If the maximum distance of normal to the ellipse x 2 4 + y 2 b 2 = 1 , b < 2 , from the origin is 1 , then the eccentricity of the ellipse is:

Step-by-step solution

Given: x 2 4 + y 2 b 2 = 1 Equation of normal to the ellipse is 2 x sec θ - b y cosec θ = 4 - b 2 Distance of normal from origin is d = 4 - b 2 4 sec 2 θ + b 2 cosec 2 θ For maximum distance, denominator must be minimum. 4 sec 2 θ + b 2 cosec 2 θ = 4 + 4 tan 2 θ + b 2 + b 2 cot 2 θ Now, 4 tan 2 θ + b 2 cot 2 θ 2 ≥ 4 tan 2 θ × b 2 cot 2 θ ⇒ 4 tan 2 θ + b 2 cot 2 θ ≥ 4 b ⇒ 4 + b 2 + 4 tan 2 θ + b 2 cot 2 θ &#8