# Problem of the week #46 - February 11th, 2013

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#### Jameson

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

$\displaystyle d=\frac{|mx_0+b-y_0|}{\sqrt{m^2+1}}$

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#### Jameson

Staff member
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.
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:

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

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

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