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 angle OPA = 90°. Then the locus of the centroid of such triangles OPA is

A locus is the set of all points satisfying a given geometric condition. The standard method for locus problems:

Step 1: Take a general point using parametric form (e.g., $(at^2, 2at)$ for parabola).
Step 2: Apply the given geometric condition to find relationships between coordinates and parameter.
Step 3: Express the coordinates of the required point (here centroid) in terms of parameter $t$.
Step 4: Eliminate $t$ between the coordinate expressions to get the locus equation.

This 4-step method works for all locus problems in JEE — centroid, circumcenter, midpoint, etc.

For parabola $y^2 = 4ax$ with $a > 0$:

Property	Value
Vertex	$(0,0)$
Focus	$(a,0)$
Directrix	$x = -a$
Axis	$x$-axis ($y=0$)
Latus Rectum length	$4a$
Parametric point	$(at^2, 2at)$
Tangent at $(at^2,2at)$	$ty = x + at^2$
Normal at $(at^2,2at)$	$y + tx = 2at+at^3$

For $y^2=12x$: $a=3$, focus=$(3,0)$, directrix $x=-3$, LR length $=12$.

Two vectors $\vec{u} = (u_1,u_2)$ and $\vec{v} = (v_1,v_2)$ are perpendicular if and only if: $$\vec{u}\cdot\vec{v} = u_1v_1 + u_2v_2 = 0$$ In this problem, $\angle OPA = 90°$ means the angle at vertex $P$ is $90°$. So vectors from P to O and from P to A must be perpendicular.

Important: The right angle is at $P$, so vectors $\vec{PO}$ and $\vec{PA}$ (both starting at $P$) must have zero dot product — not $\vec{OP}$ and $\vec{OA}$.

Equivalently, using slopes: if $m_1$ = slope of $PO$ and $m_2$ = slope of $PA$, then $m_1 \cdot m_2 = -1$.

The centroid $G$ of triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$: $$G = \left(\frac{x_1+x_2+x_3}{3},\; \frac{y_1+y_2+y_3}{3}\right)$$ Key properties of centroid:
• Divides each median in ratio $2:1$ from vertex
• Always lies inside the triangle
• Centroid of $\triangle OPA$: since $O=(0,0)$ and $A$ is on x-axis, the $y$-coordinate of centroid is simply $y_P/3$
• In this problem: $k = 6t/3 = 2t$ (only $P$ contributes to y-coordinate)

The locus $y^2 = 2x – 8 = 2(x-4)$ is a parabola:

Compare with $Y^2 = 4aX$ where $Y=y$, $X=x-4$:
$4a = 2 \implies a = 1/2$

Properties of locus parabola:
• Vertex: $(4,0)$ (shifted 4 units right from origin)
• Axis: $y=0$ (x-axis), opens rightward
• Focus: $(4+\frac{1}{2}, 0) = (4.5, 0)$
• Directrix: $x = 4 – \frac{1}{2} = 3.5$
• LR length: $4a = 2$

This means as $P$ moves along $y^2=12x$, the centroid traces a smaller parabola with vertex at $(4,0)$.

Slope of $PO$: $m_1 = \dfrac{0-6t}{0-3t^2} = \dfrac{-6t}{-3t^2} = \dfrac{2}{t}$

Slope of $PA$: $m_2 = \dfrac{0-6t}{\lambda-3t^2} = \dfrac{-6t}{\lambda-3t^2}$

For perpendicularity: $m_1 \cdot m_2 = -1$: $$\frac{2}{t} \cdot \frac{-6t}{\lambda-3t^2} = -1$$ $$\frac{-12}{\lambda-3t^2} = -1$$ $$\lambda – 3t^2 = 12 \implies \lambda = 3t^2+12$$ Same result as dot product method ✓

❌ Mistake 1: Taking angle at wrong vertex. $\angle OPA = 90°$ is the angle at $P$, not at $O$ or $A$. Always identify which vertex has the right angle before forming the perpendicularity condition.

❌ Mistake 2: Forgetting to divide by $t^2$ after applying dot product. This is valid only when $t \neq 0$ (i.e., $P \neq O$).

❌ Mistake 3: Using centroid formula incorrectly — it’s $(\sum x_i)/3$, not $(\sum x_i)/2$ (that’s midpoint for two points).

❌ Mistake 4: Writing final locus as $y^2 = 2x – 8$ without rearranging to match the options format $y^2 – 2x + 8 = 0$.

❌ Mistake 5: Computing $h$ as $(0+3t^2)/2$ — centroid involves all three vertices including $A$.

$y^2=12x$: $a=3$, parametric $(3t^2,6t)$ $\angle OPA=90°$: $\vec{PO}\cdot\vec{PA}=0$ $A=(3t^2+12,0)$ Centroid: $h=2t^2+4$, $k=2t$ $t=k/2$, eliminate: $k^2=2h-8$ Locus: $y^2-2x+8=0$