Exams › JEE Advanced › Maths
Correct answer: sqrt(2/(3 - sqrt(5)))
Let P = (r cos t, r sin t) be on the curve x² + 2y² + 2xy = 1. Substituting: r²(cos² t + 2 sin² t + 2 sin t cos t) = 1. So r² = 1/(cos² t + 2 sin² t + sin 2t) = 1/(1 + sin² t + sin 2t). To maximise r², minimise f(t) = 1 + sin² t + sin 2t = 1 + (1-cos 2t)/2 + sin 2t = 3/2 + sin 2t - (cos 2t)/2. Let u = 2t. Minimise g(u) = 3/2 + sin u - (cos u)/2. dg/du = cos u + (sin u)/2 = 0 => tan u = -2 => sin u / cos u = -2. At minimum: sin u = -2/sqrt(5), cos u = 1/sqrt(5) (picking the branch that gives minimum). Min value of g = 3/2 + (-2/sqrt(5)) - (1/sqrt(5))/2 = 3/2 - 2/sqrt(5) - 1/(2*sqrt(5)) = 3/2 - 4/(2*sqrt(5)) - 1/(2*sqrt(5)) = 3/2 - 5/(2*sqrt(5)) = 3/2 - sqrt(5)/2 = (3 - sqrt(5))/2. So max r² = 1/((3-sqrt(5))/2) = 2/(3-sqrt(5)). Max r = sqrt(2/(3-sqrt(5))).