is the abscissa of point $M$, then $\triangle AMP \sim \triangle ABO$ and

$$
\begin{gathered}
\frac{OA}{PA} = \frac{OB}{MP} = \frac{AB}{MA} \cdot \text{ But } \frac{AB}{MA} = 2 \text{ and } PA = OP, \text{ therefore} \\
\frac{OA}{OP} = \frac{OB}{MP} = 2, \text{ i.e., } OA = 2x, OB = 2y
\end{gathered}
$$

The slope of the tangent to the curve $y = f(x)$ at point $M(x, y)$ is expressed using the derivative $\frac{dy}{dx} = \tan \alpha$, where $\alpha = \angle BAX$. On the other hand, since $\tan \alpha = -\tan \angle BAO$ and $\tan \angle BAO = \frac{OB}{OA}$, then

$$
\tan \alpha = -\frac{OB}{OA} = -\frac{2y}{2x} = -\frac{y}{x}
$$

Therefore,

$$
\frac{dy}{dx} = -\frac{y}{x}
$$

Integrating this separable differential equation (see § 9.1, example 2), we get

$$
xy = C.
$$

Since the curve must pass through the point $M_0(1, 4)$, substituting its coordinates into the given equation, we find:

$$
1 \cdot 4 = C, \quad C = 4
$$

Thus, the desired curve is defined by the equation

$$
xy = 4.
$$