This value is constant for any arbitrary point $(x_0,y_0)$ on either hyperbola.

[Jameson]

Thank you to MarkFL for this week's problem!

For the hyperbolas:

$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=\pm1$

Demonstrate that the product of the perpendicular distances from a arbitrary point on either hyperbola to its asymptotes is constant, and give the value of this constant as a function of the parameters.

Hint:
[sp]The perpendicular distance $d$ between the point $(x_0,y_0)$ and the line $y=mx+b$ is given by:

$\displaystyle d=\frac{|mx_0+b-y_0|}{\sqrt{m^2+1}}$[/sp]
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[Jameson]

No one answered the question correctly this week, but honorable mention goes to Sudharaka for an answer that was very close. Here is the solution to the problem from MarkFL.
[sp] For either hyperbola, the asymptotes are:

$\displaystyle y=\pm\frac{b}{a}x$

Let $(x_0,y_0)$ be an arbitrary point on either hyperbola, and by the formula for the perpendicular distance between a point and a line, we find the product of the two distances $d_1,\,d_2$ is:

$\displaystyle d_1\cdot d_2=\frac{\left|\frac{b}{a}x_0-y_0 \right|}{\sqrt{\left(\frac{b}{a} \right)^2+1}}\cdot\frac{\left|-\frac{b}{a}x_0-y_0 \right|}{\sqrt{\left(-\frac{b}{a} \right)^2+1}}=$

$\displaystyle \frac{\left|y_0^2-\frac{b^2}{a^2}x_0^2 \right|}{\frac{b^2}{a^2}+1}=\frac{(ab)^2\left| \frac{x_0^2}{a^2}-\frac{y_0^2}{b^2} \right|}{b^2+a^2}=\frac{(ab)^2\left|\pm1 \right|}{b^2+a^2}=\frac{(ab)^2}{a^2+b^2}$[/sp]