The set of points on the axis of the parabola `y^2 =4ax`, from which three distinct normals can be drawn to theparabola `y^2 = 4ax`, is

Options

A. `{(x, 0):xgta}`
B. `{(x, 0):xgt2a}`
C. `{(x, xgt4a}`
D. `{x:altxlt2a}`

Explanation

To solve the problem of finding the set of points on the axis of the parabola \( y^2 = 4ax \) from which three distinct normals can be drawn, we will follow these steps:

### Step 1: Understand the Normal to the Parabola
The equation of the normal to the parabola \( y^2 = 4ax \) at a point \( (at^2, 2at) \) is given by:
\[ y = mx - 2am - am^3 \]
where \( m \) is the slope of the normal.

### Step 2: Set the Point from Which Normals are Drawn
We need to find points on the x-axis, which can be represented as \( (h, 0) \). This means we want the normal to pass through the point \( (h, 0) \).

### Step 3: Substitute the Point into the Normal Equation
Substituting \( (h, 0) \) into the normal equation gives:
\[ 0 = mh - 2am - am^3 \]
Rearranging this, we get:
\[ am^3 + (2a - h)m = 0 \]

### Step 4: Factor the Equation
Factoring out \( m \):
\[ m(am^2 + (2a - h)) = 0 \]
This gives us two cases: \( m = 0 \) or \( am^2 + (2a - h) = 0 \).

### Step 5: Analyze the Quadratic Equation
For three distinct normals, the quadratic equation \( am^2 + (2a - h) = 0 \) must have three distinct solutions for \( m \). This means the discriminant of the quadratic must be greater than zero.

The discriminant \( D \) of the quadratic \( am^2 + (2a - h) = 0 \) is given by:
\[ D = (2a - h)^2 - 4a \cdot 0 = (2a - h)^2 \]
For there to be three distinct normals, we need:
\[ (2a - h)^2 > 0 \]

### Step 6: Solve the Inequality
The inequality \( (2a - h)^2 > 0 \) implies:
\[ 2a - h \neq 0 \quad \text{or} \quad h \neq 2a \]
Thus, \( h \) must be such that:
\[ h < 2a \quad \text{or} \quad h > 2a \]

### Conclusion
The set of points on the x-axis from which three distinct normals can be drawn to the parabola \( y^2 = 4ax \) is:
\[ h > 2a \quad \text{or} \quad h < 2a \]