The tangent to `y=a x^2+b x+7/2a t(1,2)` is parallel to the normal at the point `(-2,2)` on the curve `y=x^2+6x+10 ,` then `a=-1` b. `a=1` c. `b=5/2` d. `b=-5/2`

Options

A. `a = 1`
B. `a = -1`
C. `b = (-15)/(2)`
D. `b = (5)/(2)`

Explanation

To solve the problem step by step, we need to find the values of \( a \) and \( b \) such that the tangent to the curve \( y = ax^2 + bx + \frac{7}{2} \) at the point \( (1, 2) \) is parallel to the normal at the point \( (-2, 2) \) on the curve \( y = x^2 + 6x + 10 \).

### Step 1: Find the slope of the normal at the point (-2, 2)

1. Differentiate the curve \( y = x^2 + 6x + 10 \) to find the slope of the tangent:
\[ \frac{dy}{dx} = 2x + 6 \]

2. Substitute \( x = -2 \) into the derivative to find the slope at that point:
\[ \frac{dy}{dx} = 2(-2) + 6 = -4 + 6 = 2 \]

3. The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{2} \]

### Step 2: Find the slope of the tangent to the curve \( y = ax^2 + bx + \frac{7}{2} \) at the point (1, 2)

1. Differentiate \( y = ax^2 + bx + \frac{7}{2} \):
\[ \frac{dy}{dx} = 2ax + b \]

2. Substitute \( x = 1 \) into the derivative:
\[ \frac{dy}{dx} = 2a(1) + b = 2a + b \]

### Step 3: Set the slopes equal

Since the tangent at \( (1, 2) \) is parallel to the normal at \( (-2, 2) \), we have:
\[ 2a + b = -\frac{1}{2} \]

### Step 4: Use the point (1, 2) to find another equation

1. Substitute \( x = 1 \) and \( y = 2 \) into the equation of the curve:
\[ 2 = a(1^2) + b(1) + \frac{7}{2} \]
Simplifying gives:
\[ 2 = a + b + \frac{7}{2} \]
Rearranging gives:
\[ a + b = 2 - \frac{7}{2} = -\frac{3}{2} \]

### Step 5: Solve the system of equations

Now we have a system of two equations:
1. \( 2a + b = -\frac{1}{2} \) (Equation 1)
2. \( a + b = -\frac{3}{2} \) (Equation 2)

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

### Step 6: Substitute \( a \) back to find \( b \)

Substituting \( a = 1 \) into Equation 2:
\[ 1 + b = -\frac{3}{2} \]
Solving for \( b \):
\[ b = -\frac{3}{2} - 1 = -\frac{5}{2} \]

### Final Answer

Thus, the values are:
\[ a = 1, \quad b = -\frac{5}{2} \]