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 solve the problem, we need to find the equation of the normal from point P on the parabola \( y^2 = 4ax \) that is not along the lines PA and PB. We will follow these steps:

### Step 1: Find the slopes \( T_1 \) and \( T_2 \)

Given points:
- \( A(4a, -4a) \)
- \( B(9a, -6a) \)

For point A, we know the normal at any point on the parabola \( y^2 = 4ax \) is given by:
\[ y + 2aT_1 = -T_1(x - 4a) \]
Substituting the coordinates of point A:
\[ -4a + 2aT_1 = -T_1(4a - 4a) \]
This simplifies to:
\[ -4a + 2aT_1 = 0 \implies 2aT_1 = 4a \implies T_1 = 2 \]

For point B, similarly:
\[ y + 2aT_2 = -T_2(x - 9a) \]
Substituting the coordinates of point B:
\[ -6a + 2aT_2 = -T_2(9a - 9a) \]
This simplifies to:
\[ -6a + 2aT_2 = 0 \implies 2aT_2 = 6a \implies T_2 = 3 \]

### Step 2: Find the slope \( T_3 \)

Since the normals at points A and B meet at point P, we have the condition:
\[ T_1 + T_2 + T_3 = 0 \]
Substituting the values of \( T_1 \) and \( T_2 \):
\[ 2 + 3 + T_3 = 0 \implies T_3 = -5 \]

### Step 3: Write the equation of the normal at point P

The equation of the normal at any point on the parabola \( y^2 = 4ax \) with slope \( T_3 \) is given by:
\[ y + 2aT_3 = -T_3(x - 8a) \]
Substituting \( T_3 = -5 \):
\[ y + 2a(-5) = -(-5)(x - 8a) \]
This simplifies to:
\[ y - 10a = 5(x - 8a) \]
Expanding this:
\[ y - 10a = 5x - 40a \]
Rearranging gives:
\[ y - 5x + 30a = 0 \]
Thus, the equation of the normal from point P on the parabola \( y^2 = 4ax \) is:
\[ y + 5x = 30a \]

### Final Answer
The equation of the normal from point P on \( y^2 = 4ax \) is:
\[ y + 5x = 30a \]