Talk:Ellipse

Finding the foci and directrices geometrically


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. &#x27A7; datumizer   &#9742;  10:47, 2 March 2021 (UTC)


 * 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)
 * Maybe I am looking at the wrong image? I don't see any similarity. &#x27A7; datumizer   &#9742;  20:45, 3 March 2021 (UTC)
 * 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. &#x27A7; datumizer   &#9742;  21:30, 3 March 2021 (UTC)

I mentioned the picture in subsection "standard equation". Any construction of a focus uses the equation $$a^2=b^2+c^2$$ and a right angular triangle, Yours too. Your construction of one point of the directrix line is based on the fact, that the points $$(c,0), (\tfrac{a^2}{c},0)$$ are inverse with respect to circle $$x^2+y^2=a^2$$. This is equivalent to a pole-polar relation (see section of this article) with respect of this circle.--Ag2gaeh (talk) 08:06, 4 March 2021 (UTC)


 * 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? &#x27A7;  datumizer   &#9742;  05:31, 6 March 2021 (UTC)


 * 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)


 * 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. &#x27A7; datumizer   &#9742;  06:00, 8 March 2021 (UTC)

With "unnecessary" I meant just "white space" to shrink the picture a little. For the hyperbola it is the same game: F and D are inverse with respect of the green circle. In both cases You should add the lines of the great main axis, on which the points F,D,A lie. Please insert Your pictures together with some explanations in a suitable subsection.--Ag2gaeh (talk) 14:34, 8 March 2021 (UTC)

Article is difficult to understand
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)

Note on the above: x and y are the most common axes in Cartesian coordinate geometry, and their nature of up/down and left/right applies here as well. You do not need to be a mathematician to understand it, but you do need early secondary school mathematics. I agree that it would be nice if the axes were labelled, but it's a very safe assumption in the absence of other data.

However, as an engineer, I find that the most common use of the pronumeral 't' regarding conic sections is for time. As such, the section on #Parametric_representation was slow to decipher. Is it acceptable to change the pronumeral to 'phi' or similar, to acknowledge that this is not the common angle 'theta', but a different angular variable? JAyerson (talk) 09:30, 23 February 2022 (UTC)

Error in formula for 'F' - the intercept in terms of semimajor and semiminor?
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)

Proposed change
''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)
 * 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)


 * 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)


 * Please, look at the diagram and the definition of an ellipse ! --Ag2gaeh (talk) 15:26, 30 November 2021 (UTC)

parameter in elliptic integral is negative???
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)


 * $$k$$ 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)

Umfang der Ellipse: Annäherungsformel
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)


 * 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)

General ellipse formulas
I consulted one of the sources referenced for the General ellipse but could not find those formulas. In particular I'm concerned with the formula for the cotangent of the angle $$\theta$$. Is it possible there are additional solutions, or that there are typos in a couple of signs? I'm referring to the first case of the last equation, for which during verification I obtain the following:

$$\theta = \arccot\dfrac{-C + A -\sqrt{(A-C)^2+B^2}}B $$

rather than

$$\theta = \arccot\dfrac{C - A -\sqrt{(A-C)^2+B^2}}B $$

--Gciriani (talk) 16:01, 25 May 2023 (UTC)
 * This additional link shows a much simpler equation, which gives results in agreement with my numerical calculations.

$$tan(2\theta) = \dfrac{B}{A-C}, with {} 0<\theta <\pi/4 $$

--Gciriani (talk) 20:43, 25 May 2023 (UTC)


 * I think you want
 * $$2\theta = \operatorname{atan2}(-B, C-A),$$
 * where atan2 is the 2-argument arctangent. Or if you want to avoid that, you can instead do:

\theta = \arctan\left(\frac{-B}{\sqrt{(C-A)^{2}+B^{2}}+C-A}\right) = \arctan\left(\frac{\sqrt{(C-A)^{2}+B^{2}}-\left(C-A\right)}{-B}\right) $$
 * With the caveat that these can break down when $$B = 0$$ (see atan2 for details).
 * Here is a Desmos plot: https://www.desmos.com/calculator/3xjo0omput –jacobolus (t) 03:12, 26 May 2023 (UTC)
 * Thank you for simplifying the theta formula with atan2. Several years ago when I added this general -> canonical formula subsection, I originally had:
 * $$\theta = \arctan \frac{C - A - \sqrt{(A-C)^2 + B^2}}{B}$$
 * with special cases for 0 and 90 degrees. Various people changed the theta formula over time, leading to the contangent form with even more cases.  This atan2 form is much simpler.
 * Now I see another simplification. Looking at the canonical to general formulae above, it would appear that a and b should depend only on A, B, & C and not on D, E or F, and in fact, it's true.  If you substitute the equation for F into the a,b equation, then substitute the x0 and y0 equations and simplify, D and E fall out and you're left with only:
 * $$a,b = \sqrt{\frac{1}{2} \left(A + C \pm \sqrt{A^2 + B^2 + C^2 - 2 A C}\right)}$$.
 * I propose changing the formula to this. I'm surprised no one else noticed this or otherwise edited the a,b formula in all the time this has been up.
 * The one question I have is what to do with the reference. There is a reference to a Wolfram document that has similar (but not exactly the same) formulae to what I wrote containing D, E & F.  I don't know of a reference giving the simpler form.  Any ideas where to look? Pdx scooter (talk) 07:59, 11 January 2024 (UTC)
 * You'll want to double check your formula. This version gives numerically different answers than the previous version. –jacobolus (t) 00:36, 14 January 2024 (UTC)
 * Note that changing any of D, E, or F (translates and) scales the ellipse, which has the effect of also changing a and b. It's plausible that there's a clearer way to write the formula though. –jacobolus (t) 00:42, 14 January 2024 (UTC)
 * I see the problem -- it's that the general formula coefficients are not unique, as you can scale them all by an arbitrary non-zero real constant without changing the result. My original a,b formulae are invariant under scaling of all of the constants, but the proposed simpler versions are not.
 * What led me to proposing the alternate versions is that if you substitute them and the theta equation back into the A,B,C equations above, then use the half-angle cosine and sine functions: $$\{\cos^2, \sin^2\}\left(\tfrac12 \operatorname{atan2}{(u, x)}\right) = \tfrac12 (1 \pm \frac{x}{\sqrt{x^2 + y^2}})$$ to eliminate the trig functions then simplify, the A,B,C equations do reduce to A=A, B=B, and C=C. But that's assuming no scaling constant. Thanks again for your input. Pdx scooter (talk) 04:05, 14 January 2024 (UTC)
 * Feel free to use https://www.desmos.com/calculator/goocxro2gg to check these. –jacobolus (t) 00:46, 14 January 2024 (UTC)

Circumference
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)


 * Where do you get this? Where is the information necessary to derive a power series? —Tamfang (talk) 07:14, 24 June 2023 (UTC)
 * 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)

Focus-to-focus reflection property
The angle bisector theorem applies to a triangle and LQF_1 is not necessarily a straight line. It seems to me that the proof should be corrected by adding a point Q' such that LQ'F_1 is a straight line and then using the fact the PQ' is a bisector to claim the same ratio for QL/QF_1. Mba14 (talk) 06:16, 3 August 2023 (UTC)


 * If $$PL \cong PF_1,$$ then the angle bisector of $$\angle LPF_1$$ is the same as the perpendicular bisector of the segment $$LF_1.$$
 * Frankly, this proof (and some others on this page) is already more detailed than it needs to be for the purpose of a Wikipedia article of such broad scope. I would cut it to just:
 * Because the tangent line is perpendicular to the normal, an equivalent statement is that the tangent is the external angle bisector of the lines to the foci (see diagram).
 * Let $$L$$ be the point on the line $$\overline{PF_2}$$ with distance $$2a$$ to the focus $$F_2$$, where $$a$$ is the semi-major axis of the ellipse. Let line $$w$$ be the external angle bisector of the lines $$\overline{PF_1}$$ and $$\overline{PF_2}.$$ Take any other point $$Q$$ on $$w.$$ By the triangle inequality and the angle bisector theorem, $$2a = \left|LF_2\right| < {}$$$$\left|QF_2\right| + \left|QL\right| = {}$$$$\left|QF_2\right| + \left|QF_1\right|,$$ therefore $$Q$$ must be outside the ellipse. As this is true for every choice of $$Q,$$ $$w$$ only intersects the ellipse at the single point $$P$$ so must be the tangent line.
 * –jacobolus (t) 17:46, 3 August 2023 (UTC)

More on the focus-to-focus section
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)


 * 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)
 * To save people the bother of clicking the diff, the text added was "Furthermore, due to the focus-to-focus reflection property of ellipses, if the reflected rays are allowed to continue propagating, they will eventually converge along a single direction along the major axis. Specifically, the rays will travel from one point on the major axis to another point on the opposite side of it." A diagram would be very helpful but I don't think incomprehensible is right. Johnuniq (talk) 03:35, 25 May 2024 (UTC)
 * It is not obvious that the focus-to-focus property causes convergence to the major axis! —Tamfang (talk) 19:22, 25 May 2024 (UTC)
 * 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)

“c^2 = a^2 - b^2”?
In my humble opinion this article seems to omit an explanation for the key point that

$$c^2 = a^2 - b^2$$

Note the following is stated w/o a preface :


 * Linear eccentricity This is the distance from the center to a focus: $$c = \sqrt{a^2 - b^2}$$.

Also e.g., In the following figure the hypotenuse is labeled as “a” with no explanation:

[[File:Ellipse-param.svg|thumb|Shape parameters:a: semi-major axis,

b: semi-minor axis,

c: linear eccentricity,

p: semi-latus rectum (usually $\ell$).]]

(It would seem to me that a simple explanation could make use of the quoted assumption that “the ellipse is the set of points such that the sum of the distances PF1+PF2 is equal to 2a”, applied to the special case in the above figure in which the hypotenuse is clearly half of PF1+PF2.)

Apologies if I am missing something obvious! 😞 JdelaF (talk) 19:29, 11 June 2024 (UTC)

JdelaF (talk) 19:40, 11 June 2024 (UTC)


 * The relation $c^2 + b^2 = a^2$ is the Pythagorean relation between sides of a right triangle. There's some explanation of the meaning of a in . Feel free to add a couple more sentences someplace if you think the current explanations are too compressed. There's some trade off between concision vs. explicitness, but we don't want to leave readers behind. –jacobolus (t) 20:07, 11 June 2024 (UTC)