Principle Lines of curvature

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In the text by Manfredo P. Do Carmo entitled Differential Geometry of Curves and Surfaces, an analysis of the principle directions is made near a non umbilic point on pp 160-161. I have followed his argument in detail myself and derived the determinant in equation (8). However, a statement is made that since the principle directions are orthogonal, it is easy to see that a necessary and sufficient condition for a parameterized curve to be a line of curvature is $F=f=0$. Im not quite sure how we easily get this from the determinant. A paraphrased derivation of the determinant:

A matrix representation of the shape operator $S_p$ is

\begin{align} &\text{First fundamental form: } I_p =\left(\begin{array}{cc}E&F\\F&G\end{array}\right)\\[2mm] &\text{Second fundamental form: } {II}_p =\left(\begin{array}{cc}e&f\\f&g\end{array}\right)\\ &\text{Shape Operator: } S_p =I_p^{-1}{II}_p = \frac{1}{EG-F^2}\left(\begin{array}{cc}G&-F\\-F&E\end{array}\right)\left(\begin{array}{cc}e&f\\f&g\end{array}\right)\\ & = \frac{1}{EG-F^2}\left(\begin{array}{cc}Ge-Ff&Gf-Fg\\Ef-Fe&Eg-Ff\end{array}\right) \end{align}

Let $\alpha(t) = \mathbb{X(u(t),v(t))}$ be a parameterized curve on some surface $M$ with no umbilic points(ie. the principle curvatures are not equal). Then the tangent vector $\alpha'(t)=\mathbb{X}_uu'+\mathbb{X}_vv'$ A necessary and sufficient condition for a curve to be a line of curvature is $S_p(\alpha') = \lambda(t)\alpha'$ so we get:

\begin{align} \frac{1}{EG-F^2}\left(\begin{array}{cc}Ge-Ff&Gf-Fg\\Ef-Fe&Eg-Ff\end{array}\right)\left(\begin{array}{c}u'\\v'\end{array}\right)= \lambda\left(\begin{array}{c}u'\\v'\end{array}\right) \end{align}

or the system of ODE's:

\begin{align} \frac{Ge-Ff}{EG-F^2}u'+\frac{Gf-Fg}{EG-F^2}v' = \lambda u'\\[2mm] \frac{Ef-Fe}{EG-F^2}u'+\frac{Eg-Ff}{EG-F^2}v' = \lambda v'\\[2mm] \end{align}

Eliminating $\lambda$ gives:

\begin{align} (fE-eF)(u')^2+(gE-Eg)u'v'+(gF-fG)(v')^2= 0 \end{align}

which can be written as:

\begin{align} \left|\begin{array}{ccc}(v')^2&-u'v'&(u')^2\\E&F&G\\e&f&g\end{array}\right| = 0 \end{align}