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  • The distance between the directrices of the hyperbola <img alt="\mathrm{x=8 \sec \theta, y=8 \tan \theta}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%3D8%20%5Csec%20%5C

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

Option: 1

16\sqrt{2}


Option: 2

\sqrt{2}


Option: 3

8 \sqrt{2}


Option: 4

4, \sqrt{2}


Answers (1)

best_answer

Equation of hyperbola is \mathrm{x=8 \sec \theta, y=8 \tan \theta \Rightarrow \frac{x}{8}=\sec \theta \quad \frac{y}{8}=\tan \theta}
\mathrm{\therefore \sec ^2 \theta-\tan ^2 \theta=1 \Rightarrow \frac{x^2}{8^2}-\frac{y^2}{8^2}=1}
\mathrm{\text { Here } a=8, b=8 \text {. Now } e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{8^2}{8^2}}=\sqrt{2}}
\mathrm{\therefore } Distance between directrices \mathrm{=\frac{2 a}{e}=\frac{2 \times 8}{\sqrt{2}}=8 \sqrt{2}}

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shivangi.bhatnagar

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