= Solution
Let $Q=(x_1\ x_2)$ for the normalized <eigenvectors> already chosen. Then $Q^TQ=I$, $(x,y)^T=Q(x',y')^T$, and <orthogonal diagonalization> gives
$$
Q^TAQ=\begin{pmatrix}2&0\\0&8\end{pmatrix},\qquad
\boxed{2(x')^2+8(y')^2=1.}
$$
The <ellipse> has semi-axis lengths $1/\sqrt2$ and $1/\sqrt8$ along $x_1$ and $x_2$, respectively. Rescale $x''=\sqrt2x'$ and $y''=\sqrt8y'$ to obtain the <unit circle> $(x'')^2+(y'')^2=1$. Since $(x',y')^T=Q^T(x,y)^T$,
$$
\boxed{\begin{pmatrix}x''\\y''\end{pmatrix}
=B\begin{pmatrix}x\\y\end{pmatrix},\qquad
B=\begin{pmatrix}\sqrt2&0\\0&\sqrt8\end{pmatrix}Q^T
=\begin{pmatrix}1&1\\2&-2\end{pmatrix}.}
$$
The identity $B^TB=A$ checks the rescaling directly. Swapping or reversing <eigenvectors> gives other valid coordinate matrices; the displayed one follows the explicit choices in part (a).
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