If β is one of the angles between the normals to the ellipse x ^ 2...

Given equation of ellipse is $\dfrac{{x}^{2}}{9} + \dfrac{{y}^{2}}{3} = 1$

Normal at \left(3cos\theta, \sqrt{3}sin\theta\right)$is</p><p>$3sec\theta . x - \sqrt{3}cosec\thetay = 6$</p><p>$⇒$Slope${m}_{1} = \dfrac{3sec\theta}{\sqrt{3}cosec\theta} = \sqrt{3}tan\theta$</p><p>Normal at \left(- 3sin \theta, \sqrt{3}cos\theta\right)$

$- 3cosec\theta . x - \sqrt{3}sec\theta . y = 6$

$⇒$Slope ${m}_{2} = \dfrac{- 3cosec\theta}{\sqrt{3}sec\theta} = - \sqrt{3}cot\theta$

Angle between normal is $\beta$.

$⇒ tan\beta = \left|\dfrac{{m}_{1} - {m}_{2}}{1 + {m}_{1}{m}_{2}}\right| = \left|\dfrac{\sqrt{3}tan\theta + \sqrt{3}cot\theta}{1 - 3}\right| = \left|\dfrac{\sqrt{3}}{2sin\theta cos\theta}\right|$

$⇒ tan\beta = \dfrac{\sqrt{3}}{\sin 2\theta} ⇒ \dfrac{2cot\beta}{\sin 2\theta} = \dfrac{2}{\sqrt{3}}$