Question

If e is the eccentricity of $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and θ be the angle between the asymptotes then sec θ/2 equals

• A. e²
• B. 1/e
• C. 2e
• D. e

Answer 1

Answer: (D)

e

Answer 2

Equation of asymptotes to $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ are given by

$\frac{x}{a} \pm \frac{y}{b} = 0$

⇒ $\frac{x}{a} + \frac{y}{b} = 0\mspace{6mu}\mspace{6mu} and\frac{x}{a} - \frac{y}{b} = 0$

⇒ $\frac{y}{b} = - \frac{1}{a}x$

y = -$\frac{b}{a}x$ ∴ m₁ = b/a

Similarly y = $\frac{bx}{a}$ ∴ m₂ = b/a

Now θ = 2tan^-1(b/a)

∴ tanθ/2 = b/a ⇒ tan²θ/2 = $\frac{b^{2}}{a^{2}} = e^{2} - 1$

∴ sec²θ/2 = e² ∴ secθ/2 = e