The slope of the line touching both the parabolas `y^2=4x and x^2=−32y` is (a) 1/2 (b) 3/2 (c) 1/8 (d) 2/3

Options

A. `1//2`
B. `3//2`
C. `1//8`
D. `2//3`

Explanation

To find the slope of the line touching both the parabolas \( y^2 = 4x \) and \( x^2 = -32y \), we will follow these steps:

### Step 1: Write the equation of the tangent to the first parabola
The first parabola is given by \( y^2 = 4x \). The equation of the tangent to this parabola at a point \( (t^2, 2t) \) is given by:
\[ yt = x + t^2 \]

### Step 2: Substitute the tangent equation into the second parabola
The second parabola is given by \( x^2 = -32y \). We can rearrange this to \( y = -\frac{x^2}{32} \). We will substitute \( y \) from the tangent equation into this equation.

From the tangent equation, we have:
\[ y = \frac{x + t^2}{t} \]
Substituting this into the second parabola's equation:
\[ x^2 = -32\left(\frac{x + t^2}{t}\right) \]

### Step 3: Rearrange the equation
Multiplying both sides by \( t \) gives:
\[ tx^2 = -32(x + t^2) \]
Rearranging this, we have:
\[ tx^2 + 32x + 32t^2 = 0 \]

### Step 4: Use the condition for tangency
For the line to be tangent to the second parabola, the quadratic equation must have equal roots. Therefore, we set the discriminant to zero:
\[ b^2 - 4ac = 0 \]
Here, \( a = t \), \( b = 32 \), and \( c = 32t^2 \). Thus, we have:
\[ 32^2 - 4(t)(32t^2) = 0 \]
This simplifies to:
\[ 1024 - 128t^3 = 0 \]

### Step 5: Solve for \( t \)
From the equation \( 128t^3 = 1024 \), we can divide both sides by 128:
\[ t^3 = 8 \]
Taking the cube root gives:
\[ t = 2 \]

### Step 6: Find the slope of the tangent
The slope of the tangent line is given by:
\[ \text{slope} = \frac{1}{t} \]
Substituting \( t = 2 \):
\[ \text{slope} = \frac{1}{2} \]

### Conclusion
Thus, the slope of the line touching both parabolas is \( \frac{1}{2} \).

### Final Answer
The correct option is (a) \( \frac{1}{2} \).

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