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: \[ y + 18 \left( \frac{x - 6}{21} \right)^{3/2} = 0. \]

Answer 2

Given the parabola:
\[ y^2 = 12x \] and points \( P \) and \( Q \) on it whose ordinates (y-coordinates) are in the ratio 1:2.
We need to find the locus of the point of intersection of the normals at \( P \) and \( Q \).

Step 1: Parameterize the parabola.
For parabola \( y^2 = 4ax \), here \(4a = 12 \Rightarrow a = 3\).
Coordinates on parabola can be expressed as:
\[ (x, y) = (a t^2, 2 a t) = (3 t^2, 6 t) \]

Step 2: Let the parameters corresponding to points \( P \) and \( Q \) be \( t_1 \) and \( t_2 \) respectively.
Since the ordinates are in ratio 1:2:
\[ \frac{y_P}{y_Q} = \frac{6 t_1}{6 t_2} = \frac{t_1}{t_2} = \frac{1}{2} \implies t_1 = \frac{t_2}{2} \]

Step 3: Equation of normal at parameter \( t \) on parabola \( y^2 = 4ax \):
\[ y + t x = 2 a t + a t^3 \] For \( a = 3 \), normal becomes:
\[ y + t x = 6 t + 3 t^3 \]

Step 4: Normals at \( P(t_1) \) and \( Q(t_2) \):
\[ y + t_1 x = 6 t_1 + 3 t_1^3 \] \[ y + t_2 x = 6 t_2 + 3 t_2^3 \]

Step 5: Find intersection of these two normals by subtracting:
\[ (y + t_1 x) - (y + t_2 x) = (6 t_1 + 3 t_1^3) - (6 t_2 + 3 t_2^3) \] \[ t_1 x - t_2 x = 6 (t_1 - t_2) + 3 (t_1^3 - t_2^3) \] \[ x (t_1 - t_2) = 6 (t_1 - t_2) + 3 (t_1^3 - t_2^3) \] Divide both sides by \( t_1 - t_2 \) (non-zero):
\[ x = 6 + 3 \frac{t_1^3 - t_2^3}{t_1 - t_2} \] Recall:
\[ t_1^3 - t_2^3 = (t_1 - t_2)(t_1^2 + t_1 t_2 + t_2^2) \] So:
\[ x = 6 + 3 (t_1^2 + t_1 t_2 + t_2^2) \]

Step 6: Substitute \( t_1 = \frac{t_2}{2} \):
\[ x = 6 + 3 \left( \left(\frac{t_2}{2}\right)^2 + \frac{t_2}{2} \times t_2 + t_2^2 \right) = 6 + 3 \left( \frac{t_2^2}{4} + \frac{t_2^2}{2} + t_2^2 \right) = 6 + 3 \times \frac{7 t_2^2}{4} = 6 + \frac{21}{4} t_2^2 \]

Step 7: Find \( y \) at intersection by substituting \( x \) in one normal, say:
\[ y + t_2 x = 6 t_2 + 3 t_2^3 \] \[ y = 6 t_2 + 3 t_2^3 - t_2 x \] Substitute \( x \):
\[ y = 6 t_2 + 3 t_2^3 - t_2 \left(6 + \frac{21}{4} t_2^2 \right) = 6 t_2 + 3 t_2^3 - 6 t_2 - \frac{21}{4} t_2^3 = 3 t_2^3 - \frac{21}{4} t_2^3 = - \frac{9}{4} t_2^3 \]

Step 8: Express \( t_2^2 \) in terms of \( x \):
\[ x = 6 + \frac{21}{4} t_2^2 \implies t_2^2 = \frac{4}{21} (x - 6) \] So: \[ t_2^3 = t_2 \cdot t_2^2 = t_2 \times \frac{4}{21} (x - 6) \] Since \( t_2 = \pm \sqrt{t_2^2} \), take the positive root for the locus.
Hence: \[ t_2^3 = \left(\frac{4}{21}\right)^{3/2} (x - 6)^{3/2} \] Ignoring constants, the locus is:
\[ y = -\frac{9}{4} t_2^3 = -18 \left( \frac{x - 6}{21} \right)^{3/2} \] Rewrite as:
\[ y + 18 \left( \frac{x - 6}{21} \right)^{3/2} = 0 \]

Therefore, the locus of the point of intersection of the normals is:
\[ \boxed{ y + 18 \left( \frac{x - 6}{21} \right)^{3/2} = 0 } \]