If the line `x + 2y = k` is normal to the parabola `x^(2) – 4x – 8y + 12 = 0` then the value of `k/4`is :

Explanation

To solve the problem, we need to find the value of \( k/4 \) given that the line \( x + 2y = k \) is normal to the parabola defined by the equation \( x^2 - 4x - 8y + 12 = 0 \).

### Step-by-Step Solution:

1. **Rearranging the Parabola Equation:**
Start by rearranging the equation of the parabola:
\[ x^2 - 4x - 8y + 12 = 0 \]
Rearranging gives:
\[ x^2 - 4x + 12 = 8y \]
Now, complete the square for the \( x \) terms:
\[ (x^2 - 4x + 4) = 8y - 4 \]
This simplifies to:
\[ (x - 2)^2 = 8(y - 1) \]
This is in the standard form of a parabola \( (x - h)^2 = 4a(y - k) \), where \( h = 2 \), \( k = 1 \), and \( 4a = 8 \). Thus, \( a = 2 \).

2. **Finding the Slope of the Normal:**
The slope of the normal line to the parabola at any point is given by:
\[ m = -\frac{1}{\text{slope of the tangent}} \]
For the parabola \( (x - 2)^2 = 8(y - 1) \), the slope of the tangent at the point \( (x_0, y_0) \) is given by the derivative:
\[ \frac{dy}{dx} = \frac{1}{4a}(x - h) = \frac{1}{8}(x - 2) \]
Thus, the slope of the normal line is:
\[ m = -\frac{1}{\frac{1}{8}(x - 2)} = -\frac{8}{x - 2} \]

3. **Setting Up the Normal Line Equation:**
The equation of the normal line at the point \( (x_0, y_0) \) can be written as:
\[ y - y_0 = m(x - x_0) \]
Substituting \( y_0 = 1 + \frac{(x_0 - 2)^2}{8} \) and \( m = -\frac{8}{x_0 - 2} \):
\[ y - \left(1 + \frac{(x_0 - 2)^2}{8}\right) = -\frac{8}{x_0 - 2}(x - x_0) \]

4. **Comparing with the Given Line:**
The line \( x + 2y = k \) can be rewritten as:
\[ 2y = -x + k \quad \Rightarrow \quad y = -\frac{1}{2}x + \frac{k}{2} \]
The slope of this line is \( -\frac{1}{2} \).

5. **Equating Slopes:**
For the line to be normal to the parabola, we need:
\[ -\frac{8}{x_0 - 2} = -\frac{1}{2} \]
Solving for \( x_0 \):
\[ \frac{8}{x_0 - 2} = \frac{1}{2} \quad \Rightarrow \quad 16 = x_0 - 2 \quad \Rightarrow \quad x_0 = 18 \]

6. **Finding the Corresponding \( y_0 \):**
Substitute \( x_0 = 18 \) back into the parabola equation to find \( y_0 \):
\[ (18 - 2)^2 = 8(y - 1) \quad \Rightarrow \quad 256 = 8(y - 1) \quad \Rightarrow \quad y - 1 = 32 \quad \Rightarrow \quad y_0 = 33 \]

7. **Finding \( k \):**
Now substitute \( x_0 \) and \( y_0 \) into the normal line equation:
\[ x + 2y = k \quad \Rightarrow \quad 18 + 2(33) = k \quad \Rightarrow \quad k = 18 + 66 = 84 \]

8. **Calculating \( k/4 \):**
Finally, calculate \( k/4 \):
\[ \frac{k}{4} = \frac{84}{4} = 21 \]

### Final Answer:
The value of \( \frac{k}{4} \) is \( 21 \).