Given equation of ellipse 16x2+7y2=1
Now finding eccentricity =1−167=43
So, foci ≡(±ae,0)≡(±3,0)
Now, hyperbola: (25144)x2−(25α)y2=1
Eccentricity will be =1+144α=121144+α
Foci ≡(±ae,0)≡(±512⋅121144+α,0)
Given foci coincide then 3=51144+α⇒α=81
Hence, hyperbola is (512)2x2−(59)2y2=1
Length of latus rectum =2ab2=2⋅5122581=1027