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(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.Here,$a^2 = 2$ and $b^2 = 1$.T

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$\frac{2}{3}$

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(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.Here,$a^2 = 2$ and $b^2 = 1$.The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x}{a} \pm \frac{y}{b} = 0$.Substituting the values,the asymptotes are $\frac{x}{\sqrt{2}} - y = 0$ and $\frac{x}{\sqrt{2}} + y = 0$,or $x - \sqrt{2}y = 0$ and $x + \sqrt{2}y = 0$.Let $P(x_1, y_1)$ be any point on the hyperbola,so $x_1^2 - 2y_1^2 = 2$.The lengths of the perpendiculars from $P(x_1, y_1)$ to the asymptotes are $p_1 = \frac{|x_1 - \sqrt{2}y_1|}{\sqrt{1^2 + (-\sqrt{2})^2}} = \frac{|x_1 - \sqrt{2}y_1|}{\sqrt{3}}$ and $p_2 = \frac{|x_1 + \sqrt{2}y_1|}{\sqrt{1^2 + (\sqrt{2})^2}} = \frac{|x_1 + \sqrt{2}y_1|}{\sqrt{3}}$.The product of the lengths is $p_1 p_2 = \frac{|x_1^2 - 2y_1^2|}{3}$.Since $x_1^2 - 2y_1^2 = 2$,the product is $\frac{2}{3}$.

What is the product of the lengths of the perpendiculars drawn from any point on the hyperbola $x^2 - 2y^2 - 2 = 0$ to its asymptotes?

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