Example 3. Determine the shape of the mirror reflecting the rays from a point source into a parallel beam.
Solution. Let the source of light be at the origin, and the axis $O x$ be parallel to the reflected rays.

Let $M(x, y)$ be an arbitrary point on the curve, which is the section of the mirror by the plane $O x y$. Draw the normal to the curve at point $M$ (Fig. 9.2) until it intersects the axis $O x$ at point $N(x, 0)$.

Denote by $\alpha$ the angle $O M N$ and by $\beta$ the angle $S M N$. According to the law of reflection, $\alpha=\beta$. Since $\gamma=\angle O N M=\beta=\alpha$, $\triangle O M N$ is isosceles, with $O M=O N$. Clearly,

$$
O M=\sqrt{x^{2}+y^{2}} .
$$

The segment $O N$ can be determined from the equation of the normal

$$
Y-y=-\frac{1}{y^{\prime}}(X-x)
$$

Setting $Y=0$ here, we get

$$
-y=-\frac{1}{y^{\prime}}(X-x)
$$

from which we find the segment

$$
X=O N=x+y y^{\prime}
$$

Substituting the expressions for $O M$ and $O N$ into the equality $O M=O N$, we obtain the homogeneous differential equation

$$
x+y y^{\prime}=\sqrt{x^{2}+y^{2}}
$$

which can be solved using the substitution $y=u x$.

To simplify the calculations, we will use an artificial trick. Let

$$
\rho=\sqrt{x^{2}+y^{2}}, \rho^{2}=x^{2}+y^{2}
$$

Differentiating the second equation with respect to $x$, we get

$$
2 \rho \frac{d \rho}{d x}=2 x+2 y y^{\prime} \text { or } \rho \frac{d \rho}{d x}=x+y y^{\prime}
$$

But

$$
x+y y^{\prime}=\sqrt{x^{2}+y^{2}}=\rho
$$

therefore

$$
\rho \frac{d \rho}{d x}=\rho \text { or } \frac{d \rho}{d x}=1
$$

from which $\rho=x+C$. Thus,

$$
\sqrt{x^{2}+y^{2}}=x+C, y^{2}=2 C x+C^{2}
$$

These curves are parabolas. The surface of the desired mirror is a paraboloid of revolution.