= Solution
{c}
Choose the rotated <Cartesian coordinates>
$$
X=\frac{x+y}{\sqrt2},\qquad Y=\frac{y-x}{\sqrt2},\qquad Z=z.
$$
Their inverse is $x=(X-Y)/\sqrt2$, $y=(X+Y)/\sqrt2$. Consequently $x^2+y^2=X^2+Y^2$ and $xy=(X^2-Y^2)/2$, giving the <canonical form of a quadric>
$$
\boxed{\left(1+\frac a2\right)X^2+\left(1-\frac a2\right)Y^2+Z^2=1.}
$$
The <orthogonal diagonalization> identifies the <eigenvalues> $1+a/2,1-a/2,1$ of the <symmetric matrix> of the <quadratic form>.
For $|a|<2$, all three coefficients are positive: the surface is an <ellipsoid> with semiaxes $(1+a/2)^{-1/2},(1-a/2)^{-1/2},1$. At $a=0$ it is a <sphere>.
For $a=2$, the equation is $2X^2+Z^2=1$, an <elliptic cylinder> parallel to the $Y$ axis. For $a=-2$, it is $2Y^2+Z^2=1$, an <elliptic cylinder> parallel to the $X$ axis. The absent coordinate is unrestricted; its disappearance does not turn the <surface> into an <ellipse>.
For $a>2$, the negative coefficient is that of $Y^2$: this is a <one-sheet hyperboloid> with axis $Y$. For $a<-2$, the negative coefficient is that of $X^2$: it is a <one-sheet hyperboloid> with axis $X$. This gives the <signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid>, including both transition values.
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