If three distinct normals can be drawn to the parabola `y^(2)-2y=4x-9` from the point (2a, 0) then range of values of a is

Options

A. no real values possible
B. `(2,oo)`
C. `(-oo,2)`
D. none of these

Answer

To solve the problem, we need to find the range of values of \( a \) such that three distinct normals can be drawn to the parabola given by the equation \( y^2 - 2y = 4x - 9 \) from the point \( (2a, 0) \).

### Step-by-Step Solution:

1. **Rearranging the Parabola Equation:**
Start with the given equation of the parabola:
\[ y^2 - 2y = 4x - 9 \]
Rearranging gives:
\[ y^2 - 2y + 9 = 4x \]
This can be rewritten as:
\[ y^2 - 2y + 1 = 4x - 8 \]
Which simplifies to:
\[ (y - 1)^2 = 4(x - 2) \]
This shows that the parabola opens to the right with vertex at \( (2, 1) \).

2. **Finding the Focus:**
The vertex form of the parabola \( (y - k)^2 = 4p(x - h) \) gives us the focus. Here, \( h = 2 \), \( k = 1 \), and \( p = 1 \) (since \( 4p = 4 \)). Thus, the focus is at:
\[ (h + p, k) = (3, 1) \]

3. **Condition for Three Distinct Normals:**
For three distinct normals to be drawn from the point \( (2a, 0) \) to the parabola, the point must lie outside the region defined by the focus. The distance from the point \( (2a, 0) \) to the focus \( (3, 1) \) must be greater than the distance from the vertex to the focus.

The distance from the vertex \( (2, 1) \) to the focus \( (3, 1) \) is:
\[ 1 \text{ unit} \]
Therefore, we require:
\[ \sqrt{(2a - 3)^2 + (0 - 1)^2} > 1 \]

4. **Squaring Both Sides:**
Squaring both sides gives:
\[ (2a - 3)^2 + 1 > 1 \]
Simplifying this leads to:
\[ (2a - 3)^2 > 0 \]
This inequality holds true for all \( a \) except when \( 2a - 3 = 0 \) or \( a = \frac{3}{2} \).

5. **Finding the Range of \( a \):**
Since we need three distinct normals, \( a \) must be such that \( 2a \) is either less than 3 or greater than 3:
\[ a < \frac{3}{2} \quad \text{or} \quad a > \frac{3}{2} \]
However, we also need to ensure that the point \( (2a, 0) \) is to the right of the vertex, which means:
\[ 2a > 2 \quad \Rightarrow \quad a > 1 \]

Combining these inequalities, we find:
\[ a > \frac{3}{2} \quad \text{and} \quad a > 1 \]
Thus, the valid range of \( a \) is:
\[ a \in \left( \frac{3}{2}, \infty \right) \]

### Final Answer:
The range of values of \( a \) is:
\[ \boxed{\left( \frac{3}{2}, \infty \right)} \]