Equation of the line joining the foci of the parabola `y^(2)=4x and x^(2) = -4y` is

Options

A. `x+y-1=0`
B. `x-y-1=0`
C. `x-y+1=0`
D. `x+y+1=0`

Answer

To find the equation of the line joining the foci of the parabolas given by the equations \( y^2 = 4x \) and \( x^2 = -4y \), we can follow these steps:

### Step 1: Identify the Foci of the Parabolas

1. **For the parabola \( y^2 = 4x \)**:
- This is a standard form of a parabola that opens to the right.
- The standard form is \( y^2 = 4ax \), where \( a \) is the distance from the vertex to the focus.
- Here, \( 4a = 4 \), so \( a = 1 \).
- The focus of this parabola is at the point \( (a, 0) = (1, 0) \).

2. **For the parabola \( x^2 = -4y \)**:
- This is a standard form of a parabola that opens downward.
- The standard form is \( x^2 = -4ay \), where \( a \) is again the distance from the vertex to the focus.
- Here, \( 4a = 4 \), so \( a = 1 \).
- The focus of this parabola is at the point \( (0, -a) = (0, -1) \).

### Step 2: Write Down the Coordinates of the Foci

- The focus of the first parabola \( y^2 = 4x \) is \( F_1(1, 0) \).
- The focus of the second parabola \( x^2 = -4y \) is \( F_2(0, -1) \).

### Step 3: Find the Equation of the Line Joining the Foci

1. **Determine the slope of the line joining the points \( F_1(1, 0) \) and \( F_2(0, -1) \)**:
- The slope \( m \) is given by the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 0}{0 - 1} = \frac{-1}{-1} = 1 \]

2. **Use the point-slope form of the line equation**:
- The point-slope form is given by:
\[ y - y_1 = m(x - x_1) \]
- Using point \( F_1(1, 0) \):
\[ y - 0 = 1(x - 1) \]
- Simplifying this gives:
\[ y = x - 1 \]

### Final Answer

The equation of the line joining the foci of the parabolas \( y^2 = 4x \) and \( x^2 = -4y \) is:
\[ \boxed{y = x - 1} \]