# maximum inscribed angle within an ellipse

Consider the figure of an ellipse below:

where the coordinate pair $$(x_e , y_e)$$ denote a point on the positive half of the ellipse, the red stars denote the foci of the ellipse (at $$x = \pm c$$) , and $$d$$ denotes a distance centered on the foci as shown.

If we draw two lines which connect the points $$(-c - \frac{d}{2}, 0)$$ and $$(-c+\frac{d}{2},0)$$ to the point $$(x_e,y_e)$$ these lines will form an angle $$\theta$$ between them.

The question is:

For a given d, for what pair $$(x_e, y_e)$$ does the angle $$\theta$$ maximize?

My work so far

I use the law of cosines to find $$\theta$$ :

$$d^2 = s_1^2 + s_2^2 - 2 \cdot s_1 \cdot s_2 \cdot cos(\theta)$$

where

$$s_1 = \sqrt{(x_e - (-c + \frac{d}{2}))^2 + y_e^2}$$ and
$$s_2 = \sqrt{(x_e - (-c - \frac{d}{2}))^2 + y_e^2}$$

so $$cos(\theta) = \frac{s_1^2 + s_2^2 - d^2}{2 \cdot s_1 \cdot s_2}$$

I get that $$s_1^2 + s_2^2 -d^2$$ becomes:

$$2\cdot \left((x_e + c)^2 + y_e^2 - (\frac{d}{2})^2 \right)$$

and the full expression becomes:

$$cos(\theta) = \frac{(x_e + c)^2 + y_e^2 - (\frac{d}{2})^2}{\sqrt{\left((x_e + c - \frac{d}{2})^2 + y_e^2\right)\left((x_e + c + \frac{d}{2})^2 + y_e^2\right)}}$$

because $$y_e = b \cdot \sqrt{1 - (\frac{x_e}{a})^2}$$ it means that (for a constant $$d$$) $$cos(\theta)$$ is a function of only $$x_e$$

so finally I got:

$$\theta(x_e) = \cos^{-1}\left(\frac{(x_e + c)^2 + (b^2 \cdot (1 - (\frac{x_e}{a})^2) - (\frac{d}{2})^2}{\sqrt{\left((x_e + c - \frac{d}{2})^2 + (b^2 \cdot (1 - (\frac{x_e}{a})^2)\right)\left((x_e + c + \frac{d}{2})^2 + (b^2 \cdot (1 - (\frac{x_e}{a})^2)\right)}}\right)$$

$$x_e$$ goes from $$[-a,a]$$ and I think that $$d \le 2\cdot (a - c)$$ so that $$-c - \frac{d}{2}$$ lies within [-a,a]

Going back to the question posed, I have two sub-questions:

1) Is this the correct equation for $$\theta (x_e)$$ ?

2) If this equation is correct, would I just have to set the derivative equal to $$0$$ to find the critical points (i.e. $$x_e$$ which gives the maximum $$\theta$$) ?

Extra plot:

I plotted the above function $$\theta(x)$$ (with the given constants for a, b, c, and d and $$\theta(x)$$ is given in degrees) in python, where the red vertical lines show the x values of the foci ($$x = \pm c$$), and I got this:

what prompted me to ask this question is that it seems the maximum inscribed angle occurs at an x position which is not the focus $$x = - c$$, even though intuitively, I would have thought it would have been at that position. Did I just mess up in the derivation of $$\theta(x)$$ ?

No you did not mess up - although I have not checked your calculation.

The maximum angle $$\theta$$ results from two possibly competing factors :

1. For a fixed direction of the point $$E(x_e, y_e)$$ from the focus $$C$$, angle $$\theta$$ is largest when the distance $$EC$$ is smallest.
2. For a fixed distance $$EC$$, angle $$\theta$$ is largest when the direction of $$E$$ from $$C$$ is perpendicular to the axis on which $$d$$ lies.

The 2nd factor favours the point which you expected. However points to the left of that are closer so they are favoured by the 1st factor. A balance between these two factors determines where $$\theta$$ is maximum.

This problem has a simple geometric interpretation. The locus of points within the upper half-plane at which the segment of width $$2d$$ is subtended by an angle $$\theta$$, is a circular arc having the same endpoints as that segment, thus with its center on the perpendicular axis of the segment. The larger is the radius, the smaller is $$\theta$$. Hence the maximum value of $$\theta$$ corresponds to the smallest such circle intersecting the ellipse, with its center on the perpendicular axis of the segment (and in the upper half plane) and internally touching the ellipse.

A quick sketch made with GeoGebra, with $$a=1$$, $$b=1/2$$, $$d=(2-\sqrt3)/4$$, gives a maximum angle of about $$38°$$, corresponding to $$x_e\approx-0.96$$, which is in good agreement with your graph.

EDIT.

The above geometric setting can also lead to a quantitative result. If we parameterise point $$E=(a\cos t,b\sin t)$$ then a normal vector to the ellipse at $$E$$ is $${\mathbf n}=(b\cos t,a\sin t)$$, and we can write: $$C=E+s{\mathbf n}$$.

Parameter $$s$$ can then be found from $$x_C=-c$$ and to find $$t$$ we can write the equation $$CE^2=CD^2$$, where $$D=(-c-d,0)$$ is an endpoint of the given segment. The resulting equation for $$x=\cos t=x_E/a$$ is quite simple: $$e^2x^3-(1+2e^2-d^2/a^2)x-2e=0,$$ where $$e=c/a$$ is the ellipse eccentricity. Solving this with the given values ($$a=1$$, $$d=(2-\sqrt3)/4$$, $$e=\sqrt3/2$$) gives $$x\approx -0.959788$$ and $$\theta=\arctan(d/y_C)\approx 38.0735$$.

EDIT 2.

I realised only after writing my answer that I gave the given segment a width of $$2d$$, whereas at the beginning of the question its width is named $$d$$. And in computing numerical results I used a value for $$d$$ which is double of that given in the question.

• @OscarLanzi I don't understand your remark above: as far as I know, $\theta=\arctan d/\sqrt{4r^2-d^2}$, which is monotonically decreasing with increasing values of $r$. – Intelligenti pauca May 1 at 12:50
• That's true if $\theta$ is acute. But $\theta$ could be obtuse and then it's $\pi-\arctan(d/\sqrt{4r^2-d^2})$ instead. Both functions are monotonic for their respective domains, but in opposite directions. The same values of $d$ and $r$ will give both an acute angle by your formula and a supplementary obtuse angle by mine. – Oscar Lanzi May 1 at 13:01
• I took for granted that the center of the circle lies in the upper half plane. I'll state that explicitly. – Intelligenti pauca May 1 at 13:06
• So, why must the center be in the upper half-plane? You should explain that the point lies outside circle centered directly on the major axis, where $\theta=\pi/2$. That's what really makes $\theta$ acute in order to reach the ellipse. – Oscar Lanzi May 1 at 13:33
• @OscarLanzi You are right, but that should be self-evident. Intelligenti pauca. – Intelligenti pauca May 1 at 13:35