If the area enclosed by `y^(2)=2x and x^(2)+4+4x=4y^(2)` is k square units, then the value of 3k is equal to

Answer

To find the area enclosed by the curves \( y^2 = 2x \) and \( x^2 + 4 + 4x = 4y^2 \), we will follow these steps:

### Step 1: Identify the curves
The first curve is given by:
\[ y^2 = 2x \]
This is a parabola that opens to the right.

The second equation can be rearranged:
\[ x^2 + 4 + 4x - 4y^2 = 0 \]
We can rewrite it as:
\[ x^2 + 4x + 4 - 4y^2 = 0 \]
This can be factored as:
\[ (x + 2)^2 - (2y)^2 = 0 \]
This represents a pair of straight lines:
\[ x + 2 - 2y = 0 \quad \text{and} \quad x + 2 + 2y = 0 \]
or:
\[ x = 2y - 2 \quad \text{and} \quad x = -2y - 2 \]

### Step 2: Find the points of intersection
To find the points of intersection, we substitute \( y^2 = 2x \) into the line equations.

For the first line \( x = 2y - 2 \):
\[ y^2 = 2(2y - 2) \]
\[ y^2 = 4y - 4 \]
\[ y^2 - 4y + 4 = 0 \]
\[ (y - 2)^2 = 0 \implies y = 2 \]
Substituting \( y = 2 \) back to find \( x \):
\[ x = 2(2) - 2 = 2 \]
So one point of intersection is \( (2, 2) \).

For the second line \( x = -2y - 2 \):
\[ y^2 = 2(-2y - 2) \]
\[ y^2 = -4y - 4 \]
\[ y^2 + 4y + 4 = 0 \]
\[ (y + 2)^2 = 0 \implies y = -2 \]
Substituting \( y = -2 \) back to find \( x \):
\[ x = -2(-2) - 2 = 2 \]
So the other point of intersection is \( (2, -2) \).

### Step 3: Set up the integral for area
The area \( A \) between the curves from \( y = -2 \) to \( y = 2 \) can be calculated as:
\[ A = \int_{-2}^{2} \left[ (2y - 2) - \frac{y^2}{2} \right] dy \]

### Step 4: Calculate the integral
Calculating the integral:
\[ A = \int_{-2}^{2} \left( 2y - 2 - \frac{y^2}{2} \right) dy \]
\[ = \int_{-2}^{2} \left( 2y - 2 - 0.5y^2 \right) dy \]

Calculating the integral:
\[ = \left[ y^2 - 2y - \frac{y^3}{6} \right]_{-2}^{2} \]

Calculating at the limits:
At \( y = 2 \):
\[ = 2^2 - 2(2) - \frac{2^3}{6} = 4 - 4 - \frac{8}{6} = -\frac{8}{6} = -\frac{4}{3} \]

At \( y = -2 \):
\[ = (-2)^2 - 2(-2) - \frac{(-2)^3}{6} = 4 + 4 + \frac{8}{6} = 8 + \frac{4}{3} = \frac{24}{3} + \frac{4}{3} = \frac{28}{3} \]

Now subtract:
\[ A = \left( -\frac{4}{3} - \frac{28}{3} \right) = -\frac{32}{3} \]
Since we are interested in the area, we take the absolute value:
\[ A = \frac{32}{3} \]

### Step 5: Find \( k \) and \( 3k \)
Thus, \( k = \frac{32}{3} \). Therefore:
\[ 3k = 3 \times \frac{32}{3} = 32 \]

### Final Answer
The value of \( 3k \) is \( \boxed{32} \).