Plane section in hyperboloid

Find equations of planes passing through the point $(a,0,0)$, which intersect a one-sheet hyperboloid $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$ in a pair of parallel lines.

My attempt: First, $(a,0,0)$ belongs to hyperboloid. Then, we can rewrite the equation of a hyperboloid this way: $(\frac{y}{b}-\frac{z}{c})(\frac{y}{b}+\frac{z}{c})=(1-\frac{x}{a})(1+\frac{x}{a})$. So, there will be two kind of generating lines of a hyperboloid. Our plane is passing through these generating lines. I don't know what to do next. How to combine the point $(a,0,0)$ with generating lines?

The answer is $\frac{y}{b}-\frac{z}{c}=0$ and $\frac{y}{b}+\frac{z}{c}=0$, but I don't understand how to get this answer.

You probably think it's harder that it actually is.
Just focus on the point $(a, 0, 0)$.
So $x=\pm a$.
The hyperboloid becomes $\frac{y^2}{b^2} - \frac{z^2}{c^2} = 0$. Rewrite it as:
$$\left(\frac{y}{b} + \frac{z}{c}\right) \left(\frac{y}{b} - \frac{z}{c}\right) = 0$$.

Which means....
$\left(\frac{y}{b} + \frac{z}{c}\right) = 0 $
$\left(\frac{y}{b} - \frac{z}{c}\right) = 0 $

Now, since $x = \pm a$
on pair of parallel lines is
$\left(\frac{y}{b} + \frac{z}{c}\right) = 0, x = a $
and
$\left(\frac{y}{b} + \frac{z}{c}\right) = 0, x = -a $

The other pair..
$\left(\frac{y}{b} - \frac{z}{c}\right) = 0, x = a $
and
$\left(\frac{y}{b} - \frac{z}{c}\right) = 0, x = -a $