Question
The maximum distance from origin of a point on the curve
$$x = a\sin t - b\left( {\frac{{at}}{b}} \right)$$
$$y = a\cos t - b\cos \left( {\frac{{at}}{b}} \right),$$ both $$a,b > 0$$ is
A.
$$a - b$$
B.
$$a + b$$
C.
$$\sqrt {{a^2} + {b^2}} $$
D.
$$\sqrt {{a^2} - {b^2}} $$
Answer :
$$a + b$$
Solution :
$$\eqalign{
& {\text{Distance}}\,{\text{of}}\,{\text{origin}}\,{\text{from}}\,\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr
& = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - \frac{{at}}{b}} \right)} ; \cr
& \leqslant \sqrt {{a^2} + {b^2} + 2ab} \left[ {{{\left\{ {\cos \left( {t - \frac{{at}}{b}} \right)} \right\}}_{\min }} = - 1} \right] \cr
& = a + b \cr
& \therefore {\text{Maximum}}\,{\text{distance}}\,{\text{from}}\,{\text{origin}} = a + b \cr} $$