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Consider the following alternative definition of a parabola:

Given two points $F$ and $O$ in the plane, the parabola having focus $F$ and vertex $O$ is the locus of points $P$ of the plane such that $$(FP - OF)(FP + 3 OF) = OP^2.$$

Using coordinates it is easy to see that the definition is equivalent to the usual one. Indeed, if we let $O = (0, 0)$, $F = (0, f)$ for some $f > 0$ and $P = (x, y)$, then the given equation simplifies to $x^2 = 4 f y$, which is precisely the equation of the parabola having focus $F$ and vertex $O$ as it is usually defined.

What I am interested in is a geometric proof that any parabola satisfies the above property, which should hopefully give some insight on why such an equality must hold. I have attempted to prove it in two ways:

  1. As it is written, the equality seems to say that a certain rectangle (or maybe parallelogram?) has the same area as the square on the line segment $OP$. I have noticed that $FP - OF$ is the distance from $P$ to the tangent line to the parabola at $O$, but I don't know what to do with $FP + 3 OF$.
  2. The equality can be rewritten as $$OP^2 + (2 OF)^2 = (FP + OF)^2.$$ Now it looks as though it could be proven using the Pythagorean theorem. But I haven't been able to draw a triangle having sides $OP$, $2OF$ and $FP+OF$ so that it can be seen that it is indeed a right triangle.

Any help would be highly appreciated.

(Background: this problem came up while trying to prove a similar property about the cissoid of Diocles, see this other question of mine. The two properties are related through inversion with respect to the unit circle centered at $O$.)

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Let $D$ be the symmetric of $F$ with respect to $O$ and let $R$ be some point on the $OF$ line, such that $O$ lies between $F$ and $R$. Let $S$ be the symmetric of $R$ with respect to $O$. If the perpendicular to $OF$ through $S$ meets the circle centered at $F$ through $R$ at $P$, $P$ lies on the wanted parabola, since $PF=FR=SD$. Let $T$ be the symmetric of $R$ with respect to $F$ and let $OR=z$. Since $PO$ is a median in the right triangle $PSR$,

$$ PO^2 = \frac{2PS^2+2PR^2-SR^2}{4}=\frac{4PS^2+SR^2}{4}=PS^2+OR^2$$ but $PS^2 = RS\cdot ST = 2 OR\cdot ST$, hence $$ PO^2 = OR\left(OR+2ST\right)=(PF-OF)(2(OS+ST)-OR)$$ and $$ PO^2 = (PF-OF)(2OT-(PF-OF))=(PF-OF)(2(PF+OF)-(PF-OF))=(PF-OF)(PF+3OF).$$

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  • $\begingroup$ Thank you. That was a bit more involved than I expected, but it certainly answers my question. Can I just ask you how you came up with this proof? I admit that I hadn't considered drawing the circle centered at $F$, but even if I had I doubt that I would've got to the desired equality if not by chance. $\endgroup$ – Luca Bressan Jan 22 '18 at 12:50
  • $\begingroup$ @LucaBressan: the elementary properties of the parabola (like the fact that $PR$ is a tangent) are usually derived by angle chasing and symmetries, and your relation looked like an identity involving the power of a point with respect to a circle. So I just started to construct a point on a parabola in the usual way, then considered some extra points in order to have segments with length $OP,OF,FP+OF$ on the same line (the axis of the parabola). After that, it is just routine inspection. $\endgroup$ – Jack D'Aurizio Jan 22 '18 at 13:11
  • $\begingroup$ Wooaahh!! As always, amazing! $\endgroup$ – Jaideep Khare Jan 22 '18 at 13:41

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