The slope of the line touching both the parabolas $y^{2}=4x$ and $x^{2}=-32y$ is :

Question

Accepted Answer

To determine the slope of the line that touches both parabolas $ y^2 = 4x $ and $ x^2 = -32y $, we need to find a common tangent to both curves.

### Explanation for Option 1: $\frac{1}{2}$

1. **Equation of the Parabola $ y^2 = 4x $:**
   - This is a standard parabola that opens to the right with vertex at the origin. The equation of the tangent to this parabola at any point $(x_1, y_1)$ on the parabola is given by:
     $$
     yy_1 = 2(x + x_1)
     $$
   - Rearranging, the slope $m$ of the tangent line is:
     $$
     m = \frac{y_1}{2}
     $$

2. **Equation of the Parabola $ x^2 = -32y $:**
   - This parabola opens downwards with vertex at the origin. The equation of the tangent to this parabola at any point $(x_2, y_2)$ on the parabola is given by:
     $$
     xx_2 = -16(y + y_2)
     $$
   - Rearranging, the slope $m$ of the tangent line is:
     $$
     m = -\frac{x_2}{16}
     $$

3. **Finding the Common Tangent:**
   - For the line to be tangent to both parabolas, the slopes must be equal:
     $$
     \frac{y_1}{2} = -\frac{x_2}{16}
     $$
   - Solving for $m$, we substitute $y_1 = 2m$ and $x_2 = -16m$ into the equations of the parabolas:
     - For $y^2 = 4x$, substituting $y_1 = 2m$ gives:
       $$
       (2m)^2 = 4x_1 \implies 4m^2 = 4x_1 \implies x_1 = m^2
       $$
     - For $x^2 = -32y$, substituting $x_2 = -16m$ gives:
       $$
       (-16m)^2 = -32y_2 \implies 256m^2 = -32y_2 \implies y_2 = -8m^2
       $$

4. **Equating the Slopes:**
   - From the slope condition $\frac{y_1}{2} = -\frac{x_2}{16}$, substituting $y_1 = 2m$ and $x_2 = -16m$ gives:
     $$
     \frac{2m}{2} = -\frac{-16m}{16} \implies m = \frac{1}{2}
     $$

Thus, the slope of the line that is tangent to both parabolas is $\frac{1}{2}$, confirming that Option 1 is correct.

### Explanation for Other Options:

- **Option 2: $\frac{3}{2}$**
  - This slope does not satisfy the condition derived from equating the slopes of the tangents to both parabolas. Substituting $\frac{3}{2}$ into the equations does not yield consistent results for both parabolas.

- **Option 3: $\frac{1}{8}$**
  - Similarly, substituting $\frac{1}{8}$ into the slope condition does not satisfy the tangent condition for both parabolas simultaneously.

- **Option 4: $\frac{2}{3}$**
  - This slope also fails to meet the derived condition for the common tangent line to both parabolas.

In conclusion, the only slope that satisfies the condition for a line to be tangent to both parabolas is $\frac{1}{2}$, making Option 1 the correct choice.