Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics IV / 3

The ellipsoid is the level set $F(x,y,z)=2x^2+y^2+z^2=1$. Its gradient is $(4x,2y,2z)$, which never vanishes on the ellipsoid. Half this gradient gives the outward normal vectorThis normal vector remains valid at $\theta =0,\pi$, where the given parametrization is singular.

The partial derivatives of the parametrization areTheir cross product isTaking its euclidean norm gives the surface area elementThe two poles and duplicated seam have zero surface area, so integrating over the stated parameter domain counts the surface area once. With the change of variables $u=\cos \theta$ and the supplied definite integral,Here $-\ln (\sqrt{2}-1)=\ln (\sqrt{2}+1)$ because $(\sqrt{2}-1)(\sqrt{2}+1)=1$.

Using the normal vector from i, the normal line isFor $0<\theta <\pi$, the $y$- and $z$-coordinates vanish together exactly when $\lambda =-1$. At that point, the $x$-coordinate isThusAt $\theta =0$ or $\theta =\pi$, both the position vector and normal vector lie along the $x$-axis, so $L$ is the whole $x$-axis in either case.

The required lines are the $x$-axis and every line through the origin in the plane $x=0$. Explicitly,The $x$-axis meets the ellipsoid at its two poles and is a normal line at each. A line of the second kind meets it at the antipodal points $(0,\pm \cos \varphi ,\pm \sin \varphi )$, whose normal vectors are parallel to that line.

To prove completeness, suppose a common normal line passes through distinct points $P=\mathbf{r}(\theta ,\varphi )$ and $Q=\mathbf{r}({\theta }^{\prime },{\varphi }^{\prime })$. If either is a pole, b forces the line to be the $x$-axis. Otherwise b says that the line meets the $x$-axis at $(-\cos \theta /\sqrt{2},0,0)$ and at $(-\cos {\theta }^{\prime }/\sqrt{2},0,0)$. A line other than the $x$-axis has at most one such intersection, so $\cos \theta =\cos {\theta }^{\prime }$.

The points $P,Q$ consequently have equal $x$-coordinates. Since they are distinct, their joining line has zero $x$-component in its direction. Its normal vector at $P$ therefore has $\sqrt{2}\cos \theta =0$. Hence $P,Q$ lie in $x=0$, and the normal line there passes through the origin. This gives exactly the listed family.