The value of a for which the area between the curves `y^(2) = 4ax` and `x^(2) = 4ay` is 1 unit is

Options

A. `sqrt3`
B. `4`
C. `4sqrt3`
D. `(sqrt3)/(4)`

Answer

To find the value of \( a \) for which the area between the curves \( y^2 = 4ax \) and \( x^2 = 4ay \) is equal to 1 unit, we can follow these steps:

### Step 1: Identify the curves
The equations given are:
1. \( y^2 = 4ax \) (a rightward-opening parabola)
2. \( x^2 = 4ay \) (an upward-opening parabola)

### Step 2: Find the points of intersection
To find the points of intersection, we can express \( y \) from the first equation:
\[ y = \sqrt{4ax} \quad \text{and} \quad y = \frac{x^2}{4a} \]
Setting these equal to each other:
\[ \sqrt{4ax} = \frac{x^2}{4a} \]
Squaring both sides:
\[ 4ax = \frac{x^4}{16a^2} \]
Rearranging gives:
\[ 16a^3 x = x^4 \]
Factoring out \( x \):
\[ x(x^3 - 16a^3) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{or} \quad x^3 = 16a^3 \implies x = 2a \]

### Step 3: Corresponding \( y \) values
Substituting \( x = 2a \) back into either equation to find \( y \):
Using \( y^2 = 4a(2a) \):
\[ y^2 = 8a^2 \implies y = 2\sqrt{2}a \]
Thus, the points of intersection are \( (0, 0) \) and \( (2a, 2\sqrt{2}a) \).

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 2a \) is given by:
\[ A = \int_0^{2a} \left( \sqrt{4ax} - \frac{x^2}{4a} \right) dx \]

### Step 5: Calculate the integral
Calculating the integral:
1. The integral of \( \sqrt{4ax} \):
\[ \int \sqrt{4ax} \, dx = \int 2\sqrt{a}\sqrt{x} \, dx = 2\sqrt{a} \cdot \frac{2}{3} x^{3/2} = \frac{4\sqrt{a}}{3} x^{3/2} \]
2. The integral of \( \frac{x^2}{4a} \):
\[ \int \frac{x^2}{4a} \, dx = \frac{1}{4a} \cdot \frac{x^3}{3} = \frac{x^3}{12a} \]

Putting it all together:
\[ A = \left[ \frac{4\sqrt{a}}{3} x^{3/2} - \frac{x^3}{12a} \right]_0^{2a} \]
Calculating at the bounds:
\[ = \left( \frac{4\sqrt{a}}{3} (2a)^{3/2} - \frac{(2a)^3}{12a} \right) - 0 \]
Calculating \( (2a)^{3/2} = 2^{3/2} a^{3/2} = 2\sqrt{2} a^{3/2} \):
\[ = \frac{4\sqrt{a}}{3} \cdot 2\sqrt{2} a^{3/2} - \frac{8a^2}{12a} \]
\[ = \frac{8\sqrt{2} a^2}{3} - \frac{2a^2}{3} = \frac{8\sqrt{2} a^2 - 2a^2}{3} \]
\[ = \frac{(8\sqrt{2} - 2)a^2}{3} \]

### Step 6: Set the area equal to 1
Setting the area equal to 1:
\[ \frac{(8\sqrt{2} - 2)a^2}{3} = 1 \]
Multiplying both sides by 3:
\[ (8\sqrt{2} - 2)a^2 = 3 \]
Solving for \( a^2 \):
\[ a^2 = \frac{3}{8\sqrt{2} - 2} \]
Taking the square root:
\[ a = \sqrt{\frac{3}{8\sqrt{2} - 2}} \]

### Step 7: Final simplification
To simplify \( 8\sqrt{2} - 2 \):
\[ = 2(4\sqrt{2} - 1) \implies a = \sqrt{\frac{3}{2(4\sqrt{2} - 1)}} = \frac{\sqrt{3}}{\sqrt{2(4\sqrt{2} - 1)}} \]

### Final Answer
Thus, the value of \( a \) for which the area between the curves is 1 unit is:
\[ a = \sqrt{\frac{3}{8\sqrt{2} - 2}} \]