Let `C_(1)` be the graph of `xy=1` and the reflection of `C_(1)` in the line `y=2x` is `C_(2)`. If the equation of `C_(2)` is expressed as `12x^(2)+bxy+cy^(2)+d=0`, then the value of `(b+c+d)` is equal to

Answer

To solve the problem, we need to find the reflection of the curve \( C_1: xy = 1 \) in the line \( y = 2x \) and express the resulting equation \( C_2 \) in the form \( 12x^2 + bxy + cy^2 + d = 0 \). We will then calculate \( b + c + d \).

### Step 1: Understand the reflection process
The reflection of a point \( (x, y) \) across the line \( y = mx \) can be found using the formula for the reflection of a point across a line. Here, the slope \( m = 2 \).

### Step 2: Find the reflection of a point on \( C_1 \)
Let’s take a point \( (x_1, y_1) \) on the curve \( C_1 \) such that \( x_1 y_1 = 1 \). The reflection of this point across the line \( y = 2x \) will give us the point \( (x', y') \).

Using the reflection formulas:
- The slope of the line \( y = 2x \) is \( m_1 = 2 \).
- The slope of the line perpendicular to this line is \( m_2 = -\frac{1}{2} \).

### Step 3: Set up the equations
The slope between the points \( (x_1, y_1) \) and \( (x', y') \) must equal \( m_2 \):
\[ \frac{y' - y_1}{x' - x_1} = -\frac{1}{2} \]
This gives us:
\[ 2(y' - y_1) = - (x' - x_1) \quad \text{(Equation 1)} \]

The midpoint of the segment connecting \( (x_1, y_1) \) and \( (x', y') \) lies on the line \( y = 2x \):
\[ \frac{y_1 + y'}{2} = 2 \cdot \frac{x_1 + x'}{2} \]
This simplifies to:
\[ y_1 + y' = 2(x_1 + x') \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
From Equation 1:
\[ 2y' - 2y_1 = -x' + x_1 \implies 2y' = -x' + x_1 + 2y_1 \implies y' = \frac{-x' + x_1 + 2y_1}{2} \]

Substituting \( y' \) into Equation 2:
\[ y_1 + \frac{-x' + x_1 + 2y_1}{2} = 2(x_1 + x') \]
Multiplying through by 2 to eliminate the fraction:
\[ 2y_1 - x' + x_1 + 2y_1 = 4x_1 + 4x' \]
Combining like terms:
\[ 4y_1 - x' = 4x_1 + 5x' \]
Rearranging gives:
\[ 4y_1 = 5x' + 4x_1 \implies x' = \frac{4y_1 - 4x_1}{5} \]

### Step 5: Substitute back to find \( y' \)
Using \( x' \) in the equation for \( y' \):
\[ y' = \frac{-\left(\frac{4y_1 - 4x_1}{5}\right) + x_1 + 2y_1}{2} \]

### Step 6: Find the equation of \( C_2 \)
Now we need to express \( C_2 \) in the form \( 12x^2 + bxy + cy^2 + d = 0 \). After substituting and simplifying, we find that:
\[ C_2: 12x^2 + 7xy - 12y^2 + 25 = 0 \]

### Step 7: Identify coefficients
From the equation \( 12x^2 + bxy + cy^2 + d = 0 \):
- \( b = 7 \)
- \( c = -12 \)
- \( d = 25 \)

### Step 8: Calculate \( b + c + d \)
Now we calculate:
\[ b + c + d = 7 - 12 + 25 = 20 \]

Thus, the final answer is:
\[ \boxed{20} \]