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 $O$.)


enter image description here

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 $$ PO^2 = (PF-OF)(2OT-(PF-OF))=(PF-OF)(2(PF+OF)-(PF-OF))=(PF-OF)(PF+3OF).$$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.