The point $\mathrm{P}(-2\sqrt{6},\sqrt{3})$ lies on the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ having eccentricity $\frac{\sqrt{5}}{2}.$ If the tangent and normal at P to the hyperbola intersect its conjugate axis at the points Q and R respectively ,then QR is equal to

$$\begin{gathered} Given hyperbola\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 — ① \\ and \mathrm{e}=\frac{\sqrt{5}}{2}\Rightarrow {\mathrm{e}}^2=\frac{5}{4} \\ \Rightarrow 1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{5}{4} \\ \Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{1}{4} \\ Since ① \mathrm{passes} \mathrm{through} (-2\sqrt{6}, \sqrt{3}) \mathrm{then} \frac{24}{{\mathrm{a}}^2}-\frac{3}{{\mathrm{b}}^2}=1 \\ \Rightarrow \mathrm{multiplay} \mathrm{both} \mathrm{sides} \mathrm{by} {\mathrm{b}}^2 \\ \left(\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right)24-3={\mathrm{b}}^2 \\ \Rightarrow \left(\frac{1}{4}\right)24-3={\mathrm{b}}^2 \\ \Rightarrow {\mathrm{b}}^2=3 \\ \mathrm{Now} {\mathrm{a}}^2=4{\mathrm{b}}^2=12 \end{gathered}$$

$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{P}(-2\sqrt{6}, \sqrt{3}) \mathrm{is} \\ \Rightarrow \frac{\mathrm{x}(-2\sqrt{6})}{12}-\frac{\mathrm{y}\left(\sqrt{3}\right)}{3}=1 \\ \mathrm{It} \mathrm{meets} \mathrm{conjugate} \mathrm{axis} \mathrm{at} \mathrm{Q}=(0, -\sqrt{3}) \\ \mathrm{Equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P} \mathrm{is} \\ \frac{12\mathrm{x}}{-2\sqrt{6}}+\frac{3\mathrm{y}}{\sqrt{3}}=15 \\ \mathrm{It} \mathrm{meets} \mathrm{conjugate} \mathrm{axis} \mathrm{at} \mathrm{R}=(0, 5\sqrt{3}) \\ \mathrm{Now} \mathrm{QR}=\sqrt{0+{(5\sqrt{3}+\sqrt{3})}^2} \\ =\sqrt{{\left(6\sqrt{3}\right)}^2} \\ =6\sqrt{3} \end{gathered}$$