x=3cost+bsint+1⇒acost+bsint=3x−1 y=3asint−bcost⇒asint−bsint=3y Squaring & adding, (3x−1)2+(3y)2=a2+b2
Locus of a centriod of the triangle whose vertices are (a cos t, a sin t),(b sin t,-b cos t) and (1,0), where t is a parameter, is
JEE Main 2003 — Mathematics Coordinate Geometry
Locus of a centriod of the triangle whose vertices are (acost,asint),(bsint,−bcost) and (1,0), where t is a parameter, is
Held on 30 Apr 2003 · Verified 6 Jul 2026.
Options
- A
(3x+1)2+(3y)2=a2−b2
- B
(3x−1)2+(3y)2=a2−b2
- C
(3x−1)2+(3y)2=a2+b2
- D
(3x+1)2+(3y)2=a2+b2
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