Let the foci of a hyperbola coincide with the foci of the ellipse $$\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$$. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

For the ellipse $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ with $$a^2 = 36$$ and $$b^2 = 16$$, we have $$c_e^2 = a^2 - b^2 = 36 - 16 = 20$$ and hence $$c_e = \sqrt{20} = 2\sqrt{5}$$. Thus the foci of the ellipse are at $$(\pm 2\sqrt{5}, 0)$$.

The hyperbola shares these foci, so $$c_h = 2\sqrt{5}$$ and its eccentricity is $$e_h = 5$$. 

Using the relation $$c_h = a_h \cdot e_h$$ for a hyperbola gives $$a_h = \frac{c_h}{e_h} = \frac{2\sqrt{5}}{5}\,.$$

Next, since $$c_h^2 = a_h^2 + b_h^2$$ for a hyperbola, we find

$$b_h^2 = c_h^2 - a_h^2 = 20 - \left(\frac{2\sqrt{5}}{5}\right)^2 = 20 - \frac{20}{25} = 20 - \frac{4}{5} = \frac{96}{5}\,.$$

The length of the latus rectum of a hyperbola is given by $$L = \frac{2b_h^2}{a_h}$$. Substituting the above values yields

$$L = \frac{2 \times \frac{96}{5}}{\frac{2\sqrt{5}}{5}} = \frac{\frac{192}{5}}{\frac{2\sqrt{5}}{5}} = \frac{192}{2\sqrt{5}} = \frac{96}{\sqrt{5}}\,, $$