Question

locus of centroid of the triangle whose vertices are (acost,asint),(bsint,-bcost)and(1,0),where t is the parameter , is

Answer 1

(acost,asint),(bsint,-bcost),(1,0)
Let locus of centroid be (h,k)
h=(acost+bsint+1)/3
3h-1=acost+bsint ------(1)
k=(asint-bcost+0)/3
3k=asint-bcost -------(2)
Squaring and adding (1) and (2) we get
(3h-1)²+(3k)²=a²+b²
Hence the locus is (3x-1)²+(3y)²=a²+b²

Answer 2

Answer:

first use the formula of centroid and then put yes co-ordinate as a h and y coordinates as a k then apply sin square theta + cos square theta is equal to 1 then we get answer 3 x minus y whole square + 3 y square is equal to a square + b square ok done