A) \[x=0\]
B) \[y=0\]
C) \[x=-a\]
D) \[y=a\]
Correct Answer: B
Solution :
Equation of normal in terms of slope, \[y=mx-2am-a{{m}^{3}}\] \[\therefore \] Normal meets in \[(h,\,k),\] then \[a{{m}^{3}}-(h-2a)m+k=0\] Let \[P(am_{1}^{2},\,\,-2a{{m}_{1}}),\] \[Q\equiv \,(am_{2}^{2},\,-2a{{m}_{2}})\] and \[R\equiv \,(am_{3}^{2},\,-2a{{m}_{3}})\] \[\therefore \] \[{{m}_{1}}+{{m}_{2}}+{{m}_{3}}=0\] \[\Rightarrow \] \[\frac{-2a{{m}_{1}}-2a{{m}_{2}}-2a{{m}_{3}}}{3}=0\] \[\therefore \] y-coordintates of centroid of \[\therefore \] Centroid lies on x-axis (y = 0).
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