The golden ratio in a parabola

Question

It is nice that the golden ratio appears automatically when we are not looking for it. This is what happened to me when I was using GeoGebra and trying to solve a different problem that occurred to me:

Data: $C$ parabola Featured F and L, Point A moving freely on L, Point B represents Muscat-based F on L, Point M represents

Intersection of a F widget with C.

Required: Find the appropriate position for a point A which makes AM=FB.

My solution:

Was it previously discovered or not? In any case, how do we prove that?

Answer 1

GeoGebra plot with polar equations for the parabola and its directrix. Parameter $p=FB$ has been taken equal to $1$.

Let us take the focus $F$ of the parabola as the origin and the horizontal line passing through the focus as the reference $x$-axis, in particular for angles.

Let us consider the following polar equations where $\theta := \angle (Fx,FP)$ ($P$ being a generic point on the parabola) :

• parabola's polar equation : $r= -\frac{p}{1-\sin \theta}$

• directrix' AP polar equation : $r= -\frac{p}{\sin \theta}$

With these equations, condition :

$$AM=FB \iff FA-FM=FB$$

can be expressed as :

$$- \frac{p}{1-\sin \theta}+\frac{p}{\sin\theta}-=p$$

Setting $x:=\sin \theta$, and simplifying by $p$ : one gets quadratic equation :

$$x^2+x-1=0$$

whose unique root in interval $[-1,1]$ is

$$x=\sin \theta = \frac{1}{\Phi} = \frac{FB}{FA} \tag{1}$$

(indeed $\theta = \angle BAF$)

Tking the inverse ratio in (1) :

$$\frac{FA}{FB}=\Phi$$

Answer 2

Draw point $D$ on $AB$ such that $MD\perp AB$.

By similiar triangles, we have

$$\frac{AF}{BF}=\frac{AM}{DM}$$

We are given that $AM=BF$. Since $F$ is the focus and $L$ is the directrix, we have $DM=FM$.

$$\frac{AF}{BF}=\frac{BF}{FM}=\frac{BF}{AF-AM}=\frac{BF}{AF-BF}=\frac{1}{\frac{AF}{BF}-1}$$

$$\left(\frac{AF}{BF}\right)^2-\frac{AF}{BF}-1=0$$

$$\therefore \frac{AF}{BF}=\frac{1+\sqrt5}{2}$$

Answer 3

Jean Marie and Dan have already proven the first question, in my answer I'll show why all those ratios are equal to each other: At first we have:

$\frac{AF}{FB}=\frac{AF}{AM}=Φ$

According to what has been proven. We could also write:

$\frac{AF}{AM}=\frac{AM}{FM}$

as defined by the Golden Ratio. So since $AM=FB$ we can write:

$\frac{FB}{FM}=\frac{AM}{FM}$

Thus, in the triangles $∆AFB,∆BFM$, we have proportional sides and the angle between them is equal ( $\angle AFB=\angle BFM$) Therefore $∆AFB,∆BFM$ They are similar.

From the similarities we write: $AM⊥BM$ Thus, we get three similar right triangles, which gives:

$\frac{AF}{FB}=\frac{FB}{FM}=\frac{AB}{BM}$

It remains only to prove that the ratio $\frac{NF}{AF}$ is equal to them.

I will prove that later.