Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 6 / ii

Use the orthogonal change of coordinatesIt preserves distances and angles and gives $x^{T}Mx=x^{\prime T}D{x}^{\prime }$. Thus the quadric surface is expressed along its principal axes asThe following three cases are the subparts of (ii) in the original PDF.

Writing the common positive eigenvalue as $\lambda$, the equation is $|x^{\prime }{|}^2=k/\lambda$. Hence it is a sphere withEvery orthonormal set of axes is a principal-axis set in this fully degenerate case.

Let ${\lambda }_1={\lambda }_2=\lambda >0$ and ${\lambda }_3=-\gamma <0$. ThenThis is a one-sheet hyperboloid of revolution around the $x_3^{\prime }$ axis. Its circular waist has radius $\sqrt{k/\lambda }$ at $x_3^{\prime }=0$, and its radius grows with $|x_3^{\prime }|$. In the original coordinates the axis is the eigenvector direction belonging to the negative eigenvalue.

The coordinate $x_1^{\prime }$ does not occur. The perpendicular cross-section isThus the surface is an elliptic cylinder extending in both directions along the zero-eigenvalue axis. Its cross-sectional semiaxes are $\sqrt{k/{\lambda }_2}$ and $\sqrt{k/{\lambda }_3}$. It becomes a circular cylinder when the two positive eigenvalues coincide.