Generating Conic Sections with Circles | Part 2. The Parabola
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- Опубліковано 10 вер 2024
- Strengthen your problem-solving skills at:
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Let C be a circle centered at F and let L denote a line on the same plane as C which doesn't intersect C. Then construct a variable circle tangent to L and C and denote its center as X. A collection of all such possible centers X is a parabola.
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You know your day is going to be good when you see that think twice uploaded
visual: 10/10
audio: 10/10
Your proofs are always a pleasure!
Looking forward for your next one. :)
Thank you!
@@ThinkTwiceLtu it’s more like 1003/10 visual and 1004/10 audio
In the special case where a=0, the centers of all circles that intersect the focus and are tangent to the directrix form a parabola too.
I want this channel to flourish to millions of math geeks.
The pauses between the steps are so well-timed, and you fill them with such neat illustrations. Nothing like educating using the best tools of story-telling! Keep up the good work.
my 9th grade math teacher used to show us your videos they were pretty great
if you're reading this, hello Mr. Savaadra
Hey, you finally uploaded after 4 months :)
yes, it took a while :D Had a very busy semester:/
@@ThinkTwiceLtu oh no! Don't worry 🥰🥰🥰
@@ThinkTwiceLtu which year r u in? just asking :)
@@ThinkTwiceLtu how do you make these videos ? What do you use ? ✨
I too have always wanted to make animations that would help explain concepts 🥺
I appreciate this as a gift to humanity. Thanks man, please keep making these.
Thanks for watching, I'm glad you like it:)
I'm likely missing something, but I don't see the purpose of L or the circles of radius a. You can start with F and d, select a point M d, construct a perpendicular to d at M, construct the segment FM, make a perpendicular bisector of FM, and label the point X as in the video.
This is exactly the same as the construction in the video, but with a radius of zero for the red circles, but it simplifies quite a bit.
In any case, excellent video!
9 years after being introduced to quadratic equations, I finally know what a parabola is!
Clear and concise. I love these visualizations.
Thank you!
so amazing...making me think of days back in my high school...when I sat in that math class, I literally didn't understand those difficult things...fortunately, math never loses its charm in my heart, and my love for it never ends for a second...and now I see this incredible video, I fell in love with it again. thanks!
I am always waiting for your math animations.
Appreciate your support
Love the background music too
Hey TT! =D
Hello:)
Hi i am Mathematican this is my number theory (1) 21-12=9 (2) 321-123 = 198 = 9 + 189 (3) 4321-1234 = 3087 = 198+ 2889 = 9 + 189 + 2889 (4) 54321 - 12345 = 41976 = 3087 + 38889 = 198 + 2889 + 38889 = 9 + 189 + 2889 + 38889 (5) 654321-123456 = 530865 = 41976 + 488889 = 3087 + 38889 + 488889 = 198 + 2889 + 38889 + 488889= 9 + 189 + 2889 + 38889 + 488889 (6)7654321-1234567=6419754=530865+ 5888889 = 530865 + 41976 + 488889 = 3087 + 38889 + 488889 + 5888889 = 198 + 2889 + 38889 + 488889 + 5888889= 9 + 189 + 2889 + 38889 + 488889 + 5888889 (7)87654321-12345678=75308643=6419754 + 68888889 = 530865+5888889 + 68888889 =41976 + 488889 + 5888889 +68888889 =3087 + 38889 + 488889 + 5888889 + 68888889= 198 + 2889 + 38889 + 488889 + 5888889 + 68888889= 9 + 189 + 2889 + 38889 + 488889 + 5888889 + 68888889 (8)987654321-123456789= 864197532 =75308643 + 788888889 =6419754 + 68888889 + 788888889= 530865+5888889 + 68888889 + 788888889 = 41976 + 488889 + 5888889 +68888889 + 788888889 = 3087 + 38889 + 488889 + 5888889 + 68888889 + 788888889= 198 + 2889 + 38889 + 488889 + 5888889 + 68888889 + 788888889 = 9 + 189 + 2889 + 38889 + 488889 + 5888889 + 68888889 + 788888889 plz tell me which website i publish this ??
@@bopaliyaharshal2399 🙄
@@bopaliyaharshal2399 can you also explain a bit?
@@ThinkTwiceLtu You should do a visual proof for the area of a regular octagon. I worked out its area is the height squared minus the side length squared. This is because if you imagine a square and you subtract 4 congruent triangles from the corners, then you have the area of the octagon left over, and it works out that the area of the 4 triangles combined is equal to the side length squared!
Beautifully explained with such clean animations. Loved it ! I want this channel to explode in growth soon :)
6:35
in case anyone missed it
3blue1brown is also a patron!!! 😊😊😊
Thank you for uploading this amazing content
Hey, youtube need more of this content, just keep it up, you are simple amazing, don't let anybody make you think otherwise, we all support you
This video is hypnotizing. Great work aa always!
I wonder who may have thought like this. Far beyond amazing
I have this really cool proof of making the circle out of circles
Aii , Atlast the part 2 is out!! Nice video!
Thanks:)
I love this channel more with each day i see the same video
The animations are always so elegant, great work
Thank you very much
this is incredible!!! Brazilian hugs for you!
Thank you:)
Man, this is sooooo satisfying
#suggestion Post more videos plz. your channel is so far the best channel in every way on YT. Plz post more
Love the videos, never stop!
Thanks for the support:)
Wow! Astonishing!!! Thank you so much!! 😃
Thanks for watching :)
I LOVE all your content but i must admit that i didn't find this result as amaizing as others or the need to introduce more circles than necessary to get the parabola
But i supose that as an introduction to the topic it's great
Can we get the background music it is really nice ☺️ as the visualisation become easy after seeing the videos of yours
Awesome video... as usual.
Thanks!
Can you do a video on 4D? It would be very interesting to see what you could make of them
Love it
Thank you
You make these animations in processing right? Can you share some part of the code. I really want to see how you it, cause I can't figure it, especially the actual animation part.
That's looking satisfying👍👍👍👍👌👌👌👌. Very good
Uhhh, what's the point of having length 'a' and circle 'c'? If a=0, you just have a circle tangent to L (which is 'd') and intersecting point 'f' and you're done. Sure it's a shorter video, but did you make this unnecessarily complicated on purpose?
True. But this series is all about generating conics from circles. :)
beautiful vid, and nice ad at the end too
Thank you:)
Hello....could you recommend some books that helps strengthen intuition like the ones which you use for your videos?
If I had you as a math teacher in high school, I probably would have been a math major
was waiting
awesome video
He’s finally back!
Wish this wasn’t so slow in places. Really dragged on during a few statements.
Also, is the path traced by the center of any circle tangent to another fixed circle and straight line always going to form a parabola? Was the initial construction even necessary?
it was a way to construct that middle circle using classical geometry only
just saying, the x1.25 button exists
How do you make these masterpieces???
What software did you use to create the animation?
Awesome keep going
Nice
:))
Which software do you use for animation?
Love the video, but you said "choose an arbitrary point X", but given the line and the focus, and xf = xd, all x's are determined
What's the equation of the parabola X in terms of F, a, and L? It's been too long since I took algebra. Also, what if we let L be formed by two arbitrary points not intersecting C? How do we generate the implicit form of the rotated parabola given F (in terms of Fx, Fy), a, and L (through L1x, L1y, L2x, and L2y)?
How were you able to do these animations??
Awesome video; very clean and smooth! What program did you generate this with?
Thank you. I am using Cinema 4d:)
Now show how that relates to: f(x) = ax^2 + bx + c for all quadratics!
What are you using to animate?
How do you make your videos, do you use 3b1b's manim library or do it by hand or some other tools?
I mainly use a 3d graphics software called cinema4d.
@@ThinkTwiceLtu new video soon?
The exact same question of locus was asked in jee mains
Nice background musictrack
5:00 wouldn't be easier let a->0 or a=0?
in the video of fermat point you gave a challenge for viewers to find fermat point of a quadrilateral and generalize it for n sided polygon can youplease please make a video on how to do that
Superb content
Thank you ☺️
hell yeah, maths
Too good 👍🏻
Why circles of radius a ? You don't need these. Take a =0, and F=d.
True. But this series is all about generating conics with circles. :)
@@Monochrome_math
Ok, but it doesn't. In the end, you take the perpendicular bisector anyway, which is what gives the parabola. You took a circle of radius a, but you could have taken 20 consecutive adjacent circles the same way, and still, in the end, needed the same exact perpendicular bisector.
@@piwi2005 Alright I carefully watched the video again. So, yes I agree that drawing circle Q would be pointless. We can simply construct perpendicular bisector and move the perpendicular line. However, lets look at where did he came up with such construction in the first case. Lets assume circle with some radius 'a' (I will come back to it why you need to do so). Now consider ALL the circles which are tangent to circle and line. And plot all the centers of such circles. Now all those points will form a parabola. Now how do we construct such a circle? Thats what this video shows. This video shows how to construct such circle. Now, looking at the steps, you may say that constructing the circle at last is useless, but that ruins the main essence which this video is trying to convey. I.e. Centers of all circles tangent to circle with focus as its center and tangent to line segment will form a parabola. Now as for why not simply take 'a' (aka radius) as 0? This is important coz it can be cleverly used to change the shape of parabola. In case of a = 0, that would mean constructing circles tangent to line and focus containing on the circumference of the circle. Yes, it would still be possible to create the parabola but you won't have that same flexibility to change the parabola as you would have if you were to change 'a'.
Correct me anywhere if I am wrong. Hope that clears things up.
@@Monochrome_math
I agree that the video shows a theorem with circles and parabolas. However, it does not construct a parabola with circles. It construct first the parabola, with the bisector, between F and d(="L-a"), then it constructs a circle from the parabola which is tangent to the first two circles. One could "construct" a parabola in such way only if you already had a set of circles that you could glide on the plane, make them tangent somehow with the circle and F, then mark the center.
4:20 interestingly, the directrix construction is just the case where a=0.
Aaaaye lets gooo!
Can you talk about the 4th dimension
💜
Is more short join F to K. Are unnecessary the circles and L is automatically the directrix.
instead of taking segment FM and perpendicularly bisect it, if you use FK, you obtain a parabola (with L directrix) and the circles are useless...
circle at X is not intersecting perpendicular bisector
maths makes cool imagery
There is only one true parabola! Gloria in x-squaris!
phonon has joined the chat
Anyone played Euclidea? It's level 13.1 there. :)
When are going to upload man? 5 months and zero videos.. we like your videos
Sorry, I had a very busy spring semester. I'm almost done with part 3, it should be out sometime this week:)
@@ThinkTwiceLtu cool! I'll be eagerly waiting to see more from you
What is the benefit of doing this way?
Just fun
@@ThinkTwiceLtu oh, though it was somehow easier or faster to calculate programmatically but couldn’t figure out how adding more powers of two would do it 🙂. Thanks for response.
this isn't the definition of a parabola that i was taught and i don't know how to convince myself that they're the same
I dont either but if you want to give It a try this may help:
ua-cam.com/video/pQa_tWZmlGs/v-deo.html
You could do it much easier...
You could build a circle tangent with f and d...
XD?
the circle method is just a really contrived and boring way of demonstrating the definition.
I find it cool tbh. Dont know abt u.
@@Monochrome_math I prefer thinking about parabolas as cross sections of the 3D surface created by revolving a line in imaginary space.
@@executive Thats why its called a conic section. Revolving a line would create a cone and cross-sections of such cone would be called as a conic section. But we need not to define it that way. Again, whatever is best for u. 👍
@@Monochrome_math thank you for re-explaining what I already said.
Wtf, already seen this one. Smh
Jk ily
@@oliot4814 :))