JEE MAIN - Mathematics (2002 - No. 16)
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
$$x = a\sin t - b\sin \left( {{{at} \over b}} \right)$$
$$y = a\cos t - b\cos \left( {{{at} \over b}} \right),$$ both $$a,b > 0$$ is
$$a-b$$
$$a+b$$
$$\sqrt {{a^2} + {b^2}} $$
$$\sqrt {{a^2} - {b^2}} $$
Explanation
Distance of origin from $$\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $$
$$ = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;$$
$$ \le \sqrt {{a^2} + {b^2} + 2ab} $$ $$\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]$$
$$=a+b$$
$$\therefore$$ Maximum distance from origin $$=a+b$$
$$ = \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;$$
$$ \le \sqrt {{a^2} + {b^2} + 2ab} $$ $$\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]$$
$$=a+b$$
$$\therefore$$ Maximum distance from origin $$=a+b$$
Comments (0)
