`P_(1):y^(2)=x,P_(2):y^(2)= - x, P_(3):x^(2) =y ,P_(4):x^(2)= -y` are four parabola, points of intersection of the parabola `P_(3) and P_(4)` with `P_(1) and P_(2)` (other than the origin ) enclose a square of area (in sq. units ).

Options

A. 2
B. 4
C. 8
D. 16

Answer

To find the area of the square formed by the intersection points of the given parabolas, we will follow these steps:

### Step 1: Identify the Parabolas
The parabolas given are:
1. \( P_1: y^2 = x \)
2. \( P_2: y^2 = -x \)
3. \( P_3: x^2 = y \)
4. \( P_4: x^2 = -y \)

### Step 2: Find Intersection Points of \( P_3 \) and \( P_4 \) with \( P_1 \) and \( P_2 \)
We will first find the intersection points of \( P_3 \) and \( P_4 \) with \( P_1 \).

**Finding intersection of \( P_3 \) and \( P_1 \):**
- From \( P_3: x^2 = y \)
- Substitute \( y \) in \( P_1: y^2 = x \):
\[ (x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0 \]
This gives \( x = 0 \) or \( x = 1 \).

- For \( x = 0 \): \( y = 0 \) (origin)
- For \( x = 1 \): \( y = 1 \) (point \( (1, 1) \))

**Finding intersection of \( P_4 \) and \( P_1 \):**
- From \( P_4: x^2 = -y \)
- Substitute \( y \) in \( P_1: y^2 = x \):
\[ (-x^2)^2 = x \implies x^4 - x = 0 \implies x(x^3 - 1) = 0 \]
This gives \( x = 0 \) or \( x = 1 \).

- For \( x = 0 \): \( y = 0 \) (origin)
- For \( x = 1 \): \( y = -1 \) (point \( (1, -1) \))

### Step 3: Find Intersection Points of \( P_3 \) and \( P_4 \) with \( P_2 \)
**Finding intersection of \( P_3 \) and \( P_2 \):**
- From \( P_3: x^2 = y \)
- Substitute \( y \) in \( P_2: y^2 = -x \):
\[ (x^2)^2 = -x \implies x^4 + x = 0 \implies x(x^3 + 1) = 0 \]
This gives \( x = 0 \) or \( x = -1 \).

- For \( x = 0 \): \( y = 0 \) (origin)
- For \( x = -1 \): \( y = 1 \) (point \( (-1, 1) \))

**Finding intersection of \( P_4 \) and \( P_2 \):**
- From \( P_4: x^2 = -y \)
- Substitute \( y \) in \( P_2: y^2 = -x \):
\[ (-x^2)^2 = -x \implies x^4 + x = 0 \implies x(x^3 + 1) = 0 \]
This gives \( x = 0 \) or \( x = -1 \).

- For \( x = 0 \): \( y = 0 \) (origin)
- For \( x = -1 \): \( y = -1 \) (point \( (-1, -1) \))

### Step 4: Identify the Points of Intersection
The points of intersection (other than the origin) are:
1. \( (1, 1) \)
2. \( (1, -1) \)
3. \( (-1, 1) \)
4. \( (-1, -1) \)

### Step 5: Determine the Area of the Square
The points \( (1, 1), (1, -1), (-1, 1), (-1, -1) \) form the vertices of a square. The side length of the square can be calculated as:
\[ \text{Length of side} = 1 - (-1) = 2 \]

The area \( A \) of the square is given by:
\[ A = \text{side}^2 = 2^2 = 4 \text{ square units} \]

### Final Answer
The area of the square formed by the intersection points is \( 4 \) square units.

---