The set of points on the axis of the parabola `y^2-4x-2y+5=0` from which all the three normals to the parabola are real , is

Options

A. `lamda,0),xgt1`
B. `(lamda,1),lamdagt3`
C. `(lamda,2),lamdagt6`
D. `(lamda,3),lamdagt8`

Answer

To solve the problem, we need to find the set of points on the axis of the parabola given by the equation \( y^2 - 4x - 2y + 5 = 0 \) from which all three normals to the parabola are real.

### Step-by-Step Solution:

1. **Rearranging the Parabola Equation**:
Start with the given equation:
\[ y^2 - 4x - 2y + 5 = 0 \]
Rearranging gives:
\[ y^2 - 2y + 5 = 4x \]

2. **Completing the Square**:
To rewrite the left side in a more usable form, complete the square for the \( y \) terms:
\[ (y - 1)^2 - 1 + 5 = 4x \]
This simplifies to:
\[ (y - 1)^2 = 4x - 4 \]
or
\[ (y - 1)^2 = 4(x - 1) \]

3. **Identifying the Vertex and Orientation**:
The equation \( (y - 1)^2 = 4(x - 1) \) indicates that the parabola opens to the right with vertex at \( (1, 1) \).

4. **Finding the Axis of the Parabola**:
The axis of the parabola is the line \( y = 1 \).

5. **Condition for Three Normals**:
For a point \( (x_0, 0) \) on the axis of the parabola from which three normals can be drawn, the point must satisfy:
\[ x_0 > 2a \]
where \( a \) is the distance from the vertex to the focus. From the standard form \( (y - k)^2 = 4a(x - h) \), we have \( a = 1 \) (since \( 4a = 4 \)). Therefore, \( 2a = 2 \).

6. **Setting Up the Inequality**:
The condition becomes:
\[ x_0 > 2 \]

7. **Conclusion**:
The set of points on the axis of the parabola from which all three normals are real is:
\[ (x, 1) \text{ where } x > 2 \]

### Final Answer:
The set of points is \( (x, 1) \) where \( x > 2 \).

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