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

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$2/3$

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(B) The given hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - y^2 = 1$.The asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x}{a} \pm \frac{y}{b} = 0$.Here,$a^2 = 2$ and $b^2 = 1$,so $a = \sqrt{2}$ and $b = 1$.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 product of the perpendiculars from $P(x_1, y_1)$ to the lines $x - \sqrt{2}y = 0$ and $x + \sqrt{2}y = 0$ is:$p_1 p_2 = \left| \frac{x_1 - \sqrt{2}y_1}{\sqrt{1^2 + (-\sqrt{2})^2}} ight| 	imes \left| \frac{x_1 + \sqrt{2}y_1}{\sqrt{1^2 + (\sqrt{2})^2}} ight|$$= \frac{|x_1^2 - 2y_1^2|}{\sqrt{3} 	imes \sqrt{3}} = \frac{|x_1^2 - 2y_1^2|}{3}$Since $x_1^2 - 2y_1^2 = 2$,the product is $\frac{2}{3}$.

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

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