Solution (source code)

= Solution

For $A'$, the centre and constant remain $\mathbf a=(-1,0,0)$ and $C=3$. Its <eigenvalues> are $3,1,-1$, so rotated coordinates put the surface in the form $3u^2+w^2-y^2=3$. This is a <one-sheet hyperboloid>. Since $3\|\mathbf x\|^2\ge\mathbf x^{\mathsf T}A'\mathbf x=3$, the nearest centre-to-surface distance is at least one, attained in the eigen-direction with eigenvalue three. The coordinate $y$ is unbounded, because arbitrarily large $|y|$ can be balanced by $u$ or $w$. Thus \b[the nearest distance is one, and there is no maximum value of $y$].

For the printed nonsymmetric $A''$, use the <symmetric part of a matrix>:
$$
S=\frac{A''+A''^{\mathsf T}}2
=\begin{pmatrix}2&0&1\\0&1&1/2\\1&1/2&2\end{pmatrix},
\qquad \mathbf r^{\mathsf T}A''\mathbf r=\mathbf r^{\mathsf T}S\mathbf r.
$$
The <skew-symmetric part> contributes zero to this scalar. The same centre solves $2S\mathbf a+\mathbf b=0$, and again $C=3$. Sylvester's positive principal minors are $2,2,5/2$, so this surface is an <ellipsoid>.

Let $\lambda_*$ be the largest <eigenvalue> of $S$. Its characteristic equation is $4\lambda^3-20\lambda^2+27\lambda-10=0$, and the <Rayleigh quotient> gives
$$
\boxed{d_{\min}=\sqrt{3/\lambda_*},\qquad
\lambda_*=\frac{5+\sqrt{19}\cos\!\left[\tfrac13\arccos\!\left(\frac{55}{19\sqrt{19}}\right)\right]}{3}.}
$$
For the maximum $y$, the <support function of a centred ellipsoid> gives the bound below. Equivalently, apply the <Cauchy-Schwarz inequality> in the $S$-weighted <inner product>: $y^2\le3\,\mathbf e_y^{\mathsf T}S^{-1}\mathbf e_y$. The inverse entry is $\mathbf e_y^{\mathsf T}S^{-1}\mathbf e_y=6/5$, and equality is attained for a centred vector proportional to $S^{-1}\mathbf e_y$. Thus
$$
\boxed{y_{\max}=\sqrt{18/5}.}
$$
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2016-resit-quadratic-surfaces.png]
{title=Ellipsoid and hyperboloid geometry after translation to the common centre}

Using the eigenvalues of the raw nonsymmetric $A''$ would be incorrect: it is its symmetric part that determines the surface and its Euclidean geometry.