Finding the normal or tangent line to an ellipse

A

a_guy_and_he_does_stuff

New member
Joined
Dec 13, 2019
Messages
6
Hello all, first time posting here. I'm currently chipping away at a rather tricky problem about mirrors and reflections, and I need help finding something - the equation of the normal/tangent line to an ellipse.

I have an ellipse with equation [MATH]k = \sqrt{a^2 + b^2} + \sqrt{a^2 + (AB - b)^2}[/MATH], where [MATH]k[/MATH] is some real number and [MATH]AB[/MATH] is the length of the line which connects the two centers of the ellipse.

Now, what I want to do is have a generalized equation that gives the vector function of the normal or tangent line to the surface of any ellipse of that specified form (OR the direction of that given line). My first attempt consisted of making the ellipse into a function of two variables and taking its gradient hoping to find the normal line to the surface (fresh out of multi-variable, first thing I thought of), and, though it produced a very interesting pattern, did not do what I wanted it to. It produced the gradient in reference to k.

I'm unsure how to proceed. My only other thought is the implicit function theorem or the simple, calculus 1 method for differentiating implicit functions, but I'm also unsure how to apply those two to this case.

I'm very tired right now, so I'll end it here.
 
I'm not sure I understand what you are doing.

First, you mention a surface; are you really talking about an ellipsoid (surface) rather than an ellipse (curve in a plane)?

Second, are your a and b coordinates (in place of x and y), or something else? Are your "two centers" the foci, or something else?

Finally, is there a reason you haven't reworked the equation into the standard form?
 
My terminology is unclear probably because I haven't actually learned about or worked with ellipses before.

a and b are coordinates in place of x and y. The two centers are the foci of the ellipse, and AB is, as follows, the distance between the foci. I'm referring to a family of ellipses described by the above function, where k and AB are semi-arbitrary values which, when both filled in, create an ellipse. When I referred to the 'surface' I mean to say the points which describe the ellipse, not an ellipsoid.

The reason that I haven't put the equation into standard form is because I don't know how to get to that point (while also keeping in mind that expressing it in the particular variables I am already using will make the final steps easier). I've built this equation from the ground up using Pythagoras - it's only by coincidence that this happens to form an ellipse.

What I meant to say, revised, is that I want to find the normal or tangent vector to the curve of form [MATH]k=\sqrt{x^2 + y^2} + \sqrt{x^2 + (AB - y)^2}[/MATH], where k is some value and AB is the distance between the foci. I want to express it in terms of those variables, but I have no idea how.
 
This is why we teach vocabulary in math classes, so it's possible to talk about something like an ellipse and be understood.

I don't have time at the moment to help directly, but I'll just ask: Have you tried looking up information on ellipses, so you can see what the equation of the curve is in a simpler form? And how much do you know about calculus that you will be able to apply to that equation? Finally, you mentioned that this is part of a larger problem involving reflections. It is entirely possible that if you told us that problem, we could suggest a quicker way to get where you want to go.

I'll try to get back to you when I can.
 
Okay. First, I will explain the problem and my approach to solve it. In simple terms, this is a problem about discovering if there exists a mirror such that all light starting from some given point will reach some other given point at the same time by reflecting off of the mirror.

What I found was that the only possible shape for such a mirror is an ellipse; this process was used to discover it:

The distance that light travels in a given time is assumed the speed of light in this problem, so the problem is now about distance as opposed to time.

Light starting at a point A reflects off of a point C and reaches a point B. Stipulating that line AB is the vertical axis, and that the line perpendicular to AB through A is the horizontal axis, we can make the following construction:

Annotation 2019-12-15 204757.png

(Variables [MATH]a, b, c, d, f, g,[/MATH] refer to the length of the lines they represent.)

On a given mirror, the length of [MATH]f+g[/MATH] is the distance light travels to reach the observer at point B. This distance must remain constant through all starting angles. As such, the equation [MATH]k=f+g[/MATH] is created to represent this relationship, where k is a constant stipulating the distance light travels from A to C to B.

Note that [MATH]a=c[/MATH], and [MATH]d=AB-b[/MATH].

Then the distance [MATH]f=\sqrt{a^2+b^2}[/MATH] and [MATH]g=\sqrt{c^2+d^2}=\sqrt{a^2+(AB-b)^2}[/MATH]By substitution, we find [MATH]k=\sqrt{a^2+b^2}+\sqrt{a^2+(AB-b)^2}[/MATH].

Graphically, [MATH]x=a[/MATH], [MATH]y=b[/MATH]. Substituting values for k and AB yields an ellipse.

However, the process which reaches this point does not guarantee that this surface is necessarily a mirror. There is no stipulation in the premise which suggests that the normal line for reflection is actually perpendicular to the tangent of the mirror. Essentially, this curve may or may not actually obey the law of reflection.

If the mirror obeys the law of reflection, the tangent vector to the curve must always be perpendicular to the vector which bisects angle ACB. Based on the construction, the direction of the bisecting vector can be determined using trigonometry (the method to do so of which I will go into upon request, as it is a bulky process).

If these vectors are orthogonal, then their dot product will be zero. However, I am not sure how to find the tangent vector to the curve. I could also find the normal vector to the curve and compare its direction to the vector which bisects ACB, which would be equally valid. However, I don't understand how to find that vector in terms of this implicit function.

In the meanwhile, I will try to put the function into standard form.
 
a_guy_and_he_does_stuff said:
Okay. First, I will explain the problem and my approach to solve it. In simple terms, this is a problem about discovering if there exists a mirror such that all light starting from some given point will reach some other given point at the same time by reflecting off of the mirror.

What I found was that the only possible shape for such a mirror is an ellipse; this process was used to discover it:

The distance that light travels in a given time is assumed the speed of light in this problem, so the problem is now about distance as opposed to time.

Light starting at a point A reflects off of a point C and reaches a point B. Stipulating that line AB is the vertical axis, and that the line perpendicular to AB through A is the horizontal axis, we can make the following construction:

View attachment 15490

(Variables [MATH]a, b, c, d, f, g,[/MATH] refer to the length of the lines they represent.)

On a given mirror, the length of [MATH]f+g[/MATH] is the distance light travels to reach the observer at point B. This distance must remain constant through all starting angles. As such, the equation [MATH]k=f+g[/MATH] is created to represent this relationship, where k is a constant stipulating the distance light travels from A to C to B.

Note that [MATH]a=c[/MATH], and [MATH]d=AB-b[/MATH].

Then the distance [MATH]f=\sqrt{a^2+b^2}[/MATH] and [MATH]g=\sqrt{c^2+d^2}=\sqrt{a^2+(AB-b)^2}[/MATH]By substitution, we find [MATH]k=\sqrt{a^2+b^2}+\sqrt{a^2+(AB-b)^2}[/MATH].

Graphically, [MATH]x=a[/MATH], [MATH]y=b[/MATH]. Substituting values for k and AB yields an ellipse.

However, the process which reaches this point does not guarantee that this surface is necessarily a mirror. There is no stipulation in the premise which suggests that the normal line for reflection is actually perpendicular to the tangent of the mirror. Essentially, this curve may or may not actually obey the law of reflection.

If the mirror obeys the law of reflection, the tangent vector to the curve must always be perpendicular to the vector which bisects angle ACB. Based on the construction, the direction of the bisecting vector can be determined using trigonometry (the method to do so of which I will go into upon request, as it is a bulky process).

If these vectors are orthogonal, then their dot product will be zero. However, I am not sure how to find the tangent vector to the curve. I could also find the normal vector to the curve and compare its direction to the vector which bisects ACB, which would be equally valid. However, I don't understand how to find that vector in terms of this implicit function.

In the meanwhile, I will try to put the function into standard form.
Click to expand...
By definition, sum of the distances of any point on the ellipse from its foci is constant. This property is used to draw an ellipse.

1576548608495.png

F1P1 + P1F2 = F1P2 + P2F2 = F3P3 + P3F2

But be aware that many mirrors are presently manufactured using this property.
 
As you may recognize from what you have looked up about ellipses, the equation simplifies more nicely if you define your coordinate system with the origin at the midpoint of AB; and the constants k and AB turn out to be what are traditionally called 2a (twice the major semiaxis) and 2c (twice the focal distance) respectively.

If we rename things so that your a becomes x, b becomes y, k becomes 2a, AB becomes 2c, the equation (still with the origin where you have it) will become after simplification

[MATH]\frac{x^2}{a^2 - c^2} + \frac{(y-c)^2}{a^2} = 1[/MATH]​

and we can define [MATH]b^2 = a^2 - c^2[/MATH]. This is easier to work with, and you can implicitly differentiate it and continue as you plan.

Or, you can look up "reflection property of ellipse" to find proofs ... this is nothing new!

Incidentally, this all works for an ellipsoid as well. (Look up "whispering gallery".) And the property can be proved without calculus; there is a long, detailed development of this here.
 
Okay, thank you! The source you provided has a good explanation of a far more clever way to prove this property. I won't finish with my differentiation, as the point of all this was to prove the property I was trying to investigate. Next time I run into a wall, perhaps I'll step back instead of pushing through it.
 
I was pondering whether it should be surprising that the ellipse has both the reflection property (all reflections from one focus pass through the other) and the equal-total-distance property, and realized that they are closely related. A reflection in a mirror takes the shortest path from one point, to the mirror, to the other point (just as light travels in straight lines, which are the shortest path). If every ray from one point, to a curve, to the other point is a reflection, then they must all be shortest paths, and therefore equal. There may be a way to prove the equivalence of these two properties directly.
 
If you mean to present the idea of proving this equivalence (as opposed to presenting the equivalence itself), then I can certainly see it being done. In fact, I decided to try myself, and I've got something here.

Trying to prove this is a bit wonky, because the properties are so closely intertwined that they trip over one another in the premises. So I had to define the properties more rigorously.

If we say that a curve has the equal-total-distance property, then the path from a fixed point to a point on the curve to another fixed point has the same length regardless of the point chosen on that curve. So, differentiating the reflection property, if a curve has the reflection property, then the path from a fixed point to any point on the curve to another fixed point is always a reflection, and therefore said path is always the shortest possible.

By using geometry, and a bit of cleverness, it should very well be possible to prove that each of these properties imply the other.

If we start with a very bold assumption, then the first half of this task is actually quite trivial! By previous result, it has already been proven that an ellipse has both the equal-total-distance property and the reflection property, meaning that, if one property implies an ellipse, and an ellipse implies both properties... well, the argument here is relatively easy to derive.

The backward direction is more complicated. Unless I'm mistaken, you can prove it through contradiction by thinking about the result: if an arbitrary point on the curve creates the shortest path between it and the fixed points, then all the paths from all the points must be equal. If we assume that they are not all equal, then there must exist some point on the curve which produces a path which is not equal to the others. But, this point cannot exist, because it must produce the shortest path - but it cannot, because if it does, then the other points do not produce the shortest path. This creates a contradiction.

In the forward direction, assume that there exists a curve and two points, [MATH]A[/MATH] and [MATH]B[/MATH], such that the equal-distance-property applies. Let [MATH]C[/MATH] be a point on the curve. Then by the equal-distance-property, the length of paths [MATH]AC[/MATH] and [MATH]BC[/MATH] sum to a fixed constant. Then the described curve is an ellipse. By previous result, the ellipse obeys the reflection property.

In the backward direction, we proceed by contradiction. Assume there exists a curve and two points, [MATH]A[/MATH] and [MATH]B[/MATH], such that the reflection property applies, but the equal-total-distance property does not apply. Let [MATH]C[/MATH] be a point on the curve. Let [MATH]k[/MATH] be equal to the length of [MATH]AC[/MATH] plus the length of [MATH]BC[/MATH]. By the reflection property, the path [MATH]A[/MATH] to [MATH]C[/MATH] to [MATH]B[/MATH] is a reflection, and [MATH]C[/MATH] is positioned such that the path [MATH]A[/MATH] to [MATH]C[/MATH] to [MATH]B[/MATH] is the shortest possible path.

Since the equal-total-distance property does not apply, then there must exist another point, [MATH]D[/MATH], on the curve, where the path [MATH]A[/MATH] to [MATH]D[/MATH] to [MATH]B[/MATH] has length [MATH]l[/MATH] equal to the length of [MATH]AD[/MATH] plus the length of [MATH]BD[/MATH], where [MATH]l\not=k[/MATH]. By the reflection property, the path [MATH]A[/MATH] to [MATH]D[/MATH] to [MATH]B[/MATH] is a reflection, and [MATH]D[/MATH] is positioned such that the path [MATH]A[/MATH] to [MATH]D[/MATH] to [MATH]B[/MATH] is the shortest possible path. Either [MATH]l>k[/MATH] or [MATH]l<k[/MATH], so we use cases.

Case 1: [MATH]l>k[/MATH]. Then the path [MATH]A[/MATH] to [MATH]D[/MATH] to [MATH]B[/MATH] is not the shortest path, because the path from [MATH]A[/MATH] to [MATH]C[/MATH] to [MATH]B[/MATH] is shorter.

Case 2: [MATH]l<k[/MATH]. Then the path [MATH]A[/MATH] to [MATH]C[/MATH] to [MATH]B[/MATH] is not the shortest path, because the path from [MATH]A[/MATH] to [MATH]D[/MATH] to [MATH]B[/MATH] is shorter.

...both of which are contradictions, giving us a grander contradiction as a result. So the equal-total-distance property must hold for any curve for which the reflection property holds (and also proven vice-versa). As long as the proof holds water, then a curve obeys the reflection property if and only if it obeys the equal-total-distance property as well, and additionally, we can extrapolate that only a curve with the properties of an ellipse can satisfy these requirements.

Thoughts?
 
You must log in or register to reply here.
Top