Product of distances to asymptotes in a rectangular hyperbola I need to prove that in a rectangular hyperbola, then the product of the distances of any point in the hyperbole to the asymptotes is a constant. 
So I started by using the definition of a rectangular hyperbole.
this being $$b=a$$
So then the equations of the asymptotes are: $$y=x$$ and $$y=-x$$
From there I used the general formula of a hyperbole to get the formula for any point given an x coordinate, which was $$\left(x,\frac{\sqrt{x^2b^2}}{a}\right)$$
Then I applied the formula of distance from line to point using both asymptotes, and now I proceed to multiplying them. But I don't know if I'm going the right direction. Any help would be appreciated! (:
 A: The asymptotes, as you’ve noted, are the lines $x\pm y=0$. The distances to these lines from an arbitrary point $(x,y)$ are given by the usual point-to-line distance formulas: ${|x \pm y| \over \sqrt2}$; their product is $\frac12|x^2-y^2|$. For a point on the hyperbola, $x^2-y^2=a^2$, therefore the product of the distances for any point on the hyperbola is $\frac{a^2}2$.  
This generalizes to any hyperbola: Given a pair of intersecting lines $ax+by+c=0$ and $px+qy+r=0$, the family of hyperbolas with those asymptotes is given by the equations $(ax+by+c)(px+qy+r)=k$ for various values of the parameter $k$. However, the left-hand side of this equation is just the product of the numerators of the formulas for the distances of a point to these lines, and so the product of these distances for a point on the hyperbola is $$\left|{k \over \sqrt{(a^2+b^2)(p^2+q^2)}}\right|.$$ Indeed, we can drop the absolute value signs: the product of the signed distances for a fixed choice of orientation of the normals to the asymptotes is constant.
