A) \[16\sqrt{2}\]
B) \[\sqrt{2}\]
C) \[8\sqrt{2}\]
D) \[4\sqrt{2}\]
Correct Answer: C
Solution :
Equation of hyperbola is \[x=8\,\sec \theta ,\,y=8\tan \theta \] Þ \[\frac{x}{8}=\sec \theta ,\,\frac{y}{8}=\tan \theta \] \[\because \,\,{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\] Þ \[\frac{{{x}^{2}}}{{{8}^{2}}}-\frac{{{y}^{2}}}{{{8}^{2}}}=1\]. Here, \[a=8,\,\,b=8\] Now, \[e=\sqrt{1+\frac{{{b}^{2}}}{{{a}^{2}}}}=\sqrt{1+\frac{{{8}^{2}}}{{{8}^{2}}}}=\sqrt{1+1}\]Þ \[e=\sqrt{2}\] \ Distance between directrices \[=\frac{2a}{e}\]\[=\frac{2\times 8}{\sqrt{2}}=8\sqrt{2}.\]
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