There was an interesting conversation on twitter today about the net of an oblique cylinder. I misunderstood the question and produced a net of a right cylinder sliced by a plane with the equation $ax+by+cz+d=0$. (The example I printed uses particular planes, $z=2-y$ and $z=-2-y$ and a cylinder with radius 1 around the z-axis)
The ends of my cylinder are non-circular ellipses. It even works in construction.
But, an oblique cylinder has circles on each end.
If I have an oblique cylinder can I trim it down into a right cylinder?
I suspect the answer is no. A cut at any angle other than the one that produces a circle will produce a non-circular ellipse.
What confuses me is the image of two cylinders of infinite height. One is formed by extending the oblique cylinder. The other is formed by extending the right cylinder. How can I tell them apart?
If you mean by an oblique cylinder a surface obtained by shearing a cylinder, then the answer is in general NO. Its cross-section perpendicular to the 'axial direction' will be an ellipse, which means that this surface is an elliptic cylinder.
Indeed, consider a cylinder $x^2 + y^2 = 1$ of radius $1$ and let us shear it to $x$-direction. Then with the shearing angle $\alpha$, the resulting surface is given by the equation $(x-z\tan\alpha)^2 + y^2 = 1$. Applying the rotation $(x, y, z) = (x'\cos\alpha + z'\sin\alpha, y', -x'\sin\alpha + z'\cos\alpha)$ so that the axial direction is now the $z'$-direction, the equation reduces to
$$ \frac{(x')^2}{\cos^2\alpha} + (y')^2 = 1, $$
which is indeed an elliptic cylinder. It is even obvious from the graph if we apply extreme degree of shearing: If $\tan\alpha = 3$, then