JEE Main 2002Annual, Paper 1MathematicsEasyLimit, Continuity and Differentiability420× since 2002The maximum distance from origin of a point on the curve x=asint−bsin(atb)x = a\sin t - b\sin \left( {{{at} \over b}} \right)x=asint−bsin(bat) y=acost−bcos(atb),y = a\cos t - b\cos \left( {{{at} \over b}} \right),y=acost−bcos(bat), both a,b>0a,b > 0a,b>0 isAOption A. a−ba-ba−bBOption B. a+ba+ba+bCOption C. a2+b2\sqrt {{a^2} + {b^2}}a2+b2DOption D. a2−b2\sqrt {{a^2} - {b^2}}a2−b2CheckSkip · show solution
