Parabolic Mirrors - Numberphile

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[@TomRocksMaths]

I'm going to claim we definitely planned to release this at the same time as Steve Mould's mirror video...

[@bergerniklas6647]

Don't forget alphaphoenix, also a very nice channel, released also a video at almost the same times about mirrors

[@deliciousrose]

So you guys..... _mirrored_ each other?

[@TomRocksMaths]

@@bergerniklas6647 thanks for the tip - subscribed :)

[@BrianRousseau]

Let's see your proofs.

[@aryst0krat]

@@bergerniklas6647 AlphaPhoenix and Steve Mould were actually collaborating though

[@phizc]

Is it the International Day of Reflection or something? Mirror videos from Numberphile, AlphaPhoenix, and Steve Mould, all within 12 minutes. 😃

[@user-yv5mt9rm3d]

Came to the comments to see whether it was just me cracking up

[@katakana1]

They all decided to mirror each other

[@swordfishxd-]

@@katakana1 hehe

[@darealpoopster]

Steve and Alpha both had a collab. Dunno about numberphille.

[@kjdude8765]

What a Convergence!

[@AttilaAsztalos]

One thing to note that the parabola maintains this focusing property for rays that are NOT coming in horizontally but at some angle - as long as the rays are parallel, the focus point continues to exist, but it moves off the zero axis in the opposite direction. Which is why those "satellite dishes" that have their receiver very obviously not on their centerline still work just fine - they're just not looking where they are apparently pointing... :)

[@replicaacliper]

where can i find a proof?

[@davidgillies620]

They're sections of a paraboioid. It's advantageous to have the LNB off-axis so the dish can be mounted vertically.

[@cedv37]

As a first approach, only relying on visualization; Parallel beams which subtend an angle with respect to the center axis of a bounded parabola greater than the tangent at one of the endpoints, will be covered out an arc of the parabola by the parabola's outside shape ("not hitting it from the inside"). I wonder what can be said about the focus in that case. I suspect no satellite dish (since they are quite wide, and since there would be wasted material) is designed to operate that way.

[@koenth2359]

This is only by approximation, and will only work in practice for small deviation angles. So it is not a mathematical fact.

[@tamasdhgebrq5968]

The greater the deviation from the axis of the parabola, the less true this is.

[@sameadslighting]

Most modern spotlights (especially in theatre) use ellipsoidal reflectors. Parabolic reflectors are still used in fixtures called PAR Cans, but ellipsoidal reflectors allow fixtures to become more light projectors, not just a source of light. Awesome video!

[@agmessier]

When Brady asked about a distant star, he and Tom remarked that the light rays are *almost* parallel. I always see it described this way, but I think it's a missed opportunity. The ideal shape for focusing a point (rather than a truly collimated) light source is an ellipse. Instead of saying the light rays are nearly parallel, you could say that the paraboloid is nearly elliptical, as the eccentricity approaches 1.

[@theinsanitypenguin]

Lots of mirror videos today

[@benwyckoff4165]

Right? My feed is mirror vids from here, Mould, and Alpha Phoenix back to back to back

[@amankadian1722]

Yessss. That's what I thought too. Weird.😅

[@dhayes5143]

Secret mirror collab 😮

[@banderi002]

Minor correction at 8:40: The vector from point A to point B is, counter-intuitively, point B minus point A, not the reverse! You can think of it in terms of "what displacement does A need to have to reach B" and you can keep in mind even in 1 dimensions, e.g.: To go from 3 to 2 ---> 2 - 3 = -1 ---> decrease by one!

[@tristanridley1601]

He said "doesn't matter if we're going in or out" so I don't think it's a correction. As long as you only care about the absolute value, it's moot.

[@ROMEMoussolini]

@@tristanridley1601 That claim is false in general when computing angles between vectors. Because in the video, he also mistakes the direction of the tangent vector, which he defined by (t,1) is actually pointing upwards and to the right since both x and y elements are positive. Considering that the drawing in the image shown at 9:28 illustrates a tangent vector pointing in the opposite direction, you would then compute a different angle between the two vectors than theta2 if the direction of vector L was actually going from point P to the Focus point (as Tom himself is shown drawing in the video). What is in fact computed is the vertical angle of Theta2, and not Theta2 itself, which can be confusing for someone not as well known with vectors and geometry. Small details like these can easily slip by, and if Tom did this intentionally or not, it is easy to make a mistake when not being aware of these things, or not having it explicated for the more unaware viewer.

[@tamasdhgebrq5968]

Conic sections and reflection (waves of light, sound, etc.) are an interesting topic. 1. The light of a point source in the center of the circle is reflected back into itself. 2. Light rays coming from one focus of the ellipse converge at the other focus. 3. Parallel rays of light going inside the parabola meet at the focus of the parabola. 4. Rays of light reaching the outer surface of the hyperbola and going towards one focus meet at the other focus.

[@obansrinathan]

1,2 and 3 ( maybe 4 too) all follow the same rule. Rays from one focus go to the other. It's most obvious with the elipse, but when you think about it, a circle is just an ellipse with both foccii on top of each other, and a parabola is just an ellipse with one foccii at infinity, which is the same as Hella far away

[@MagicGonads]

indeed they are all the same rule if you consider a general conic in the projective plane

[@unspeakablevorn]

Correction at 14:20: the focus of the parabola is actually much further in, about 1/4 of the way from the vertex to the shown point. this is why the angles are coming out unequal. Really, the focus is on the line that connects the two points where the mirror makes a 1/8 circle angle with the axis of symmetry.

[@michaelgibbons7014]

Tom is such a great math communicator. We love Tom in this house

[@wheresmyoldaccount]

The "Whispering Wall" dam in South Australia is an amazing example of the parabolic effect, in this case using sound, that I have experienced first hand. You can have a conversation over 100m as if the other person is standing right next to you - as in really next to you like maybe 1 or 2 metres away.

[@Sathrandur]

Thank you for that information. If I'm ever nearby in SA I would love to visit it.

[@jonathanross6260]

An interesting additional thing to prove, is that the path length of all the light beams to the focus is the same for all paths, so the signal retains coherence.

[@danielshamano3613]

Spot on! That's actually the most important thing he should have shown and not proving that those angels are the same, Physics has already told us so.

[@Fogmeister]

1:13 “All possible parabolas” There is only one true parabola! StandUpMaths would take issue with that statement. 😂

[@Zejgar]

Gloria in X-squaris!

[@flamencoprof]

I thought this, thanks for saving me from looking it up.

[@arcuscotangens]

I remember reading about this in my maths text book. I spent that day proving to myself that it is actually true. If memory serves, I used the derivative of the parabola to calculate the reflection of an incoming beam of light at any point, and found the intersection of all the reflected rays. Not a terribly elegant solution, but that's what I came up with.

[@aguyontheinternet8436]

Ah, that's quite clever. You think it would be easier to prove that all beams of light coming out of the focus will become a horizontal line when reflected, or that any horizontal beam of light will pass through the focus when reflected? They both effectively say the same thing, but perhaps one is easier to prove than the other.

[@toaster4693]

​@@aguyontheinternet8436That actually sounds harder than what OP did.

[@arcuscotangens]

@@aguyontheinternet8436 They sound about equally difficult, tbh. I believe by far the difficult part was to actually work out the reflection itself, and you'd have to do that in either case. I think I made things harder than it needed to be, because I didn't use the position of the focus point. I wanted to work it out as if I didn't have any prior knowledge of the solution.

[@adityavardhanjain]

I read the title, immediately paused all my work and grabbed my earphones. Some things just sound that interesting!

[@TomRocksMaths]

^^this

[@rkalle66]

It's not only the bouncing that angles into the focal point ... but the distance the wavefront of the beam is travelling is equal, too.

[@Bludgeoned2DEATH2]

That ink is so awesome

[@TomRocksMaths]

❤

[@alphamath3851]

Great video!! Now I can show my students why we learn about parabolas & where it is applied in real life

[@Graham_Rule]

I have vague memories of seeing a much simpler proof based on the construction of the parabola using a directrix and the focus. I think the reflective properties are almost part of the definition. But it is a long time since I've done any geometry.

[@tuppb]

Nice video! However, that LED flashlight (torch) is using *refractive* optics -- not reflective optics. It is not utilizing a parabolic reflector to focus its beam. Note that the LED emitter is located in the back of the "reflector" and not forward at a focal point. Also, notice that the reflector isn't even specular. Due to the structure and size of LED modules, there are few LED lighting fixtures that utilize parabolic reflectors.

[@obansrinathan]

Isn't the backing acting as a rough focus, to make the rays close enough to parallell that they can be focused neatly by the lense?

[@tuppb]

@@obansrinathan No. As I mentioned above, the "backing" is not specular, and the LED emitter is set in the back of the "reflector" -- not anywhere near where a focal point should be located. All of the focusing is achieved with the refractive lens in front of the LED.

[@bogdanieczezbyszka6538]

Exactly. This particular torch (as most LED torches do) uses a lens, not a parabolic reflector like incandescent bulb torches.

[@jondoolio]

What I love about math is that I was able to prove the same thing but through a different method. I feel like my method was a bit more complex compared to this, required differential equations, but it proved the same thing

[@Jodabomb24]

So this kind of derivation is used in optics to demonstrate that the optimal shape for the surface of a lens is also a parabola, for the exact same reason: all of the light rays coming in horizontally will be focused to a point, or equivalently a point source of light would be perfectly collimated. However, it's a lot easier to make lenses (and mirrors) that have a spherical shape instead of a parabolic one. That's usually ok, because a sphere looks like a parabola (or the 3d version, a paraboloid) as long as you're close to the center, which we of course know thanks to Taylor series and, every physicist's friend, approximating everything as a quadratic. However, if you start to reach the edges of the lens, which you need to do to create a really tight focus, then you start to see what we call spherical aberrations: that is, places where the shape of the beam no longer has the ideal shape you would expect from a parabolic optic because the optic is actually spherical. There are also special types of lenses called aspheres, which are manufactured with a different shape specifically to avoid these spherical aberrations.

[@yashrawat5071]

Interesting property of parabola , one of the most fundamental conic sections ! Also there is a nice result for ellipses which they have covered before !

[@UncleKennysPlace]

I made my first parabolic mirror more than half-a-century ago, and have made many since! I have three in progress now.

[@EveryWayWorks]

Always a good day when numberphile uploads :)

[@gleedads]

Very nice video, as always. This is one of my favourite channels on KZbin. Just a quick point on the connections to real-life devices. Yes, radio dishes, acoustic mirrors, solar ovens and so on often use parabolic dishes, for all of the reasons that Tom explains. But, the reflector in a typical torch ("flashlight" for colonials like me ;-) ) is probably not a parabola. First of all you don't actually want the outgoing rays of light to be parallel, because you want the beam to widen with distance. Also, the size of the bulb relative to the dish is rather large so it can't be well approximated as a point source and so even if the dish was a parabola you wouldn't achieve parallel rays. But most importantly, it is easier and cheaper to make the dish some other shape and so most manufacturers do (some are cones, some are "slightly rounded cones", some might be portions of spheres).

[@tuppb]

A lighting fixture with a parabolic reflector does not necessarily emit a collimated beam. There are plenty of light fixtures utilizing parabolic reflectors yield a range of beam angle focus from spot to flood. One notable example is the Maglite flashlight.

[@koenth2359]

The principle of least time, together with the definition of a paraboloid as the collection of points that have the same distance to a (focus) point and a plane, gives the result right away.

[@johnchessant3012]

Awesome! It would also be cool to do this derivation without already knowing that the answer is a parabola, i.e. instead of just verifying that this is true for the parabola, what if you had this focal point application in mind and you needed to find the curve that had that property?

[@martijn8554]

I imagine you'd approach it as: start with the focal point and a point on the curve. The derivative at that point should be half the angle of the line between the focal point and that point. Then integrate. You might need to handle the part to the left and right of the focal point separately.

[@elementalsheep2672]

Looks like Numberphile got Dereked by Steve Mould and AlphaPhoenix

[@miroslavzikic]

For someone who ground and polished his own parabolic mirror to make a complete Dobsonian telescope, this episode was quite familiar and satisfying to watch.

[@markherbert4723]

Always love a Tom video

[@TomRocksMaths]

❤

[@El_Girasol_Fachero]

¡Excellent animations!

[@Nethershaw]

Every time I heard "torches," what I processed what I heard was "tortures," and I had to keep reassuring myself that the subject of discussion was beams of light.

[@user-uy8yt7ku4w]

Another cool thing about parabolas: if you spin a liquid (for a example a glass with water), the surface of the liquid will form a paraboloid. That's actually how big lenses for telescopes are manufactured

[@annaclarafenyo8185]

A much better (and easier) proof is by using the 'infinite length ellipse' formulation, then the distance from focus minus the distance to the y-axis is constant to get this property, the extra distance to travel from the focus to a point P on the parabola is equal to the distance from P to a point at infinity x, or, equivalently, to a line parallel to the y-axis at infinite x. The least-time formulation of optics gives the equal angle law from this alternate formulation, this is Pascal's principle. The distance relation, in equations, says C + a x^2 = sqrt(x^2 + (f-ax^2)^2) where C is constant for all x. Setting x=0 gives C=f, and the relation becomes (f+ax^2) = sqrt(x^2 + (f-ax^2)^2). So we want to get (f+ax)^2 inside the square-root, so the two sides are equal, which means that we need to flip the sign of the middle term in expanding (f-ax^2)^2 so it becomes (f+ax^2)^2. That means f*a=1/4. The middle term in expanding (f-ax^2)^2, i.e. 2fax^2, becomes - x^2/2, and, adding back the x^2, it flips sign to +x^2/2, and then the thing inside the square root is (f+ax^2)^2, and the two sides are equal. This gives the focal length as 1/4a. This is the book proof.

[@Einyen]

The vectors L at 8:47 - 8:48 and T at 9:22 - 9:23 are drawn in the opposite direction on the drawing, since the y-values of the vectors are positive "2at" (with a and t positive) and 1, but the y-values of the drawn vectors are negative.

[@stevenpurtee5062]

It doesn't matter. The angle between the vectors and one vector and the opposite of the other is the same.

[@Einyen]

@@stevenpurtee5062 Yeah sure, the proof is fine, I just meant they were pointing the wrong way on the drawing compared to the vector coordinates.

[@prosamis]

Very cool!

[@pjoden]

Just a comment, but you don't get a focused beam out, you get a collimated beam. The rays of a focused beam will intersect at some distance.

[@Phaust94]

Did you sync it up with Steve Mould?

[@strixking]

And alpha phoenix?

[@oaxUK]

Years later and he's still at it.

[@tombouie]

Well-Done

[@frankharr9466]

That is very interesting indeed.

[@macronencer]

The observant will have spotted that taking the arccos of that value gives you another angle as well, which is the negative of theta 2 - if you look at the geometry, you will see that this represents a hypothetical ray that travels straight to the focus as if the reflector were not an obstacle.

[@stephendavies5968]

A simpler proof: Draw y=x^2, draw tangent from [x,x^2] to [0,-x^2] (because dy/dx=2x), draw reflection from [x,x^2] to [0,a], spot isosceles triangle (because 2 equal angles), and use equal sides to show a=1/4.

[@aamiddel8646]

Nice. The next time also explain that the length of the signal path to the focal point is the same (s arrive in phase). Or not? BTW if the focal point also receives a signal from the back this is slightly out of phase. (NB the LED in the flashlight is the wrong way..)

[@dvilardi]

This approach “only” proved that the parabola is one function that has this property, but how would you answer the generalized question “find all families of functions f(x) that have this parallel beam focusing property”?

[@wadehines9971]

There can be only one.

[@soranuareane]

There's a result that proves all parabolas are linear transformations of each other. There is only one "true" parabola. Standupmaths did a video proving this; search for "one true parabola" and you'll find it.

[@dvilardi]

@@wadehines9971 i know but what’s the proof for that statement?

[@wadehines9971]

@@dvilardi I haven't done a proof in 45 years but I imagine you establish the fact that the derivative of the a shape y=f(x) must point to a common focal point for all x and that the solution is the formula for parabola. I leave the details as an exercise for the student.

[@archivist17]

That was a lot of fun! 😀

[@adrianf.5847]

Of course, you can also use a standard parabola, like y = x²/b, and the point (0, b/4). The slope vector is simply the derivative of the parametrised curve (x, x²/b) by x, and then you can form the vector from the curve point to the focal point, and the downward vector you compare it to may then be the absolute value of the previous vector (which is an expression that doesn't need a square root) multiplied by (0, 1).

[@adrianf.5847]

I'm pretty sure that's how I'm going to memorise it.

[@TerranIV]

I always like to take the angle from the normal, not the angle from the tangent, just so you can keep your angle on the same side that the incoming and reflected wave are in. When you use the tangent you have to "measure" the angle THROUGH the parabola - which would be very difficult in a real-world situation!

[@markoo134]

Great vid

[@subliminalvibes]

Steve Mould with a video on parabolic saddle-shaped mirrors today too. 👍😎🇦🇺

[@jameswyatt1304]

It'd be interesting to see what happens with phase coherence as radius becomes some smaller fraction of a wavelength. More an audio than RF or light thought, but I'm hoping that makes sense.

[@midhunnellikattu]

Awesome

[@Nethershaw]

If there is a focusing error on parabolic surfaces due to parallax, what is the minimum distance an object would have to be from the parabolic mirror for that focusing error to be less than the diffraction limit of the mirror?

[@NyznTvfk]

this is really cool....

[@michaeldunkerton3805]

The question Brady brings up at 5:15 got me thinking (and apologies if this is either wrong or obvious). But the fact that it only works on beams coming straight at the parabola (ie, parallel to the x axis in this example) is actually a useful feature. He mentioned that a star is so far away that everything is basically coming in parallel. But if your intent is to amplify something close, you might *want* to avoid capturing beams that come from other angles because they come from another source. The example of a listening device: sound waves from your target are coming in parallel to x and get reflected onto a. But waves from another target aren't parallel and will scatter to a point other than a. So the parabola also serves to filter out sound from places other than your intended target, when dealing with close targets. Right?

[@SirBrian_]

This is exactly how rotating dish radars work. The parabolic antenna serves as a sort of "spatial filter," in that you only receive a signal from the direction you are pointing. There is a caveat; the filter is not perfect, and you do get some leakage from directions that are nearby to where you are pointing (otherwise known as beamwidth in radar and other contexts.)

[@gideonk123]

@Michael Dunkerton The solution is to use a shape of part of an ellipsoid, rather than part of a paraboloid. That is because an ellipse has 2 focal points, rather that a parabola which has only 1. To use it for filtering, as you propose, you place your sensor at one focal point, and the source of your signal at the 2nd focal point of the ellipse (actually an ellipsoid). Of course this assumes you know both the distance from source to sensor, as well as having control of the ellipse (ellipsoid) dish… If these assumptions are not met, it won’t work.

[@topilinkala1594]

You can prove it the other way around also. If there is no mirror the light will travel the same length horizontally and if it the case that they travel the same total length to the focus they must have same total lenght even withouth the mirror. So they will hit a plane. So to figure out what is the shape of the mirror where the distance of a fixed plane is same as the reflected beam to the focus. If you solve that you'll found out that itr is a parabola. By the way just like there is only one circle, there is only one parabola.

[@danitajaye7218]

You are super cool AND super brilliant.

[@joshwilliams7692]

You missed a great opportunity to actually build a parabolic mirror and set things on fire!

[@1st2nd2]

Wrong channel for such content. Steve Mould or ElectroBOOM might do such a thing.

[@la6mp]

But it is also of great importance whether the signals arrive in phase at the focal point or not, in principle they could all cancel out.

[@Veptis]

a TV satellite dish is also a parabola. But off axis. meaning it's just a section of the 3D paraboloid - where the receiver (and LNA) are in the focal point.

[@General12th]

Hi Dr. Crawford! Hi Brady!

[@mr_truex]

please do more video with Dr James Grime

[@tamasdhgebrq5968]

Conic slices also play an important role in celestial mechanics, especially regarding the movement of planets and other celestial bodies. There is a deep connection between reflection and the operation of gravitational fields, which in both cases may be related to the optimization of certain distances.

[@andrewdjang2080]

What would be really cool is a proof of this fact using synthetic geometry instead of analytic geometry and calculus. Archimedes cites some of these tangent properties in his work so they were clearly known prior to calculus.

[@skilz8098]

Are there any other curves such as hyperbolic, elliptical, or conic sections that would be satisfied by this as well?

[@milliams]

I'd be interested to see the reverse calculation: "we want a function which focusses light, what does it have to be?" - we should end up with a parabola.

[@aliam555]

It feels a bit backwards to assume a seemingly arbitrary solution (the parabola) and prove its validity by checking that the laws of physics, in this case the law of reflection or Snell's law, are correct. It would make more sense if this analysis came in as a second (verification) step where the first step is to answer the question "what is the shape that would focus all straight lines into a point" and arrive to the parabola as the answer.

[@Graeme_Lastname]

What about the phase of the signal? 🙂

[@romanburtnyk]

Wow. That's great. But now want to go vice versa, build curve which will focus all in single point :)

[@KyleDB150]

Is this the only shape that does this? If so, how is it proved? What if you allow multiple reflections before reaching the focus?

[@petrospaulos7736]

we need number theory videos!!!

[@jms019]

Should say something about the path remaining length constant keeping stuff in phase.

[@Firebird1005]

Lovely proof, but for maximum amplification the phase of the light or sound should be the same. Could you also prove that the path length is the same?

[@z4zuse]

Compound Parabolic Concentrators would have been a nice topic for the 2nd channel

[@stevenpurtee5062]

The result for cos(theta) was independent of a. That is very interesting.

[@renyhp]

well, that's only because the dependence on a was hidden in the x coordinate of the point of reflection

[@txikitofandango]

Just a slight thing troubles me here, you started out by saying you're going to show that if you want the curve to have the ray collecting property, then that curve must be a parabola. But you ended up doing it the other way, where you showed that if you have a parabola, And it's collecting the rays as you would like it to, then it must also be obeying the incidence law of optics. So you pulled a little switch there, am I wrong?

[@pratikgurjar7578]

Can someone explain me why proving the angles are same would prove that all the horizontal parallel rays will pass through the focus? Even if we dont take a parabolic mirror, any mirror has the property of reflecting light (rays) at the same angle with the tangent/normal line of the curve at that point.

[@jursamaj]

This is why parabolic mirrors are preferred for telescopes, over lenses. *All* light follows the same reflection rule, angle in equals angle out. Lenses suffer chromatic aberration, because different wavelength of light are refracted differently (which is what makes rainbows).

[@imkharn]

How to compensate if the focus is not a point but a sphere circle radius R?

[@schitlipz]

Nice to have this explained. Unfamiliar with the hat thing. And is it just a British thing to have the vectors written with the bar under them?

[@doomdoot6731]

Not just British, as we also do that in Germany. Bar (above or below, doesn't really matter) for vectors, two bars for matrices. Unless it is already predefined that you're working with vectors and matrices only, in which case you typically use lower case letters for vectors and upper case letters for matrices: a = b*C. If I'm not entirely mistaken, if you need to include a scalar with those notations, you then typically use a greek letter like e.g. lower case lambda.

[@schitlipz]

@@doomdoot6731 Yeah, thanks, kinda got that I guess. Not sure about the hat thing and its context here. Internet says something like "ball park figure"... I'm confused about the hat thing. I've been called a something-hat before, but I think I just feel like one now.

[@andrybak]

ISO 31-11 recommends using a right arrow on top of a letter for the notation.

[@user-gd8no2jr7l]

Is there something similar to this for the two foci of an ellipse? How does this concept apply to other conics?

[@y.herstein2199]

Yes, there is. If you put a point source in one focus of an elliptical mirror, it will focus in the other. This has several applications in optics as well.

[@thrillscience]

How does a retroreflector work?

[@leolennox1209]

Confused me at 1:00 he introduces a position at^2. Where does t come from?

[@daltongrowley5280]

but what about saddle shaped mirrors

[@hoschi49]

You can Show this in a Short way using fermats principle that light takes the shortest path.

[@GreggDurishan]

HOLY moly, there has GOT to be an easier way. I'm no mathematician, but can't you just do something like: a parabola is an ellipse with a focal point at infinity. ergo effectively parallel light (infinity) will reflect to the other focal point. done.

[@illesizs]

Is this a crossover episode with Steve Mould?

[@sebas9174]

I'm not a math expert but I think what was shown is that the angle of incidence is equal to the reflected angle. But I don't see that it has been shown that all reflected vectors point to the same point on the x axis. 🤔

[@DrLogical987]

Yeah... "Horizontal" if the x axis is horizontal. Buy, in general, the incoming rays should be *parallelle* to the x axis.

[@bahrammehrandish7699]

the proof does not need such a messy algebra and calculus, just consider a parabola as a ellipse where one of its pole lies in infinity then any rays comes parallel to parabola's axis crossing the corresponding elipse pole in infinity ,then it is enough to say that the ray reflects to another ellipse pole (which is correspond to parabola's pole or the focal point), because it is known as a main characteristic of ellipse, any ray starting from one pole encountering the ellipse reflects to the other pole as the tangent to ellipse in the encountering point is perpendicular to the angle bisector of ray's deflected angle and it has a completely simple geometric proof.

[@DaniMakes]

at^2 all over the equations, and you'd think this was a physics problem for something falling.

[@pink_plasticbag]

so that WWI giant-ears actually make sense?? that's amazing

[@pzelact4328]

i though parabola will be the line which will reflect a function like when u switch x and y but relatively about parabola line

[@simonf8370]

Tom, amazing delivery as always. I just love the fact we know the answer, but we had to wade through a bunch if equations that looked like a minefield.

[@TomRocksMaths]

[@scowell]

So, for a deep optical parabola, your sensor needs to be spherical? And the shallower your parabola, the more the sensor can be planar? For a spherical mirror, like Arecibo, your sensor should be linear, not a point.

[@rowanrobinson]

How can you measure if something is a parabola or paraboloid?

[@user-jp1qt8ut3s]

Next step: how do zernikes change the focus point?

[@krabkit]

this is the secret part 0 to the Steve mould and alpha phenix collaboration

[@thermidorthelobster4645]

Tom’s parabola drawing seems to have a pointy end.