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!

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. 
