Is it correct, that the number of intersections of two polynomials with degree 2 is at most 2?
I would argue that the intersection points can be determined by using the quadratic formula, which provides at most 2 different values.
Update: I consider two functions of form $f(x) = \frac{d}{ax^2+bx+c}$.
You can try $$ax^2+bx+c=dx^2+ex+f$$ and we get $$(a-d)x^2+x(b-e)+(c-f)=0$$ If $a=d$ the the equation is linear, if not then we get by the quadratic formula $$x_1=1/2\,{\frac {-b+e+\sqrt {-4\,ac+4\,af+{b}^{2}-2\,be+4\,dc-4\,df+{e}^{2 }}}{a-d}} $$ $$x_2=-1/2\,{\frac {b-e+\sqrt {-4\,ac+4\,af+{b}^{2}-2\,be+4\,dc-4\,df+{e}^{2 }}}{a-d}} $$ Also your case can reduced to this case: $$\frac{d}{ax^2+bx+c}=\frac{e}{fx^2+gx+h}$$ is equivalent to $$d(fx^2+gx+h)=e(ax^2+bx+c)$$ if $$ax^2+bx+c\ne 0$$ and $$fx^2+gx+h\neq 0$$