= Solution
If $\mathbf b=0$, the objective is zero at every feasible point, so \b[every point of the ellipsoid is a minimizer]. Suppose now $\mathbf b\ne0$. A minimum cannot be interior: a sufficiently small step in direction $-\mathbf b$ would remain feasible and decrease the objective. The boundary is active.
For
$$
\mathcal L(\mathbf x,\lambda)=\mathbf b^{\mathsf T}\mathbf x+
\lambda(\mathbf x^{\mathsf T}A\mathbf x-1),
$$
stationarity gives $\mathbf b+2\lambda A\mathbf x=0$. A <positive-definite symmetric matrix> is invertible, and the minimizing sign has $\lambda>0$. Enforcing the boundary gives
$$
\lambda=\frac12\sqrt{\mathbf b^{\mathsf T}A^{-1}\mathbf b},\qquad
\boxed{\mathbf x_*=
-\frac{A^{-1}\mathbf b}{\sqrt{\mathbf b^{\mathsf T}A^{-1}\mathbf b}}.}
$$
For global optimality, use the <inner product> $\langle u,v\rangle_A=u^{\mathsf T}Av$. The <Cauchy-Schwarz inequality> gives
$$
|\mathbf b^{\mathsf T}\mathbf x|
=|\langle A^{-1}\mathbf b,\mathbf x\rangle_A|
\le\sqrt{\mathbf b^{\mathsf T}A^{-1}\mathbf b}
\sqrt{\mathbf x^{\mathsf T}A\mathbf x}
\le\sqrt{\mathbf b^{\mathsf T}A^{-1}\mathbf b}.
$$
The proposed point attains the negative of this bound. Thus \b[the minimum is $-\sqrt{\mathbf b^{\mathsf T}A^{-1}\mathbf b}$], and equality in the <Cauchy-Schwarz inequality>, together with the sign and boundary conditions, makes this minimizer unique when $\mathbf b\ne0$.
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