TITLE: A line intersects a hyperbola at the points $(-2,-6)$ and $(4,2)$ and one of the asymptotes of the hyperbola at $(1,-2)$. Find the centre.
QUESTION [5 upvotes]: A line intersects a hyperbola at the points $(-2,-6)$ and $(4,2)$. It also intersects one of the asymptotes of the hyperbola at the point $(1,-2)$.
  Find the centre of the hyperbola. 

My approach:
The mid-point of the two points of contact $(-2,-6)$ and $(4,2)$ is the point $(1,-2)$ and we are given that the line cuts the asymptote at this point. Hence the answer is $(1,-2)$. But this seems intuitive and I am not sure whether my reasoning is correct. I am looking for another approach to the problem.

REPLY [3 votes]: Let $A$ symmetric matrix and let $u_1^TAu_1=u_2^TAu_2=1$. Then
\begin{align}
\underbrace{\left(\frac{u_1-u_2}{2}\right)^TA\left(\frac{u_1-u_2}{2}\right)}_{P_1}+\underbrace{\left(\frac{u_1+u_2}{2}\right)^TA\left(\frac{u_1+u_2}{2}\right)}_{P_2}=1~~,\tag{1}
\end{align}
which implies that if $P_2$ above is zero, then $P_1=1$.
Any hyperbola (rotated or not) centered in the origin can be defined as the points $\vec{x}$ such that $x^TAx=1$ for A symmetric$^*$, and the asymptotes as those point such that $x^TAx=0$. In Equation (1), $u_1$ and  $u_2$ are in the hyperbola. Midpoint is $(u_1+u_2)/2$, which being in the asymptote means $P_2=0$. Therefore, $P_1=1$ or $(u_1-u_2)/2$ is in the hyperbola. But if $u_2$ is in the hyperbola (centered in origin) so it is $-u_2$. Therefore, we have that $u_1$, $-u_2$, and  $(u_1-u_2)/2$ are in the hyperbola, which is impossible unless $(u_1-u_2)/2=u_1$ or $=-u_2$ (a line intersects hyperbola at most in two points). In either case, $u_1+u_2=0$ or, what is the same, the midpoint between these two points is the origin. 
Reasoning above can be taken outside origin taking the equation for the hyperbola as $(x-x_0)^TA(x-x_0)=1$ and the asymptotes $(x-x_0)^TA(x-x_0)=0$. 
$^*$ $A$ must have one eigenvalue positive and one negative, but it is not necessary for this proof.