The area bounded by `y^(2) = 4x ` and `x^(2) = 4y` is `A_(1)` and the area bounded by `x^(2) = 4y , x = 4` and x-axis is `A_(2)` . If `A_(1) : A_(2) = K : 1` then K is _______

Answer

To solve the problem, we need to find the areas \( A_1 \) and \( A_2 \) as described in the question, and then determine the ratio \( K \) such that \( A_1 : A_2 = K : 1 \).

### Step 1: Find the intersection points of the curves

The curves given are:
1. \( y^2 = 4x \) (which can be rewritten as \( y = 2\sqrt{x} \))
2. \( x^2 = 4y \) (which can be rewritten as \( y = \frac{x^2}{4} \))

To find the intersection points, we set \( 2\sqrt{x} = \frac{x^2}{4} \).

Squaring both sides:
\[ (2\sqrt{x})^2 = \left(\frac{x^2}{4}\right)^2 \]
\[ 4x = \frac{x^4}{16} \]
Multiplying through by 16 to eliminate the fraction:
\[ 64x = x^4 \]
Rearranging gives:
\[ x^4 - 64x = 0 \]
Factoring out \( x \):
\[ x(x^3 - 64) = 0 \]
Thus, \( x = 0 \) or \( x^3 = 64 \) which gives \( x = 4 \).

Now substituting back to find \( y \):
- For \( x = 0 \): \( y = 2\sqrt{0} = 0 \)
- For \( x = 4 \): \( y = 2\sqrt{4} = 4 \)

The intersection points are \( (0, 0) \) and \( (4, 4) \).

### Step 2: Calculate area \( A_1 \)

The area \( A_1 \) is bounded by the curves from \( x = 0 \) to \( x = 4 \):
\[ A_1 = \int_{0}^{4} \left( 2\sqrt{x} - \frac{x^2}{4} \right) dx \]

Calculating the integral:
\[ A_1 = \int_{0}^{4} 2\sqrt{x} \, dx - \int_{0}^{4} \frac{x^2}{4} \, dx \]

Calculating each integral:
1. \( \int 2\sqrt{x} \, dx = \frac{2}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}} \)
- Evaluating from 0 to 4:
\[ = \frac{4}{3} (4^{\frac{3}{2}} - 0) = \frac{4}{3} \cdot 8 = \frac{32}{3} \]

2. \( \int \frac{x^2}{4} \, dx = \frac{1}{4} \cdot \frac{x^3}{3} = \frac{x^3}{12} \)
- Evaluating from 0 to 4:
\[ = \frac{4^3}{12} - 0 = \frac{64}{12} = \frac{16}{3} \]

Putting it all together:
\[ A_1 = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} \]

### Step 3: Calculate area \( A_2 \)

The area \( A_2 \) is bounded by the curve \( x^2 = 4y \), the line \( x = 4 \), and the x-axis. We need to find the area from \( x = 0 \) to \( x = 4 \):
\[ A_2 = \int_{0}^{4} \frac{x^2}{4} \, dx \]

Calculating the integral:
\[ A_2 = \frac{1}{4} \int_{0}^{4} x^2 \, dx = \frac{1}{4} \cdot \frac{x^3}{3} \bigg|_{0}^{4} = \frac{1}{4} \cdot \frac{64}{3} = \frac{16}{3} \]

### Step 4: Find the ratio \( A_1 : A_2 \)

Now we have:
\[ A_1 = \frac{16}{3}, \quad A_2 = \frac{16}{3} \]
Thus, the ratio \( A_1 : A_2 \) is:
\[ \frac{16/3}{16/3} = 1 : 1 \]
This means \( K = 1 \).

### Final Answer
The value of \( K \) is \( \boxed{1} \).