Parabola
Locus of Centroid — Right Angle Condition
nta_pyq_2026_jan
Grade 11
Question:
Let $y^2=12x$ 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 $\angle OPA=90°$. Then the locus of the centroid of such triangles $OPA$ is:
$y^2-2x+8=0$
$y^2-9x+6=0$
$y^2-4x+8=0$
$y^2-6x+4=0$
Step-by-Step Solution
Key Concept: $P=(3t^2,6t)$, $A=(h,0)$. $\angle OPA=90°\Rightarrow\vec{PO}\cdot\vec{PA}=0\Rightarrow(-3t^2)(h-3t^2)+(-6t)(-6t)=0\Rightarrow h=3t^2+12$. Centroid $G=\left(\tfrac{3t^2+h}{3},2t\right)=(2t^2+4,2t)$.
Locus: $y^2-2x+8=0$.
Correct Answer: 1