If `y^2 = ax` and `x^2 = ay` intersect at A & B. Area bounded by both curves is bisected by line x = b(given `a gt b gt 0`). Area of triangle formed by line AB, x = b and x-axis is `1/2` . Then

Options

A. `a^6-12a^3-4 = 0`
B. `a^6+12a^3-4 = 0`
C. `a^6-12a^3+4 = 0`
D. `a^6+12a^3+4 = 0`

Explanation

To solve the problem step by step, we will analyze the given equations and conditions systematically.

### Step 1: Identify the curves and their intersections
We have two curves given by the equations:
1. \( y^2 = ax \)
2. \( x^2 = ay \)

To find the points of intersection, we can express \( y \) in terms of \( x \) from the first equation:
\[ y = \sqrt{ax} \]

Substituting this expression for \( y \) into the second equation:
\[ x^2 = a(\sqrt{ax}) \]
Squaring both sides gives:
\[ x^4 = a^2x \]
Rearranging this, we have:
\[ x^4 - a^2x = 0 \]
Factoring out \( x \):
\[ x(x^3 - a^2) = 0 \]

Thus, the solutions for \( x \) are:
1. \( x = 0 \)
2. \( x^3 = a^2 \) which gives \( x = a^{2/3} \)

Now substituting these values back to find \( y \):
- For \( x = 0 \):
\[ y = \sqrt{a \cdot 0} = 0 \]
So one intersection point is \( (0, 0) \).

- For \( x = a^{2/3} \):
\[ y = \sqrt{a \cdot a^{2/3}} = a^{1/3} \]
So the other intersection point is \( (a^{2/3}, a^{1/3}) \).

### Step 2: Find the area between the curves
The area \( A \) between the curves from \( x = 0 \) to \( x = a^{2/3} \) can be calculated using the integral:
\[ A = \int_0^{a^{2/3}} \left( \sqrt{ax} - \frac{x^2}{a} \right) dx \]

### Step 3: Calculate the integral
Calculating the integral:
1. The integral of \( \sqrt{ax} \):
\[ \int \sqrt{ax} \, dx = \frac{2}{3} \cdot \frac{2}{3} a^{1/2} x^{3/2} \]
Evaluating from \( 0 \) to \( a^{2/3} \):
\[ = \frac{2}{3} a^{1/2} \left( a^{2/3} \right)^{3/2} = \frac{2}{3} a^{1/2} a = \frac{2}{3} a^{3/2} \]

2. The integral of \( \frac{x^2}{a} \):
\[ \int \frac{x^2}{a} \, dx = \frac{1}{3a} x^3 \]
Evaluating from \( 0 \) to \( a^{2/3} \):
\[ = \frac{1}{3a} \left( a^{2/3} \right)^3 = \frac{1}{3a} a^2 = \frac{a}{3} \]

Thus, the area \( A \) becomes:
\[ A = \frac{2}{3} a^{3/2} - \frac{a}{3} \]

### Step 4: Area of triangle formed by line AB, x = b, and x-axis
The area of the triangle formed by the line \( AB \) (which has the equation \( y = x \)), the line \( x = b \), and the x-axis can be calculated as:
\[ \text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times b = \frac{b^2}{2} \]

Given that this area is \( \frac{1}{2} \), we have:
\[ \frac{b^2}{2} = \frac{1}{2} \]
Thus, \( b^2 = 1 \) and therefore \( b = 1 \).

### Step 5: Area bisected by the line \( x = b \)
The area to the left of \( x = b \) is half of the total area:
\[ \int_0^b \left( \sqrt{ax} - \frac{x^2}{a} \right) dx = \frac{1}{2} A \]

Substituting \( b = 1 \) into the area integral gives:
\[ \int_0^1 \left( \sqrt{ax} - \frac{x^2}{a} \right) dx = \frac{1}{2} \left( \frac{2}{3} a^{3/2} - \frac{a}{3} \right) \]

### Step 6: Solve for \( a \)
Setting up the equation and solving for \( a \) will yield the required relationship.