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) \[{{(3x+1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}-{{b}^{2}}\]
Correct Answer: B
Solution :
Let coordinate of the centroid of the triangle be (h k), then \[h=\frac{a\cos t+b\sin t+1}{3}\] \[\Rightarrow \] \[3h-1=a\cos t+b\sin t\] ...(i) and \[k=\frac{a\sin t-b\cos t}{3}\] \[\Rightarrow \] \[3k=a\sin t-b\cos t\] ...(ii) On squaring and adding Eqs. (i) and (ii), \[{{(3h-1)}^{2}}+{{(3k)}^{2}}={{a}^{2}}{{\cos }^{2}}t+{{b}^{2}}{{\sin }^{2}}t\] \[+2ab\cos t\sin t+{{a}^{2}}{{\sin }^{2}}t+{{b}^{2}}{{\cos }^{2}}t\] \[-2ab\cos t\sin t\] \[\Rightarrow \] \[{{(3h-1)}^{2}}+3{{k}^{2}}={{a}^{2}}+{{b}^{2}}\] \[\therefore \]Required locus is \[{{(3x-1)}^{2}}+{{(3y)}^{2}}={{a}^{2}}+{{b}^{2}}\]
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