Exams › JEE Advanced › Maths › Applications of Derivatives / Coordinate Geometry
1 questions with worked solutions.
Q1. Find the shortest distance between the curve y² = x³ and the circle 9x² + 9y² - 30y + 16 = 0.
Answer: (sqrt(13))/3 - 1
Circle: 9(x² + y² - (10/3)y) = -16 => x² + (y-5/3)² = 1. Centre (0, 5/3), radius 1. Curve: parametrise as (t², t³). Distance² from centre: D² = t⁴ + (t³ - 5/3)². Minimising: d(D²)/dt = 4t³ + 6t²(t³ - 5/3) = 0 => t=0 or 3t³ + 2t - 5 = 0. t=1 solves this (3+2-5=0). Point (1,1). Distance = sqrt(1 + (1-5/3)²) = sqrt(1+4/9) = sqrt(13)/3. Shortest distance from curve to circle = sqrt(13)/3 - 1.