Let $0<\theta<\frac{\pi}{2}$. If the eccentricity of the hyperbola $\frac{x^{2}}{\cos ^{2} \theta}-\frac{y^{2}}{\sin ^{2} \theta}=1$ is greater than 2, then the length of its latus rectum lies in the interval:

(B) $\left(\frac{3}{2}, 2\right]$

(D) $\left(1, \frac{3}{2}\right]$

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Hyperbola

Let

0 < θ < \frac{π}{2}

. If the eccentricity of the hyperbola

\frac{x ^{2}}{c o s ^{2} θ} - \frac{y ^{2}}{s i n ^{2} θ} = 1

is greater than 2, then the length of its latus rectum lies in the interval:

Answer: (A)

With

a^{2} = cos^{2} θ

and

b^{2} = sin^{2} θ

, the eccentricity is

e = 1 + \frac{b ^{2}}{a ^{2}} = sec θ

. The requirement

e > 2

means

sec θ > 2

, i.e.

θ \in (\frac{π}{3}, \frac{π}{2})

. The latus rectum is

\frac{2 b ^{2}}{a} = \frac{2 s i n ^{2} θ}{c o s θ}

, an increasing function on that interval; at

θ = \frac{π}{3}

it equals

3

and it grows without bound as

θ \to \frac{π}{2}

. So the latus rectum lies in

(3, \infty)

JEE Main · 2019 · 09 Jan 2019 · Shift 1 · Single correct option