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  • If e is the eccentricity of <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D-%5Cfrac%7By%5E2

If e is the eccentricity of \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} and \mathrm{\theta} is the angle between the asymptotes, then \mathrm{\cos \frac{\theta}{2}} is equal to

Option: 1

\mathrm{\frac{1}{2 e}}


Option: 2

\mathrm{\frac{1}{3 e}}


Option: 3

\mathrm{\frac{1}{4 e}}


Option: 4

\mathrm{\frac{1}{e}}


Answers (1)

best_answer

The asymptotes to \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} are \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=0}

Or \mathrm{\quad \frac{x}{a}-\frac{y}{b}=0}

And \mathrm{\frac{x}{a}+\frac{y}{b}=0}

\therefore \quad The slope of \mathrm{\frac{x}{a}-\frac{y}{b}=0} is

\mathrm{ \begin{aligned} & \quad \tan \frac{\theta}{2}=\frac{b}{a} \\\\ & \Rightarrow \quad \tan ^2 \frac{\theta}{2}=\frac{b^2}{a^2}=\left(e^2-1\right) \\\\ & \text { As } b^2=a^2\left(e^2-1\right) \\\\ & \text { Hence, } \sec ^2 \frac{\theta}{2}=1+\left(e^2-1\right)=e^2 \\\\ & \text { Or } \cos \frac{\theta}{2}=\frac{1}{e} \end{aligned} }

The answer is (d) 

Posted by

Divya Prakash Singh

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