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

We are given the parabola \( y^2 = 12x \) with its vertex at the origin \( O(0, 0) \). A point \( P(x_1, y_1) \) lies on this parabola, so it satisfies the equation \( y_1^2 = 12x_1 \).

Let \( A(a, 0) \) be a point on the \( x \)-axis. We are given that \(\angle OPA = 90^\circ\). The condition of perpendicularity suggests that the dot product of vectors \( \overrightarrow{OP} \) and \( \overrightarrow{PA} \) is zero. Let's calculate these vectors and their dot product:

• \( \overrightarrow{OP} = (x_1, y_1) \) 
• \( \overrightarrow{PA} = (a - x_1, -y_1) \)

The dot product is:

\(x_1(a - x_1) + y_1(-y_1) = 0\)

Which simplifies to:

\(x_1a - x_1^2 - y_1^2 = 0\)

Substituting \( y_1^2 = 12x_1 \) into the equation gives:

\(x_1a - x_1^2 - 12x_1 = 0\)

Rearranging terms, we have:

\(x_1(a - x_1 - 12) = 0\)

Since \( x_1 \neq 0 \) because \( P \) is on the parabola (not at the vertex), we have:

\(a = x_1 + 12\)

Now, let's find the centroid of triangle \( OPA \). The coordinates of the centroid \( G \) are given by:

\( G \left( \frac{0 + x_1 + a}{3}, \frac{0 + y_1 + 0}{3} \right) \)

Substitute \( a = x_1 + 12 \):

\( G \left( \frac{x_1 + x_1 + 12}{3}, \frac{y_1}{3} \right) = \left( \frac{2x_1 + 12}{3}, \frac{y_1}{3} \right) \)

Now using \( y_1^2 = 12x_1 \), we can express \( y_1 \) in terms of \( x_1 \) as \( y_1 = \sqrt{12x_1} \). Thus, the coordinates of \( G \) become:

\( G \left( \frac{2x_1 + 12}{3}, \frac{\sqrt{12x_1}}{3} \right) \)

The locus of the centroid \( (X, Y) \) is:

\(

From \( Y = \frac{\sqrt{12x_1}}{3} \), we have \( Y^2 = \frac{12x_1}{9} \) or \( x_1 = \frac{3}{4}Y^2 \).

Substituting \( x_1 \) in the expression for \( X \):

\(X = \frac{2 (\frac{3}{4} Y^2) + 12}{3}\)

Simplifying gives:

\(X = \frac{\frac{3}{2} Y^2 + 12}{3} = \frac{3}{2*3}Y^2 + \frac{12}{3} = \frac{1}{2}Y^2 + 4\)

Therefore, the locus equation is derived from:

\(X = \frac{1}{2}Y^2 + 4 \to 2X = Y^2 + 8 \Rightarrow Y^2 = 2x - 8\)

This corresponds to the equation \( y^2 - 2x + 8 = 0 \). Hence, the correct answer is:

Option: \( y^2 - 2x + 8 = 0 \)