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(B) The equations of the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \f

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$6$

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(B) The equations of the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.Let $P(x_1, y_1)$ be any point on the hyperbola. The product of the perpendicular distances from $P$ to the asymptotes is given by:$p_1 p_2 = \left| \frac{\frac{x_1}{a} - \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| 	imes \left| \frac{\frac{x_1}{a} + \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| = \frac{|\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2}|}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{a^2 + b^2}{a^2 b^2}} = \frac{a^2 b^2}{a^2 + b^2}$.Given $e = \sqrt{3}$,we have $b^2 = a^2(e^2 - 1) = a^2(3 - 1) = 2a^2$.Substituting $b^2 = 2a^2$ into the product formula:$p_1 p_2 = \frac{a^2(2a^2)}{a^2 + 2a^2} = \frac{2a^4}{3a^2} = \frac{2a^2}{3}$.Given $p_1 p_2 = 6$,we have $\frac{2a^2}{3} = 6$ $\Rightarrow a^2 = 9$ $\Rightarrow a = 3$.The length of the transverse axis is $2a = 2(3) = 6$.

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ of eccentricity $e = \sqrt{3}$ from its asymptotes is equal to $6$,then the length of the transverse axis of the hyperbola is

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