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Let there be two 2nd degree curves: $$f(x,y)=ax^2+by^2+cx+dy+e=0$$ and $$g(x,y)=fx^2+gy^2+hx+iy+j=0,$$ how is it possible to determine if these two curves intersect in some region, say $x \le 1 , y \ge 1$, without actually calculating the roots of these two curves.

Alternatively what are the conditions on the coefficients for these two curves to intersect in the region $x \le 1 , y \ge 1$.

Any help would be appreciable

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Translate your curves so that the region boundaries align with coord axes; comparing coords to $0$ is easier. Use Sylvester's Eliminant (or resultants) to eliminate $y$ from your system and get a quartic $p(x)$ whose roots are the $x$-coords of the intersections; likewise, get $q(y)$. Then, the Descartes Rule of Signs can give insights into how many roots of $p(x)$ and $q(y)$ lie on either side of $0$. Of course, Sylvester is cumbersome, and Descartes often inconclusive --and not knowing how the $p$-roots pair-up with $q$-roots is problematic-- but this approach could help in certain cases. –  Blue May 1 '12 at 9:27
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1 Answer 1

This is not what you want:

It is hard to say what concrete conditions on the coefficients such that $F,G$ intersect in $\mathbb{C}^2$.

However, by Chevalley's theorem, we know some informations of the set of coefficients such that $F,G$ intersect in $\mathbb{C}^2$.

Consider the map $\mathbb{C}[a,b,c,d,e,f,g,h,i,j]\to \mathbb{C}[a,b,c,d,e,f,g,h,i,j][x,y]/(F,G)$. Let $X,Y$ denote the affine schemes respectively corresponding to the rings. So we have a map from $Y\to X$, by Chevalley's theorem, we know that the image of $Y$ is a constructible set of $X$.

If we only consider the closed points $X_0$ of $X$ and $Y_0$ of $Y$. Then $F,G$ intersect if and only if $(a,b,c,d,e,f,g,h,i,j)$ in the image of $Y_0$.

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I just want to say it is not easy even for $\mathbb{C}^2$. –  wxu May 1 '12 at 8:21
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