Prove that the locus of the centroid of the triangle whose vertices are `(acost ,asint),(bsint ,-bcost),` and `(1,0)` , where `t` is a parameter, is circle.

Find the locus of the centroid of a triangle whose vertices are (a cos t,a sin t),(b sin t,-b cos t) and (1,0) where t' is the parameter.

Locus of centroid of the triangle whose vertices are (a cos t, a sin t), (b sin t, - b cos t) and (1, 0), where is a

A locus is the curve traced out by a point which moves under certain geomatrical conditions: To find the locus of a point first we assume the co-ordinates of the moving point as (h,k) and then try to find a relation between h and k with the help of the given conditions of the problem. If there is any variable involved in the process then we eliminate them. At last we replace h by x and k by y and get the locus of the point which will be an equation in x and y. On the basis of above information, answer the following questions Locus of centroid of the triangle whose vertices are (acost, asint), (bsint,-b cost) and (1, 0) where t is a parameter is -

Locus of centroid of the triangle whose vertices are (a cos t ,a s in t ), (b s in t ,-b cos t ) and (1, 0), where t is a parameter, is (3x-1)^2+(3y)^2=a^2-b^2 (3x-1)^2+(3y)^2=a^2+b^2 (3x+1)^2+(3y)^2=a^2+b^2 (3x+1)^2+(3y)^2=a^2-b^2

Locus of centroid of the triangle whose vertices are (acost, asint), (bsint,-bcost) and (1,0) where t is a parameter is : (A) (3x+1)^2 + (3y)^2 = a^2 - b^2 (B) (3x-1)^2 + (3y)^2 = a^2 - b^2 (C) (3x-1)^2 + (3y)^2 = a^2 + b^2 (D) (3x+1)^2 + (3y)^2 = a^2 + b^2

The centroid of the triangle whose vertices are (3, 10), (7, 7), (-2, 1) is

Locus of centroid 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: (3x-1)^(2)+(3y)^(2)=a^(2)-b^(2)(3x-1)^(2)+(3y)^(2)=a^(2)+b^(2)(3x+1)^(2)+(3y)^(2)=a^(2)+b^(2)(3x+1)^(2)+(3y)^(2)=a^(2)-b^(2)

The co-ordinates of the centroid of a triangle whose vertices are (0,6),(8,12) and (8,0).