Question

Let \(O\) be the vertex of the parabola \(y^{2}=16x\). The locus of the centroid of \(\triangle OPA\), when point \(P\) lies on the parabola and point \(A\) lies on the \(x\)-axis such that \(\angle OPA = 90^\circ\), is:

• A. \( y^{2} = 8(3x - 16) \)
• B. \( 9y^{2} = 8(3x - 16) \)
• C. \( y^{2} = 8(3x + 16) \)
• D. \( 9y^{2} = 8(3x + 16) \)

Answer 1

The Correct Option is B

Concept:

• Take a general point \( P \) on the parabola using parametric form.
• Use the condition \( \angle OPA = 90^\circ \) via dot product.
• Find coordinates of the centroid and eliminate the parameter.

Step 1: Coordinates of Points

The equation of the parabola is \( y^2 = 16x \). The general point \( P \) on the parabola can be expressed as: \[ P(4t^2, 8t) \] Where \( t \) is the parameter, and the vertex of the parabola is at \( O(0, 0) \). Let point \( A \) lie on the x-axis at \( A(a, 0) \).

Step 2: Apply the Right-Angle Condition \( \angle OPA = 90^\circ \)

Vectors from point \( O \) to point \( P \) and from point \( P \) to point \( A \) are: \[ \overrightarrow{PO} = (-4t^2, -8t) \] \[ \overrightarrow{PA} = (a - 4t^2, -8t) \] Since the angle between vectors \( \overrightarrow{PO} \) and \( \overrightarrow{PA} \) is \( 90^\circ \), their dot product is zero: \[ \overrightarrow{PO} \cdot \overrightarrow{PA} = 0 \] Expanding the dot product: \[ (-4t^2)(a - 4t^2) + (-8t)(-8t) = 0 \] Simplifying: \[ -4t^2a + 16t^4 + 64t^2 = 0 \] \[ -a + 4t^2 + 16 = 0 \] Solving for \( a \): \[ a = 4t^2 + 16 \] Thus, the coordinates of point \( A \) are \( A(4t^2 + 16, 0) \).

Step 3: Coordinates of the Centroid

The centroid \( G(x, y) \) of triangle \( OPA \) is given by the average of the coordinates of points \( O(0, 0) \), \( P(4t^2, 8t) \), and \( A(4t^2 + 16, 0) \): \[ x = \frac{0 + 4t^2 + (4t^2 + 16)}{3} = \frac{8t^2 + 16}{3} \] \[ y = \frac{0 + 8t + 0}{3} = \frac{8t}{3} \]

Step 4: Eliminate the Parameter \( t \)

From \( y = \frac{8t}{3} \), we can solve for \( t \): \[ t = \frac{3y}{8} \] Substitute this value of \( t \) into the equation for \( x \): \[ x = \frac{8}{3} \left(\frac{3y}{8}\right)^2 + 16 \] Simplifying: \[ x = \frac{9y^2}{64} + 16 \] Rearranging the terms: \[ x = \frac{9y^2}{64} + 2 \] Now, multiplying through by 64 to simplify: \[ 64x = 9y^2 + 128 \] Finally, rearranging: \[ 9y^2 = 8(3x - 16) \]

Final Answer: The locus of the centroid is:

\[ 9y^2 = 8(3x - 16) \]