Intersection of two parabolas Given $a>0$ and $b>0$, I want to find the points of intersection of the two parabolas 
\begin{align}
y&=1-ax^2   \\x&=1-by^2 
\end{align}
Clearly I can just eliminate one of the variables, and I'll get a quartic equation, whose general solutions will be an enormous mess (according to Mathematica, anyway).
I also tried using this approach, but again got stuck in a quagmire of algebra.
Or, I could just use numerical methods, but that's what I'm trying to avoid.
The general problem of intersecting two conic section curves is well understood, and can only be solved by the techniques I described above (as far as I know). But my problem is not the general one, it's a very specific special case, and I'm wondering if someone can see some clever shortcut.
According to this question, the intersection points all lie on a circle, but I don't know if that helps.

Perhaps this is an improvement.
$$
y = 1 + a x^{2}
\tag{1}
$$
$$
x = 1 + b y^{2}
\tag{2}
$$
Substitute $(2)$ into $(1)$ to obtain
$$
  y = 1 + a \left(b y^2+1\right)^2
$$
and solve for $y$.

$$
y = \color{blue}{\pm} \frac{1}{2} \sqrt{-\frac{4 \sqrt[3]{2} (4 a+3)}{3 \sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}-\frac{\sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}{3 \sqrt[3]{2} a b^2}-\frac{2 \sqrt{6}}{a b^2 \xi }\color{red}{\pm} \frac{8}{3 b}}
\color{red}{\pm} \frac{\xi }{2 \sqrt{6}}
$$
where
$$
\alpha = a^2 b^4 (27-32 a b (8 a (b+1)+8 b+9))
$$
$$
\beta = a b^2 (27-16 a (8 a+9) b)
$$
$$
\xi = \sqrt{\frac{\frac{2^{2/3} \sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}{a}+\frac{8 \sqrt[3]{2} (4 a+3) b^2}{\sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}-8 b}{b^2}}
$$
There are a total of $4$ cases. The $\color{blue}{blue}$ and $\color{red}{red}$ signs are independent.

Intrigued by your comment about the intersection points, a few cases were plotted.




