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 equation of the line that touches both parabolas \( y^2 = 4x \) and \( x^2 = -32y \), we can follow these steps:

### Step 1: Identify the equations of the parabolas
The first parabola is \( y^2 = 4x \), which opens to the right. The second parabola is \( x^2 = -32y \), which opens downwards.

### Step 2: Assume the equation of the tangent line
We can assume the equation of the tangent line in slope-intercept form as:
\[ y = mx + c \]
where \( m \) is the slope and \( c \) is the y-intercept.

### Step 3: Find the condition for tangency with the first parabola
To find the condition for tangency with the first parabola \( y^2 = 4x \), we substitute \( y = mx + c \) into the equation of the parabola:
\[ (mx + c)^2 = 4x \]
Expanding this gives:
\[ m^2x^2 + 2mcx + c^2 = 4x \]
Rearranging, we have:
\[ m^2x^2 + (2mc - 4)x + c^2 = 0 \]
For this quadratic equation in \( x \) to have exactly one solution (tangency), the discriminant must be zero:
\[ (2mc - 4)^2 - 4m^2c^2 = 0 \]

### Step 4: Simplify the discriminant condition
Expanding the discriminant:
\[ (2mc - 4)^2 = 4m^2c^2 \]
This simplifies to:
\[ 4m^2c^2 - 16mc + 16 = 4m^2c^2 \]
Thus, we have:
\[ -16mc + 16 = 0 \implies mc = 1 \quad \text{(1)} \]

### Step 5: Find the condition for tangency with the second parabola
Next, we find the condition for tangency with the second parabola \( x^2 = -32y \). Substituting \( y = mx + c \) into the parabola gives:
\[ x^2 = -32(mx + c) \]
Rearranging this gives:
\[ x^2 + 32mx + 32c = 0 \]
For tangency, the discriminant must also be zero:
\[ (32m)^2 - 4 \cdot 1 \cdot 32c = 0 \]
This simplifies to:
\[ 1024m^2 - 128c = 0 \implies 128c = 1024m^2 \implies c = 8m^2 \quad \text{(2)} \]

### Step 6: Substitute equation (2) into equation (1)
Now, we substitute \( c = 8m^2 \) into \( mc = 1 \):
\[ m(8m^2) = 1 \implies 8m^3 = 1 \implies m^3 = \frac{1}{8} \implies m = \frac{1}{2} \]

### Step 7: Find the value of \( c \)
Substituting \( m = \frac{1}{2} \) back into equation (2):
\[ c = 8\left(\frac{1}{2}\right)^2 = 8 \cdot \frac{1}{4} = 2 \]

### Step 8: Write the equation of the tangent line
Now we have \( m = \frac{1}{2} \) and \( c = 2 \). Thus, the equation of the tangent line is:
\[ y = \frac{1}{2}x + 2 \]
Rearranging this into the standard form \( ax + by + c = 0 \):
\[ -\frac{1}{2}x + y - 2 = 0 \implies x - 2y + 4 = 0 \]
Thus, \( a = 1 \), \( b = -2 \), and \( c = 4 \).

### Step 9: Calculate \( a + b + c \)
Now we calculate:
\[ a + b + c = 1 - 2 + 4 = 3 \]

### Final Answer
The value of \( a + b + c \) is \( \boxed{3} \).