Homework: Problem concerning first fundamental form Here's a strange problem in our differential geometry textbook.

At a point on surface $\mathbf{r}=\mathbf{r}(u,v)$, the equation $Pdudu+2Qdudv+Rdvdv=0$ determines two tangential directions. Prove that these two tangential directions are normal iff $$ER-2FQ+GP=0$$
  where $E=\langle\mathbf{r}_u,\mathbf{r}_u\rangle,F=\langle\mathbf{r}_u,\mathbf{r}_v\rangle,G=\langle\mathbf{r}_v,\mathbf{r}_v\rangle$

I don't understand what this problem wants me to do. Since the statement :"the equation $Pdudu+2Qdudv+Rdvdv=0$ determines two tangential directions" is a bit ambiguous. Hope to find some good understanding of this problem, no need for solutions, thanks!
 A: Consider $(1,a),\ (1,b)$ vectors on $uv$-plane. And ${\bf x}$ is a parametrication. 
$$ d{\bf x}\ (1,a) \perp d{\bf x}\ (1,b) \Leftrightarrow ({\bf x}_u+a{\bf x}_v )\cdot ({\bf x}_u+b{\bf x}_v)=0\Leftrightarrow E+(a+b)F+abG =0 $$
And if $R=1,\ Q=-1/2(a+b),\ P=ab$, then note that $$ ( Pdudu +2 Qdudv+ Rdvdv )((1,a),(1,a)) = P+aQ+a^2R=0$$ and $$ ( Pdudu +2 Qdudv+ Rdvdv )((1,b),(1,b)) = P+bQ+b^2R=0$$
So with these observations, we have the desired result. 
A: Well we know the first fundamental form gives: $$\begin{align} & {} \quad \mathrm{I}(aX_u+bX_v,cX_u+dX_v) \\ & = ac \langle X_u,X_u \rangle + (ad+bc) \langle X_u,X_v \rangle + bd \langle X_v,X_v \rangle \\ & = Eac + F(ad+bc) + Gbd. \end{align}$$ 
And we have the unit normal vector to a tangent plane at a point as: $$n = \frac{{\bf x}_u\times {\bf x}_v}{ |{\bf x}_u\times {\bf x}_v|}.$$
So if you combine your two equations $ER-2FQ+GP=0$ and $Pdudu+2Qdudv+Rdvdv=0$ and replace your $E,F,G$ by there respective inner products then I think you will get your desired outcome? I hope I am going in the right direction. 
I will delete this answer if it is wrong. Hopefully someone can answer this question, seems interesting to me. 
A: My interpretation can be wrong too.
For me statement "the equation $Pdudu+2Qdudv+Rdvdv=0$ determines two tangential directions" means the following. You have a quadratic form which is defined on tangent vector $(du, dv)$ at point $(u, v)$ with a matrice 
$A = \left ( \begin{matrix} P & Q \\ Q & R \end{matrix} \right )$ and you are looking for solutions of $(du\; dv) \cdot A \cdot (du\; dv)^{T} = 0$. In fact, if matrice $A$ has eigenvalues of different signs this equation really determines two family of solutions (there is one family of solutions when $A \leqslant 0$ or $A \geqslant 0$ and no solutions at all when $A > 0 $ or $A < 0$), which gives rise to a tangential directions on surface. You can measure the angle between these directions using first fundamental form and the orthogonality condition must be expressed in terms of quadratic forms' coefficients. 
Were there any conditions on $P$, $Q$ and $R$, like I've supposed?
