The tangent to `y = ax ^(2)+ bx + (7 )/(2) at (1,2)` is parallel to the normal at the point `(-2,2)` on the curve `y = x ^(2)+6x+10.` Then the vlaue of `a/2-b` is:

Options

A. 2
B. 0
C. 3
D. 1

Answer

To solve the problem, we need to find the value of \( \frac{a}{2} - b \) given the conditions about the tangent and normal lines to the curves.

### Step-by-Step Solution:

1. **Identify the first curve and find its derivative:**
The first curve is given by:
\[ y = ax^2 + bx + \frac{7}{2} \]
To find the slope of the tangent at the point \( (1, 2) \), we differentiate:
\[ \frac{dy}{dx} = 2ax + b \]
At \( x = 1 \):
\[ \text{slope of tangent (MT)} = 2a + b \tag{1} \]

2. **Verify that the point \( (1, 2) \) lies on the first curve:**
Substitute \( x = 1 \) and \( y = 2 \):
\[ 2 = a(1)^2 + b(1) + \frac{7}{2} \]
Simplifying gives:
\[ 2 = a + b + \frac{7}{2} \]
Rearranging:
\[ a + b = 2 - \frac{7}{2} = -\frac{3}{2} \tag{2} \]

3. **Identify the second curve and find its derivative:**
The second curve is given by:
\[ y = x^2 + 6x + 10 \]
Differentiate to find the slope of the tangent:
\[ \frac{dy}{dx} = 2x + 6 \]
At \( x = -2 \):
\[ \text{slope of tangent} = 2(-2) + 6 = -4 + 6 = 2 \]
Thus, the slope of the normal (MN) is:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{2} \tag{3} \]

4. **Set the slopes equal:**
Since the tangent to the first curve is parallel to the normal of the second curve:
\[ MT = MN \]
From equations (1) and (3):
\[ 2a + b = -\frac{1}{2} \tag{4} \]

5. **Solve the system of equations:**
We now have two equations:
- \( a + b = -\frac{3}{2} \) (from (2))
- \( 2a + b = -\frac{1}{2} \) (from (4))

Subtract equation (2) from equation (4):
\[ (2a + b) - (a + b) = -\frac{1}{2} + \frac{3}{2} \]
This simplifies to:
\[ a = 1 \]

6. **Substitute \( a \) back to find \( b \):**
Substitute \( a = 1 \) into equation (2):
\[ 1 + b = -\frac{3}{2} \]
Thus:
\[ b = -\frac{3}{2} - 1 = -\frac{5}{2} \]

7. **Calculate \( \frac{a}{2} - b \):**
Now we can find \( \frac{a}{2} - b \):
\[ \frac{a}{2} - b = \frac{1}{2} - \left(-\frac{5}{2}\right) = \frac{1}{2} + \frac{5}{2} = \frac{6}{2} = 3 \]

### Final Answer:
The value of \( \frac{a}{2} - b \) is \( \boxed{3} \).