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(D) The equations of the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $bx - ay = 0$ and $bx + ay = 0$.Let $P(a \sec 	heta, b \

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$12$

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(D) The equations of the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $bx - ay = 0$ and $bx + ay = 0$.Let $P(a \sec 	heta, b 	an 	heta)$ be any point on the hyperbola.The perpendicular distance from $P$ to the asymptote $bx - ay = 0$ is $PQ = \frac{|b(a \sec 	heta) - a(b 	an 	heta)|}{\sqrt{b^2 + a^2}} = \frac{ab(\sec 	heta - 	an 	heta)}{\sqrt{a^2 + b^2}}$.The perpendicular distance from $P$ to the asymptote $bx + ay = 0$ is $PR = \frac{|b(a \sec 	heta) + a(b 	an 	heta)|}{\sqrt{b^2 + a^2}} = \frac{ab(\sec 	heta + 	an 	heta)}{\sqrt{a^2 + b^2}}$.The product of these distances is given as $6$:$\frac{a^2 b^2(\sec^2 	heta - 	an^2 	heta)}{a^2 + b^2} = 6$Since $\sec^2 	heta - 	an^2 	heta = 1$,we have $\frac{a^2 b^2}{a^2 + b^2} = 6$.Given $e = \sqrt{3}$,we know $b^2 = a^2(e^2 - 1) = a^2(3 - 1) = 2a^2$.Substituting $b^2 = 2a^2$ into the equation:$\frac{a^2(2a^2)}{a^2 + 2a^2} = 6$ $\Rightarrow \frac{2a^4}{3a^2} = 6$ $\Rightarrow \frac{2}{3}a^2 = 6$ $\Rightarrow a^2 = 9$.Then $b^2 = 2(9) = 18$,so $b = \sqrt{18} = 3\sqrt{2}$.The length of the conjugate axis is $2b = 2(3\sqrt{2}) = 6\sqrt{2}$.

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

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