Equation of a section plane in hyperbolic paraboloid

Solution 1:

A generic plane containing the $x$ axis ($Ox$) has a unit normal vector

$ n = (0, \cos \theta , \sin \theta ) $

Two vectors (unit and mutually perpendicular) that span this plane are

$v_1 = (1, 0, 0) $ and $v_2 = (0, \sin \theta, - \cos \theta ) $

Hence, on this plane, the position vector of any point is $r = [v_1, v_2] u $

where $u = [u_1, u_2]^T $ is the coordinate vector of $r$ with respect to $v_1, v_2$.

Hence,

$x = u_1$

$ y = \sin \theta u_2$

$ z = -\cos \theta u_2$

Plug this into the equation of the hyperbolic paraboloid,

$\dfrac{u_1^2}{p} - \dfrac{\sin^2 \theta u_2^2 }{q} = - 2 \cos \theta u_2 $

which is an equation of hyperbola in $u_1$ and $u_2$. If the semi-axes are equal then

$ p = \dfrac{q}{\sin^2 \theta} $

Hence, the required angle $\theta$ satisfies $ | \sin \theta | = \sqrt{\dfrac{q}{p}} $

This determines four possible values for $\theta$ and correspondingly four planes.

Solution 2:

The last equation you got is that of a hyperbola, which is the projection of the hyperboloid section onto $xy$ plane. The projection has semi-axes $\sqrt p$ (parallel to $x$) and $\sqrt q$ (parallel to $y$).

The tilted plane forms an angle $\alpha$ (with $\tan\alpha=-B/C$) with $xy$ plane, and is parallel to $x$ axis. Hence the hyperbola on the plane has semi-axes $\sqrt p$ (parallel to $x$) and $\sqrt q/|\sin\alpha|$ (parallel to $y$). You then get a right hyperbola if $|\sin\alpha|=\sqrt{q/p}$.