Talk:Ellipse

message: Here are some ways to find the focus (point F) and directrix (line through D) using compass and straight edge. Didn't see them mentioned in the article. I would create an animation but my software is not that good. ➧datumizer  ☎  10:47, 2 March 2021 (UTC)
replies: text: Your method is just a variation of the method shown in the picture of subsection "standard equation" and the fact, that a directrix and the corresponding focus are a pair pole-polar.--Ag2gaeh (talk) 13:21, 2 March 2021 (UTC)
replies: text: Maybe I am looking at the wrong image? I don't see any similarity. ➧datumizer  ☎  20:45, 3 March 2021 (UTC)

text: The article says, "C2 is called the circular directrix (related to focus {\displaystyle F_{2}}F_{2}) of the ellipse.[1][2] This property should not be confused with the definition of an ellipse using a directrix line below." So the two images describe different things. ➧datumizer  ☎  21:30, 3 March 2021 (UTC)

text: The image doesn't actually show how to plot the points geometrically though. Yeah it is possible to infer this. But is it maybe better to show instead of tell? Or is there a length limit on Wikipedia articles?  ➧datumizer  ☎  05:31, 6 March 2021 (UTC)
replies: text: If You agree, I add Your picture to section "pole-polar relation". But, it would be suitable to cut off not necessary parts and to add the lines containing the main axes. The usual construction of focus F (see picture) is more easy, so You should concentrate on the construction of point D and the directrix (orthogonal to line OD). --Ag2gaeh (talk) 08:48, 6 March 2021 (UTC)
replies: text: What parts are unnecessary? I don't understand. I have created a similar image for the hyperbola (see to the right) and would like to keep as many elements as possible the same between them. I also can create two separate sets of images for the ellipse and hyperbola: one set for constructing the focus, another set for constructing the directrix. ➧datumizer  ☎  06:00, 8 March 2021 (UTC)
thread_title:  Finding the foci and directrices geometrically 

message: This article is very difficult to understand. It seems to be written by mathematicians for mathematicians, on the basis that nobody else needs to understand it.  (Or perhaps the people who wrote it copied from books by mathematicians, but could not explain it simply because the editors did not understand it.)
:Analytically, the equation of a standard ellipse centered at the origin with width 2a and height 2b is:
: \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 .
It says what a and b are.  But it leaves x and y undefined.
The lead needs to explain simple concepts in simple language.  When concepts such a focal points (focii) are introduced they need to be explained clearly.  Equations should have their terms defined.-- Toddy1 (talk) 11:29, 6 August 2021 (UTC)
thread_title:  Article is difficult to understand 

message: Formula reads
F = A x_0^2 + B x_0 y_0 + C y_0^2 + a^2 b^2
But the a and b here are different from the a and b in the other equations - they should be doubled, no?
I found the formula to end with  4 a^2 b^2 .— Preceding unsigned comment added by 88.109.89.43 (talk) 23:21, 23 August 2021 (UTC)
thread_title:  Error in formula for 'F' - the intercept in terms of semimajor and semiminor? 

message: The section In Cartesian coordinates seems to be assuming b² = a² - c² without proving it. Should it be replaced with:
Ellipse as conic section
Let the ellipse be centered at origin with directrix x =  k, semi-major axis a, semi-minor axis b, eccentricity e and the focus at (c,0).
From the definition of conic section, 
\sqrt{(x - c)^2 +y^2} = e(k - x)
At (a, 0), 
a - c = e(k - a)
At (-a, 0), 
a + c = e(k + a)
These yield a = ek and 
c = ae = e^2k
At (0, b), \sqrt{c^2 +b^2} =ek
Or, b^2 =a^2(1-e^2)
Plugging back and squaring,
x^2 + e^4 k^2 - 2e^2kx + y^2 = e^2(k^2 +x^2 - 2kx)
x^2(1 - e^2) + y^2 = e^2 k^2(1-e^2)
\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 
Shubjt (talk) 06:20, 30 November 2021 (UTC)
replies: text: No, we do not need to derive the Pythagorean theorem here. Readers can be expected to know it and a proof is off-topic. —David Eppstein (talk) 08:19, 30 November 2021 (UTC)

text: I mean it isn't proven that the distance between covertex and focus is equal to the semi-major axis.  Shubjt (talk) 13:59, 30 November 2021 (UTC)
replies: text: Please, look at the diagram and the definition of an ellipse ! --Ag2gaeh (talk) 15:26, 30 November 2021 (UTC)
thread_title:  Proposed change 

message: In the section entitled "Arc length", E{z,1-a^2/b^2} is compared to E{z,m}, where m=k^2. Because a > b, 1-a^2/b^2 <0, which would require purely imaginary k??? I think something is wrong there...  — Preceding unsigned comment added by 2A02:908:1C10:F8C0:2933:DA59:F60:5BDE (talk) 19:51, 13 March 2022 (UTC)
replies: text: k is purely imaginary, but there's nothing wrong with that. See e.g. https://dlmf.nist.gov/19.7.E5 (Imaginary-Modulus Transformation). Also k\in i\mathbb{R} is just a matter of flipping the sign in \operatorname{E}(\varphi,k)=\int_0^\varphi\sqrt{1-k^2 \sin^2\theta}\, \mathrm d\theta, so you "work with" real quantities anyway. A1E6 (talk) 20:06, 13 March 2022 (UTC)
thread_title:  parameter in elliptic integral is negative??? 

message: the most easiest derivation for calculating the circumference of the ellipse you find in the apendix and leads to my new formula, which uses the "YNOT" constant:
 
All "YNOT"-formulas: 
 
first formula:     
P  ≈ 4 (a^y + b^y)^(1/y)  for   y=ln (2) / ln (pi/2) = 1.5349285.... (YNOT constant)
 
the second is my  formula: 
P ≈ 4a(1 + (b/a)^y)^(1/y)
 
and the third is: 
P ≈ 4a(1 + (1-e²)^(y/2))^(1/y)   
 
These three formulas lead to the same result. 
 
In my opinion all of these three fomulas should be published in wikipedia.
 
 
The most simple derivation of the astroid approximation for the circumference of the ellipse:
If you want to find out a formula of approximation for the circumference of the ellipse now, you need an elementary function, 
which proceeds as exact as possible on the frequency polygon. 
The frequency polygon develops when a' as x and b' as y of the ellipses which all have the circumference C=2pi are stripped of. 
By comparing this formula with other elementary formulas, you will recognize the similarity of the frequency polygon with 
an outside curved “bugly astroid”.
The general formula of the astroid with a length of the branch of the astroid of π/2 is:
y=((π/2)^z - x^z)^(1/z)             at which the “kurtosis“  is determined by z.
If you search the special astroid, which proceeds through the point (1/1), the kurtosis z must be ascertained at which the function y takes the value 1 although x=1. 
You get the value „bex“ for the kurtosis by inserting the values and solving to “Y”:
1=((π/2)^y - 1^y )^(1/y)      ->    1^y=((π/2)^y - 1^y)) ->    1^y=(π/2)^y - 1^y   ->     2^y=(π/2)^y  -> 
ln(2) = y * ln(π/2)    ->    y =  ln(2)/ ln(π/2)    ->    y = 1,5349285…  = konstant  
The equation for the astroid then is:
f(x)=((π/2)^y - x^y)^(1/y)      because    (π/2)^y  = 2  you get f(x)=(2-x^y )^(1/y)      
This is a nice elementary provisional result. 
The line  f(x) = b/a x strips on  the whole length the ratio of b/a
If you equalize the both equations, in the intersection you get the x- and  the y-value for the desired ratio for b/a  of the  given ellipse.
The formulas  are:
astroid-formula             =       line-formula
f(x)=((π/2)^y - x^y )^(1/y) =        b/(a ) x
(π/2)^y - x^y               =        (b/a)^y x^y
(π/2)^y                   =        (b/a)^y x^y + x^y
(π/2)^y                   =        x^y ( (b/a)^y+1 )
x^y                       =        (π/2)^y/((b/a)^y+1)
x                         =        (((π/2)^y / ((b/a)^y+1 )))^(1/y)   =    a'
f(x)                      = b/a * a'                                  =    b'
The values for x and f(x) are a` and b`of the ellipse with the U = 2π at the same time.
The circumference of the searched ellipse results by U = 2π a/a`.
By inserting of a’ results: 
U = 2π a /((  (π/2)^y  /  ( (b/a)^y+1 )))^(1/y) =   2π a /  π/2  / ((( (b/a)^y+1 )))^(1/y)
By reducing results:
U ≈  4 a (((b/a))^y+1)^(1/y)        for  the 2. solution you have to change a and b: 
U ≈  4 b (((a/b))^y+1)^(1/y)
as second „Y-NOT“-circumference-formula for the ellipse
with a relative deviation of max. 1,54%.
	
quod erat demonstrandum 
Greetings
Jürgen Beck, Germany
Becksjuergen (talk) 17:27, 13 June 2022 (UTC)
replies: text: Thanks for your efforts, but Wikipedia is not a venue to spread original research, see WP:NOR. - MrOllie (talk) 18:10, 13 June 2022 (UTC)
thread_title:  Umfang der Ellipse: Annäherungsformel 

message: I like to point out to people who say that the circumference of an ellipse is hard to calculate the following:
There is a formula for circumference of an ellipse in form C=p(a/b)(a+b) where a & b are the semiaxis. The function p(a/b) has its minima when a=b and it's value then is pi. So p(a/b) is easy to calculate for every value as pi is for the case a=b. So in fact the circumference of an ellipse is as easy to calculate as the circumference of a circle. It's just that for the one special case the value of the function has been calculated, but everyone who knows something about series can get a powerseries for p(a/b) to calculate it's values. I always also demonstrate this by setting b=1 so the circumfence is C=p(a)(a+1). It is easy from this to see that p(a) has maximum when a=0 or infinite and then p(a)=4.
Linkato1 (talk) 09:37, 4 June 2023 (UTC)
replies: text: Where do you get this?  Where is the information necessary to derive a power series? —Tamfang (talk) 07:14, 24 June 2023 (UTC)
replies: text: Take the first integral you see on the page for circumference. Substitute the e as expressed by a & b. Divide the integral by (a+b). Rearrange and you get an integral where the parameter is a/b or its inverse, this is p(a/b). Then just look up how integral can be set as a power series. Insert your integrand and you have your powerseries. Linkato1 (talk) 10:10, 29 June 2023 (UTC)
thread_title:  Circumference 

message: The new edit https://en.wikipedia.org/w/index.php?title=Ellipse&diff=next&oldid=1220981169 is practically incomprehensibly worded. I haven't undone it, because I don't want to discourage its author, who, sad to say, doesn't use a chosen user name. If the wording is not made clearly comprehensible soon, I will undo it. I would ask its editor to choose and use a Wikipedia user name. Generally, he should choose it to give himself privacy and security. It is just good manners to do so.Chjoaygame (talk) 23:55, 24 May 2024 (UTC)
replies: text: It might be better to make a whole separate section about dynamical billiards in an ellipse. There are numerous sources about this topic, e.g. see google scholar. I'm not an expert though, and don't really feel motivated to write it. –jacobolus (t) 02:39, 25 May 2024 (UTC)

text: To save people the bother of clicking the diff, the text added was "" A diagram would be very helpful but I don't think incomprehensible is right. Johnuniq (talk) 03:35, 25 May 2024 (UTC)
replies: text: It is not obvious that the focus-to-focus property causes convergence to the major axis! —Tamfang (talk) 19:22, 25 May 2024 (UTC)
replies: text: Billiard paths passing outside the foci end up indefinitely bouncing around and around with their envelope as a confocal ellipse. In the limit as you move the path toward passing through one focus, this ellipse approaches the straight line segment between the foci. –jacobolus (t) 21:34, 25 May 2024 (UTC)
thread_title: More on the focus-to-focus section