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:
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:
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For parabola problems, use the normal equation and substitution method to find the required locus.
Updated On: Mar 24, 2025
- \( y + 18 \left( \frac{x - 6}{21} \right)^{3/2} = 0 \)
- \( y - 18 \left( \frac{x - 6}{12} \right)^{3/2} = 0 \)
- \( y + 12 \left( \frac{x - 6}{14} \right)^{1/2} = 0 \)
- \( y - 12 \left( \frac{x - 6}{18} \right)^{1/2} = 0 \)
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The Correct Option is A
Solution and Explanation
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:
\[
y + 18 \left( \frac{x - 6}{21} \right)^{3/2} = 0.
\]
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