The point P ( - 2 6 , 3 ) lies on the hyperbola x ^ 2 a ^ 2 - y ^ 2...

As point $P ( - 2\sqrt{6}, \sqrt{3} )$ lies on hyperbola.
$\dfrac{24}{{a}^{2}} - \dfrac{3}{{b}^{2}} = 1 . . . \left(1\right)$

Given, eccentricity $e = \dfrac{\sqrt{5}}{2}$

$⇒ e = \sqrt{\dfrac{{a}^{2} + {b}^{2}}{{a}^{2}}} = \dfrac{\sqrt{5}}{2}$

$⇒ 4{a}^{2} + 4{b}^{2} = 5{a}^{2}$
$⇒ {a}^{2} = 4{b}^{2} . . .$\left(2\right)$</p><p>On solving equations \left(1\right)$ and \left(2\right)$, we get<br />$⇒ \dfrac{6}{{b}^{2}} - \dfrac{3}{{b}^{2}} = 1$<br />$⇒ {b}^{2} = 3 \text{ and } {a}^{2} = 12$<br />Hyperbola$\dfrac{{x}^{2}}{12} - \dfrac{{y}^{2}}{3} = 1$<br />$\dfrac{2x}{12} - \dfrac{2y \cdot {y}^{'}}{3} = 0$</p><p>$y ' {|}_{( - 2\sqrt{6}, \sqrt{3} )} = - \dfrac{1}{\sqrt{2}}$</p><p>Equation of tangent is$( y - \sqrt{3} ) = - \dfrac{1}{\sqrt{2}} ( x + 2\sqrt{6} )$<br />at$x = 0, y = - \sqrt{3}, Q ( 0, - \sqrt{3} )$<br />Equation of normal is$( y - \sqrt{3} ) = \sqrt{2} ( x + 2\sqrt{6} )$<br />at$x = 0, y = 5\sqrt{3}$<br />$R ( 0, 5\sqrt{3} )$</p><p>Apply distance formula,<br />$QR = 6\sqrt{3}