Three points $A$, $B$, $C$ and a line $p$ (that does not pass through the said points) are given. Let $H$ be a hyperbola that passes through points $A$, $B$ and $C$, and the direction $p$ one of its asymptotes. Construct the line through the point $A$ that is perpendicular to the other asymptote of the hyperbola.
If another point $D$ of the hyperbola $H$ was given, the solution would be relatively simple. By applying Pascal's theorem to the hexagon $AABCDP$, where $P$ is the infinitely far point of the asymptote $p$, tangent $t_1$ of hyperbola $H$ at point $A$ is easily found. Analogously, by applying Pascal's theorem to the hexagon $ABCCDP$, we can easily find tangent $t_2$ to hyperbola $H$ at point $C$. Now, let $p \cap t_1 =S_1$ and $c_1=c(A, |AS_1|)$ (circle of radius $|AS_1|$ centered at $A$ ). Let $G_1$ denote the intersection $c_1 \cap t_1$. By the same logic, $p \cap t_2=S_2$ and $c_2=c(C, |CS_2|)$. Now $G_2=c_2 \cap t_2$. $G_1G_2$ is the other asymptote of the hyperbola. Now the construction of the perpendicular from $A$ to $G_1G_2$ is trivial. However, point $D$ is not given. I am wondering if there is a way to construct point $D$ using the given points $A$, $B$ and $C$?