Let 0 < θ < (π)/(2) . If the eccentricity of the hyperbola x ^ 2 ^...

For given hyperbola the eccentricity is given as

${e}^{2} = 1 + \dfrac{\sin^{2}\theta}{\cos^{2}\theta} = 1 + \tan^{2}\theta{= sec}^{2}\theta$

$⇒ e = sec\theta$

$∴$ Length of latus rectum $l = \dfrac{2\sin^{2}\theta}{cos\theta} = \dfrac{2\tan^{2}\theta}{sec\theta}$

$⇒ l = \dfrac{2 ( {e}^{2} - 1 )}{e} = 2\left(e - \dfrac{1}{e}\right)$
On differentiating w.r.t. $e$, we get

$\dfrac{dl}{de} = 2\left(1 + \dfrac{1}{{e}^{2}}\right) > 0$

$∴$ $l$ is an increasing function

∴ {l}_\min = 2\left(2 - \dfrac{1}{2}\right) = 3

$∴$ Range of latus rectum is \left(3, \infty\right)$.