A right circular cone, with the apex angle $\alpha=60^{o}$, is thoroughly cut with a smooth plane inclined at an acute angle $\theta=70^{o}$ with its geometrical axis to generate an elliptical section (As shown in the diagram) .

How to calculate the eccentricity of the elliptical section generated?

Is there any set formula to calculate eccentricity for the similar case i.e. eccentricity, $e$ as the function $\alpha$ & $\theta$ i.e. $e=f(\alpha, \theta)$?

elliptical section of right circular cone

  • $\begingroup$ Where's the diagram? It might not be that important but, since you said it... $\endgroup$ – user228113 May 12 '15 at 16:07
  • $\begingroup$ Hint The eccentricity of an ellipse is $\sqrt{1-\left({b\over a}\right)^2}$, where $a\ge b$ are the lengths of its axes... I think $b\over a$ can most likely be found knowing only $\alpha$ and $\theta$. $\endgroup$ – user228113 May 12 '15 at 16:14
  • $\begingroup$ As far as i remember there really was a formula for e in terms of $\theta$ and $\alpha $ only. I recall reading like this somewhere. $\endgroup$ – Mann May 12 '15 at 16:38

Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $\theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $\beta$ so that $\beta = \alpha/2$ (this will help simplify the calculations later).

Clearly in $\Delta AOC$ we have $$\frac{OA}{\sin \angle OCA} = \frac{AC}{\sin \angle AOC}$$ so that $$b = AC = \dfrac{\sin \theta}{\sin(\theta + \beta)}\tag{1}$$ Similarly $$c = AB = \frac{\sin \theta}{\sin(\theta - \beta)}\tag{2}$$ and $$a = BC = \frac{b\sin 2\beta}{\sin(\theta - \beta)} = \frac{\sin\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}\tag{3}$$ Let the incircle of $\Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $\Delta ABC$ then we know that $$\tan \frac{C}{2} = \frac{r}{CD}$$ so that $$CD = \frac{r}{\tan(C/2)} = \dfrac{r}{\tan\left(\dfrac{\pi - \theta - \beta}{2}\right)} = r\tan\left(\frac{\theta + \beta}{2}\right)\tag{4}$$ Note further that the inradius $r$ is given by the formula \begin{align} r &= \frac{\Delta}{s}\notag\\ &= \frac{bc\sin 2\beta}{a + b + c}\notag\\ &= \dfrac{\dfrac{\sin^{2}\theta\sin 2\beta}{\sin(\theta + \beta)\sin(\theta - \beta)}}{\dfrac{\sin\theta\{\sin(\theta + \beta) + \sin(\theta - \beta) + \sin 2\beta\}}{\sin(\theta + \beta)\sin(\theta - \beta)}}\notag\\ &= \frac{\sin\theta\sin 2\beta}{2\sin\theta\cos\beta + 2\sin\beta\cos\beta}\notag\\ &= \frac{2\sin\theta\sin\beta\cos\beta}{2\cos\beta\{\sin\theta + \sin\beta\}}\notag\\ &= \frac{\sin\theta\sin\beta}{\sin\theta + \sin\beta}\tag{5} \end{align} If $e$ is the eccentricity then we know that $$e = \frac{\text{distance of focus from center}}{\text{length of semi major axis}} = \frac{(a/2) - CD}{a/2} = \frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as \begin{align} e &= \dfrac{a - 2r\tan\left(\dfrac{\theta + \beta}{2}\right)}{a}\notag\\ &= 1 - \frac{2r}{a}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \frac{2\sin\theta\sin\beta}{\sin\theta + \sin\beta}\cdot\frac{\sin^{2}\theta - \sin^{2}\beta}{2\sin\theta\sin\beta\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \frac{\sin\theta - \sin\beta}{\cos\beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \dfrac{2\cos\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\tan\left(\dfrac{\theta + \beta}{2}\right)\notag\\ &= 1 - \dfrac{2\sin\dfrac{\theta + \beta}{2}\sin\dfrac{\theta - \beta}{2}}{\cos \beta}\notag\\ &= 1 - \frac{\cos \beta - \cos \theta}{\cos \beta}\notag\\ &= \frac{\cos \theta}{\cos \beta} = \frac{\cos\theta}{\cos(\alpha/2)}\tag{6} \end{align} Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < \beta = \alpha / 2 < \theta \leq \pi/2$. If $\beta = \alpha/2 = \theta$ then the section becomes a parabola and if $\beta = \alpha/2 > \theta$ then the section becomes a hyperbola. For the angles given in question we have $e = \cos 70^{\circ}/\cos 30^{\circ} = 0.39493084\ldots$.

Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = \cos\theta/\cos\beta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.

  • $\begingroup$ Thanks for your effort. Could you please reduce your result to $$e=\frac{\cos\theta}{\cos\alpha}$$ ? Or I will check it soon. $\endgroup$ – Harish Chandra Rajpoot Aug 23 '15 at 6:46
  • $\begingroup$ @HarishChandraRajpoot; Your result looks problematic because if $\alpha > \pi/2$ then eccentricity comes out to be negative. $\endgroup$ – Paramanand Singh Aug 23 '15 at 6:49
  • $\begingroup$ But let me tell you the restriction of $\alpha$ & $\theta$ which must be held true to make $e$ always positive as follows $$0<\alpha<\theta\leq\frac{\pi}{2}$$ $\endgroup$ – Harish Chandra Rajpoot Aug 23 '15 at 6:52
  • $\begingroup$ Your individual restrictions are true but in addition there must always be $\alpha<\theta$. OK, for instance, kindly check it for any values such that $\alpha>\theta\leq \frac{\pi}{2}$ will such a case exist $\alpha>\theta$.? $\endgroup$ – Harish Chandra Rajpoot Aug 23 '15 at 7:02
  • $\begingroup$ @HarishChandraRajpoot: check the figure you have drawn. Clearly $\alpha > \theta$. Please read my previous comment. Your comments assume $\alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $\alpha$ is the full angle at the vertex of the cone. there is no further constraint on $\alpha, \theta$ apart from the one I mention. $\endgroup$ – Paramanand Singh Aug 23 '15 at 7:05

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