= Solution
For $\mathbf r\ne0$, the <dot product> equation says that the angle $\theta$ between $\mathbf r$ and the oriented <unit vector> $\mathbf u$ satisfies $\cos\theta=\alpha$. Thus $\theta=\arccos\alpha$ is fixed. More explicitly, using the <orthogonal projection> from the introductory solution,
$$
s=\alpha\sqrt{s^2+|\mathbf v|^2}
\quad\Longrightarrow\quad
|\mathbf v|=|s|\frac{\sqrt{1-\alpha^2}}{|\alpha|},\qquad \operatorname{sgn}s=\operatorname{sgn}\alpha.
$$
Consequently \b[the surface is one infinite nappe of a <right circular cone>, with apex at the origin], including the apex itself. The cone's axis points along $\operatorname{sgn}(\alpha)\mathbf u$, and its acute semi-angle is $\arccos|\alpha|$. Equivalently, $\alpha$ is the cosine of the angle from the oriented $\mathbf u$ axis. Squaring the original equation without retaining the sign of $s$ would incorrectly add the opposite nappe of a <double circular cone>. No base disk is included in this infinite lateral surface.
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