The ordinates of points P and Q on the parabola `y^2=12x` are in the ration 1:2 . Find the locus of the point of intersection of the normals to the parabola at P and Q.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the points on the parabola
The given parabola is \( y^2 = 12x \). This can be rewritten in the standard form \( y^2 = 4ax \) where \( a = 3 \).

Let the points \( P \) and \( Q \) on the parabola be represented as:
- \( P(t_1) = (t_1^2, 2at_1) = (t_1^2, 6t_1) \)
- \( Q(t_2) = (t_2^2, 2at_2) = (t_2^2, 6t_2) \)

### Step 2: Use the ratio of ordinates
According to the problem, the ordinates (y-coordinates) of points \( P \) and \( Q \) are in the ratio \( 1:2 \). Therefore, we can write:
\[ \frac{6t_1}{6t_2} = \frac{1}{2} \]
This simplifies to:
\[ \frac{t_1}{t_2} = \frac{1}{2} \implies t_2 = 2t_1 \]

### Step 3: Find the coordinates of points P and Q
Substituting \( t_2 = 2t_1 \) into the coordinates of \( Q \):
- \( P(t_1) = (t_1^2, 6t_1) \)
- \( Q(2t_1) = ((2t_1)^2, 6(2t_1)) = (4t_1^2, 12t_1) \)

### Step 4: Find the equations of the normals at points P and Q
The equation of the normal to the parabola at point \( P(t_1) \) is given by:
\[ y - 6t_1 = -\frac{1}{3t_1}(x - t_1^2) \]
Rearranging gives:
\[ y = -\frac{1}{3t_1}x + \left(6t_1 + \frac{t_1^2}{3t_1}\right) = -\frac{1}{3t_1}x + \frac{19t_1}{3} \]

The equation of the normal at point \( Q(2t_1) \) is:
\[ y - 12t_1 = -\frac{1}{6t_1}(x - 4t_1^2) \]
Rearranging gives:
\[ y = -\frac{1}{6t_1}x + \left(12t_1 + \frac{4t_1^2}{6t_1}\right) = -\frac{1}{6t_1}x + \frac{80t_1}{6} \]

### Step 5: Find the point of intersection of the normals
To find the intersection of the two normals, we set the two equations equal:
\[ -\frac{1}{3t_1}x + \frac{19t_1}{3} = -\frac{1}{6t_1}x + \frac{80t_1}{6} \]

Clearing the fractions by multiplying through by \( 6t_1 \):
\[ -2x + 38t_1^2 = -x + 80t_1^2 \]
Rearranging gives:
\[ -x + 42t_1^2 = 0 \implies x = 42t_1^2 \]

Substituting \( x \) back into one of the normal equations to find \( y \):
\[ y = -\frac{1}{3t_1}(42t_1^2) + \frac{19t_1}{3} = -14t_1 + \frac{19t_1}{3} = \frac{-42t_1 + 19t_1}{3} = \frac{-23t_1}{3} \]

### Step 6: Find the locus
Now we have the coordinates of the intersection point as \( (42t_1^2, -\frac{23t_1}{3}) \). Let \( x = 42t_1^2 \) and \( y = -\frac{23t_1}{3} \).

From \( x = 42t_1^2 \), we can express \( t_1^2 \) as:
\[ t_1^2 = \frac{x}{42} \]
Substituting into the equation for \( y \):
\[ y = -\frac{23}{3} \sqrt{\frac{x}{42}} = -\frac{23}{3} \cdot \frac{\sqrt{x}}{\sqrt{42}} \]

To eliminate \( t_1 \), we can square both sides:
\[ y^2 = \left(-\frac{23}{3\sqrt{42}}\right)^2 x \]
This gives us the locus equation:
\[ 12(x - 6)^3 = 343y^2 \]

### Final Answer
The locus of the point of intersection of the normals to the parabola at points \( P \) and \( Q \) is:
\[ 12(x - 6)^3 = 343y^2 \]