Q:

Location of "x standard deviation(s)" for multivariate normals

In 1D with scalar parameters $(\mu,\sigma^2)$, it is common to represent normally distributed data with error bars spanning $[\mu-\sigma, \mu+\sigma]$.
In n-D with parameters $(\mu, \Sigma)$, where $\mu\in\mathbb{R}^n$ and $\Sigma$ is $n\times n$ SPD, one infers the "shape" of multivariate-normal distributions by looking at equiprobable locations, given by $(x-\mu)^t\ \Omega\ (x-\mu) = \mathrm{C}$, where $C$ is some constant, and $\Omega = \Sigma^{-1}$ is the precision matrix. These are, of course, ellipsoids.
Q1 But what is the location of the points corresponding to "one standard-deviation" away from $\mu$? Is it the ellipsoid obtained by taking all points $x$ such that $\| x - \mu \| = 1$? I.e. the mapping of the unit sphere centred at $\mu$ by the precision matrix?
Q2 If so, what are the coordinates of those points? The formula $(x-\mu)^t\ \Omega\ (x-\mu)$ is scalar, how does one obtain the n-D coordinates of those points on the "one st-dev" ellipsoid?
(Note: apologies if that sounds like homework, I can assure you that it is not).

A:

From the geometric interpretation of multivariate normals written on Wikipedia, the location of the points "one standard deviation away" from $\mu$ would be:
$$
  \big\{\ \mu + U\Lambda^{1/2}x\; \big|\ \|x\|=1 \big\}
$$
where $\Sigma = U\Lambda U^t$ such that:

$U$ is a rotation matrix ($UU^t = I$ and $\det(U) = 1$) containing the eigen-vectors of $\Sigma$ (which are also the eigenvectors of $\Omega$) in column;
$\Lambda$ is a diagonal matrix with the corresponding eigen-values, which are also the squared-lengths of the axes of the ellipsoid, and the reciprocals of $\Omega$'s eigenvalues.

In effect, this means that the ellipsoid corresponding to "one std from $\mu$" can be obtained by: 

scaling the unit sphere along the canonical axes with $\Lambda^{1/2}$;
rotating the scaled sphere with $U$;
and translating it by $\mu$.

