I have the hyperbolic paraboloid of equation $x^2-y^2=16z$ and I have to find rectilinear generators that pass through $M(1,1,0)$.
The form for the equations that I found on https://encyclopediaofmath.org/wiki/Hyperbolic_paraboloid#:~:text=A%20hyperbolic%20paraboloid%20is%20a,p−y0√q does not apply since the last term has a denominator of $0$.
I was also able to find a somehow different form for the equations in a geometry book by Pogorelov (photo snippet here: https://i.stack.imgur.com/YAzsg.jpg) , but this form only yields valid generator for the second family mentioned, namely $\begin{cases} z = 0 \\ x+y=128 \end{cases}$. Is it correct to infer this is the only one in this case?
You can apply those formulas for the $x,y$ equations, to get: $$ x=y\quad\text{and}\quad x+y=2. $$ Substitute these into the equation of the paraboloid to obtain the equations for $z$: $$ z=0\quad\text{and}\quad z={1\over4}(x-1) $$ respectively.