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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

Vedclass · Accepted Answer

$4$

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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 $x = \pm y$,or $x + y = 0$ 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 $(x, y)$ lies on the hyperbola $x^2 - y^2 = 8$,we have $d_1 	imes d_2 = \frac{8}{2} = 4$.

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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