JEE Main 2005Annual, Paper 1MathematicsEasyLimit, Continuity and Differentiability420× since 2002Area of the greatest rectangle that can be inscribed in the ellipse x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1a2x2+b2y2=1AOption A. 2ab2ab2abBOption B. abababCOption C. ab\sqrt {ab}abDOption D. ab{a \over b}baCheckSkip · show solution