Equation of hyperboloid of one sheet resulting from rotating a (skew) line about an axis

Suppose I have a line in $3D$ given in parametric form as $ L(t) = P_0 + t V $, with $t$ being the parameter, and I rotate it about an axis passing through the origin whose direction is specified by the unit vector $A$. The vector equation of the surface resulting from rotating $L(t)$ about $A$ is found as follows:

$X(t, \theta) = R_A(\theta) L(t) $

where $R_A(\theta)$ is the rotation matrix about the axis $A$ by an angle $\theta$, and is given by the Rodrigues formula as

$R_A(\theta) = A A^T + (I - A A^T) \cos \theta + S_A \sin \theta $

with the skew-symmetric matrix $S_A$ being the matrix representation of the linear transformation $A \times P$ (the cross product between $A$ and a vector $P$ )

I want to show that the surface specified by all points $X(t,\theta)$ is indeed a hyperboloid of one sheet (under certain conditions imposed on $P_0$, $V$ and $A$), and I also want to derive the algebraic equation of this hyperboloid, which is of the form

$ (X - X_0)^T Q (X - X_0) = 1 $

Any pointers on these two questions are highly appreciated.

I used the following construction to prove that the surface is a hyperboloid of one sheet under certain conditions imposed on $P_0, V, A$.

Let $p(t) = P_0 + t V$ be the parametric equation of the line, and $ q(s) = s A $ be the equation of the axis, where we can assume without any loss of generality that $A$ and $V$ are unit vectors.

The first part of the construction is to find the point $p_1$ and $q_1$ on $p(t)$ and $q(s)$ respectively that result in the minimum distance between $p(t)$ and $q(s)$.

This can be done by minimizing the function of the squared distance between $p(t)$ and $q(s)$ defined as follows

$f(t, s) = (P_0 + t V - s A)^T (P_0 + t V - s A) $

By finding the partial derivative of $f(t, s)$ with respect to $t$ and $s$ and setting them to zero, we get the following equations in $t$ and $s$

$ (P_0 + t V - s A)^T V = 0 $

$ (P_0 + t V - s A)^T A = 0 $

which is a simple $2 \times 2$ linear system in the unknowns $t$ and $s$, and it will have a unique solution if the determinant is nonzero. The determinant is given by

$ \Delta = (V^T V) (A^T A) - (A^T V)^2 = 1 - cos^2 \psi $

where $\psi$ is the angle between $A$ and $V$, which means that there will a unique solution if and only if the vectors $A$ and $V$ are non-parallel, which is already assumed because we're talking about the skew lines $p(t)$ and $q(s)$. One can see that if $V$ is parallel to $A$ then the surface resulting will be a right circular cylinder.

Now assuming there is a unique solution for the minimum distance between $p(t)$, and $q(s)$ then there is a unique pair $t^*, s^*$ as a solution to the linear system above, and we can compute $p^* = P_0 + t^* V $ and $q^* = s^* A $ as the unique pair of points satisfying the minimum distance between the line and the axis.

Now equations of the two lines become

$p(t) = P_0 + t V = P_0 + t^*V + (t - t^*) V = p^* + t' V $

$q(s) = s A = s^* A + (s - s^*) A = q^* + s' A $

Shifting the coordinate system by $-q^*$ , the equations in the new system are

$p'(t') = (p^* - q^*) + t' V $

$q'(s') = s' A $

We now make one more assumption that the vector $(p^* - q^*)$ is nonzero, which means the two lines do not intersect. With this assumption define the unit vector

$u_1 = \dfrac{ (p^* - q^*) }{\| p^* - q^* \| } $

Note that $u_1$ is perpendicular to both $A$ and $V$ (this can be seen from the linear system above that we solved). Also define $u_2 $ by

$u_2 = A \times u_1 $

The triple $u_1, u_2, A$ define a coordinate system, thus define the rotation matrix

$R = [ u_1, u_2, A ] $

To find the coordinates of $p'(t') $ in this new frame, we have to premultiply by $R^T$, and this results in

$p''(t') = d_\text{min} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + t' \begin{bmatrix} 0 \\ \sin \phi \\ \cos \phi \end{bmatrix} = \begin{bmatrix} d_\text{min} \\ t' \sin \phi \\ t' \cos \phi \end{bmatrix}$

where $d_{min} = \| p^* - q^* \| $, and since $V$ is orthogonal to $u_1$ then it can be expressed as $V = \cos \phi A + \sin \phi u_2 $ , thus $\phi$ is the angle that satisfies: $\cos \phi = A^T V $ and $\sin \phi = u_2^T V $.

Applying $R^T$ to $q'(s')$ results in

$q''(s') = s' \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $

Now we note that rotating $p(t)$ about the axis $A$ is equivalent to rotating $p''(t')$ about the unit vector along the $z''$ axis, the rotation matrix for that is

$ R_z (\theta) = \begin{bmatrix} \cos \theta && - \sin \theta && 0 \\ \sin \theta && \cos \theta && 0 \\ 0 && 0 && 1 \end{bmatrix} $

Applying this matrix to $p''(t') $ results in

$R_z p''(t') = \begin{bmatrix} \cos \theta d_\text{min} - t' \sin \theta \sin \phi \\ \sin \theta d_\text{min} + t' \cos \theta \sin \phi \\ t' \cos \phi \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}$

Finally, from this expression, we have

$ {x'}^2 + {y'}^2 = {d_\text{min}}^2 + {t'}^2 \sin^2 \phi $

and

${z'}^2 = {t'}^2 \cos^2 \phi $

Thus,

${x'}^2 + {y'}^2 = {d_\text{min}}^2 + {z'}^2 \tan^2 \phi $

Hence the coordinates of $p''$ satisfy

$p''^T Q_1 p'' = 1 $

where

$ Q_1 = \dfrac{1}{d_\text{min}^2 } \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && - \tan^2 \phi \end{bmatrix} $

Recalling that $p'' = R^T p' $ and $ p' = p - q^* $ then $p'' = R (p - q^*) $

Hence, the equation of the hyperboloid is

$ (X - X_0)^T Q (X - X_0) = 1 $

where $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix} $ , $X_0 = q^* $ and

$Q = R Q_1 R^T $