A doubly ruled surface which is not a plane must be quadratic I want to show that a doubly ruled surface which is not plane must be quadratic.
Any help will be appreciated. 
 A: A proof using affine differential geometry can be found in Exercises 14 and 16 of Volume 3, Chapter 4 of Spivak's A Comprehensive Introduction to Differential Geometry. 
However, one can give an elementary argument just using basic facts about quadrics (see, for example, section 4 of Chapter 8 of my book Abstract Algebra, A Geometric Approach). The outline is simple: If $L_1, L_2, L_3$ are three lines in $\Bbb P^3$ in general position, then there is a unique quadric surface $S$ containing them. It follows that any line intersecting all three of them lies on $S$, and hence the conclusion.
A: I think the only way to do this that makes sense is to try to show that all second derivatives of the surface are 0. But to get there seems non-trivial.
I think you want to begin by starting at a point on the surface, $p=(x, y, z)$ and its two lines, with directions $v$ and $w$, and we may assume $|v|=|w|=1$. Now, for every point on the line $\{p+tv\}_{t\in\mathbb{R}}$ there is precisely one other line through that point, with its associated direction, and we call them w(t), and we may choose them so that they vary continuously, rather than switching signs. The surface is then, at least locally, the set of points $\{p + tv + sw(t)\}_{s,t\in \mathbb{R}}$. You can do this in the other direction too, and build the surface as $\{p + sw + tv(s)\}_{s,t\in \mathbb{R}}$.
Using these two representations of the surface, you can find a quadrilateral of lines, two through p, the other two in the two families of lines already discussed, $\{p + t_0v + sw(t_0)\}_{s\in\mathbb{R}}$ and $\{p + s_0w + tw(s_0)\}_{t\in\mathbb{R}}$. These two lines must intersect somewhere, put some variables on that too, say $p + t_0v + s_1w(t_0) = p + s_0w + t_1w(s_0)$. Fix one more line that's parallel to the $w$'s but intersecting the $v$'s and you now have exactly enough information to have closed forms for all the points "inside" the quadrilateral. With that closed form, proving that all the second derivatives vanish should be possible.
My apologies for the partial answer.
