For hyperbola 10x2−10cos2θy2=1
The eccentricity of the hyperbola eH=1+a2b2=1+cos2θ
For ellipse 5cos2θx2+5y2=1
The eccentricity of an ellipse eE=1−a2b2=1−cos2θ=sinθ
Given, eH=5eE
⇒1+cos2θ=5sin2θ⇒cos2θ=32
Now, the length of the latus rectum of an ellipse =a2b2=510cos2θ=3520=345