Find the rectilinear generatrices of the hyperbolic paraboloid $ P_{h}: \frac{x^{2}}{p}-\frac{y^{2}}{q}=2z $, where $ p,q > 0 $, which are parallel to the plane $ (\pi ):\frac{x}{\sqrt{p}}+\frac{y}{\sqrt{q}}=0 $.
Here's my attempt:
We get the family of lines:
$ d_{\lambda }:\begin{cases}
\frac{x}{\sqrt{p}} - \frac{y}{\sqrt{q}} = \lambda \\
\lambda \cdot (\frac{x}{\sqrt{p}} + \frac{y}{\sqrt{q}}) = 2z
\end{cases} $
$ d_{\mu }:\begin{cases} \frac{x}{\sqrt{p}} + \frac{y}{\sqrt{q}} = \mu \\ \mu \cdot (\frac{x}{\sqrt{p}} - \frac{y}{\sqrt{q}}) = 2z \end{cases} $
Afterwards, I computed the directors for these families of lines:
$ \vec{d_{\lambda}} = \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k}\\
\frac{1}{\sqrt{p}}& -\frac{1}{\sqrt{q}} & 0 \\
\frac{\lambda }{\sqrt{p}}& \frac{\lambda }{\sqrt{q}} & -2
\end{vmatrix} = \frac{2}{\sqrt{q}}\cdot \vec{i} + \frac{2}{\sqrt{p}}\cdot \vec{j} + \frac{2\lambda }{\sqrt{pq}}\cdot \vec{k}\Rightarrow \vec{d_{\lambda}} =\left ( \frac{2}{\sqrt{q}}, \frac{2}{\sqrt{p}}, \frac{2\lambda }{\sqrt{pq}} \right ) $.
Analoguous, $ \vec{d_{\mu }} =\left ( -\frac{2}{\sqrt{q}}, -\frac{2}{\sqrt{p}}, -\frac{2\mu }{\sqrt{pq}} \right ) $.
A straight line $ d: \frac{x-x_{0}}{p} = \frac{y-y_{0}}{q} = \frac{z-z_{0}}{r} $ is parallel to a plane $ (\pi ): Ax + By + Cz + D = 0 $ if and only if $ Ap + Bq + Cr= 0 $.
Thus, $ d_{\lambda } || \pi \Leftrightarrow \frac{1}{\sqrt{p}}\cdot \frac{2}{\sqrt{q}} + \frac{1}{\sqrt{q}}\cdot \frac{2}{\sqrt{p}} + 0\cdot \frac{2\lambda }{\sqrt{pq}} = 0 \Leftrightarrow \frac{4}{\sqrt{pq}} = 0 $, which is obviously false.
Hence, I got stuck here.