Let the latus rectum of the hyperbola x ^ 2 9 - y ^ 2 b ^ 2 = 1...

Given,

Equation of hyperbola $\dfrac{{x}^{2}}{9} - \dfrac{{y}^{2}}{{b}^{2}} = 1$

And latusrectum $LR$ subtends ${60}^{°}$ at centre

Now, plotting the diagram we get,

Now, from above diagram we get $A\left(ae, \dfrac{{b}^{2}}{a}\right) \text{ and } B\left(ae, \dfrac{- {b}^{2}}{a}\right)$

$⇒ \tan 30 ° = \dfrac{\dfrac{{b}^{2}}{a}}{ae} = \dfrac{{b}^{2}}{{a}^{2}e} = \dfrac{1}{\sqrt{3}}$

$⇒ e = \dfrac{\sqrt{3}{b}^{2}}{9}$ \left\{\text{as}{a}^{2} = 9\right\}$</p><p>Also,${e}^{2} = 1 + \dfrac{{b}^{2}}{9}$</p><p>$⇒ 1 + \dfrac{{b}^{2}}{9} = \dfrac{3{b}^{4}}{81}$</p><p>$⇒ {b}^{4} = 3{b}^{2} + 27$</p><p>$⇒ {b}^{4} - 3{b}^{2} - 27 = 0$</p><p>$⇒ {b}^{2} = \dfrac{3 + \sqrt{117}}{2}${ignoring the negative sign as it is a square function}</p><p>$⇒ {b}^{2} = \dfrac{3}{2} ( 1 + \sqrt{13} )$</p><p>Hence, on comparing with$\dfrac{l}{m}$\left(1 + \sqrt{n}\right)$ we get,

$⇒ l = 3, m = 2, n = 13$

$⇒ {l}^{2} + {m}^{2} + {n}^{2} = 182$