how to find ellipse's minor axis and BC segment?

Question

the picture depicts a cone and the red line is ellipse's major axis.

I found the major axis to be $\frac{rh^2}{l^3}\sqrt{l^2+3h^2}$ which is calculated using Pythagorean rule, Thales' theorem and Trigonometry.

but I don't know how to find the minor axis.

Is there any way?

Also, I need to find $BC$ segment in order to find the difference between the ellipse's center and the point that the cone's height goes through the ellipse. How come?

See at the end of this answer: https://math.stackexchange.com/questions/3102248/is-the-right-intersection-of-an-oblique-circular-cone-an-ellipse/3103027#3103027 — Intelligenti pauca, Mar 07 '20 at 14:39
so here, $BD$ and $AE$ are the two parallel lines in my picture like in your answer in that question. true? — , Mar 07 '20 at 14:51
Yes, exactly so. Take also into account that $b$ is half the minor axis. — Intelligenti pauca, Mar 07 '20 at 15:47
@Aretino I have edited the picture to ask another question too. Are $AB$ and $AC$ the same in length? ($C$ isn't the ellipse's center and $AC$ is on the height). — , Mar 07 '20 at 16:25
in order to find the difference between ellipse's center and the point that cone's height goes through the ellipse — , Mar 07 '20 at 18:31
can you please answer these two questions too? https://math.stackexchange.com/questions/3572719/are-semi-minor-axes-the-same-when-we-have-these-different-major-axes https://math.stackexchange.com/questions/3566483/how-to-find-the-ellipse-perimeter-and-equation-using-this — , Mar 07 '20 at 18:48
Please edit your question with the new request to find $BC$. — Intelligenti pauca, Mar 07 '20 at 19:23

Intelligenti pauca · Accepted Answer · 2020-03-13T13:54:09.387

0

Notice that triangles $ARH$, $RHE$, $HEM$, $EMB$, $MBN$ are all similar among them. Hence: $$ HE={h\over l}r,\quad ME={h^2\over l^2}r,\quad BM={h^3\over l^3}r,\quad BN={h^4\over l^4}r. $$ Semi-minor axis is then (see here for a proof): $$ b=\sqrt{EM\cdot BN}={h^3\over l^3}r=BM. $$

Triangles $CMD$ and $CNB$ are similar too, which entails: $$ {BC\over CD}={BN\over DM}={BN\over ME}={h^2\over l^2}. $$ From that, and using your result for $BD$, it is easy to find $BC$.

edited Mar 13 '20 at 13:54

answered Mar 07 '20 at 20:36

Intelligenti pauca

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thank you for your answer. in your picture, the triangle $ABC$ is more likely to be equilateral. aren't $AC$ and $AB$ really the same? and from your last fractions, it is stated that I have to have the length of $CD$ in order to have $BC$. is it true? – Mar 07 '20 at 20:50
It doesn't matter if $ABC$ is isosceles or not. And you know the sum $BC+CD$, hence can compute $CD$. – Intelligenti pauca Mar 07 '20 at 21:21
If $ABC$ is isosceles, then we can compute $BC$ using law of cosines. and, yes, I know $BC+CD$; but when I don't know $BC$, how can I compute $CD$? my goal is to compute $BC$. – Mar 07 '20 at 21:28
If one knows both $BC+ CD$ and $BC/CD$ it's easy. – Intelligenti pauca Mar 07 '20 at 21:44
$${BC+CD\over BC}=1+{CD\over BC}=1+{l^2\over h^2}.$$ – Intelligenti pauca Mar 07 '20 at 21:50
is the same things applicable to the following question? (let the intersection of semi-major axes be $O$. https://math.stackexchange.com/questions/3572719/are-semi-minor-axes-the-same-when-we-have-these-different-major-axes – Mar 07 '20 at 22:44

how to find ellipse's minor axis and BC segment?

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