If the line `x+by+c=0, (bc ne 0)` , touches both the parabolas `y^(2)=4x and x^(2)= -32y`, then `(b+c)/(8)` is equal to

Answer

To solve the problem, we need to find the value of \((b+c)/8\) given that the line \(x + by + c = 0\) touches both the parabolas \(y^2 = 4x\) and \(x^2 = -32y\).

### Step 1: Identify the properties of the parabolas
The first parabola \(y^2 = 4x\) opens to the right and has its vertex at the origin. The standard form is \(y^2 = 4ax\) where \(a = 1\).

The second parabola \(x^2 = -32y\) opens downwards and can be rewritten in the standard form as \(x^2 = 4by\) where \(b = -8\).

### Step 2: Find the equation of the tangent to the first parabola
For the parabola \(y^2 = 4x\), the equation of the tangent line at a point \((at^2, 2at)\) is given by:
\[ yy_1 = 2a(x + x_1) \]
Substituting \(a = 1\):
\[ yy_1 = 2(x + t^2) \]
This can be rearranged to give:
\[ y = mx + \frac{1}{m} \]
where \(m\) is the slope of the tangent.

### Step 3: Find the equation of the tangent to the second parabola
For the parabola \(x^2 = -32y\), the equation of the tangent line at a point \((2b, -8b^2)\) is:
\[ yy_1 = -16(x + x_1) \]
Substituting \(b = -8\):
\[ y = mx - 8m^2 \]

### Step 4: Set the equations equal to each other
Since the line touches both parabolas, we equate the two tangent equations:
\[ mx + \frac{1}{m} = mx - 8m^2 \]
This simplifies to:
\[ \frac{1}{m} + 8m^2 = 0 \]

### Step 5: Solve for \(m\)
Multiplying through by \(m\) (assuming \(m \neq 0\)):
\[ 1 + 8m^3 = 0 \]
Thus, we have:
\[ 8m^3 = -1 \implies m^3 = -\frac{1}{8} \implies m = -\frac{1}{2} \]

### Step 6: Substitute \(m\) back to find \(b\) and \(c\)
Using \(m = -\frac{1}{2}\) in the tangent equation for the first parabola:
\[ y = -\frac{1}{2}x - 2 \]
This gives us \(b = -2\) and \(c = -2\).

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

### Step 8: Calculate \((b+c)/8\)
Finally, we compute:
\[ \frac{b+c}{8} = \frac{2}{8} = \frac{1}{4} \]

### Final Answer
Thus, the value of \((b+c)/8\) is \(\frac{1}{4}\).