Showing positive definiteness in the projection of ellipsoid I would like to show that the projection onto the $xoy$ plane of the centered ellipsoid given by the definition $$\mathbf{x'}\mathbf{A}\mathbf{x}=1$$ where we have a positive definite $$\mathbf{A}=
 \left[\begin{array}{rrr}
   a & d & e \\
   d & b & f \\
   e & f & c \\ 
   \end{array} \right]
$$ is an ellipse.
I know this has been covered in posts such as this, however, the answers there end by showing that the figure we need is the shadow of an ellipsoidal section (by a plane).  Then two statements are made that I know from intuition are true, but are not proved, namely.

*

*An ellipsoidal section is an ellipse

*The shadow of an ellipse is an
ellipse.

Is there a way instead to show from the definition in 2-d space that it is indeed an ellipse in 2-D, the set of all $\mathbf{x}=(x,y)$ satisfying $$\mathbf{(x-v)'}\mathbf{B}\mathbf{(x-v)}=1$$ where $B$ is a positive definite $2 \times 2$ matrix and $\mathbf{v} \in \mathbb{R}^2$ is the center of the ellipse?
Here is what I have so far.  Similar to Andrew Hwang's answer and Christian Blatter's answer, I need that the z-component of $\nabla f$ should vanish on the curve I seek, where $f(\mathbf{x})=\mathbf{x'}\mathbf{A}\mathbf{x}-1$.  Solving this along with the original equation of the ellipsoid, I get that the equation of the curve I seek is $$\mathbf{x'}\mathbf{B}\mathbf{x}=1$$ where $$\mathbf{B}=
 \left[\begin{array}{rr}
   (a-\frac{e^2}{c}) & (d-\frac{ef}{c}) \\
   (d-\frac{ef}{c}) & (b-\frac{f^2}{c}) \\
   \end{array} \right]$$
EDITED: The matrix B previously had a mistake in it - the off-diagonal element was stated as $(\frac{d}{2}-\frac{ef}{c})$ instead of $(d-\frac{ef}{c})$. Now fixed.
My question is, how do I show that this $B$ is positive definite?  The info I have is that from Sylvester's criterion applied to $A$, I know

*

*$a > 0$

*$ab - d^2 >0$

*$a(bc-f^2) + d(ef-cd) + e(df-eb) >0$
I also know that $b$ and $c$ have to be positive.
Note: As an example of the kind of answer I was trying: I was able to prove for a different problem (that the section of the ellipsoid by a plane $z=l$ is an ellipse).  Here the curve is given by $$\mathbf{(x-v)'}\mathbf{B}\mathbf{(x-v)}=1-cl^2+\frac{l^2(af^2+be^2)}{ab-d^2}+\frac{efl^2d^3}{(ab-d^2)^2}$$ with $\mathbf{v}=(\frac{l(fd-eb)}{ab-d^2},\frac{l(ed-af)}{ab-d^2})$.In this case, $B$ was
$$\mathbf{B}=
 \left[\begin{array}{rr}
   a  & d \\
   d & b \\
   \end{array} \right]$$ and I know that is positive definite by Sylvester's criterion applied to the original $A$ matrix.
 A: First of all, a graphical illustration (Matlab program given below).

Answer to the question "Why is $B$ an spd (symmetric positive definite) matrix ?"
It is because $\mathbf{B}$ can be written, in a very natural way, as a "Schur complement"  (http://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/IFEM.AppP.d/IFEM.AppP.pdf) with respect to matrix $\mathbf{A}$. 
Let us partition matrix $\mathbf{A}$ into the following way:
$$\tag{1}\mathbf{A}=
 \left(\begin{array}{rr|r}
   a & d & e \\
   d & b & f \\ 
   \hline 
   e & f & c \\ 
   \end{array} \right)$$
yielding the following Schur complement:
$$\tag{2}\begin{pmatrix}a&d\\d&b\end{pmatrix}-\tfrac{1}{c}\begin{pmatrix}e\\f\end{pmatrix}\begin{pmatrix}e&f\end{pmatrix}=
 \left(\begin{array}{rr}
   a-\frac{e^2}{c} & d-\frac{ef}{c} \\
   d-\frac{ef}{c} & b-\frac{f^2}{c} \\
   \end{array} \right)$$
which coincides with the given matrix $\mathbf{B}$.
Now use the fact that a Schur complement of a spd matrix is itself spd (see "Properties" in  (https://en.wikipedia.org/wiki/Schur_complement)).
Edit: a complementary view (see (https://stats.stackexchange.com/q/30588/147896)) is to interpret matrix $\mathbf{A}$ as a covariance matrix of a certain multivariate normal random variable $(X,Y,Z) \in \mathbb{R^3}$ and $\mathbf{B}$  (https://stats.stackexchange.com/q/30588/147896) as the covariance matrix associated with marginal distribution of $(X,Y) \in \mathbb{R^2}.$
Matlab program for the "cigar and smoke ring" figure:

clear all;close all;hold on;axis equal;
p=0.05;
[X, Y, Z] = meshgrid(-1:p:1, -1:p:1, -1.5:p:1.5);
r=1;a=3;b=1;c=2;d=2;e=1;f=2;
A=[a b d
   b c e
   d e f];% caution: the coefficients' order has been changed
isosurface(X, Y, Z, a*X.^2 + 2*b*X.*Y + c*Y.^2 +...
  f*Z.^2 + 2*d*X.*Z+ 2*e*Y.*Z, r^2);
camlight(60,0);shading flat;view([30,15]);
S=A(1:2,1:2)-(1/f)*A(1:2,3)*A(3,1:2);%Schur complement
u=S(1,1);v=S(2,2);w=S(2,1);
ff = @(x,y) (u*x.^2 + v*y.^2 + 2*w*x.*y-1);% please note the -1
ep=ezplot(ff);set(ep,'linecolor',[0.5,0.5,0.5],'linewidth',3);


