Suppose we have a family of 1-parameter curves that foliates a subset of the plane, defined by a function that is a polynomial at the coordinates $ (x, y) $ and a parameter that identifies the individual curves.

For example, the family of curves defined by:

$$ P (x, y, a) = 16 a ^ 2 left (x ^ 2 + y ^ 2-a ^ 2 right) – left (2 x_f x + y_f left (2 y-y_f right ) -x_f ^ 2 right) left (8 a ^ 2 + 2 x_f x + 2 y_f y-x_f ^ 2-y_f ^ 2 right) = 0 $$

is the set of confocal ellipses with foci in $ (0,0) $ and $ left (x_f, y_f right) $, with the parameter $ to $ giving the semi-major axis of each ellipse.

Associated with this family of curves is a field of tangents to the curves: at each point $ (x, y) $ a single curve passes through the point, and if we choose an orientation we can assign a unit vector to each point that is tangent to that curve.

We can derive a polynomial equation satisfied by the components of these tangent vectors, $ (x & # 39 ;, y & # 39;) $ along with the coordinates $ (x, y) $assuming we have parameterized each curve as $ left (x_a (t), y_a (t) right) $ then taking the derivative of $ P (x_a (t), y_a (t), a) $ with respect to $ t $, to produce a polynomial $ D (x, y, x & # 39 ;, y & # 39 ;, a) $. In our example, we have:

$$ D (x, y, x & # 39 ;, y & # 39 ;, a) = 16 a ^ 2 left (2 xx & # 39; + 2 y y & # 39; -x_f x & # 39; -y_f y & # 39; right) -4 left (2 x_f x + 2 y_f y-x_f ^ 2-y_f ^ 2 right) left (x_f x & # 39; + y_f y & # 39; right) $$

When $ P $ and $ D $ they are zero their resultant with respect to $ to $ will be zero, and we have:

$$ Res_ {a} (P, D) = left (y & # 39; right) ^ 4 left (2 x ^ 3 x_f left (2 y-y_f right) y_f + x ^ 2 left (x_f ^ 2 left (-2 y y_f + y_f ^ 2-4 y ^ 2 right) + y_f ^ 2 left (y_f-2 y right) {} ^ 2 right) +2 xy ^ 2 x_f left (2 x_f ^ 2 + y_f left (y_f-2 y right) right) -y ^ 2 x_f ^ 2

left (x_f ^ 2 + y_f left (y_f-2 y right) right) right) {} ^ 2+ left (x & # 39; right) ^ 4 left (2 x ^ 3 x_f left (2 y-y_f right) y_f + x ^ 2 left (x_f ^ 2 left (-2 y y_f + y_f ^ 2-4 y ^ 2 right) + y_f ^ 2 left (y_f-2 y right) {} ^ 2 right) +2 xy ^ 2 x_f

left (2 x_f ^ 2 + y_f left (y_f-2 y right) right) -y ^ 2 x_f ^ 2 left (x_f ^ 2 + y_f left (y_f-2 y right) right) right) {} ^ 2-4 left (x & # 39; right) ^ 3 y & # 39; left (x ^ 3 left (2 y-y_f right) + x ^ 2 x_f left (y_f-3 y right) + xy left (x_f ^ 2 + 3 y

y_f-y_f ^ 2-2 y ^ 2 right) + y ^ 2 x_f left (y-y_f right) right) left (2 x ^ 2 x_f y_f-x left (x_f ^ 2 left ( y_f + 2 y right) + y_f ^ 2 left (y_f-2 y right) right) + y x_f left (x_f ^ 2 + y_f left (y_f-2 y right) right) right ) {} ^ 2 + 4 x & # 39;

left (y & # 39; right) ^ 3 left (x ^ 3 left (2 y-y_f right) + x ^ 2 x_f left (y_f-3 y right) + xy left (x_f ^ 2 + 3 y y_f-y_f ^ 2-2 y ^ 2 right) + y ^ 2 x_f left (y-y_f right) right) left (2 x ^ 2 x_f y_f-x left (x_f ^ 2 left (y_f + 2 y right) + y_f ^ 2

left (y_f-2 y right) right) + y x_f left (x_f ^ 2 + y_f left (y_f-2 y right) right) right) {} ^ 2 + 2 left (x & # 39; right) ^ 2 left (y & # 39; right) ^ 2 left (-4 x ^ 3 x_f + x ^ 2 left (2 x_f ^ 2 + 8 y y_f-y_f ^ 2- 8 y ^ 2 right) + y ^ 2 left (2

left (y-y_f right) {} ^ 2-x_f ^ 2 right) +2 xy x_f left (4 y-3 y_f right) +2 x ^ 4 right) left (2 x ^ 2 x_f y_f-x left (x_f ^ 2 left (y_f + 2 y right) + y_f ^ 2 left (y_f-2 y right) right) + y x_f left (x_f ^ 2 + y_f left (y_f-2

y right) right) right) {} ^ 2 $$

So we have removed the parameter $ to $ to get a homogeneous polynomial in $ (x & # 39 ;, y & # 39;) $ will be zero for any field of tangents to the curves, since any multiple of $ (x & # 39 ;, y & # 39;) $ it will also give zero.

My question is: when and how can this construction be reversed? Given a polynomial $ R (x, y, x & # 39 ;, y & # 39;) $ Like this one that is satisfied with a field of tangents, is there any proof that can determine if it originated from a family of polynomial curves? And can the defining polynomial be reconstructed?

If we write a polynomial $ P $ of a sufficiently high degree, with unknown coefficients, we can carry out all this construction to produce a resulting polynomial, which we could then try to match with our $ R (x, y, x & # 39 ;, y & # 39;) $ solving a potentially very large number of polynomial equations at unknown coefficients.

So my question is: can we do better than that brute force approach?

For the specific case I am trying to solve, I have a polynomial satisfied by the tangent vectors that is degree 8 in the derivatives and degree 6 in the coordinates.