Conic Sections Question 379
Question: The distance between the directrices of the hyperbola $ x=8\sec \theta ,\y=8\tan \theta $ is
[Karnataka CET 2003]
Options:
A) $ 16\sqrt{2} $
B) $ \sqrt{2} $
C) $ 8\sqrt{2} $
D) $ 4\sqrt{2} $
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Answer:
Correct Answer: C
Solution:
Equation of hyperbola is $ x=8\sec \theta ,y=8\tan \theta $
therefore $ \frac{x}{8}=\sec \theta ,\frac{y}{8}=\tan \theta $
$ \because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 $
therefore $ \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} $
therefore $ e=\sqrt{2} $ Distance between directrices $ =\frac{2a}{e} $
$ =\frac{2\times 8}{\sqrt{2}}=8\sqrt{2}. $
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