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Locus of a centriod of the triangle whose vertices are (a t, a t),(b t,-b t) and (1,0) , where t is a parameter, is

Options

  1. A(3 x+1)^2+(3 y)^2=a^2-b^2
  2. B(3 x-1)^2+(3 y)^2=a^2-b^2
  3. C(3 x-1)^2+(3 y)^2=a^2+b^2
  4. D(3 x+1)^2+(3 y)^2=a^2+b^2

Correct answer

C. (3 x-1)^2+(3 y)^2=a^2+b^2

Step-by-step solution

x= t+b t+1 3 a t+b t=3 x-1y= a t-b t 3 a t-b t=3 y Squaring & adding, (3 x -1)^2+(3 y )^2= a ^2+ b ^2

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