Normals drawn to `y^(2)=4ax` at the points where it is intersected by the line `y = mx +c` intersected at P. Coordinates of foot of the another normal drawn to the parabola from the point 'P' is

Options

A. `((a)/(m^(2)), -(2a)/(m))`
B. `(9)/(m), - (6a)/(m)`
C. `(a m^(2), -2a m)`
D. `((4a)/(m^(2)), - (4a)/(m))`

Answer

To solve the problem, we need to find the coordinates of the foot of the normal drawn to the parabola \( y^2 = 4ax \) from the point of intersection \( P \) of the line \( y = mx + c \) with the parabola.

### Step 1: Find the points of intersection of the line and the parabola.

The equation of the parabola is given by:
\[ y^2 = 4ax \]
Substituting \( y = mx + c \) into the parabola's equation:
\[ (mx + c)^2 = 4ax \]
Expanding this:
\[ m^2x^2 + 2mcx + c^2 = 4ax \]
Rearranging gives us a quadratic equation in \( x \):
\[ m^2x^2 + (2mc - 4a)x + c^2 = 0 \]
Let the roots of this quadratic equation be \( T_1 \) and \( T_2 \).

### Step 2: Use the roots to find the coordinates of intersection points.

The coordinates of the points of intersection \( A \) and \( B \) can be expressed as:
\[ A = (aT_1^2, 2aT_1) \quad \text{and} \quad B = (aT_2^2, 2aT_2) \]

### Step 3: Determine the slope of the line joining points A and B.

The slope \( m \) of the line joining points \( A \) and \( B \) is given by:
\[ m = \frac{2aT_2 - 2aT_1}{aT_2^2 - aT_1^2} = \frac{2a(T_2 - T_1)}{a(T_2^2 - T_1^2)} = \frac{2(T_2 - T_1)}{T_2 + T_1} \]
From this, we can express \( T_1 + T_2 \) in terms of \( m \):
\[ T_1 + T_2 = \frac{2}{m} \]

### Step 4: Find the coordinates of the foot of the normal from point P.

The foot of the normal from point \( P \) to the parabola can be expressed as:
\[ C = (aT_3^2, 2aT_3) \]
where \( T_3 \) is related to \( T_1 \) and \( T_2 \) by:
\[ T_3 = -(T_1 + T_2) = -\frac{2}{m} \]

### Step 5: Substitute \( T_3 \) to find the coordinates of the foot of the normal.

Substituting \( T_3 \) into the coordinates of point \( C \):
\[ C = \left(a\left(-\frac{2}{m}\right)^2, 2a\left(-\frac{2}{m}\right)\right) \]
Calculating this gives:
\[ C = \left(a \cdot \frac{4}{m^2}, -\frac{4a}{m}\right) \]

### Final Result

Thus, the coordinates of the foot of the normal drawn to the parabola from point \( P \) are:
\[ \left(\frac{4a}{m^2}, -\frac{4a}{m}\right) \]