Find the equation of a plane passing through $Ox$ and intersecting a hyperbolic paraboloid $\frac{x^2}{p}-\frac{y^2}{q}=2z$ $(p>0, q>0)$ along a hyperbola with equal semi-axes.
My attempt: The equation of a plane passing through $Ox$ is $By+Cz=0$. So $z=\frac{-By}{C}$. Now
substitute $z=\frac{-By}{C}$ into the equation of hyperbolic paraboloid $\frac{x^2}{p}-\frac{y^2}{q}=\frac{-2By}{C}$ and transform to $\frac{x^2}{p}-\frac{(y-\frac{Bq}{C})^2}{q}=-\frac{B^2q}{C^2}$. What to do next? I don't understand how to find B and C if we assume $\frac{x^2}{p}-\frac{(y-\frac{Bq}{C})^2}{q}=-\frac{B^2q}{C^2}$ is a hyperbola with equal semi-axes.
Best Answer
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.