Question

If $e$ is the eccentricity of the hyperbola $x²/a² - y²/b² = 1$ and $θ$ is the angle between the asymptotes, then $\cos(θ/2)$ is equal to

• A. $1/e$
• B. $-1/e$
• C. $e$
• D. $1$

Hint:

If $e$ is the eccentricity of the hyperbola $x/a - y/b = 1$ and $θ$ is the angle between the asymptotes, then $\cos(θ/2)$ is equal to

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The asymptotes of a hyperbola are $y = \pm(b/a)x$.
Step 2: Evaluation
The angle $\alpha$ an asymptote makes with the transverse axis satisfies $\tan(\theta/2) = b/a$.
Step 3: Final Calculation
Using the identity $\sec^2(\theta/2) = 1 + \tan^2(\theta/2) = 1 + b^2/a^2 = (a^2+b^2)/a^2$.
Step 4: Conclusion
Since $e^2 = (a^2+b^2)/a^2$, we have $\sec^2(\theta/2) = e^2 \Rightarrow \cos(\theta/2) = 1/e$.
Hence, the Answer is: (a)