The maximum distance from origin of a point on the curve x = a\sin t - b\sin \left( {{{at} \over b}} \right) y = a\cos t - b\cos \left( {{{at} \over b}} \right), both a,b 0 is

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The maximum distance from origin of a point on the curve x = a\sin t - b\sin \left( {{{at} \over b}} \right) y = a\cos t - b\cos \left( {{{at} \over b}} \right), both a,b > 0 is

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Answer: B. a+b Distance of origin from \left( {x,y} ight) = \sqrt {{x^2} + {y^2}} = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} ight)} ; \le \sqrt {{a^2} + {b^2} + 2ab} \left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} ight)} ight\}}_{\min }} = - 1} ight] =a+b 	herefore Maximum distance from origin =a+b

The maximum distance from origin of a point on the curve $$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$ $$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is

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