Proof that any conjugate diameters of rectangular hyperbolas are reflections across an asymptote. (Hyperbolic Orthogonality) I came across the Wikipedia page on conjugate diameters of ellipses, circles and hyperbolas, stating

[T]wo diameters of a conic section are said to be conjugate if each chord parallel to one diameter is bisected by the other diameter.

There is a statement that the conjugate of a diameter of a rectangular hyperbola is a reflection across an asymptote. The entry further links to this page that describes it as the hyperbolic orthogonality property of two lines. Visually, it is easy and intuitive to make this deduction. However, the pages do not include any proofs - perhaps it is too lengthy. Where can I find a proof of this property?
 A: Rectangular hyperbola is given by $x^2-y^2=1$. Consider a diameter given by $y=kx$ (for relevant $k$). In order to describe its conjugate we need to find the midpoints of the chords parallel to $y=kx$. Such a chord is given by $y=kx+n$, so let $(x_1,y_1)$ and $(x_2,y_2)$ be the endpoints of this chord. Note that we are interested only in the midpoint of the chord, i.e. in $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$, hence we don't really need to calculate the coordinates but only $x_1+x_2$ and $y_1+y_2$. 
Endpoints of the chord are in the intersection of $x^2-y^2=1$ and $y=kx+n$. If we solve it for $x$, we get $x^2-(kx+n)^2=1$, i.e. $(1-k^2)x^2-2knx-n^2-1=0$. The solutions of this equation are $x_1$ and $x_2$, so by Vieta's formulae we have $x_1+x_2=\frac{2kn}{1-k^2}$.
If we solve the system for $y$, we have $(\frac{y-n}{k})^2-y^2=1$, i.e. $(y-n)^2-k^2y^2=k^2$, and we get $(1-k^2)y^2-2ny+n^2-k^2=0$. As before, $y_1+y_2=\frac{2n}{1-k^2}$.
So the midpoints of the chords are $\{(\frac{kn}{1-k^2},\frac{n}{1-k^2})\mid n\in\mathbb R\}$. This is obviously a line given by $y=\frac{1}{k}x$, which is obviously symmetric to $y=kx$ with respect to an asymptote. 
