## Task B-4.4.

The line $t_{1}$ touches the left, and the line $t_{2}$, which is parallel to it, touches the right branch of the hyperbola $x^{2}-y^{2}=$ 1. If the lines $t_{1}$ and $t_{2}$ intersect the $x$-axis at an angle of $60^{\circ}$, calculate their mutual distance.
## Solution.

We are looking for tangents of the form $y=k x+l$.

From $k= \pm \tan 60^{\circ}$ we get $k= \pm \sqrt{3}$.

For the given hyperbola, $a=b=1$.

We use the tangency condition $k^{2} a^{2}-b^{2}=l^{2}$, which gives $l^{2}=2, l= \pm \sqrt{2}$.

The tangents are:

$$
\begin{aligned}
& t_{1} \ldots y=x \sqrt{3}+\sqrt{2} \\
& t_{2} \ldots y=x \sqrt{3}-\sqrt{2}
\end{aligned}
$$

or

$$
\begin{aligned}
& t_{1} \ldots y=-x \sqrt{3}+\sqrt{2} \\
& t_{2} \ldots y=-x \sqrt{3}-\sqrt{2}
\end{aligned}
$$

Let $O$ be the origin of the coordinate system. The distance between the tangents is $2 \cdot d\left(O, t_{1}\right)=2 \cdot d\left(O, t_{2}\right)=2 \frac{\sqrt{2}}{2}=\sqrt{2}$.

Note: The required distance can also be calculated as follows.

If we choose, for example, the point $T(0, \sqrt{2})$ on the line $t_{1}$, we calculate the distance to the line $t_{2}$ using the formula:

$$
d\left(T, t_{2}\right)=\frac{|0-\sqrt{2}-\sqrt{2}|}{\sqrt{3+1}}=\sqrt{2}
$$

The distance is the same regardless of which pair of tangents we choose.

Therefore, a student can write down either one or both pairs of tangents and receive all the points allocated for the problem.