The locus of point of intersection of two normals drawn to the parabola `y^2 = 4ax` are perpendicular to each other is

Options

A. `y^2 = 2a (x-a)`
B. `y^2 = a (x-4a)`
C. `y^2 = a (x-3a)`
D. `y^2 = 4a (x+a)`

Answer

To find the locus of the point of intersection of two normals drawn to the parabola \( y^2 = 4ax \) that are perpendicular to each other, we can follow these steps:

### Step 1: Equation of the Normal
The equation of the normal to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ y + tx = 2at + at^3 \]
Rearranging gives:
\[ y = -tx + 2at + at^3 \]

### Step 2: Points of Intersection
Let’s consider two normals at parameters \( t_1 \) and \( t_2 \). The equations of the normals can be written as:
1. \( y = -t_1 x + 2a t_1 + a t_1^3 \)
2. \( y = -t_2 x + 2a t_2 + a t_2^3 \)

To find the point of intersection of these two normals, we set the right-hand sides equal:
\[ -t_1 x + 2a t_1 + a t_1^3 = -t_2 x + 2a t_2 + a t_2^3 \]

### Step 3: Rearranging the Equation
Rearranging gives:
\[ (t_2 - t_1)x = 2a(t_2 - t_1) + a(t_2^3 - t_1^3) \]
If \( t_1 \neq t_2 \), we can divide through by \( t_2 - t_1 \):
\[ x = 2a + a \frac{t_2^3 - t_1^3}{t_2 - t_1} \]

### Step 4: Simplifying the Expression
Using the identity \( t_2^3 - t_1^3 = (t_2 - t_1)(t_2^2 + t_2 t_1 + t_1^2) \), we can simplify:
\[ x = 2a + a(t_2^2 + t_2 t_1 + t_1^2) \]
Thus,
\[ x = 2a + a(t_2^2 + t_2 t_1 + t_1^2) \]

### Step 5: Finding the y-coordinate
Substituting \( x \) back into one of the normal equations to find \( y \):
\[ y = -t_1\left(2a + a(t_2^2 + t_2 t_1 + t_1^2)\right) + 2a t_1 + a t_1^3 \]

### Step 6: Condition for Perpendicular Normals
For the normals to be perpendicular, the product of their slopes must equal -1:
\[ t_1 t_2 = -1 \]

### Step 7: Substituting the Condition
Substituting \( t_2 = -\frac{1}{t_1} \) into the expression for \( x \) and simplifying will yield the locus equation.

### Step 8: Final Locus Equation
After substituting and simplifying, we find the locus of the point of intersection of the normals:
\[ y^2 = 4a(x - 3a) \]

### Conclusion
The locus of the point of intersection of two normals drawn to the parabola \( y^2 = 4ax \) that are perpendicular to each other is given by:
\[ y^2 = 4a(x - 3a) \]