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(C) The given parametric equations are $x = 8 \sec 	heta$ and $y = 8 	an 	heta$.We know that $\sec^2 	heta - 	an^2 	heta = 1$.Substituting the v

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$8 \sqrt{2}$

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(C) The given parametric equations are $x = 8 \sec 	heta$ and $y = 8 	an 	heta$.We know that $\sec^2 	heta - 	an^2 	heta = 1$.Substituting the values,we get $\left(\frac{x}{8}ight)^2 - \left(\frac{y}{8}ight)^2 = 1$,which simplifies to $\frac{x^2}{64} - \frac{y^2}{64} = 1$.This is a rectangular hyperbola where $a^2 = 64$ and $b^2 = 64$,so $a = 8$ and $b = 8$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{64}{64}} = \sqrt{2}$.The distance between the directrices of a hyperbola is given by $\frac{2a}{e}$.Substituting the values,we get $\frac{2 	imes 8}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 8 \sqrt{2}$.

The distance between the directrices of the hyperbola $x = 8 \sec \theta, y = 8 \tan \theta$ is

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