The set of a points on the axis of the parabola `y^(2)-4x-4y+12=0` from which all three normals to the parabola are real is

Options

A. `(k,2),k gt 4`
B. `(k,0),kgt5`
C. `(k,1),kgt4`
D. none

Answer

To solve the problem, we need to find the set of points on the axis of the parabola given by the equation \( y^2 - 4x - 4y + 12 = 0 \) from which all three normals to the parabola are real.

### Step 1: Rewrite the Equation of the Parabola
First, we will rearrange the equation of the parabola into a standard form. The given equation is:
\[ y^2 - 4y - 4x + 12 = 0 \]
We can complete the square for the \(y\) terms:
\[ (y^2 - 4y + 4) - 4x + 12 - 4 = 0 \]
This simplifies to:
\[ (y - 2)^2 = 4(x - 2) \]
This is the standard form of a parabola that opens to the right, where the vertex is at the point \((2, 2)\).

### Step 2: Identify the Axis of the Parabola
The axis of the parabola is the line \(y = 2\). We are looking for points on this line from which all three normals to the parabola are real.

### Step 3: Equation of the Normal to the Parabola
The equation of the normal to the parabola at a point \((x_1, y_1)\) is given by:
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope of the normal. The slope of the normal can be expressed in terms of the parameter \(t\) (where \(t\) is related to the point on the parabola).

### Step 4: Condition for Real Normals
For all three normals to be real, the discriminant of the equation formed by substituting the point \((\alpha, 2)\) into the normal equation must be non-negative.

### Step 5: Finding the Discriminant
Substituting \(y = 2\) into the normal equation derived from the parabola:
\[ 2 - 2 = m(\alpha - 2) - 2m - m^2 \]
This simplifies to:
\[ 0 = m(\alpha - 2) - 2m - m^2 \]
Rearranging gives:
\[ m^2 + (2 - \alpha)m = 0 \]
Factoring out \(m\):
\[ m(m + (2 - \alpha)) = 0 \]
This means \(m = 0\) or \(m = \alpha - 2\).

### Step 6: Condition for Real Values of \(m\)
For \(m\) to have real solutions, we need:
\[ \alpha - 2 \geq 0 \implies \alpha \geq 2 \]
However, we also need to ensure that the condition for three real normals is satisfied, which leads us to the requirement:
\[ \alpha - 4 \geq 0 \implies \alpha \geq 4 \]

### Final Answer
Thus, the set of points on the axis of the parabola from which all three normals are real is:
\[ \alpha \geq 4 \]