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 \( \angle OPA = 90^\circ \). Then the locus of the centroid of such triangles \( OPA \) is :

• A. \( y^2 - 4x + 8 = 0 \)
• B. \( y^2 - 6x + 4 = 0 \)
• C. \( y^2 - 9x + 6 = 0 \)
• D. \( y^2 - 2x + 8 = 0 \)

Hint:

For locus problems, expressing the coordinates in terms of a single parameter \( t \) and then eliminating it is usually the most efficient method.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We use parametric coordinates for \( P \) on the parabola and satisfy the perpendicularity condition using slopes. Finally, we find the centroid and eliminate parameters to get the locus.

Step 2: Detailed Explanation:
Parabola: \( y^2 = 12x \implies a = 3 \).
Vertex: \( O(0, 0) \). Let \( P = (3t^2, 6t) \).
Let \( A = (x_0, 0) \) be on the \( x \)-axis.
Since \( \angle OPA = 90^\circ \), \( \text{slope of } OP \times \text{slope of } PA = -1 \).
\[ \frac{6t}{3t^2} \times \frac{0 - 6t}{x_0 - 3t^2} = -1 \]
\[ \frac{2}{t} \times \frac{-6t}{x_0 - 3t^2} = -1 \implies \frac{-12}{x_0 - 3t^2} = -1 \implies x_0 = 3t^2 + 12 \]
Let centroid \( G \) be \( (h, k) \):
\[ h = \frac{0 + 3t^2 + (3t^2 + 12)}{3} = 2t^2 + 4 \]
\[ k = \frac{0 + 6t + 0}{3} = 2t \implies t = k/2 \]
Substitute \( t \) in the expression for \( h \):
\[ h = 2(k/2)^2 + 4 = \frac{k^2}{2} + 4 \implies 2h = k^2 + 8 \]
Converting back to \( x, y \): \( y^2 - 2x + 8 = 0 \).

Step 3: Final Answer:
The locus is \( y^2 - 2x + 8 = 0 \).