Let $l_1$, $l_2$ be the hyberbolic lines in $\mathbb{H}$ as in the following picture. $l_1$ has end points at infinity $0$ and $\infty$, $l_2$ has endpoints at infinity $1$ and $x$, where $x>0$.
From Hyperbolic Geometry, James W. Anderson (page 110-111) Prop 3.22 and Prop. 3.23 , I have constructed a hyperbolic line $c_r$ with radius $r$ and $1<r<1$. Assume that for any $r$, $c_r$ intersects $l_1$ perpendicularly. Since $r \in (0,x)$, $c_r$ intersects $l_2$ too as well. Construct the Euclidean circle in $\mathbb{H}$ with vertices $0, \frac{1}{2}(x+1)$ and $r.e^{i\theta}$. We determine $\theta$ by considering the Euclidean triangle above (note that $\frac{1}{2}(x+1)$ is the Euclidean center of the Euclidean circle containing $l_2$. The Euclidean lengths of the two sides of this Euclidean triangle adjacent to the vertex $0$, which has angle $θ$, are $r$ and $\frac{1}{2}(x + 1)$, and the length of the opposite side is $\frac{1}{2}(x -1)$. Calculating, we see that $$\left[\frac{1}{2}(x-1)\right]^{2}=\left[\frac{1}{2}(x+1)\right]^{2}+r^{2}-2 r\left[\frac{1}{2}(x+1)\right] \cos (\theta)$$ by the law of cosines.
Moreover, since $l_1$ and $l_2$ are ultraparalel hyperbolic lines, we know that there exists a hyperbolic line $l$ in $\mathbb{H}$ that is perpendicular to both $l_1$ and $l_2$. So to determine the values of $r$ for which $c_{r}$ is perpendicular to $l_2$, we apply the Pythagorean theorem to the Euclidean triangle with vertices $0, \frac{1}{2}(x+1),$ and $re ^{i\theta}$. The angle between $c_{r}$ and $l_2$ is $\frac{\pi}{2}$ if and only if $$\left[\frac{1}{2}(x+1)\right]^{2}=\left[\frac{1}{2}(x-1)\right]^{2}+r^{2}$$ which occurs if and only if $r=\sqrt{x}$.
By doing these steps, I'm able to find that $l=c_r$ that intersecting both $l_1$ and $l_2$ perpendicularly. I wish to show that there exists no other hyperbolic line, $l^{*}$ such that $l^*$ intersects both $l_1$ and $l_2$. However, I couldn't manage to find a way to prove uniqueness of $l$.
