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. 
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.  
 A: Your intuition can be stated as the proposition below:

If $A$ and $B$ are two different points on a hyperbola and the midpoint of $AB$ lies on the asymptotes, then this midpoint is the center of the hyperbola.

Proof: Note that this proposition is irrelevant to coordinate systems, thus without loss of generality, the hyperbola can be assumed to be $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.
Suppose the coordinates of $A$ and $B$ are $(x_1, y_1)$ and $(x_2, y_2)$, respectively, then $x_1, x_2 ≠ 0$ and$$
\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1, \quad \frac{x_2^2}{a^2} - \frac{y_2^2}{b^2} = 1. \tag{1}
$$
Because the asymptotes are $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0$ and $\left( \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \right)$ lies on the asymptotes, then$$
\frac{(x_1 + x_2)^2}{a^2} - \frac{(y_1 + y_2)^2}{b^2} = 0. \tag{2}
$$
Now, suppose $x_1 + x_2 ≠ 0$, then $y_1 + y_2 ≠ 0$ by (2). From (1) there is$$
\frac{x_1^2 - x_2^2}{a^2} = \frac{y_1^2 - y_2^2}{b^2},
$$
and from (2) there is$$
\frac{(x_1 + x_2)^2}{a^2} = \frac{(y_1 + y_2)^2}{b^2},
$$
thus\begin{align*}
&\mathrel{\phantom{\Longrightarrow}}{} \frac{x_1 - x_2}{x_1 + x_2} = \frac{\dfrac{x_1^2 - x_2^2}{a^2}}{\dfrac{(x_1 + x_2)^2}{a^2}} = \frac{\dfrac{y_1^2 - y_2^2}{b^2}}{\dfrac{(y_1 + y_2)^2}{b^2}} = \frac{y_1 - y_2}{y_1 + y_2}\\
&\Longrightarrow \frac{2x_1}{x_1 + x_2} = \frac{x_1 - x_2}{x_1 + x_2} + 1 = \frac{y_1 - y_2}{y_1 + y_2} + 1 = \frac{2y_1}{y_1 + y_2}\\
&\Longrightarrow \frac{y_1}{x_1} = \frac{y_1 + y_2}{x_1 + x_2} \Longrightarrow \frac{y_1}{x_1} = \frac{y_2}{x_2} := c.
\end{align*}
Note that $x_1 + x_2 ≠ 0$, plugging $y_1 = cx_1$ and $y_2 = cx_2$ into (2) to get $\dfrac{1}{a^2} - \dfrac{c^2}{b^2} = 0$, then by (1) there is$$
1 = \frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = \left( \frac{1}{a^2} - \frac{c^2}{b^2} \right) x_1^2 = 0,
$$
a contradiction. Therefore, $x_1 + x_2 = 0$, which implies $y_1 + y_2 = 0$ by (2). Hence the midpoint of $AB$ is the center $(0, 0)$ of the hyperbola.
A: 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. 
