Let the normals at points `A(4a, -4a)` and `B(9a, -6a)` on the parabola `y^(2)=4ax` meet at the point P. The equation of the nornal from P on `y^(2)=4ax` (other than PA and PB) is

Options

A. `5x+y-135a=0`
B. `5x-y+115a=0`
C. `5x+y+115=0`
D. `5x-y-115a=0`

Answer

To find the equation of the normal from point P on the parabola \( y^2 = 4ax \) that is not PA or PB, we will follow these steps:

### Step 1: Identify the points A and B on the parabola
Given points A and B are \( A(4a, -4a) \) and \( B(9a, -6a) \).

### Step 2: Determine the parameters \( T_1 \) and \( T_2 \)
For point A:
- The coordinates can be expressed in terms of the parameter \( T_1 \) as \( (2aT_1, aT_1^2) \).
- Setting \( 2aT_1 = 4a \) gives \( T_1 = 2 \).
- Setting \( aT_1^2 = -4a \) gives \( T_1^2 = -4 \) (not applicable since \( T_1 \) must be real).

For point B:
- The coordinates can be expressed in terms of the parameter \( T_2 \) as \( (2aT_2, aT_2^2) \).
- Setting \( 2aT_2 = 9a \) gives \( T_2 = 4.5 \).
- Setting \( aT_2^2 = -6a \) gives \( T_2^2 = -6 \) (not applicable since \( T_2 \) must be real).

### Step 3: Calculate the values of \( T_1 \) and \( T_2 \)
From the coordinates:
- For point A, \( T_1 = -2 \) (as derived from the y-coordinate).
- For point B, \( T_2 = -3 \) (as derived from the y-coordinate).

### Step 4: Find \( T_3 \) using the condition \( T_1 + T_2 + T_3 = 0 \)
Substituting the values:
\[ -2 - 3 + T_3 = 0 \implies T_3 = 5 \]

### Step 5: Write the equation of the normal at point P
The coordinates of the point corresponding to \( T_3 \) are:
\[ (2aT_3, aT_3^2) = (10a, 25a) \]

The equation of the normal at point \( (10a, 25a) \) is given by:
\[ y - 25a = -T_3(x - 10a) \]
Substituting \( T_3 = 5 \):
\[ y - 25a = -5(x - 10a) \]
This simplifies to:
\[ y - 25a = -5x + 50a \]
Rearranging gives:
\[ y + 5x - 75a = 0 \]

### Final Answer
The equation of the normal from point P on the parabola \( y^2 = 4ax \) (other than PA and PB) is:
\[ y + 5x - 75a = 0 \]