7.54. The plane $\Pi$ passes through the vertex of the cone perpendicular to its axis; point $A$ lies in the plane $\Pi$. Let $M$ be a point on the cone such that a light ray going from $A$ to $M$, after being reflected off the surface of the cone, becomes parallel to the plane $\Pi$. Find the geometric locus of the projections of points $M$ onto the plane $\Pi$.

## Problems for Independent Solving
7.54. Let's introduce a coordinate system with the origin $O$ at the vertex of the cone and the axis $O x$ passing through the point $A$ (Fig. 55). Let $\overrightarrow{O M}=(x, y, z)$, then $\overrightarrow{A M}=(x-a, y, z)$, where $a=A O$. If

![](https://cdn.mathpix.com/cropped/2024_05_21_3f2bc5f5a1af26514cfag-155.jpg?height=493&width=434&top_left_y=337&top_left_x=153)

Fig. 55 $\alpha$ is the angle between the axis of the cone $O z$ and its generatrix, then $x^{2}+y^{2}=$ $=k^{2} z^{2}$, where $k=\operatorname{tg} \alpha$. Consider the vector $\overrightarrow{P M}$, perpendicular to the surface of the cone, with the origin $P$ on the axis of the cone. This vector has coordinates $(x, y, t)$, and $0=(\overrightarrow{O M}, \overrightarrow{P M})=x^{2}+y^{2}+t z=$ $=k^{2} z^{2}+t z$, i.e., $t=-k^{2} z$.

Under symmetry relative to the line $P M$, the vector $\mathbf{a}=\overrightarrow{A M}$ transforms into the vector $2 \mathbf{b} \frac{(\mathbf{a}, \mathbf{b})}{(\mathbf{b}, \mathbf{b})}-\mathbf{a}$, where $\mathbf{b}=\overrightarrow{P M}$ (see problem 7.38). The third coordinate of this vector is

$$
-2 k^{2} z \frac{x^{2}-a x+y^{2}-k_{i}^{2} z^{2}}{x^{2}+y^{2}+k^{4} z^{2}}-z=\frac{2 a k^{2} x z}{\left(x^{2}+1-y^{2}\right)\left(1+k^{2}\right)}-z
$$

it should be equal to zero. Therefore, the desired locus of points is given by the equation $x^{2}+y^{2}-2 a k^{2} x /\left(1+k^{2}\right)=0$; it represents a circle of radius $a k^{2} /\left(1+k^{2}\right)=a \sin ^{2} \alpha$, passing through the vertex of the cone.