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(C) The equation of the hyperbola is $x^2-y^2=8$. The asymptotes of the hyperbola $x^2-y^2=a^2$ are given by $x^2-y^2=0$,which implies $x-y=0$ and $x+

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$4$

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(C) The equation of the hyperbola is $x^2-y^2=8$. The asymptotes of the hyperbola $x^2-y^2=a^2$ are given by $x^2-y^2=0$,which implies $x-y=0$ and $x+y=0$. Let $P(x, y)$ be any point on the hyperbola. The length of the perpendicular from $P(x, y)$ to the line $x-y=0$ is $d_1 = \frac{|x-y|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y|}{\sqrt{2}}$. The length of the perpendicular from $P(x, y)$ to the line $x+y=0$ is $d_2 = \frac{|x+y|}{\sqrt{1^2+1^2}} = \frac{|x+y|}{\sqrt{2}}$. The product of the lengths of the perpendiculars is $d_1 	imes d_2 = \frac{|x-y|}{\sqrt{2}} 	imes \frac{|x+y|}{\sqrt{2}} = \frac{|x^2-y^2|}{2}$. Since the point lies on the hyperbola $x^2-y^2=8$,we substitute the value: $\frac{8}{2} = 4$.

The product of the lengths of the perpendiculars from any point on the hyperbola $x^2-y^2=8$ to its asymptotes is

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