# Proof the Focus of a Parabola Rolling Traces a Catenary

As discussed in the Hyperbolic Functions page, when a parabola rolls on the *x*-axis without slipping, its focus traces the catenary curve. The focus of the parabola \(y=ax^2\) traces the curve \(\frac{1}{4a}\cosh(4ax)\). We will prove this here in easy to understand steps. First, let’s become familiar with the pieces that are necessary with the image below.

The image above shows an arbitrary point P on the parabola \(y=ax^2\) at (*t*, *at*²). We will be using this variable *t* as the parameter to find the parametric equation that defines the location of the focus. The initial position of the focus of this parabola is at \(F\left(0,\frac{1}{4a}\right)\). The tangent line has been drawn at point P. The angle BPA is equal to *α*. Per the reflective property of parabolas, angle APF is also equal to *α*. Segment AB is equal to \(\frac{t}{2}\), a property of parabola tangents.

The distance from the focus to the point on the parabola, FP, is equal to *d*. The arc length of the parabola from O to P is *s*.

Our goal is to express *d*, *α*, and *s* as functions of *t* so we can parameterize the position of the focus when P lies on the *x*-axis. So, at some point as the parabola rolls, point P will coincide with the *x*-axis. At that instance, the tangent line of point P will be the *x*-axis. The image will look like the following:

After rotation, all of the lengths stay intact for the same point given by *t*. The arc length, *s*, is OP' = V'P'. The distance is F'P'. The segment F'V' is the distance of the focus from the vertex and it remains 1/(4*a*). Angle *α* also remains the same. Therefore, we need to express *d*, *α*, and *s* as a parameter of *t* so that we can find the coordinates of F' as a function of *t*. After the parabola rolls, the horizontal position of F' is \(x = s - d\cos\alpha\) and the vertical position of F' is \(y = d\sin\alpha\).

(1) \(x = s - d\cos\alpha\) and \(y = d\sin\alpha\).

Let’s find *α* in terms of *t* first. From Figure 1, we have a right triangle ABP, which has angle BPA as the angle *α*. From the right triangle, we can state the tangent of *α*:

(2) \(\tan\alpha = \dfrac{t/2}{at^2} = \dfrac{1}{2at}\).

From (2), we can find the sine and cosine of *α* in terms of *t*:

(3) \(\cos\alpha = \dfrac{2at}{\sqrt{4a^2t^2+1}} = \dfrac{t}{\sqrt{t^2+\frac{1}{4a^2}}}\)

(4) \(\sin\alpha = \dfrac{1}{\sqrt{4a^2t^2+1}} = \dfrac{1}{2a\sqrt{t^2+\dfrac{1}{4a^2}}}\)

We do not need to evaluate the actual meansure of *α* in terms of *t*. Finding sine and cosine of *α* is sufficient for our purpose.

From Figure 1 again, we can find *d* in terms of *t* using the distance formula: \(d(t) = \sqrt{\left(at^2-\frac{1}{4a}\right)^2 + t^2} = \) \(\sqrt{a^2t^4+\frac{t^2}{2}+\frac{1}{16a^2}} =\) \(at^2+\frac{1}{4a}\). The distance reduces quite nicely to a square that we can take the square root of.

(5) \(d(t) = at^2+\dfrac{1}{4a}=\) \(a\left(t^2+\dfrac{1}{4a^2}\right)\)

Actually, this result is not at all surprising, since we know that this distance is equal to the distance of point P to the directrix of the parabola. This is discussed here: Parabola Properties.

Now, we need to find the arc length of the parabola from the vertex to point P. Our limits are 0 to *t*. The arc length formula is \(\int_{0}^{t}\sqrt{1+(dy/dx)^2}\text{ }dx\). For our parabola, whose derivative is 2*ax*, this becomes \(\int_{0}^{t}\sqrt{1+(2ax)^2}\text{ }dx\) and evaluates to \(s(t) = \frac{1}{2}t\sqrt{4a^2t^2+1} + \frac{\sinh^{-1}(2at)}{4a} = \) \( at\sqrt{t^2+\frac{1}{4a^2}} + \frac{\sinh^{-1}(2at)}{4a}\).

(6) \(s(t) = at\sqrt{t^2+\dfrac{1}{4a^2}} + \dfrac{\sinh^{-1}(2at)}{4a}\)

Now that we have *d*, *s*, and *α* in terms of *t*, we can find the coordinates of F' in terms of *t*. Let’s recall Figure 2 here in a mini form:

We already know that from (1) the formulas for the position of the focus. Let’s find the *x* and *y* position of the focus in terms of *t* separately.

The *x* position is:

(i) \(x(t) = s(t) - d(t)\cos\alpha \)

(ii) \(x(t) = at\sqrt{t^2+\frac{1}{4a^2}} + \frac{\sinh^{-1}(2at)}{4a} - a\left(t^2+\frac{1}{4a^2}\right)\cdot\frac{t}{\sqrt{t^2+\frac{1}{4a^2}}}\).

(iii) \(x(t) = \frac{\sinh^{-1}(2at)}{4a}\)

This simplified to a wonderful simple equation:

(7) \(x(t) = \frac{\sinh^{-1}(2at)}{4a}\)

In terms of logarithmic functions, (7) can be written as \(x(t) = \frac{1}{4a}\ln\left(2at+2a\sqrt{t^2+\frac{1}{4a^2}}\right)\).

Now we find the *y* position:

(i) \(y(t) = d(t)\sin\alpha \)

(ii) \(y(t) = a\left(t^2+\frac{1}{4a^2}\right)\cdot \frac{1}{2a\sqrt{t^2+\frac{1}{4a^2}}}\).

This also reduces the friendly expression for the focus.

(8) \(y(t) = \frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}}\)

The focus of a parabola \(y = at^2\) is located at \(x(t) = \frac{\sinh^{-1}(2at)}{4a}\) and \(y(t) = \frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}}\) as the parabola rolls on the *x*-axis without slipping.

The last step is solving for *t* in (vii) and substituting that into (8) to give our position of the focus in terms of *x* and *y*

(i) \(x = \frac{\sinh^{-1}(2at)}{4a}\)

(ii) \(4ax = \sinh^{-1}(2at)\)

(iii) \(\sinh(4ax) = 2at\)

(iv) \(t = \frac{\sinh(4ax)}{2a}\)

Now, substitute *t* from (iv) into (8):

(v) \(y = \frac{1}{2}\sqrt{\left(\frac{\sinh(4ax)}{2a}\right)^2 + \frac{1}{4a^2}}\)

(vi) \(y = \frac{1}{2}\sqrt{\frac{\sinh^{2}(4ax) + 1}{4a^2}}\)

(vii) \(y = \frac{1}{4a}\cosh(4ax)\)

This completes the proof.

The focus of a parabola \(y = at^2\) traces a catenary as the parabola rolls on the *x*-axis without slipping. The curve traced by the focus is \(y = \frac{1}{4a}\cosh(4ax)\).

The location of the focus is: \(x(t) = \frac{\sinh^{-1}(2at)}{4a}\) and \(y(t) = \frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}}\).

The location of the vertex is: \(x(t) = \frac{\sinh^{-1}(2at)}{4a} - \frac{t}{4a\sqrt{t^2+\frac{1}{4a^2}}}\) and \(y(t) = \frac{t^2}{2\sqrt{t^2+\frac{1}{4a^2}}}\).

Note that the equation of the catenary is usually written as \(y = a\cosh(x/a)\). Also, we will find the location of the vertex stated above next.

## Finding the Equation of the “Rolled” Parabola

### Rectangular Equation of a Rolled Parabola

Now, we want to find the equation of the parabola after it has rolled on the *x*-axis based on the parameter *t*. In order to do this, we need to find the angle of rotation and the translation of the parabola because the parabola is rolling and not simply rotating. Let’s find the angle of rotation, then we will find the translation distances.

First, let’s quickly review the relationship between the focus and the tangent of a parabola.

From Figure 4, we can see that a segment from the focus to the point on the *x*-axis where the tangent intersects the *x*-axis creates a right angle. Angle APB is equal to *α* (a property of parabolas). PAB is complementary to *α*. This makes OAF equal to *α* and OFA complementary to *α*. Moreover, AFP is also complementary to *α*.

Now, let’s look at a rolled parabola and draw the similarities from the parabola in standard position. Our goal is to find the values of *m* and *n*. These values will give us the position of V', the vertex. These are the translation values.

The segment F'A' is the similar to the FA in Figure 4 because both intersect their tangents at point P and P', respectively. Therefore, F'A'P' is a right angle, just like FAP is a right angle. Moreover a line trough A' and tangent to the parabola (at point V', which is the vertex) must be the *x*-axis of the rolled parabola. Hence, F'A'V' is equal to *α*.

This means that OA'V' is complementary to *α* and this is our angle of rotation. This is the solution to our first problem.

The angle of rotation, *θ*, of the parabola as it rolls on the *x*-axis is given by \(\theta = \pi/2 - \alpha\).

Next, we need to find *m* and *n*, which will give us the translation distances. We will substract these from the position of the focus.

A'V'F' is a right triangle because line through A'V' is the *x*-axis and V'F' is the line trough the origin and the focus. This establishes JV'F' is also equal to *α* by similar triangles.

We can find *m* and *n* in terms of *t* because we know the sine and cosine of *α*, and we know the length of V'F' is equal to 1/(4*a*) and it does not change.

(i) \(m = \frac{1}{4a}\cos\alpha = \frac{t}{4a\sqrt{t^2+\frac{1}{4a^2}}}\)

(ii) \(n = \frac{1}{4a}\sin\alpha = \frac{1}{8a^2\sqrt{t^2+\frac{1}{4a^2}}}\)

We need to subtract these values from the position of the focus. The focus located at:

\(x(t) = \frac{\sinh^{-1}(2at)}{4a}\) and \(y(t) = \frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}}\)

Therefore, our rotated parabola in (i) needs to be translated by a horizontal distance of \(h = \frac{\sinh^{-1}(2at)}{4a} - \frac{t}{4a\sqrt{t^2+\frac{1}{4a^2}}}\) and a vertical distance of \(k = \frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}} - \frac{1}{8a^2\sqrt{t^2+\frac{1}{4a^2}}} = \) \(\frac{t^2}{2\sqrt{t^2+\frac{1}{4a^2}}}\).

Now, let’s rotate the standard parabola first. The steps are not shown but the resulting equation is shown below. This is the equation of a parabola rotated by *π*/2 – *α*. The vertex remains at (0, 0).

(1) \(a(\cos^2 \alpha)x^2 - a\sin(2\alpha)xy + a(\sin^2\alpha)y^2 - \) \( (\sin \alpha)x - (\cos\alpha)y=0\)

Substituting *h* and *k* would give us our rolling parabola based on *t*.

(2) \(a(\cos^2 \alpha)(x-h)^2 - a\sin(2\alpha)(x-h)(y-k) + \) \( a(\sin^2\alpha)(y-k)^2 - (\sin \alpha)(x-h) - (\cos\alpha)(y-k)=0\)

Below is a Geogebra activity that rolls the parabola based on the equations above. Drag the parameter *t* to roll the parabola. You can increase or decrease the coefficient *a* to widen or narrow the parabola.

### Parametric Equations of a Rolled Parabola

The above method does not work for *t* < 0. No matter what I tried, the parabola just becomes inverted or rolls off the *x*-axis when *t* is less than 0. I assume it has to do with the arctangent of the rotation angle.

However, we did rotate the rectangular form of the parabola \(y = ax^2\) when we have been using the parametric equations. So, the rotation presented above was based on the *x* and *y* relationship and did not correlate with *t*.

As I thought about this, I finally realized we need to rotate the parametric form of the parabola given by the coordinates \((t, at^2)\). Since we cannot confuse the original parameter *t*, we will let the *u* be the parameter for the parabola \((u, au^2)\) and let *t* be the parameter for the angle of rotation. Applying the rotation formulas with a rotation angle of *π*/2 – *α* to the *x* and *y* values of the parametric form \((u, au^2)\), we get the following new coordinates:

(i) \(x(u) = u\cos(\pi/2-\alpha) + au^2\sin(\pi/2 - \alpha) =\) \(u\sin(\alpha) + au^2\cos(\alpha)\)

(ii) \(y(u) = au^2\cos(\pi/2-\alpha) - u\sin(\pi/2 - \alpha) =\) \(au^2\sin(\alpha) - u\cos(\alpha)\)

The sine and cosine of *α* in terms of *t* are given in equations (3) an (4). After substituting these into equation (ii), our rotated parabola has the equations:

(iii) \(x(u) = \frac{2a^2tu^2+u}{2a\sqrt{t^2+\frac{1}{4a^2}}}\) and \(y(u) = \frac{u^2-2tu}{2\sqrt{t^2+\frac{1}{4a^2}}}\)

Last step is to translate the parabola based on the coordinates of the vertex we found earlier. Therefore, the equation of our whole parabola that has rolled based on the parameter *t* is given here. Note that the simplification steps have not been shown and left to the reader.

The equation of a parabola that has rolled on the *x*-axis by the parameter *t* is given by the parametric equations:

\(x(u) = \dfrac{4a^2tu^2+2u-t}{4a\sqrt{t^2+\dfrac{1}{4a^2}}} + \dfrac{\sinh^{-1}(2at)}{4a}\)

\(y(u) = \dfrac{(u-t)^2}{2\sqrt{t^2+\dfrac{1}{4a^2}}} = \dfrac{a(u-t)^2}{\sqrt{4a^2t^2+1}}\)

Now, this parametric equation allows us to animate the rolling parabola by increasing or decreasing *t*. For graphing purposes, \(y(u) = \frac{a(u-t)^2}{\sqrt{4a^2t^2+1}}\) is the better option because if *a* is negative, then this allows us to graph an inverted parabola. (And noticing this fact allowed me to modify the rectangular form of the parabola above. This solved part of the problem.)

In the activity above, you can increase or decrease *a*, the coefficient of the parabola and the shape of the parabola will widen or narrow. To roll the parabola, increase or decrease *t*.

## Arc Length of Catenary

An interesting observation is that the arc length of the catenary from its vertex, V, to the focus of the parabola, F', is equal to *t*/2, which is the length of the segment V'A' as can be seen below.

Since V' is the origin of the axes of the parabola that has rolled, line containing V'A' is the *x*-axis of this parabola because V' is the tangent point. Also, a tangent of a parabola always crosses the *x*-axis at the midpoint of its *x* distance. Therefore, V'A' equal *t*/2. (This is more apparent in Figure 4 where OA is similar to V'A' here.

Now, we need to prove that the arc length of the catenary from V to F' is equal to *t*/2.

The arc length, *s*, of the catenary \(\frac{1}{4a}\cosh(4ax)\) from *x* = 0 to *b* is given by:

(i) \(s = \int_{0}^{b} \sqrt{1+\sinh^{2}(4ax)}\text{ }dx =\) \(\int_{0}^{b} \cosh(4ax)\text{ }dx =\) \(\left[\frac{1}{4a}\sinh(4ax) \right]_{0}^{b} = \frac{1}{4a}\sinh(4ab)\).

The *x* position of the focus is \(\frac{\sinh^{-1}(2at)}{4a}\). If we let \(b = \frac{\sinh^{-1}(2at)}{4a}\), we can find the arc length in terms of *t*.

(ii) \(s = \frac{1}{4a}\sinh\left(4a\cdot \frac{\sinh^{-1}(2at)}{4a}\right)\)

(iii) \(s = \frac{1}{4a}\sinh\left( \sinh^{-1}(2at)\right)\)

(iv) \(s = \frac{2at}{4a} = \frac{t}{2}\)

Therefore, the length of V'A' is equal to the arc length of the catenary from V to F'.

## Rolling Parabola Tangent to the *Y*-Axis

You may notice from the Geogebra activities that, when the parabola is rolled, it will be tangent to the *y*-axis at one specific positive value of *t* and one negative value of *t*. What is this value of *t*?

I tried so many methods to find the relationship between *t* and *α* and found lengths of many segments in these sketches with the goal of finding the length OT, believing that that would help me find *t*. But that lead nowhere. Then something simple just hit me in the image below. Let’s try to find this value using the image below.

As you can see in the image, the parabola is tangent to the *y*-axis at point T when it rolls toward the right for some positive value of *t*. The parabola is always tangent to the *x*-axis. But now, we have two tangents that are perpendicular. As we know, ** tangents to the parabola that are perpendicular always intersect on the directrix of the parabola.** This has been discusses here: Properties of Parabolas. And this is a major fact that allows us to find the relationship we seek.

Therefore, segment OD is equal to V'F', which is equal to \(\frac{1}{4a}\). We already know that OA' is equal to \(\frac{\sinh^{-1}(2at)}{4a}\). The angle A'OD is equal to *α* because of similar triangles. Therefore, we have the following trigonometric relationship:

(i) \(\cos\alpha = \frac{1/(4a)}{\frac{\sinh^{-1}(2at)}{4a}} = \frac{1}{\sinh^{-1}(2at)}\)

This is the equation we need to solve. We have already derived the cosine value of *α*, so equation (i) becomes:

(ii) \(\frac{t}{\sqrt{t^2+\frac{1}{4a^2}}} = \frac{1}{\sinh^{-1}(2at)}\) or \(\frac{2at}{\sqrt{(2at)^2+1}} = \frac{1}{\sinh^{-1}(2at)}\)

As it turns out, this equation is not solvable by any means. However, by graphing the two functions on either side of the equation and finding their intersection points for the various values of *a*, we can find their approximate values.

The graph above shows the solutions for *a* = 1. We should have a tangent parabola when *t* ≈ ±0.754. As a matter of fact, Figure 7 above shows exactly the parabola \(y = x^2\) when rolled with the parameter *t* = 0.754.

As an aside, it turns out that segments V'A and DA' are equal in length. Since V'A' is equal to *t*/2, that makes DA also equal to *t*/2.

Also, triangles A'F'A and A'OA are congruent triangles because they are both right triangles with equal angle measures. That means AF' = OA' and we have a rectangle when the parabola is tangent to the *y*-axis.

We can make a table of the value of *t* for particular *a*’s of the parabola.

a |
t |

0.125 | 6.0355182461533 |

0.25 | 3.0177591230766 |

1/3 | 2.2633193423705 |

0.5 | 1.5088795615383 |

1 | 0.7544397807693 |

2 | 0.3772198906872 |

3 | 0.2514799270938 |

4 | 0.1886099409664 |

Examining the values shows an inverse relationship between *t* and *a*. So if we know the value of *t* for *a* = 1, we can find any value of *a*.

The equation is basically: \(t \approx \frac{0.7544397807693}{a}\). An easy way to determine that the relationship is inverse is to let 2*at* = *x* in equation (ii) and solve for *x* by using a graphing utility. Our equation would be \(\frac{x}{\sqrt{x^2+1}} = \frac{1}{\sinh^{-1}x}\), and solution is *x* = 1.5088... Which means 2*at* = 1.5088... or *at* = 0.7544...

### Taking the *t* Out of the Equation

Since we know that *at* is constant and equal to about 0.7544, we can let *at* = *k* and replace that in our parametric equation to remove the parameter *t*. Our equations become:

(iii) \(x(u) = \frac{4ka^2u^2+2au-k}{2a\sqrt{4k^2+1}} + \frac{\sinh^{-1}(2k)}{4a}\) \(y(u) = \frac{(au-k)^2}{a\sqrt{4k^2+1}}\)

The above parametric equation will give a parabola that is always tangent to both axes, no matter what *a* is equal to. You can play with the *a* in the following Geogebra activity and see that the parabola remains tangent to both axes.

### The Other Tangent

Let’s bring back Figure 7 with another segment added.

Since OAF'A' is a rectangle and AA' is one of the diagonals, the other diagonal has to be OF'. But notice that OF' seems to be tangent to the catenary in green. And that is exactly the case when the rolling parabola is tangent to the *y*-axis.

We can find the slope of the tangent to the catenary by using its derivative.

(iii) \(y = \frac{1}{4a}\cosh(4ax)\)

(iv) \(\frac{dy}{dx} = \sinh(4ax)\)

Now, we let *x* equal the length of OA' , which is still the same as shown in Figure 8, to find the slope, *m*, at F':

(iv) \(m = \sinh\left(4a\cdot\frac{\sinh^{-1}(2at)}{4a}\right)\)

(v) \(m = \sinh(\sinh^{-1}(2at))\)

(vi) \(m = 2at\)

But we have a special relationship between *a* and *t*, which we stated is an inverse relationship. The product of *a* and *t* is constant and equal to approximately 0.7544. So twice that is 1.5088. Therefore, the slope of line OF' is approximately 1.5088. We can restate (vi) as:

(vi) \(m = 2at \approx 1.5088\)

Now, we know that the length of segment A'F' is the *y* value of the focus, which is \(\frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}}\). Therefore, the slope should also equal F'A'/OA'.

(vii) \(m = \dfrac{\frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}}}{\frac{\sinh^{-1}(2at)}{4a}} = \frac{\sqrt{(2at)^2+1}}{\sinh^{-1}(2at)}\)

If we substitute 1.5088 into (vii), we should get 1.5088 as the result. And as a matter of fact, this is correct. This is expected because if we let equations (vi) and (vii) equal each other, we get:

(viii) \(2at = \dfrac{\frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}}}{\frac{\sinh^{-1}(2at)}{4a}}\)

With some manipulation, this equals:

(ix) \(\frac{\sqrt{1+4a^2t^2}}{2at} = \sinh^{-1}(2at)\)

Equation (ix) is exactly the same equation as (ii) written in a different way.

Therefore, OF' is tangent to the catenary only when the parabola is tangent to the *y*-axis.

### The Angle of Rotation

With the above discussions about the parabola being tangent to both axes and finding the *t* at which this occurs, there’s one more to piece to this puzzle: the angle of rotation of this parabola that is tangent to both axes.

This piece of the puzzle is easier to solve than one may think because the product of *a* and *t* is constant when the parabola is tangent to both axes. Recall equation (2) in the beginning: \(\tan\alpha = \frac{1}{2at}\). The product of *a* and *t* seems to be everywhere; and it is here in this simple formula.

If the product is constant, then the angle of rotation is constant also, no matter what *a* is. Therefore, the measure of *α* is: \(\alpha = \tan^{-1}\frac{1}{2at} = \tan^{-1}\frac{1}{1.5088} \approx 33.534^{\circ}\).

The angle or rotation is complementary to *α*, so our angle of rotation is: \(\frac{\pi}{2} - 33.534^{\circ} \approx 56.466^{\circ}\). The following Geogebra activity shows how increasing or decreasing *a* does not affect the angle of rotation. The activity also shows that the line containing OF' remains in place. The focus of the parabola seems to travel along the line OF'.

### The Secant Line

Let’s look at Figure 8 with both tangent points in view. From the Parabola Properties page, it is evident that if two tangents form a right triangle, then the segment connecting the two tangent points always passes through the focus of the parabola. This segment is the secant line of the parabola, TP, which is shown in Figure 9.

Knowing this, we can easily find the length of TO. Remember angle TPO is equal to *α* and \(\tan\alpha = \frac{1}{2at}\). The ratio TO/OP is also equal to tan(*α*). Therefore, \(\text{TO} = \text{OP}\tan\alpha\).

We also know that OP is equal to the arc length of the parabola and its length was given in equation (6): \(s(t) = at\sqrt{t^2+\frac{1}{4a^2}} + \frac{\sinh^{-1}(2at)}{4a}\). After substitution, this makes TO equal to \(\text{TO} = \frac{1}{2}\sqrt{t^2+\frac{1}{4a^2}} + \frac{\sinh^{-1}(2at)}{8a^{2}t}\).

### Is a Square Possible?

If Figure 7, we identified a rectangle AF'A'O. Would a square be possible at any point?

In order for AF'A'O to be a square, the angle of rotation must be *π*/4. That means \(\tan\alpha = 1\). Howerver, established an inverse relationship between *t* and *α* above: \(t \approx \frac{0.7544397807693}{a}\). So, the product of *a* and *t* has to be approximately 0.7544... That means AF'A'O will never be a square.