The equation of the line touching both the parabolas `y^(2)=4xandx^(2)=-32y` is `ax+by+c=0`. Then the value of `a+b+c` is ___________ .

Explanation

To find the value of \( a + b + c \) for the equation of the line touching both parabolas \( y^2 = 4x \) and \( x^2 = -32y \), we will follow these steps:

### Step 1: Determine the equation of the tangent to the first parabola
The equation of the first parabola is \( y^2 = 4x \). The general equation of the tangent to this parabola is given by:
\[ y = mx + \frac{1}{m} \]
where \( m \) is the slope of the tangent.

### Step 2: Determine the equation of the tangent to the second parabola
The equation of the second parabola is \( x^2 = -32y \). The general equation of the tangent to this parabola is given by:
\[ y = mx - 8m^2 \]
where \( m \) is again the slope of the tangent.

### Step 3: Set the two tangent equations equal
Since the line is a common tangent to both parabolas, we can set the two equations equal to each other:
\[ mx + \frac{1}{m} = mx - 8m^2 \]
This simplifies to:
\[ \frac{1}{m} = -8m^2 \]

### Step 4: Solve for \( m \)
Multiplying both sides by \( m \) (assuming \( m \neq 0 \)):
\[ 1 = -8m^3 \]
Thus,
\[ m^3 = -\frac{1}{8} \quad \Rightarrow \quad m = -\frac{1}{2} \]

### Step 5: Substitute \( m \) back into the tangent equation
Substituting \( m = -\frac{1}{2} \) into the tangent equation for the first parabola:
\[ y = -\frac{1}{2}x + \frac{1}{-\frac{1}{2}} = -\frac{1}{2}x - 2 \]
Multiplying through by -1 gives:
\[ y = \frac{1}{2}x + 2 \]
Rearranging this gives:
\[ \frac{1}{2}x - y + 2 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ x - 2y + 4 = 0 \]

### Step 6: Identify coefficients \( a, b, c \)
From the equation \( x - 2y + 4 = 0 \), we have:
- \( a = 1 \)
- \( b = -2 \)
- \( c = 4 \)

### Step 7: Calculate \( a + b + c \)
Now, we can find \( a + b + c \):
\[ a + b + c = 1 - 2 + 4 = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]