Two parabolas, perpendicular axes: Can we prove the intersection points are concyclic?

Question

In this answer it is shown that when a pair of parabolas have perpendicular axes and intersect at four points, the four points are concyclic (belong to the same circle).

It's easy to show that this is true in general, if we have coordinate equations in the form

$ax^2+bxy+cy^2+dx+ey+f=0, a'x^2+b'xy+c'y^2+d'x+e'y+f'=0$

for each parabola, by taking a linear combination of the two equations you can make the quadratic component a multiple of $x^2+y^2$ (from the hypothesis you can render $a'=c, b'=-b, c'=a$).

But is there a way to prove the claim by purely geometric techniques? It smells like something that would have a purely geometric proof, but I can't see exactly how.

Answer 1

I don't forget that you desire a geometrical proof.

I thought I had one ; @Blue at the same time has found a previous identical question, and indicated me that my initial "solution" is flawed.

Having anew worked a lot on this issue, I propose to come back first to the analytic geometry solution, and after that to consider an interesting geometrical property of this rich figure, but let it be clear, it isn't a proof.

A) Analytic geometry part :

Let us take the parabolas' axes as coordinates axes.

The interest of this analytic geometry proof is that it brings information about the circle in terms of the two parabolas' parameters (the word "parameter" is classical for coefficient $p$ in $y^2=2px$ ; its geometrical interpretation is the distance from the focus to the directrix).

Fig. 1: The (red) circle passing through the 4 intersection points has its center in $(p,q)$ [Notation of formulas (1) ; here we have taken parameters $p=\frac12, q=1$ and shifts $a=-3, b=-2$]. Besides (see part $B$), the three common tangents to the parabola determine a triangle $ABC$ whose circumscribed (blue) circle is the circle with diameter the line segment joining the foci.

Here it goes.

We can give to the parabolas the following equations:

$$\begin{cases}y^2&=&2p(x-a)& \ \ (a)\\x^2&=&2q(y-b)& \ \ (b)\end{cases}\tag{1}$$

System (1a)+(1b) is equivalent to the following one (2a)+(2c): the first equation is preserved ; the second equation is the sum of the two previous ones:

$$\begin{cases}y^2&=&2p(x-a)& \ \ (a)\\x^2+y^2&=&2px+2qy-2(pa+qb)& \ \ (c)\end{cases}\tag{2}$$

(2c) is clearly the equation of a circle.

Remark: the equivalence of the two systems means that the common points of the two parabolas are exactly the same as the common points of, in fact any of, the two parabolas with the circle. This proves the co(n)cyclicity of the intersection points.

Equation (2c) can be written under the following form:

$$(x-p)^2+(y-q)^2=\underbrace{(p-a)^2+(q-b)^2-(a^2+b^2)}_{R^2}$$

It is noticeable that the coordinates of center $C$ of this circle $(p,q)$ are the two parameters of the parabolas. Moreover, the expression of radius $R$ gives the condition

$$(p-a)^2+(q-b)^2 \ge a^2+b^2$$

under which the circle exists as a "real" circle.

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B) A geometrical property : (once again, for proofs see the reference given by @Blue) I think that this property, even if it does not give a solution is interesting and complementary to the solutions given there.

Consider on the figure the three common tangents to parabolas $(P_1)$ and $(P_2)$. They form a triangle $ABC$. There is a theorem (Thm 3.1 p.207 in this reference) saying that the curcumscribed circle to $ABC$ passes through the foci $F_1, \ F_2$ of the parabolas ; this theorem doesn't assume that the axes of the parabolas are orthogonal. If we add this fact, one can establish that this circle passes through the origin, which amounts to say (right angle property of the diameter) that this circle has line segment $F_1F_2$ as its diameter (see Addendum at the end).

A second geometrical remark, important to my eyes.

Let us consider the (ill-known) "reciprocal" polar transform with respect to the unit circle (duality between a pole and its polar line). See for that this page with an illustration of the fact that, by this transformation, the image of a parabola whose axis passes through the center is an ellipse passing through the origin. In this way, we transform the problem into a dual one (see Fig. 2 and its legend).

Historical note: "reciprocal" polar transform (born in the 1820s) has historically been important as a transition step towards a more abstract version of duality.

Fig. 2: Polar reciprocation: the two parabolas are transformed by duality into two ellipses featuring the set of their tangents. In particular their intersection points are to be understood as the three common tangents of Fig. 1 + the tangent at infinity (nothing surprizing: in projective geometry, a parabola is tangent to the line at infinity).

Addendum: Sketch of the algebraic developments used for proving the property of the circumscribed circle to $ABC$.

Starting point: the generic equations of the tangents to the 2 parabolas:

$$\begin{cases}\text{in} \ (x_1,y_1) : \ \ yy_1&=&p(x+\frac{y_1^2}{2p}-a)\\ \text{in} \ (x_2,y_2) : \ \ xx_2&=&q(y+\frac{x_2^2}{2q}-b)\end{cases}\tag{3}$$

A tangent is shared by the two parabolas if the coefficients in the two equations (3) are proportionnal :

$$\dfrac{p}{x_2}=\dfrac{y_1}{q}=- \ \dfrac{y_1^2-2ap}{x_2^2-2qb}=:c$$

giving a third degree equation for the value of the common ratio denoted by $c$:

$$\dfrac{2ap-q^2c^2}{\tfrac{p^2}{c^2}-2qb}=c\tag{4}$$

Out of its three solutions, one gets $y_1=qc$ and $x_2=\frac{p}{c}$, allowing to compute the equations of the three common tangents, therefore to determine points $A,B,C$, allowing finally to give access at the (cyan colored) circle represented on Fig. 1.