I cannot find my mistake in the statement below, neither computational nor conceptual. I cannot prove that the velocity vector $\dot{x}$ is constant along the acclaimed geodesic. Could you review it for me? Briefly, both verbose and symbolic argumentation are below:
Verbose argument: We may represent an ellipsoid $\mathcal{S}$ submerged in $\mathbb{R}^3$ in ellipsoidal coordinates by equality $u_3 = 1$, for coordinate chart $(u, \eta)$ corresponding to horizontal and vertical meridian angles and radius and aforementioned map. It means, in ellipsoidal coordinates, a geodesic curve is a straight line. The geodesic curve $\gamma: [0, 1] \to \mathbb{R}^3$ in Euclidean coordinates corresponds to this line evaluated in the ellipsoidal transformation map. Since the velocity vector $\dot{\gamma}$ along a geodesic has constant value, the distance between two coordinates $p, \, q \in \mathcal{S}$ corresponds to the Euclidean norm of velocity vector $\dot{x}$ given by $J_\eta \dot{u}$ evaluated at any coordinate within geodesic parametrization $[0, 1]$. It means that the expression $(\dot{u}^\intercal J_\eta^\intercal)(J_\eta \dot{u})(t)$, for $u(t) = u(0) + t \, (u(1) - u(0))$ must not depend on variable $t$ and is equal to distance map $d(q, \, p)$.
Mathematical argument: Given length of each eigen axis to origin are equal to $\frac{1}{\sqrt{\lambda_k}}$, the ellipsoid parametrization emerges naturally from ellipsoidal coordinate chart $([0, \, 2 \pi] \times [-\pi, \, \pi] \times 1, \eta)$ given by map. , its vectorial form corresponds to equation below.
\begin{equation} \label{eq:spherical_map} \eta(u_1, u_2, u_3) = \Lambda^{-\frac{1}{2}} \, \underbrace{ \begin{pmatrix} u_3 \, \cos{(u_1)} \, \sin{(u_2)} \\ u_3 \, \sin{(u_1)} \, \sin{(u_2)} \\ u_3 \, \cos{(u_2) } \end{pmatrix} }_{\sigma(u)} \end{equation}
Ellipsoidal Cartesian representation $\lambda_1 x_1^2 + \lambda_2 x_2^2 + \lambda_3 x_3^2 - 1 = 0$ turns into equality $u_3 - 1 = \, 0$. Therefore, geodesic curves in chart coordinate $([0, 2 \, \pi] \times [-\pi, \pi] \times 1, \eta)$ are curvature-free curves whose differential behaviour respects equality $\ddot{c}(t) = 0$ for every instant $t \in [0, 1]$ and 3-tuple $u \, \in \, 1 \times [0, 2 \, \pi] \times [-\pi, \pi]$. The deviation $\delta$ is equal to difference $u(1) - u(0)$ and represents the velocity vector $\dot{u}(t)$.
\begin{equation} \label{eq:line} u(t) = p + t \, \delta \end{equation}
The distance map $d(p, \, q)$ between two coordinates $p$ and $q$ on a smooth surface is given by equation below, for the geodesic velocity vector norm along a geodesic curve.
\begin{equation} \label{eq:geodesic} d(p, q) \, = \int_0^1 \sqrt{\langle \dot{c}, \, \dot{c} \rangle} \, dt \end{equation}
By substitution of velocity vector $\dot{c}$ on equation below as product $J_\eta(u(t)) \, \dot{u}(t)$, the tensor $J(t)$ appears naturally, equal to integral of tensor given by product $J_\eta^\intercal \, J_\eta$ i.e. $J_\sigma^\intercal \, \Lambda^{-1} \, J_\sigma$ evaluated along curve $u(t)$ from $0$ to $1$. Hence, the distance map $d(p, \, q)$ is equal to expression $\delta^\intercal \, J \, \delta$.
\begin{equation} \label{eq:eta_jacobian} \begin{aligned} \frac{\partial \, \eta}{\partial \, u} & = J_\eta = \Lambda^{-\frac{1}{2}} \, J_\sigma \\ & = \Lambda^{-\frac{1}{2}} \, \begin{pmatrix} - u_3 \, \sin{(u_1)} \, \sin{(u_2)} & u_3 \, \cos{(u_1)} \, \cos{(u_2)} & \cos{(u_1)} \, \sin{(u_2)} \\ u_3 \, \cos{(u_1)} \, \sin{(u_2)} & u_3 \, \sin{(u_1)} \, \cos{(u_2)} & \sin{(u_1)} \, \sin{(u_2)} \\ 0 & - u_3 \, \sin{(u_2)} & \cos{(u_2)} \end{pmatrix} \end{aligned} \end{equation}