The angle of intersection between the curves `x^(2) = 4(y +1)` and `x^(2) =-4 (y+1)` is

Options

A. `(pi)/(6)`
B. `(pi)/(4)`
C. 0
D. `(pi)/(2)`

Explanation

To find the angle of intersection between the curves \( x^2 = 4(y + 1) \) and \( x^2 = -4(y + 1) \), we will follow these steps:

### Step 1: Find the points of intersection of the curves
We start by equating the two equations since both equal \( x^2 \):

\[ 4(y + 1) = -4(y + 1) \]

### Step 2: Simplify the equation
Rearranging gives us:

\[ 4y + 4 = -4y - 4 \]

Adding \( 4y \) to both sides:

\[ 8y + 4 = -4 \]

Subtracting \( 4 \) from both sides:

\[ 8y = -8 \]

Dividing by \( 8 \):

\[ y = -1 \]

### Step 3: Substitute \( y \) back to find \( x \)
Now, substitute \( y = -1 \) back into either of the original equations to find \( x \). Using \( x^2 = 4(y + 1) \):

\[ x^2 = 4(-1 + 1) = 4(0) = 0 \]

Thus, \( x = 0 \). Therefore, the point of intersection is:

\[ (0, -1) \]

### Step 4: Calculate the derivatives to find slopes
Next, we need to find the slopes of the tangents to both curves at the point of intersection.

For the first curve \( x^2 = 4(y + 1) \):

Differentiating implicitly:

\[ 2x = 4\frac{dy}{dx} \]

Thus,

\[ \frac{dy}{dx} = \frac{2x}{4} = \frac{x}{2} \]

At the point \( (0, -1) \):

\[ \frac{dy}{dx} = \frac{0}{2} = 0 \]

For the second curve \( x^2 = -4(y + 1) \):

Differentiating implicitly:

\[ 2x = -4\frac{dy}{dx} \]

Thus,

\[ \frac{dy}{dx} = -\frac{2x}{4} = -\frac{x}{2} \]

At the point \( (0, -1) \):

\[ \frac{dy}{dx} = -\frac{0}{2} = 0 \]

### Step 5: Determine the angle of intersection
Both curves have the same slope of \( 0 \) at the point of intersection. When two curves have the same slope at the point of intersection, the angle of intersection is \( 0 \) degrees.

Thus, the angle of intersection between the curves is:

\[ \text{Angle of intersection} = 0 \text{ degrees} \]

### Summary
The angle of intersection between the curves \( x^2 = 4(y + 1) \) and \( x^2 = -4(y + 1) \) is \( 0 \) degrees.

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