<s>[INST] Prove that the intersection of a cylinder and a plane forms an ellipse I've seen the nice proof of this using spheres, but I'm looking for a way to prove it parametrically if possible. Using a cylinder $x^2+y^2=r^2$ and a plane $ax+by+cz+d=0$ I got:
$x=r\cos(\theta), y=r\sin(\theta), z=\dfrac{-ar\cos(\theta)+br\sin(\theta)+d}{c}$
But after this I'm stuck trying different projections and messing with ellipse definitions 
 [/INST] Without loss of generality let's assume the cylinder has radius $1$, so that it has equation $x^2 + y^2 = 1$. Assuming that the plane is not parallel to the cylinder, we can always rearrange the coordinate system so that the plane goes through the origin, or even better, make the plane go through the $x$-axis after a suitable rotation. Now the plane should have equation $z = y \tan\alpha$ (with $\alpha$ being the slope in the $yz$-plane).
Now your parametrization gives the curve
$$x = \cos\theta, \quad y = \sin\theta, \quad z=\sin\theta\tan\alpha.$$
We think of a rotation (of the whole space) around the $x$-axis, through an angle of $-\alpha$ in the $yz$-plane:
$$\begin{aligned}
\begin{bmatrix} x \\ y \\ z \end{bmatrix}
& \mapsto
\begin{bmatrix}
  1 & 0 & 0 \\
  0 & \cos(-\alpha) & -\sin(-\alpha) \\
  0 & \sin(-\alpha) & \cos(-\alpha)
\end{bmatrix}
\begin{bmatrix} x \\ y \\ z \end{bmatrix} \\
& =
\begin{bmatrix}
  x \\
  y\cos\alpha + z\sin\alpha \\
  -y\sin\alpha + z\cos\alpha
\end{bmatrix}.
\end{aligned}$$
Under this rotation the curve becomes
$$\begin{aligned}
\begin{bmatrix} \cos\theta \\ \sin\theta \\ \sin\theta\tan\alpha \end{bmatrix}
& \mapsto
\begin{bmatrix}
  \cos\theta \\
  \sin\theta\cos\alpha + \sin\theta\tan\alpha\sin\alpha \\
  -\sin\theta\sin\alpha + \sin\theta\tan\alpha\cos\alpha
\end{bmatrix} \\
& = 
\begin{bmatrix}
  \cos\theta \\
  \sin\theta\cos\alpha + \sin\theta\sin^2\alpha/\cos\alpha \\
  -\sin\theta\sin\alpha + \sin\theta\sin\alpha
\end{bmatrix} \\
& = 
\begin{bmatrix}
  \cos\theta \\
  \sin\theta/\cos\alpha \\
  0
\end{bmatrix}.
\end{aligned}$$
This is
$$x = \cos\theta, \quad y = \frac{1}{\cos\alpha}\sin\theta, \quad z\equiv0,$$
which is an ellipse.
</s>