Let the foci of the ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{7}=1$ and the hyperbola $\frac{{\mathrm{x}}^2}{144}-\frac{{\mathrm{y}}^2}{\mathrm{\alpha }}=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:-

$\begin{array}{l}{a}_1=4,b_1=\sqrt{7},e_1=\sqrt{1-\frac{b_1^2}{a_1^2}}=\sqrt{1-\frac{7}{16}}=\frac{3}{4}\\ \text{ Hyperbola }\to \frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\frac{\alpha }{25}}=1\\ a_2^2=\frac{144}{25},b_2^2=\frac{\alpha }{25}\\ a_1{e}_1=a_2{e}_2\\ \Rightarrow 4\cdot \frac{3}{4}=\frac{12}{5}{e}_2\Rightarrow e_2=\frac{5}{4}\\ \sqrt{1+\frac{b_2^2}{a_2^2}}=\frac{5}{4}\\ \begin{array}{r}\Rightarrow 1+\frac{\alpha }{144}=\frac{25}{16}\\ \therefore \alpha =81\\ \text{ L.R. }=\frac{2b_2^2}{a_2}=\frac{27}{10}\end{array}\end{array}$