Step 1: Find the foci of the ellipse.

The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$. Here, $a^2 = 36$ and $b^2 = 16$.

The eccentricity of the ellipse, $e_E$, is given by $b^2 = a^2(1 - e_E^2)$, so $16 = 36(1 - e_E^2)$.

Thus, $1 - e_E^2 = \frac{16}{36} = \frac{4}{9}$, and $e_E^2 = 1 - \frac{4}{9} = \frac{5}{9}$. Therefore, $e_E = \frac{\sqrt{5}}{3}$.

The foci of the ellipse are at $(\pm ae_E, 0) = (\pm 6 \cdot \frac{\sqrt{5}}{3}, 0) = (\pm 2\sqrt{5}, 0)$.

Step 2: Determine the value of 'a' for the hyperbola.

Since the hyperbola is confocal with the ellipse, its foci are also at $(\pm 2\sqrt{5}, 0)$.

For the hyperbola, the foci are at $(\pm ae_H, 0)$, where $e_H$ is the eccentricity of the hyperbola, given as $e_H = 5$.

So, $ae_H = 2\sqrt{5}$, which means $a \cdot 5 = 2\sqrt{5}$. Therefore, $a = \frac{2\sqrt{5}}{5}$.

Step 3: Determine the value of 'b' for the hyperbola.

For the hyperbola, $b^2 = a^2(e_H^2 - 1)$.

Substituting the values, $b^2 = (\frac{2\sqrt{5}}{5})^2 (5^2 - 1) = \frac{4 \cdot 5}{25} (24) = \frac{20}{25} \cdot 24 = \frac{4}{5} \cdot 24 = \frac{96}{5}$.

Step 4: Calculate the length of the latus rectum of the hyperbola.

The length of the latus rectum of the hyperbola is $\frac{2b^2}{a}$.

Substituting the values, the length of the latus rectum is $\frac{2 \cdot \frac{96}{5}}{\frac{2\sqrt{5}}{5}} = \frac{\frac{192}{5}}{\frac{2\sqrt{5}}{5}} = \frac{192}{2\sqrt{5}} = \frac{96}{\sqrt{5}}$.