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.
 A: 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.




A: Re-writing the second equation to be in terms of $y$, we have the two equations $y = 1 − a x^2$ and $y = \sqrt{\frac{1 − x}{b}}$.
Set them equal to each other, giving $1 − a x^2 = \sqrt{\frac{1 − x}{b}}$.
Square both sides and expand to get $a^2 x^4 − 2 a x^2 + 1 = \frac{1}{b} − \frac{1}{b} x$.
Move everything to the left to get $a^2 x^4 − 2 a x^2 + \frac{1}{b} x + 1 − \frac{1}{b} = 0$.
Comparing that to the standard form of a quartic equation $A x^4 + B x^3 + C x^2 + D x + E = 0$, we see that:

*

*$A = a²$,

*$B = 0$,

*$C = −2 a$,

*$D = \frac{1}{b}$, and

*$E = 1 − \frac{1}{b}$.

The next steps use the equations for the quartic formulae found on this webpage, simplified due to the fact that $B = 0$ and thus any terms containing $B$ vanish.
Define the new variables:

*

*$p = \left(128 − \frac{144}{b}\right) a^3 + 27 \frac{a^2}{b^2}$,

*$q = \left(16 − \frac{12}{b}\right) a^2$, and

*$s = \frac{1}{3 a^2} \left(\frac{q \sqrt[3]{2}}{\sqrt[3]{p + \sqrt{p^2 − 4 q^3}}} + \frac{\sqrt[3]{p + \sqrt{p^2 − 4 q^3}}}{\sqrt[3]{2}} + 4 a\right)$
Then the $x$-values for the four potential intersections are:

*

*$x_1 =  \frac{1}{2} \sqrt{s} + \frac{1}{2} \sqrt{\frac{4}{a} − \frac{2}{a² b \sqrt{s}} − s}$

*$x_2 =  \frac{1}{2} \sqrt{s} − \frac{1}{2} \sqrt{\frac{4}{a} − \frac{2}{a² b \sqrt{s}} − s}$

*$x_3 = −\frac{1}{2} \sqrt{s} + \frac{1}{2} \sqrt{\frac{4}{a} + \frac{2}{a² b \sqrt{s}} − s}$

*$x_4 = −\frac{1}{2} \sqrt{s} − \frac{1}{2} \sqrt{\frac{4}{a} + \frac{2}{a² b \sqrt{s}} − s}$
These can be plugged into the first of the original pair of equations, $y = 1 − a x^2$, to get the corresponding $y$-values.
Depending on the values of $a$ and $b$, anywhere from none to all-four intersections may be complex-valued.
Notably, all of the real-valued intersections lie on a circle with radius $r = \sqrt{\frac{a^2 + 4 a^2 b + 4 a b^2 + b^2}{4 a^2 b^2}}$ and center $\left(\frac{−1}{2 b}, \frac{−1}{2 a}\right)$, including for cases with negative values of $a$ and/or $b$.
