Semi-minor axis of any diagonal cylindrical section equal to radius of circular base

Question

Question: Is the semi-minor axis of any diagonal cylindrical section always equal to the radius of the circular base of that cylinder?

The picture depicts a cylinder which is cut diagonally.

Answer 1

Let the equations of the cylinder and the cutting plane be

$$ x^2+y^2=r^2,\>\>\>\>\>z=y\tan\theta\tag 1$$

where $r$ is the radius of the cylinder and $\theta $ is angle the cutting plane forms with the $y$-axis. Then, the equation for the slant ellipse is implicitly given by the intersection of the two surfaces.

To see the elliptical equation explicitly, make the $\theta$-degree rotation clockwise around the $x$-axis, i.e.

$$x’=x,\>\>\>\>\>y’= y\cos\theta+z\sin\theta,\>\>\>\>\> z’=y\sin\theta -z\cos\theta\tag 2$$

From (1) and (2), we have the explicit equation below for the ellipse

$$(x’)^2+(y’\cos\theta)^2=r^2, \>\>\>\>\>z’=0$$

which shows that it’s semi-minor axis $r$ remains the same, while the semi-major axis becomes $\frac r{\cos\theta}$.

Answer 2

The cutting plane intersects the axis $a$ of the cylinder in the point $O$. Consider an arbitrary point $P$ on the ellipse, and project $P$ orthogonally onto $a$, resulting in the point $P'$. One then has $$|OP|^2=|OP'|^2+|P'P|^2=|OP'|^2+r^2\ ,$$ where $r$ is the radius of the cylinder. This shows that when $P$ runs through the ellipse then $|OP|$ is minimal when $P'=O$.