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?
 A: 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.

A: Assume that $(x,y)$ is a point common to both parabolas. If we add the two equations together and complete the square we get the circle equation
\begin{equation*}
\left( x - (c/2) \right)^2 + \left(y - (a/2) \right)^2  = (a/2)^2 + (c/2)^2 + b + d
\end{equation*}
So one condition which is necessary for a solution is that the right-hand side be $\geq 0$.
If we instead subtract the second equation from the first we get
\begin{align*}
(x-y)(x+y) &= ay - xc + b - d \newline
&= \frac{a-c}{2}(x+y) - \frac{a+c}{2}(x-y) + b - d
\end{align*}
Let $w=(x+y)$, $z=(x-y)$, $k_1 = (a-c)/2$, $k_2=(a+c)/2$. Then
\begin{equation*}
wz = k_1w - k_2z + b-d
\end{equation*}
\begin{equation*}
(w+k_2)(z-k_1) = (b-d) -k_1 k_2
\end{equation*}
So the point $(w,z)$ lies on a certain hyperbola. Unfortunately I can't think of anything else that one can say with this line of reasoning, but I thought it was interesting so I decided to post this incomplete answer anyway.
