Parabola y2=12x, a=3. Let P=(3t2,6t), O=(0,0), A=(h,0).
∠OPA=90∘⇒PO⋅PA=0:
(−3t2)(h−3t2)+(−6t)(−6t)=0⇒h=3t2+12.
Centroid G=(30+3t2+h,30+6t+0)=(2t2+4, 2t).
Let G=(X,Y): t=Y/2, X=Y2/2+4.
Locus: y2−2x+8=0.
Let y2=12x be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that ∠OPA=90∘. Then the locus of the centroid of such triangles OPA is :
Held on 21 Jan 2026 · Verified 6 Jul 2026.
y2−2x+8=0
y2−9x+6=0
y2−4x+8=0
y2−6x+4=0
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