If `t` is parameter, `A=(aSect,b tan t)` and `B=(-atant,b,Sect)`, `O=(0,0)` then the locus of the centroid of `Delta OAB` is `x^(2)+10xy+25y^(2)+ax+by=0` then `b-5a=`

If t is parameter, A=(a sec t,b tan t) and B=(-a tan t,b sec t) ,O=(0,0), then the locus of the centroid of Delta OAB is

If t is parameter,A=(a sec t,b tan t) and B=(-a tan t,b sec t),O=(0,0) then be locus of the centroid of Delta OAB is

If t is parameter,A=(a sec t,b tan t) and B=(-a tan t,b sec t),O=(0,0) then be locus of the centroid of Delta OAB is

If theta is parameter, A=(a cos theta,a sin theta) and B=(b sin theta,-b cos theta)C=(1,0) then the locus of the centroid of Delta ABC

A=(1,2),B(3,4) and the locus of C is 2x+3y=5 then the locus of the centroid of Delta ABC is

A=(2,5),B=(4,-11) and the locus of ^ prime C' is 9x+7y+4=0 then the locus of the centroid of ABC is

circle of radius 3k passes through (0,0) and cuts the axes in A and B then the locus of centroid of triangle OAB is

A circle of radius 3k passes through (0.0) and cuts the axes in A and B then the locus of centroid of triangle OAB is

A line from (-1,0) intersects the parabola x^(2)= 4y at A and B. Then the locus of centroid of DeltaOAB is (where O is origin)

A variable straight line passes through a fixed point (a, b) intersecting the co-ordinates axes at A & B. If 'O' is the origin, then the locus of the centroid of the triangle OAB is (A) bx+ ay-3xy=0 (B) bx+ay-2xy=0 (C) ax+by-3xy=0 (D) ax+by-2xy=0.