The hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ has its conjugate axis 5 and passes through the point (–2, 1) if length of latusrectum is k then $4 {\mathrm{k}}^2$ is equal to$\mathrm{\_\_\_\_\_}$

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \mathrm{and} 2\mathrm{b}=5 \Rightarrow \mathrm{b}=\frac{5}{2} \\ \mathrm{It} \mathrm{passes} \mathrm{through} (-2, 1) \mathrm{then} \\ \frac{4}{{\mathrm{a}}^2}-\frac{1}{{\displaystyle \frac{25}{4}}}=1 \\ \Rightarrow \frac{4}{{\mathrm{a}}^2}=1+\frac{4}{25}=\frac{29}{25} \\ \Rightarrow {\mathrm{a}}^2=\frac{100}{29} \\ \Rightarrow \mathrm{a}=\frac{10}{\sqrt{29}} \\ \mathrm{Now} \mathrm{L}.\mathrm{L}.\mathrm{R}=\frac{2{\mathrm{b}}^2}{\mathrm{a}} \\ \mathrm{K}=2\left(\frac{25}{4}\right)\times \frac{\sqrt{29}}{10} \\ \Rightarrow \mathrm{K}=\frac{5\sqrt{29}}{4} \\ \mathrm{Now} 4{\mathrm{k}}^2=4\times \left(\frac{25\times 29}{16}\right) \\ =181.25 \end{gathered}$$