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

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/e) (B) (2/e)-e (C) (1/e) (D) (2/e)

Tardigrade Admin · Accepted Answer

θ=2 tan -1 (b/a) ⇒ tan (θ/2)=(b/a) ⇒ cos (θ/2)=(a/√a2+b2) =(1/√1+(b2)a2)=(1/e)

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

$\frac{1 - e}{e}$

$\frac{2}{e} - e$

$\frac{1}{e}$

$\frac{2}{e}$

$θ = 2 tan^{- 1} \frac{b}{a}$
$\Rightarrow tan \frac{θ}{2} = \frac{b}{a}$
$\Rightarrow cos \frac{θ}{2} = \frac{a}{a ^{2} + b ^{2}}$
$= \frac{1}{1 + \frac{b ^{2}}{a ^{2}}} = \frac{1}{e}$

Q. If e is the eccentricity of the hyperbola a2x2​−b2y2​=1 and θ is the angle between the asymptotes, then cos2θ​ is equal to

e1−e​

e2​−e

e1​

e2​