For some θ ∈ 0 , π 2 , if the eccentricity of the hyperbola, x 2 - y 2 sec 2 θ = 10 is 5 times the eccentricity of the ellipse, x 2 sec 2 θ + y 2 = 5 , then the length of the latus rectum of the ellipse, is

Step-by-step solution

For hyperbola x 2 10 - y 2 10 cos 2 θ = 1 The eccentricity of the hyperbola e H = 1 + b 2 a 2 = 1 + cos 2 θ   For ellipse x 2 5 cos 2 θ + y 2 5 = 1 The eccentricity of an ellipse e E = 1 - b 2 a 2 = 1 - cos 2 θ = sin θ Given, e H = 5   e E ⇒ 1 + cos 2 θ = 5 sin 2 θ ⇒ cos 2 θ = 2 3   Now, the length of the latus rectum of an ellipse = 2 b 2 a = 10 cos 2 θ 5 = 20 3 5 = 4 5 3