# two polynomial equations

Let $f:\mathbb R^2\rightarrow\mathbb R$ be a smooth function such that for every point $(x,y)\in\mathbb R^2$ the system $$f_{11}+2tf_{12}+t^2f_{22}=0$$ $$f_{111}+3tf_{112}+3t^2f_{122}+t^3f_{222}=0$$ has a solution in $t$. In other words, $f$ satisfies a certain complicated partial differential equation (think of the resultant: http://en.wikipedia.org/wiki/Resultant)

(Here $f_{ij}$ and $f_{ijk}$ are second and third partial derivatives computed at $(x,y)$.)

Is there any chance that the above conditions imply the following statement:

"For a given constant $c$, the set of points $(x,y)$ that give a solution $t=c$, if nonempty, is a line, or at least must contain a line."

Or, if it doesn't hold, do you see a way to add some natural generically satisfied condition on $f$ so that it holds?

This is not my specialty, but I would be surprised if it's true.

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This is an extremely fancy way to say that at each point you have a direction $(1,t)$ in which the second and the third derivatives of $f$ vanish. Why don't you speak plain English? Formally the answer is "No" and the most trivial counterexample is $f(x,y)=x^2-y^2$ with the direction $(1,1)$ serving all points. The only lines in the level sets occur here when $c=0$. However, I suspect you meant something more interesting. So, what is this really about? – fedja Nov 9 '12 at 4:42
Not so much fancy as convoluted. – Noam D. Elkies Nov 9 '12 at 4:50
@fedja: why is $f(x,y)=x^2-y^2$ a counterexample? I want the set of $(x,y)$ for which $t$ is equal to a constant $c$ to be either empty or contain a line. in this case, for $c=1$ we get the whole plane, for $c\neq1$ it's empty. – filipm Nov 9 '12 at 5:10
maybe that part was equally convoluted, but i need the following: fix a point $(x_0,y_0)$; that point will give us some $t$, say $t=15$. now we are looking at the set of all other points $(x,y)$ that would give us also $t=15$, and i would like to claim that that set must contain a line through $(x_0,y_0)$. – filipm Nov 9 '12 at 5:13
Ah, sorry: I misread $t$ for $f$. It was getting late and I was getting stupid... Then "there is a chance" :). I'll have to check my computations before I post them though... – fedja Nov 9 '12 at 13:19