Example 6. Find the orthogonal trajectories of the family of parabolas $y=a x^{2}$.
Solution. We form the differential equation of the family of parabolas. For this, we differentiate both parts of the given equation with respect to $x: y^{\prime}=2 a x$. Eliminating the parameter $a$, we find $\frac{y^{\prime}}{y}=\frac{2}{x}$, or $y^{\prime}=\frac{2 y}{x}$ - the differential equation of the given family. Replacing $y^{\prime}$ with $-\frac{1}{y^{\prime}}$ in the equation, we obtain the differential equation of the orthogonal trajectories

$$
-\frac{1}{y^{\prime}}=\frac{2 y}{x}, \quad \text { or } \quad \frac{d y}{d x}=-\frac{x}{2 y}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-057.jpg?height=376&width=454&top_left_y=1422&top_left_x=796)

Fig. 17

Integrating, we find $y^{2}=-\frac{x^{2}}{2}+C$ or $\frac{x^{2}}{2}+y^{2}=C, C>0$. The orthogonal family is a family of ellipses (Fig. 17).