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:
\( 16 \)
\( \frac{96}{\sqrt{5}} \)
\( 24\sqrt{5} \)
\( 12 \)