Angle between the plane of an ellipse and the axis of the conic Is there a way to determine the angle between the plane of the ellipse and the axis of the conic?
 
 A: Assuming a cone at $45^\circ$, we get:
A circle is obviously at $90^\circ$. 
An ellipse is between $45^\circ$ and $90^\circ$. 
A parabola is at exactly $45^\circ$. 
A hyperbola will be between $0^\circ$ and $45^\circ$. 
Of course, if the angle of your cone is other than $45^\circ$, then these answers change. 
Assuming a cone at $\theta$ (the angle from the vertical axis to a line intersecting the vertex and on the surface of the cone), we get:
A circle is still at $90^\circ$. 
An ellipse is between $\theta$ and $90^\circ$. 
A parabola is at exactly $\theta$. 
A hyperbola will be between $0^\circ$ and $90^\circ-\theta$. 
The above can be seen when you think about how a conic section can intersect the cone. A line less than $\theta$ will intersect both the upper and lower cone, and a line with greater than $\theta$ will only intersect either the upper or lower cone. 
To determine the exact angle for a particular conic, let's work backwards. We will have to work in $3$D space here.
Take the cone $z^2=x^2+y^2$ as what we will takes slices of to generate the conic sections. Take planes parallel to $z=ax$ to generate our conics for $a\geq 0$ (we don't need to consider lines with negative slope). The plane makes angle $\phi=90^\circ-\arctan a$ with the axis of the cone. Of course the vertical location of our plane will determine the relative size of the conic. A large number of "similar" conics of various sizes will have the same angle with the cone axis.
We will generate a conic section by taking the plane $z=ax \ $ ($0 \leq a$) and shifting it up by $k$ to generate a conic. So plugging $z=ax+k$ into $z^2=x^2+y^2$ and performing some algebraic simplifications gives the equation:
$$
\frac{\left(x-\frac{ak}{1-a^2}\right)^2}{\frac{k^2}{1-a^2}}+\frac{y^2}{k^2}=1.
$$
So we can see that $a=0$ generates a circle, and $0<a<1$ generates an ellipse.
If you have an ellipse with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then it will have angle 
$$\phi=\pi/2-\arctan\left(\sqrt{1-\frac{b^2}{a^2}} \ \right)$$
with the vertical axis of the cone. 
Letting $a=1$ initially and going through the same calculation shows that we get a parabola.
Letting $a>1$ we can see that we get a hyperbola.
Carefully working out each case will show that this coincides with the eccentricity calculations given in the other answer.
A: The eccentricity of the conic.
$e = \frac {sin \beta}{sin\alpha}$
Where $\beta$ is the angle of the plane, and $\alpha$ is the slant of the cone.
If the conic is an circle, $e = 0$
If the conic is an parabola, $e = 1$
If the conic is an ellipse, $e = \sqrt{1-\frac{b^2}{a^2}}$
If the conic is an hyperbola, $e = \sqrt{1+\frac{b^2}{a^2}}$
