Area bounded by `x ^(2) y ^(2)+ y ^(4)-x ^(2)-5y ^(2)+4=0` is equal to :

Options

A. `(4pi)/(2) + sqrt2`
B. ` (4pi)/(3) -sqrt2`
C. `(4pi)/(3) -sqrt2`
D. none of these

Answer

To find the area bounded by the curve given by the equation \( x^2 y^2 + y^4 - x^2 - 5y^2 + 4 = 0 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^2 y^2 + y^4 - x^2 - 5y^2 + 4 = 0 \]
We can rearrange this to isolate \(x^2\):
\[ x^2 y^2 - x^2 = -y^4 + 5y^2 - 4 \]
Factoring out \(x^2\) gives us:
\[ x^2 (y^2 - 1) = -y^4 + 5y^2 - 4 \]

### Step 2: Factor the right-hand side
Now we will factor the right-hand side:
\[ -y^4 + 5y^2 - 4 = -(y^4 - 5y^2 + 4) \]
We can factor \(y^4 - 5y^2 + 4\) as:
\[ y^4 - 5y^2 + 4 = (y^2 - 4)(y^2 - 1) \]
Thus, we have:
\[ x^2 (y^2 - 1) = -(y^2 - 4)(y^2 - 1) \]

### Step 3: Simplify the equation
This simplifies to:
\[ x^2 = -\frac{(y^2 - 4)(y^2 - 1)}{(y^2 - 1)} \]
This gives us:
\[ x^2 = - (y^2 - 4) \quad \text{for } y^2 \neq 1 \]
Thus, we can write:
\[ x^2 = 4 - y^2 \]

### Step 4: Identify the curves
The equation \(x^2 + y^2 = 4\) represents a circle with radius 2 centered at the origin. The lines \(y = 1\) and \(y = -1\) are horizontal lines that intersect the circle.

### Step 5: Find points of intersection
To find the points where the circle intersects the lines \(y = 1\) and \(y = -1\):
1. For \(y = 1\):
\[ x^2 + 1^2 = 4 \implies x^2 = 3 \implies x = \pm \sqrt{3} \]
Points of intersection: \((\sqrt{3}, 1)\) and \((- \sqrt{3}, 1)\).

2. For \(y = -1\):
\[ x^2 + (-1)^2 = 4 \implies x^2 = 3 \implies x = \pm \sqrt{3} \]
Points of intersection: \((\sqrt{3}, -1)\) and \((- \sqrt{3}, -1)\).

### Step 6: Calculate the area
The area bounded by the circle and the lines can be calculated using integration. We will calculate the area above the line \(y = -1\) and below the line \(y = 1\).

1. The area of the circle segment above \(y = -1\):
\[ A_1 = 4 \int_{0}^{\sqrt{3}} \sqrt{4 - x^2} \, dx \]
This integral can be solved using the formula for the area of a circle segment.

2. The area of the rectangle formed by the lines \(y = 1\) and \(y = -1\):
\[ A_2 = \text{length} \times \text{breadth} = 2\sqrt{3} \times 2 = 4\sqrt{3} \]

### Step 7: Combine areas
The total area bounded by the curves is:
\[ \text{Total Area} = 4 \left( \frac{2\pi}{3} - 2 \right) + 4\sqrt{3} \]
This simplifies to:
\[ \text{Total Area} = \frac{8\pi}{3} + 4\sqrt{3} \]

### Final Answer
Thus, the area bounded by the curve is:
\[ \frac{4\pi}{3} + 2\sqrt{3} \]