You can start from the disc representation and gradually make the examples more complicated. Example: Let $p$ a point in the two dimensional plane
$$
x^2+y^2 \leq 1
$$
can be rewritten as
$$
\pmatrix{x&y}\pmatrix{1&0\\0&1}\pmatrix{x\\y} \leq 1
$$
This is the simplest case where you can see the resulting shape and read off the eigenvalues at the same time. Smashing the circle from the top and bottom would result with
$$
\pmatrix{x&y}\pmatrix{1&0\\0&4}\pmatrix{x\\y} \leq 1
$$
We can see that points satisfy this can at most have a $y$ component with magnitude $0.5$ etc. We can further generalize this to circles that is centered somewhere other than the origin and also we can rotate these eccentric ellipsoids. When rotated, instead of plain $x$ and $y$ axes being principal axes you get the eigenvectors working for you and the lengths are associated with the magnitude of the eigenvalues. You can show this by eigenvalue decomposition or simply let your variables $x,y$ be the eigenvector of the largest eigenvalue and see them as rotated $(x,y)$ pairs that we have started with.
In general, we are left with inequalities in the form of
$$
\pmatrix{p-p_0}C\pmatrix{p-p_0} \leq s
$$
where $C$ is a positive definite matrix, $p_0$ is the center of the ellipsoid and $s$ is, roughly, the size of the ellipsoid. The positive definiteness is not crucial but often physically meaningful. If the matrix is not positive definite you allow for the reflections along some axis.
Define the position vector of a point on the ellipsoid as ${r} = [x, y, z]^T$
Then the equation of the ellipsoid is:
$ r^T M r + 2 r^T n + d = 0$
Define $s = -M^{-1} n$ , then the above quadratic equation becomes,
$ (r - s)^T M (r - s) - s^T M s + d = 0$
i.e.
$ (r - s)^T M (r - s) = s^T M s - d $
Dividing by the right side,
$ (r - s)^T G (r - s) = 1 $
where $G = \dfrac{1}{s^T M s - d} M$
The last equation is the standard equation of an ellipsoid with center at $s$. The semi-axes lengths of this ellipsoid are the square root of the receiprocals of the eigenvalues of $G$. The eigenvalues of $G$ are a scaled version of the eigenvalues of matrix $M$ by a factor of $\dfrac{1}{s^T M s - d }$, hence,
$ \lambda_{Gi} = \dfrac{\lambda_{Mi}}{s^T M s - d} $
And therefore the semi-axes lengths are
$ L_i = \dfrac{1}{\sqrt{\lambda_{Gi}}} = \sqrt{ \dfrac{s^T M s - d}{ \lambda_{Mi} }}$
Finally we note that $s^T M s - d = (-M^{-1} n)^T M (- M^{-1} n ) - d = n^T M^{-1} n - d$
Best Answer
Consider the yellow and orange convex figures, which have the same bounding ellipsoid but different bounded ellipsoids.
Therefore, given just the bounding ellipsoid you cannot determine the bounded ellipsoid. (Nor vice versa.) You must know the figure $K$.
A better, limiting, example: Suppose $K$ is an ellipsoid. Then the bounding ellipsoid and the bounded ellipsoid are the same!