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The quadratic expression is ${\mathbf{x}}^{T}A\mathbf{x}$. Normalize the preceding eigenvectors to obtain an orthonormal basisWith the orthogonal matrix $U=({\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3)$ and coordinates $\mathbf{y}=U^T\mathbf{x}$, the surface equation becomes $y_1^2+2y_2^2+4y_3^2=1$. All coefficients are positive, so it is an ellipsoid, with principal semiaxes $1,1/\sqrt{2},1/2$ along the respective eigenvectors. This is principal-axis reduction of a quadric.

A map to the unit sphere is $\mathbf{x}\mapsto \mathrm{diag}(1,\sqrt{2},2)U^T\mathbf{x}$. One explicit matrix isIndeed $B^{T}B=A$, so $|B\mathbf{x}{|}^2={\mathbf{x}}^{T}A\mathbf{x}$. Since $B$ is invertible, every point on the unit sphere has a preimage on the ellipsoid; the map is onto, not merely into. The requested $B$ is not unique, since an additional orthogonal matrix on the left preserves the unit sphere.

Represent the quadratic form by the symmetric matrixThe $xy$ block has characteristic polynomial $(2-\mu )(5-\mu )-4=(\mu -1)(\mu -6)$. Corresponding orthonormal eigenvectors, together with the $z$ direction, areThe eigenvalues are $1,6,-1$. To eliminate the linear term, solve $H\tilde{O}=-(0,3\sqrt{5},0{)}^T$, obtainingIndeed the first equation gives $x=y$, and the second then gives $3y=-3\sqrt{5}$. Define the new coordinates by $X=\tilde{O}+\tilde{x}{\tilde{e}}_1+\tilde{y}{\tilde{e}}_2+\tilde{z}{\tilde{e}}_3$. The original polynomial takes the value $-15$ at the center, so its translated and orthogonally diagonalized form isThusThe surface is an elliptic one-sheet hyperboloid, centered at $\tilde{O}$, with axis parallel to ${\tilde{e}}_3$. Its waist at $\tilde{z}=0$ is an ellipse with semiaxes $\sqrt{15}$ and $\sqrt{5/2}$. Every constant-$\tilde{z}$ section is a nonempty ellipse, so the two signs of $\tilde{z}$ belong to the same connected sheet. This is an application of principal-axis reduction of a quadric.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.