Verified Solution • 2025–26 Session moderate

QJEEM02Maceb3: JEE Main 2002 Mathematics - Class 12 JEEM02Maceb3 Solution: The maximum distance from origin of a point on the curve both is

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

Multiple Choice Options

  • A a−ba-b
  • B a+ba+b
  • C a2+b2\sqrt {{a^2} + {b^2}}
  • D a2−b2\sqrt {{a^2} - {b^2}}

Step-by-Step Verified Solution

Distance of origin from (x,y)=x2+y2\left( {x,y} \right) = \sqrt {{x^2} + {y^2}}

=a2+b2−2abcos⁡(t−atb);= \sqrt {{a^2} + {b^2} - 2ab\cos \left( {t - {{at} \over b}} \right)} ;

≤a2+b2+2ab\le \sqrt {{a^2} + {b^2} + 2ab} [{cos⁡(t−atb)}min⁡=−1]\left[ {{{\left\{ {\cos \left( {t - {{at} \over b}} \right)} \right\}}_{\min }} = - 1} \right]

=a+b=a+b

∴\therefore Maximum distance from origin =a+b=a+b

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring