Intersection of two parabolas

Question

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.

Answer 1

This is too long for a comment.

If we consider the equations $$\begin{align} y&=1-Ax^2 \\x&=1-By^2 \end{align}$$ eliminating $y$ to get the quartic in $x$ and then using the procedure given in this page, we have $$\Delta=A^4B^2(256 A^2 B^2-256 A^2 B-256 A B^2+288 A B-27)$$ $$P=-16 A^3 B^2 <0$$ $$Q=8 A^4 B^2 >0$$ $$\Delta_0=4 A^2 B (4 B-3)$$ $$D=-64 A^6 B^3<0$$ So, there will be four real roots if $\Delta >0$ and two real roots if $\Delta <0$.

That is to say that, for a given value of $A$, we should have only two real roots if $B$ is between the two roots $$B_1=\frac{A(9-8A)-\sqrt{A (4 A-3)^3}}{16 A(1-A)}$$ $$B_2=\frac{A(9-8A)+\sqrt{A (4 A-3)^3}}{16A(1-A)}$$ and four roots otherwise (this assumes $A\neq 0$).

For example, using $A=5$ and $B=0.9$ leads to two real roots while $A=5$ and $B=1.1$ leads to four real roots.

Looking at the particular case where $B=\frac 1A$ $$\Delta=-A \left(256 A^2-517 A+256\right)$$ which is positive if $$\frac{517-7 \sqrt{105}}{512}< A < \frac{517+7 \sqrt{105}}{512} $$ which represents a very narrow range.

Using $A=1.1$ leads to four real roots while $A=1.2$ leads to two real roots.

The problem seems to be quite sensitive to the values of the parameters.

Looking at the case where $A=1$, the problem simplifies a lot since $\Delta=B^2 (32 B-27)$. So, if $B > \frac{27}{32}$ four real roots and only two real roots otherwise.

Let us try for $B = \frac{26}{32}$ $${x= -1.67794}\,,{x= 0.338968 -0.150441 i}\,,{x= 0.338968 +0.150441 i}\,,{x= 1.}$$ while for $B = \frac{28}{32}$ $${x=-1.65597}\,,{x= 0.182018}\,,{x= 0.473952}\,,{x= 1.}$$

Edit

Interesting is the case where $B=A$; in such a case $\Delta=A^6 (4 A-3)^3 (4 A+1)$ and then four roots if $A >\frac 34$. If this is the case, the coordinates of the intersections are $$\left( \begin{array}{cc} x & y \\ -\frac{\sqrt{4 A-3}-1}{2 A} & \frac{\sqrt{4 A-3}+1}{2 A} \\ \frac{\sqrt{4 A-3}+1}{2 A} & -\frac{\sqrt{4 A-3}-1}{2 A} \\ \frac{-\sqrt{4 A+1}-1}{2 A} & -\frac{\sqrt{4 A+1}+1}{2 A} \\ \frac{\sqrt{4 A+1}-1}{2 A} & \frac{\sqrt{4 A+1}-1}{2 A} \end{array} \right)$$

These points are along a circle centered at $\left(-\frac{1}{2 A},-\frac{1}{2 A}\right)$ with a radius equal to $R=\sqrt{\frac{1+4A}{2A^2}}$

Answer 2

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 $y$.

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$$ y = \color{blue}{\pm} \frac{1}{2} \sqrt{-\frac{4 \sqrt[3]{2} (4 a+3)}{3 \sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}-\frac{\sqrt[3]{\beta -3 \sqrt{3} \sqrt{\alpha }}}{3 \sqrt[3]{2} a b^2}-\frac{2 \sqrt{6}}{a b^2 \xi }\color{red}{\pm} \frac{8}{3 b}} \color{red}{\pm} \frac{\xi }{2 \sqrt{6}} $$ 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.

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

Answer 3

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 $b$.