Question

If the ordinates of points \( P \) and \( Q \) on the parabola \[ y^2 = 12x \] are in the ratio 1:2, then the locus of the point of intersection of the normals to the parabola at \( P \) and \( Q \) is:

• A. \( y + 18 \left( \frac{x - 6}{21} \right)^{3/2} = 0 \)
• B. \( y - 18 \left( \frac{x - 6}{12} \right)^{3/2} = 0 \)
• C. \( y + 12 \left( \frac{x - 6}{14} \right)^{1/2} = 0 \)
• D. \( y - 12 \left( \frac{x - 6}{18} \right)^{1/2} = 0 \)

Hint:

For parabola problems, use the normal equation and substitution method to find the required locus.

Answer 1

The Correct Option is A

Step 1: Finding the normals at given points The given equation of the parabola is: \[ y^2 = 12x. \] For points \( P \) and \( Q \), we use the normal equation for a parabola: \[ y = m(x - 3m^2) + 6m. \] Solving for the locus of intersection of the normals, we get: \[ y + 18 \left( \frac{x - 6}{21} \right)^{3/2} = 0. \]