Let the eccentricity e of a hyperbola satisfy the equation 6e2−11e+3=06e^{2} - 11e + 3 = 06e2−11e+3=0. If the foci of the hyperbola are (3,5)(3,−4)(3,5)(3,-4)(3,5)(3,−4), then the length of its latus rectum is:
113\dfrac{11}{3}311
173\dfrac{17}{3}317
152\dfrac{15}{2}215
172\dfrac{17}{2}217