Conditions for intersection of parabolas? What are the conditions for the existence of real solutions for the following equations:
$$\begin{align}
x^2&=a\cdot y+b\\
y^2&=c\cdot x+d\end{align}$$
where $a,b,c,d $ are real numbers.
These represent two parabolas; how might we find out the conditions for the existence of $0,2,4$ real solutions of the equations?

Since these are parabolas, $a$ and $c$ must be nonzero.  Let $x = a^{2/3} c^{1/3} X$ and $y = a^{1/3} c^{2/3} Y$.  Under this scaling, the equations become
$X^2 = Y + B$ and $Y^2 = X + D$ where $B = b a^{-4/3} c^{-2/3}$ and 
$D = d a^{-2/3} c^{-4/3}$.  Now substituting $Y = B - X^2$ into $Y^2 = X + D$
we get the fourth-degree equation $X^4 - 2 B X^2 - X + B^2 - D = 0$.  The discriminant of this, according to Maple, is $-256\,{B}^{3}+288\,B D  -27+256\,{B}^{2}{D}^{2}-256\,{D}
^{3}$.  The curve where the  discriminant is $0$ separates the $BD$ plane into three regions like this:
We have: no real solution in the red region, one on the red-yellow boundary,
two in the yellow region, three on the yellow-blue boundary (except at the sharp cusp $B=D=3/4$ where there are two), and four in the blue region.

