TITLE: Evaluate the eccentricity of the elliptical section of a right circular cone
QUESTION [1 upvotes]: 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. $e=f(\alpha, \theta)$?

REPLY [1 votes]: 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.


text:
\section{I-MMSE relation}\label{app:gaussianchannels}

In this appendix we prove the I-MMSE relation of \cite{GuoShamaiVerdu_IMMSE,guo2011estimation} for the convenience of the reader.

\begin{lemma}[I-MMSE formula]\label{app:I-MMSE}
 Consider a signal $\bX\in \mathbb{R}^n$ with $\bX\sim P_X$ that has finite support, and gaussian corrupted data $\bY\sim {\cal N}(\sqrt{R}\,\bX,{\rm I}_n)$ and possibly additional generic data $\bW\sim P_{W|X}(\cdot\,|\bX)$ with $H(\bW)$ bounded. The \emph{I-MMSE formula} linking the mutual information and the MMSE then reads
\begin{align}\label{I-MMSE-relation}
	\frac{d}{dR}I\big(\bX;(\bY,\bW)\big)=\frac{d}{dR}I(\bX;\bY|\bW) = \frac12 {\rm MMSE}(\bX|\bY,\bW)= \frac12 \EE\| \bX -\langle \bx\rangle \|^2\,,
\end{align}
where the Gibbs-bracket $\langle - \rangle$ is the expectation acting on $\bx\sim P(\cdot\,|\bY,\bW)$.
\end{lemma}

\begin{proof}
First note that by the chain rule for mutual information $I(\bX;(\bY,\bW))=I(\bX;\bY|\bW)+I(\bX;\bW)$, so the derivatives in \eqref{I-MMSE-relation} are equal.
We will now look at $\frac{d}{dR}I(\bX;(\bY,\bW))$. Since, conditionally on $\bX$, $\bY$ and $\bW$ are independent,  we have

\begin{align*}
I\big(\bX;(\bY,\bW)\big) = H(\bY,\bW)-H(\bY,\bW|\bX) = H(\bY,\bW) - H(\bY|\bX) - H(\bW|\bX)\, .
\end{align*}

With gaussian noise contribution $H(\bY|\bX)=\frac n2\ln(2\pi e)$. Therefore only $H(\bY,\bW)$ depends on $R$.
Let us then compute, using the change of variable $\bY=\sqrt{R}\,\bX+\bZ$,

\begin{align}
	&\frac{d}{dR}I\big(\bX;(\bY,\bW)\big)=\frac{d}{dR}H(\bY,\bW)&\nn
	&\ =-\frac{d}{dR} \int dP_X(\bX)d\bY d\bW P_{W|X}(\bW|\bX) \frac{e^{-\frac12\|\bY-\sqrt{R}\bX\|^2}}{(2\pi)^{n/2}}\nn
	&\qquad\qquad\times\ln \int dP_X(\bx)P_{W|X}(\bW|\bx)\frac{e^{-\frac12\|\bY-\sqrt{R}\bx\|^2}}{(2\pi)^{n/2}}\nonumber\\
	&\ =- \int dP_X(\bX)d\bZ d\bW P_{W|X}(\bW|\bX)\frac{e^{-\frac12\|\bZ\|^2}}{(2\pi)^{n/2}}\nn
	&\qquad\qquad\times\frac{d}{dR}\ln \!\int dP_X(\bx)P_{W|X}(\bW|\bx)\frac{e^{-\frac12\|\bZ-\sqrt{R}(\bx-\bX)\|^2}}{(2\pi)^{n/2}}\nonumber\\
	&\ =\frac1{2\sqrt{R}}\EE_{\bX,\bZ,\bW|\bX}\big\langle (\bZ+\sqrt{R}(\bX-\bx))\cdot (\bX-\bx)\big \rangle
	\label{IPPtoapply}
\end{align}

where $\bZ\sim{\cal N}(0,{\rm I}_n)$ and the bracket notation is the expectation w.r.t. the posterior proportional to

$$
dP_X(\bx)dP_{W|X}(\bW|\bx)d\bZ \exp\Big\{-\frac12\|\bZ-\sqrt{R}(\bx-\bX)\|^2\Big\}\, .
$$
In \eqref{IPPtoapply} the interchange of derivative and integrals is permitted by a standard application of Lebesgue's dominated convergence theorem in the case where the support of $P_X$ is bounded.

Now we use the following gaussian integration by part formula: for any bounded function ${\bm  g} :\mathbb{R}^n\mapsto \mathbb{R}^n$ of a standard gaussian random vector $\bZ\sim{\cal N}(0,{\rm I}_n)$ we obviously have

\begin{align}
\EE [\bZ\cdot {\bm  g}(\bZ)]=\EE [ \nabla_{\bZ} \cdot {\bm  g}(\bZ)]\,.
\label{SteinLemma}
\end{align}

This formula applied to a Gibbs-bracket associated to a general Gibbs distribution with hamiltonian ${\cal H}(\bx, \bZ)$ (depending on the Gaussian noise and possibly other variables) yields

\begin{align}
	\EE [\bZ\cdot \langle {\bm  h}(\bx)\rangle]&=	\EE\, \nabla_\bZ \cdot \frac{\int dP(\bx)e^{-{\cal H}(\bx,\bZ)} {\bm  h}(\bx)}{\int dP(\bx') e^{-{\cal H}(\bx',\bZ)}}\nn
	&= -\EE\,\frac{\int dP_X(\bx) e^{-{\cal H}(\bx,\bZ)} {\bm  h}(\bx)\cdot \nabla_\bZ {\cal H}(\bx,\bZ)}{\int dP_X(\bx') e^{-{\cal H}(\bx',\bZ)}}\nn
	&\hspace{2cm}+ \EE\Big[\frac{\int dP_X(\bx) e^{-{\cal H}(\bx,\bZ)} {\bm  h}(\bx)}{\int dP_X(\bx') e^{-{\cal H}(\bx',\bZ)}} \cdot \frac{\int dP_X(\bx) e^{-{\cal H}(\bx,\bZ)}\nabla_\bZ {\cal H}(\bx,\bZ) }{\int dP_X(\bx') e^{-{\cal H}(\bx',\bZ)}}\Big]\nn
	&=-\EE \big\langle {\bm  h}(\bx)\cdot \nabla_\bZ  {\cal H}(\bx,\bZ)\big\rangle+\EE \big[\big\langle {\bm  h}(\bx)\big\rangle \cdot\big\langle\nabla_\bZ  {\cal H}(\bx,\bZ)\big\rangle\big]\,.\label{GaussIPP}
	\end{align}

Applied to \eqref{IPPtoapply}, where the ``hamiltonian'' is $\mathcal{H}(\bx, \bZ)=-\ln P_{W|X}(\bW|\bx)+\frac12\|\bZ-\sqrt{R}(\bx-\bX)\|^2$, this identity gives

\begin{align*}
\frac{d}{dR}I\big(\bX;(\bY,\bW)\big)
&=\frac12\EE\big[\big\langle  \|\bX-\bx\|^2\big \rangle + \frac{1}{\sqrt{R}} \nabla_\bZ\cdot \langle \bX-\bx\rangle\big]\nonumber\\
&=\frac12\EE\big[\big\langle  \|\bX-\bx\|^2\big \rangle - \frac{1}{\sqrt{R}}\big\langle (\bX-\bx)\cdot(\bZ+\sqrt{R}(\bX-\bx))\big\rangle\nonumber \\
&\qquad\qquad+ \frac{1}{\sqrt{R}}\big\langle (\bX-\bx)\big\rangle \cdot\big\langle \bZ+\sqrt{R}(\bX-\bx)\big\rangle\big]\nonumber\\
&=\frac12\EE \|\bX-\langle \bx\rangle \|^2\,.
\end{align*}

\end{proof}

The MMSE cannot increase when the SNR increases. This translates into the concavity of
the mutual information of gaussian channels as a function of the SNR.

\begin{lemma}[Concavity of the mutual information in the SNR] \label{lemma:Iconcave}Consider the same setting as Lemma~\ref{app:I-MMSE}. Then the mutual informations $I(\bX;(\bY,\bW))$ and $I(\bX;\bY|\bW)$ are concave in the SNR of the gaussian channel:
\begin{align*}
	\frac{d^2}{dR^2}I\big(\bX;(\bY,\bW)\big)&=\frac{d^2}{dR^2}I(\bX;\bY|\bW) \nn
	&= \frac12 \frac{d}{dR}{\rm MMSE}(\bX|\bY,\bW)= -\frac{1}{2n}\sum_{i,j=1}^n\mathbb{E}\big[(\langle x_ix_j\rangle-\langle x_i\rangle\langle x_j\rangle)^2\big]\le 0
\end{align*}
where the Gibbs-bracket $\langle - \rangle$ is the expectation acting on $\bx\sim P(\cdot\,|\bY,\bW)$.
\end{lemma}
\begin{proof}
Set $Q\equiv \bx\cdot \bX/n$ where $\bx\sim P(\cdot\,|\bY,\bW)$. From a Nishimori identity
${\rm MMSE}(\bX|\bY,\bW) = \EE_{P_X}[X^2]-\EE\langle Q\rangle$. Thus by the I-MMSE formula we have, by a calculation similar to \eqref{GaussIPP},

\begin{align}
-2\frac{d^2}{dR^2}I\big(\bX;(\bY,\bW)\big)=\frac{d\,\EE\langle Q\rangle}{dR}=n\EE [\langle Q\rangle\langle {\cal L}\rangle-\langle Q {\cal L}\rangle]\label{tocombine}
\end{align}
where we have set $${\cal L}\equiv \frac{1}{n}\Big(\frac{1}{2}\|\bx\|^2 - \bx\cdot \bX - \frac{1}{2\sqrt{R}}\bx\cdot \bZ\Big)\,.$$ Now we look at each term on the right hand side of this equality.

The calculation of appendix \ref{proof:remarkable_id} shows that
$$
-\EE\langle Q{\cal L}\rangle= \EE\langle Q^2\rangle-\frac12\EE[\langle Q\rangle^2]\,
$$
so it remains to compute

\begin{align*}
\EE[\langle Q\rangle \langle {\cal L}\rangle]=\mathbb{E}\Big[\langle Q \rangle\frac{\big\langle \|\bx\|^2 \big\rangle}{2n} -  \langle Q \rangle^2 - \langle Q \rangle\frac{\bZ \cdot \langle \bx\rangle}{2n\sqrt{R}}  \Big]	\,.
\end{align*}
By formulas \eqref{SteinLemma} and \eqref{GaussIPP} in which the Hamiltonian is \eqref{Ht} we have
\begin{align*}
-\frac{1}{2n\sqrt{R}}\EE\big[ \bZ \cdot \langle \bx\rangle\langle Q \rangle \big]	&= -\frac{1}{2n\sqrt{R}}\EE\big[\langle Q\rangle \nabla_{\bZ}\cdot \langle \bx\rangle + \langle \bx\rangle\cdot \nabla \langle Q\rangle\big]\nn

&=-\frac{1}{2n}\EE\big[\langle Q\rangle \big(\big\langle \|\bx\|^2\big\rangle-\|\langle\bx\rangle\|^2\big) + \langle \bx\rangle\cdot \big(\langle Q\bx\rangle-\langle Q\rangle \langle \bx\rangle\big)\big]\nn

&\overset{\rm N}{=}-\frac{1}{2n}\EE\big[\langle Q\rangle \big\langle \|\bx\|^2\big\rangle\big]+\frac{1}{n}\EE\big[\langle Q\rangle \|\langle \bx \rangle \|^2\big]-\frac12\EE[\langle Q\rangle^2]\,.
\end{align*}
In the last equality we used the following consequence of the Nishimori identity. Let $\bx,\bx^{(2)}$ be two replicas, i.e., conditionally (on the data) independent samples from the posterior \eqref{tpost}. Then
$$
\frac1n\EE\big[\langle \bx\rangle\cdot \langle Q \bx\rangle\big]=\frac1{n^2}\EE\big\langle (\bx^{(2)}\cdot \bx) (\bx\cdot \bX)\big\rangle\overset{\rm N}{=}\frac1{n^2}\EE\big\langle (\bx^{(2)}\cdot \bX) (\bX\cdot \bx)\big\rangle=\EE[\langle Q\rangle^2]\,.
$$
Thus we obtain
\begin{align*}
\EE[\langle Q\rangle \langle {\cal L}\rangle -\langle Q {\cal L}\rangle] &=	\EE\langle Q^2\rangle-2\EE[\langle Q\rangle^2]+\frac{1}{n}\EE\big[\langle Q\rangle \|\langle \bx \rangle \|^2\big]\nn
&=\frac{1}{n^2}\EE\big\langle(\bx\cdot\bX)^2-2(\bx\cdot\bX)(\bx^{(1)}\cdot\bX)+(\bx\cdot\bX)(\bx^{(2)}\cdot\bx^{(3)})\big\rangle\nn

&\overset{\rm N}{=}\frac{1}{n^2}\EE\big\langle(\bx\cdot\bx^{(0)})^2-2(\bx\cdot\bx^{(0)})(\bx^{(1)}\cdot\bx^{(0)})+(\bx\cdot\bx^{(0)})(\bx^{(2)}\cdot\bx^{(3)})\big\rangle
\end{align*}
where $\bx^{(0)},\bx,\bx^{(1)},\bx^{(2)}, \bx^{(3)}$ are replicas and the last equality again follows from a Nishimori identity. Multiplying this identity by $n$ and rewriting the inner products component-wise we get

\begin{align}
\frac{d\,\EE\langle Q\rangle}{dR} &=\frac{1}{n}\sum_{i,j=1}^n\EE\big\langle x_ix_i^{(0)} x_jx_j^{(0)} -2x_ix_i^{(0)}x_j^{(1)}x_j^{(0)}+x_ix_i^{(0)}x_j^{(2)}x_j^{(3)}\big\rangle\nn

&=\frac{1}{n}\sum_{i,j=1}^n\EE\big[\langle x_ix_j\rangle^2 -2\langle x_i\rangle \langle x_j\rangle \langle x_ix_j\rangle+ \langle x_i\rangle^2\langle x_j\rangle^2\big] \label{secondDer_f_pos}
\end{align}
Using \eqref{tocombine} this ends the proof of the lemma. Note that we have also shown the positivity claimed in \eqref{jacPos} of section \ref{sec:adapInterp_XX}.
\end{proof}

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