Do carmo's exercise 3.2-11: conjugate directions on a surface I'm trying to find a satisfying answer to this problem, so I would appreciate some help. 
Let $p\in S$ be an eliptic point and let r and r' be conjugate directions on p. Varying r in $T_pS$, show that the minimum value for the angle between r and r' is satisfied by a single pair of vector in $T_pS$ wich are symmetric with respect to the principal directions.
My ideia was simply consider unitary vectors on r and r', say $w=cos(\theta)e_1+sin(\theta)e_2$ and $w'=cos(\phi)e_1+sin(\phi)e_2$ so that the angle between these two vectors would be given by $\displaystyle cos(\theta)cos(\phi)+sin(\theta)sin(\phi)$. 
Now, remember that $\theta$ is varying and, since r and r' are conjugate, $\phi=\phi(\theta)$. Taking the derivative of that last expression with respect to $\theta$ and making it equal to 0, we should be able to find the answer. 
I'm not very confident of this approach, though. 
Thank you!
 A: By the geometric construction of conjugate lines (seen on page 152-153) via the Dupin Indicatrix (DI), for a direction $r$, we know that its conjugate direction $r'$ is parallel to the tangent of the ellipse ($k_1x^2+k_2y^2=1$ for $x$ and $y$ along $e_1$ and $e_2$ the principal directions) forming the DI at the point where $r$ and the DI ellipse meet. I.E., if $r$ is given by a vector $w\in T_p(S)$ on the DI ellipse, then (WOLG $k_1,k_2>0$) $w=\frac{1}{\sqrt{k_1}}\cos\theta e_1+\frac{1}{\sqrt{k_2}}\sin\theta e_2$, and $r'$ lies along the direction given by $w'=v=-\frac{1}{\sqrt{k_1}}\sin\theta e_1+\frac{1}{\sqrt{k_2}}\cos\theta e_2$. Taking their inner product gives $\langle w,v\rangle=|w||v|\cos\phi=m(t)(\frac{1}{k_2}-\frac{1}{k_1})\cos\theta\sin\theta$, for $m=|w||v|$. Note that minimizing $\cos\phi$ is the same thing as maximizing $-\cos\phi$, so as the angle between $r$ and $r'$ is both $\phi$ and $\pi-\phi$ as $r=\text{span}\{w\}$ and $r'=\text{span}\{v\}$, finding the minimal angle between $r$ and $r'$ is the same as finding the maximal. So computing to find the maximal angle via Rolle's theorem, we differentiate and find through calculation that $\theta$ for maximal $\phi$ is $\frac{\pi}{4}$. Hence we have found the unique directions upon which the angle between conjugate lines are minimized, notably $r$ and $r'$ are given by the spans of $w=\frac{1}{\sqrt{2k_1}}e_1+\frac{1}{\sqrt{2k_2}}e_2$, and $v=-\frac{1}{\sqrt{2k_1}}e_1+\frac{1}{\sqrt{2k_2}}e_2$. Note their symmetry with respect to $e_1$ and $e_2$.
A: Your approach does work well with one modification, although it is probably not what do Carmo intended. 
As you note, the objective is to minimize $\theta-\phi$ or to maximize $\cos(\theta-\phi) = \cos\theta\cos\phi + \sin\theta\sin\phi$ subject to the constraint $k_1 \cos\theta\cos\phi + k_2 \sin\theta\sin\phi = 0$. While it may work if you think of $\phi$ as a function of $\theta$, I think it is better to use Lagrange multipliers. Eliminating $\lambda$ via a short, nifty calculation with lots of cancellation (which I will leave up to you) leads to the condition $\tan^2\theta = \tan^2\phi$ or to $\theta=\pm \phi$. Using the constraint it is easy to dispense with $\theta=\phi$, so you can conclude that $\theta=-\phi$ at the maximum, which establishes the first part of the problem, the symmetry with respect to the axes. Now substitute that relation back into the constraint to obtain $\tan\theta = \sqrt{k_1/k_2}$ which establishes the second part of the problem, uniqueness. This gives essentially the same answer that the other poster provided. (You should think about what happens in the question of maximizing $\theta-\phi$ in this way, which seems to have been neglected in this method. We know $\theta-\phi$ must have a maximum (by a compactness argument) but why does the Lagrange multiplier method not seem to find it?)
Alternatively you can write the problem in terms of vectors $u=(u_1,u_2)$ and $v=(v_1,v_2)$. The objective is to maximize $u\cdot v$ subject to the constraints $u^T \left( \matrix{k_1 & 0 \\ 0 & k_2} \right) v = 0$, $u\cdot u = 1$, $v\cdot v=1$. Lagrange multipliers isn't so nice (there are now seven variables and seven equations), but there is a nice theory of quadratic optimization that you might be able to use. (I'm not absolutely sure where to go with this, but it must work out one way or another.)
