Consider hyperboloid $z^2 = x^2 − 4y^2 + 5$. Find an equation of the line which belongs to this hyperboloid and the point $(0, 1, 1)$. Consider hyperboloid $z^2 = x^2 − 4y^2 + 5$. Find an equation of the line which belongs to this hyperboloid and the point $(0, 1, 1)$.
This is what I have so far:
$I(s) = OP + sv$
Letting $v = (a, b, c) $
$I(s) = (0, 1, 1) + s(a, b, c)$
$x = as,  y = bs + 1, z = cs + 1$
I then plugged the parameterization into the equation $z^2 = x^2 − 4y^2 + 5$, giving me $s^2(a^2-4b^2-c^2)-2s(4b+c)=0$.
However, how do I proceed from here to find $(a, b, c)$?
 A: We'll assume $s \neq 0$ to simplify things since we know $I(0) = (0,1,1)$. Notice then that
\begin{align*}
(cs+1)^2=(as)^2 -4(bs+1)^2+5  &\implies s \left(sa^2  - 4 b^2 s - 8 b - c^2 s - 2 c\right) = 0\\
& \implies s\left(a^2 -4b^2 -c^2\right) = 8b+2c \tag{1}
\end{align*}
From here, since it should hold that $I(s)$ is in the hyperboloid for any $s$, choosing some arbitrary $s_1, s_2 \in \mathbb{R} \setminus \{0\}$ we get that
$$
(s_1-s_2) \left(a^2 -4b^2 -c^2\right) = 0
$$
but since $s_1$ need not be equal to $s_2$, to guarantee that the above equation holds in general it must be the case that
$a^2 -4b^2 -c^2 = 0$. Plugging the previous condition in $(1)$ we get $c = - 4b$. So combining the last two conditions we get that
\begin{align}
a^2 = 4\left(- \frac{c}{4} \right)^2 +c^2 \implies c = \pm \frac{2a}{\sqrt{5}}\\
a^2 = 4b^2 +(-4b)^2 \implies b = \mp \frac{a}{2\sqrt{5}}\\
\end{align}
noticing that $c = - 4b$ imposes $b$ and $c$ to have different signs. So your line has a free parameter $a$, and choosing $a=1$ for simplicity you get your solutions to be
$$
\boxed{I(s) = (0,1,1) + s\left(1, \pm \frac{1}{2\sqrt{5}}, \mp \frac{2}{\sqrt{5}}\right)}
$$
Here's a graph of how the solutions look like:

where the dotted line is the solution with the $y$ coordinate $+ \frac{1}{2\sqrt{5}}$ and the filled-out line is the solution with the $y$ coordinate $- \frac{1}{2\sqrt{5}}$.
A: Here is how I would approach it - use the fact that the line is in hyperbola and hence it must also be  perpendicular to the normal vector to the tangent plane at $(0, 1, 1)$.
Hyperbola surface $S: -x^2 + 4y^2 + z^2 - 5 = 0$
Equation of line through point $(0, 1, 1)$ is $(0, 1, 1) + s(a, b, c)$
Taking partial derivatives, gradient vector is $(-2x, 8y, 2z)$. At point $(0, 1, 1)$, the gradient vector is $(0, 8, 2)$ or $(0, 4, 1)$.
As the direction vector of the line is $(a, b, c)$, we have $(a, b, c) \cdot (0, 4, 1) = 0$. That gives us $c = - 4b$
As the line is in the hyperbola, the points on the line must satisfy the equation of hyperbola. So as you rightly said,
$s^2(a^2-4b^2-c^2)-2s(4b+c)=0 \implies a^2 = 20b^2, a = \pm2 \sqrt5 \ b$
So equation of line is $(0, 1, 1) + s (\pm2\sqrt5, 1, -4)$
