Intersection of a plane with an infinite right circular cylinder by means of coordinates So, I started studying analytic geometry and I must say I'm finding it much harder than "classic" geometry, because of the equations without help from diagrams... Still, I wanted to see how to use it to get alternative proofs of geometric theorems.
The theorem that "the intersection of a plane not parallel nor perpendicular to the axis of an infinite right circular cylinder is an ellipse" looked appropriate to be proved via analytic geometry because imagining in three dimensions is harder than in the plane and with equations one does not have to imagine, just compute (at least this is what I'm thinking right now).
Some texts take that as the definition of an ellipse, but then one needs to show that it defines the same shape as the definition "the set of points in the plane such that the sum of the distances from the point to each of two fixed points is constant".
There is that famous synthetic proof in many texts with the auxiliary spheres, as can be found, for example, in Hilbert, D. and Cohn-Vossen, Geometry and the Imagination.
But I wanted to handle the problem by means of equations without having to imagine too much, but also without too long calculations :-)
So, since the equation of an infinite right circular cylinder is x² + y² - R² = 0 and the equation of a plane is Ax + By + Cz + D = 0, the points that satisfy both equations should form an ellipse, right? I was not sure it was correct reasoning, but I tried solving the system composed by these two equations by substitution of x for sqrt(R²-y²) on the second equation, but how to show it's the same as the equation of an ellipse (y/T)² + (z/U)² - 1 = 0 I do not know... 
I'm confused, sorry if my question does not make sense.
 A: The slick analytic way to go about this is to (prove and then) appeal to the general theorem classifying second-degree curves in the plane:

Let $P(x,y)$ be a polynomial of degree 2. The locus of points $(x,y)$ with $P(x,y)=0$ is similar to the solution set of exactly one of
  $$ \tag{ellipse or circle} x^2 + (y/a)^2 = 1 $$
  $$ \tag{hyperbola} x^2 - (y/b)^2 = 1 $$
  $$ \tag{parabola} x^2 = y $$
  $$ \tag{intersecting lines} x^2 - (y/a)^2 = 0 $$
  $$ \tag{parallel lines} x^2 = 1 $$
  $$ \tag{one line} x^2 = 0 $$
  $$ \tag{one point} x^2 + y^2=0 $$
  $$ \tag{nothing} x^2 = -1 $$
  for some $a\in (0,1]$ or $b\in(0,\infty)$.

In the case of a plane cutting a circular cylinder, if you write the plane as a parametric equation you get a coordinate system of the plane for free, and a bit of algebra easily shows that the equation of the intersection is a quadratic one.
It is also clear on geometric grounds that the solution set is bounded, and contains at least two points, which means that the only possible among the above cases is $x^2+(y/a)^2 = 1$.
We can then prove once and for all that this equation defines an ellipse in the plane. 
