How to get ellipse cross-section of an ellipsoid I'm trying to get the major and minor radius of an ellipse which represents the cross-section of a given ellipsoid. This is particularly of interest in the field of RF propagation in terms of Fresnel zones and how they interact with a ground surface.
In my particular use, I am taking an ellipsoid which represents the RF Fresnel zone and attempting to determine the cross-sectional area which comes in contact with the Earth's surface. This link provides an example of what I'm describing.
The cross-section of the ellipsoid will be an ellipse. And the major radius is determined by knowing the where the Earth's surface begins and ends it's intersection with the ellipsoid - I can do this part just fine. But I am struggling to determine the minor radius at that intersection.
I am looking for how to determine the minor radius of the intersection of the ellipsoid so that I can draw the intersection onto a map in this fashion:

 A: You have sections with planes perpendicular to a principal axis of the ellipsoid. These ellipses are all similar. If $r$ is the radius of the Fresnel ellipsoid and $h$ is the Earth's height ( or bulge) at the midpoint between the transmitter and the receiver then the cross section perpedicular to the radius $r$ at distance $r-h$ from the center of the ellipsoid is an ellipse with semiaxes $r'$ and $d'$ where 
$$\frac{r'}{r} = \frac{d'}{d} = \sqrt{1 - \left(\frac{r-h}{r}\right)^2}$$
This all follows from the fact that the (Fresnel) ellipsoid has semiaxes $d$, $r$, $r$ and so for every point on the ellipsoid we have 
$$\frac{x^2}{d^2} + \frac{y^2}{r^2} + \frac{z^2}{r^2} =1$$
At distance $r-h$ from the center we have $z$ constant $r-h$ ( perhaps with $-$ sign) - the coordinate system is center in the centered of the ellipsoid. 
${\bf Added:}$ Now for the slanted sections of the ellipsoid. Recall the system of coordinates is centered at the center of the ellipsoid and the $z$-axis is coming out of the earth (maybe not perpendicular, depending on the elevations of the transmitter and receiver). 

Let $E$, $E'$ be the ends of the slice that the Earth makes on the ellipsoid. So $EE'$ will be one of the axis of the slice. The other axis, coming out of the paper, passes through the middle $M$ of the segment $EE'$. Let $M$ have the coordinate $(x_M,0,z_M)$, the average of the coordinates of $E$, $E'$ ($x_M$, $z_M$ the signed green segments, horizontal and vertical). If $w$ is the half-axis of the section then the points of coordinates $(x_M, \pm w, z_M)$ are on the ellipsoid and so 
$$\frac{x_M^2}{d^2} + \frac{w^2}{r^2} + \frac{z_M^2}{r^2} =1$$
and from this equality you get $w$, the other semi-axis of the slice.
