[Math] Define the Intersection Points of Polynomials

Question

I am facing the following problem.
Let’s consider that there are 2 points that are not known.
$${(x_0,y_0) (x_1,y_1)}$$
I know that from these 2 unknown points a set of quadratics passes
$$f_i(x)=a_2x^2+a_1x+a_0 $$ $$0<i<n$$
i.e. the quadratics intersect at these two unknown points.
For each one of the quadratics:
I know the leading coefficient $$a_2$$ and a point $$(x_i,y_i)$$
Is it possible to find the intersection points of those polynomials? $${(x_0,y_0) (x_1,y_1)}$$
I think that if I know the leading coefficient and a point from four quadratics (n=4) I can define their intersection.

Is that right?

In this image two quadratics intercept at two unknown points.For this quadratics I only know a point and their leading coefficient.
By using Newton Interpolation's Method:
$$a_2=\frac{y_2-(y_0+\frac{y_1-y_0}{x_1-x_0}(x_2-x_0))}{(x_2-x_0)(x_1-x_0)}$$

In the previous equations the only unknowns are the 2 intersection points $$x_0,y_0 x_1,y_1$$. Having 4 equations like the previous one, does not solve the problem?

Answer 1

If I have interpreted your question correctly, then you don't have enough information to find the intersection points. As a simple example, consider when $a_2=0$ for all of the "quadratics" (i.e. we have straight lines). Then all you know is that you have a straight line passing through a particular point in each case, but you don't know their gradients so you have no way of knowing where they intersect. Now yes, these aren't really quadratics anymore, but the situation for $a_2\not = 0$ is the same- we know where the quadratics pass through, but we don't know enough about their gradient to determine exactly what quadratics they are and therefore can not find their intersections.