What does the focus point of ellipse signify?

Question

If one focus of the ellipse is the point where the axis line of the conics and the intersecting plane intersect what is geometrically at the other focus point

ellipse as a conic section:

Focus point as the point where axis of conic and plane intersect:

I know the defination of ellipse but I am just curious to ask that how this defination came from the fact of ellipse being a conic section, focus point as the point where the plane intersects one cone out of the two double napped cone(as shown in second image)

But if we are defining focus point in this way, we are defining just one focus point. So where did the other focus point came from?--1

Where do we show the other focus point in the conic section diagram of ellipse?--2

And then how can we prove the defination of ellipse from the known facts about focus point and ellipse as a conic section, that the sum of distances from the focus points to any point on ellipse is constant?--3

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. — Community, Dec 19 '24 at 08:34
Take a look at the concept of Dandelin spheres. See also this ancient answer of mine that uses a Dandelin sphere on the way to relating the eccentricity of a conic with angles associated with the cone and cutting plane. — Blue, Dec 19 '24 at 08:56
Plucker defined foci generally for plane algebraic curves, to be the points at which lines from the circular points at infinity are tangent there. There are $(d^)^2$ of which $d^$ are real, where the class $d^*=d(d-1)$ ($2$ for conics). See this answer of mine. — Jan-Magnus Økland, Dec 19 '24 at 09:07
Like @Blue I wanted to bring up Dandelin spheres. Observe that neither focal point is on the axis of symmetry. The focal points are where the Dandelin spheres graze the intersecting plane. — Jyrki Lahtonen, Dec 19 '24 at 10:00
@Blue, The concept of Dandelin spheres is the most mesmerizing concept I have seen this month. While reading your ancient answer I didn't understand how PT in the image is equal to twice of c. Can you please clarify it for me? https://i.sstatic.net/q0sSU.png — Yug Ahuja, Dec 19 '24 at 10:47
@YugAhuja: Dandelin spheres are truly remarkable (hence, my remark)! :) ... To your question about my previous answer: In this image of mine, we know $|PF|=|PQ|$ and $|P'F|=|P'Q"|$ (the two tangent segments from a given point to a circle have equal lengths). That these lengths are $a-c$ and $a+c$ follow from declaring that $|PP'|=2a$ and $|MP|=c$ (where $M$ is the midpoint of $PP'$). Also, $|P'Q'|=|TQ|$ "by symmetry" in the figure. Consequently, $$|PT|=|TQ|-|PQ|=|P'Q'|-|PQ|=(a+c)-(a-c)=2c$$ Does that help? — Blue, Dec 19 '24 at 20:31
@Blue In the image, the constant distance(sum of distances from the both focii is a+c) but shouldn't it be 2a. If the movable point is fixed at one vertex then the distance will be( a-c) + (a+c) =2a. I didn't understand this. Can you please clarify it to me? — Yug Ahuja, Dec 23 '24 at 09:59
@YugAhuja: My answer (and that figure) makes no use of the constant-sum-of-dists property. To see that property, you need both Dandelin spheres, and the argument from Richard's answer. Restating: The line through $P$ and the cone vertex meets the "horizontal" circles $k_1$ & $k_2$ (where the spheres meet the cone) at $P_1$ and $P_2$. Now, $|P_1P_2|$ is "obviously" constant for all $P$. Since it breaks into $|P_1P|+|P_2P|$, and since $|P_iP|=|F_iP|$ (as tangent segments from $P$ to each sphere), we have that $|F_1P|+|F_2P|$ is constant. Done! — Blue, Dec 23 '24 at 11:41
@Blue Thanks for the answer. I understand it now. I have another query. You wrote $$\frac{\sin(S)}{\sin(Q)} = \frac{TP}{PP'}$$ But for that the triangle TPP' should have been right angled triangle other wise you should have drawn a perpendicular to define sin cos ratios. You haven't proved for TPP' to be right angled. — Yug Ahuja, Dec 26 '24 at 19:41
@YugAhuja: The relation $$\frac{\sin S}{\sin Q}=\frac{TP}{PP'}$$ is an application of the Law of Sines in $\triangle TPP'$ (noting that $\angle P'=\angle S$ and $\angle T=\angle Q$ in that figure). It applies to any triangle, right or otherwise. ... Cheers! — Blue, Dec 27 '24 at 08:43
@Blue Sorry to put up Another question. I know I am going too far away from my original one. But these are little queries that I don't think would be reasonable to ask as new questions on SE. — Yug Ahuja, Dec 28 '24 at 18:01

score 1 · Accepted Answer · answered Dec 19 '24 at 10:13

1

First of all, your assumption that the intersection of the axis of the cone with the intersection plane would be one of the foci, clearly is wrong!

It was the French mathematician Germinal Pierre Dandelin, who outlined the correct geometry. In fact the foci are the orthogonal projections each of the cone-inscribed sphere's centers, which are tangent to the intersection plane, cf. Dandelin spheres (wikipedia).

When it comes to your 3rd question, i.e. to the well-known ellipse property of Appolonius (sum of point-to-foci-distances are constant) then you can do that by means of the Dandelin spheres too:

First you note that $d(P_1,P_2)$ is constant (parallel conncetion circles of these spheres and the cone). Next you note that $d(P,P_k)=d(P,F_k)$ because both $P_k$ and $F_k$ are situated on the same sphere and both these lines are according tangents. Thence you get $$d(P,F_1)+d(P,F_2)=d(P,P_1)+d(P,P_2)=d(P_1,P_2)=const.$$

--- rk

answered Dec 19 '24 at 10:13

Dr. Richard Klitzing

5,757

Thanks for reply. I realised that my assumption is wrong when I read about Dandelin spheres as suggested by a user in comment. I made this assumption while I was studing parabola as conic section and pre assumed it for ellipses and hyperbolas. I want to ask if this assumption in case of hyperbolas would be true or not. I made a little proof for myself that I want to share with you. Check this image. I am continuing this in next comment as I am running out of characters – Yug Ahuja Dec 19 '24 at 17:13
Here, we have the plane intersecting the conic. The angle Beta of plane and alpha of conic are equal to make the conic section a parabola. By definition we have that any point on parabola would be equidistant from the focus and directrix. I assumed directrix to be the line where plane perpendicular to axis of conic that passes through its vertex intersects the plane on which the parabola lies. I tried to prove that the point and line are respectively focus and directrix by using defination of parabola. – Yug Ahuja Dec 19 '24 at 17:18
Consider the triangle ABC perpendicular at C. Angle DAC and ACD are equal by defination of parabola. So triangle DAC is isosceles and AD=DC. Angle DCB would be 90- angle ACD And angle DBC would be 90- angle DAC. since angle ACD And DAC are equal, angle DCB and DBC are also equal and side DC = BD. DC = AD as previously proved. So BD = AD which by defination of parabola should also have been equal. So by this I assumed that the assumed point would be focus and the line would be directrix. I just want to ask if this assumption is right in the case of parabola or not? – Yug Ahuja Dec 19 '24 at 17:26

What does the focus point of ellipse signify?

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